manuscripta math 138, 141–160 (2012) © Springer-Verlag 2011 Phan Hồng Cho’n, Lê Minh Hà On May spectral sequence and the algebraic transfer Received: 30 September 2010 / Revised: 27 July 2011 Published online: October 2011 Abstract The algebraic transfer is an important tool to study the cohomology of the Steenrod algebra In this study, we will construct a version of the algebraic transfer in E -term of May spectral sequence and use this version to study the image of the algebraic transfer By this method, we obtain the description of the image of ϕs in some degrees Introduction This article explores the May spectral sequence as a tool for understanding the algebraic transfer, defined by Singer [26] We work exclusively at the prime 2, and let A denote the mod Steenrod algebra [27,19] The cohomology algebra, Ext∗,∗ A (F2 , F2 ), is a central object of study in algebraic topology because it is the initial page of the Adams spectral sequence converging to stable homotopy groups of the spheres [1] This cohomology algebra has been intensively studied, see Lin [15] and Bruner [4] for the most recent results, but its structure remains largely mysterious One approach to better understand the structure of Ext∗,∗ A (F2 , F2 ) was proposed by Singer [26] where he introduced an algebra homomorphism from a certain subquotient of the divided power algebra to the cohomology of the Steenrod algebra, which can be seen as an algebraic formulation of the the stable transfer B(Z/2)s+ / S0 Let Vs denote an s-dimensional F2 -vector space Its mod cohomology is a polynomial algebra on s generators, each in degree H ∗ (BVs ) admits a left action of the Steenrod algebra as well as a right action of the automorphism group G L s = G L(V ), and the two actions commute For each s ≥ 1, Singer [26] constructed an F2 -linear map: A ϕs : Tor s,s+∗ (F2 , F2 ) → [F2 ⊗ A H ∗ (BVs )]G L s , from the homology of the Steenrod algebra to the elements of F2 ⊗ A H ∗ (BVs ) invariant under the group action This is called the rank s algebraic transfer Dually, let P H∗ (BVs ) Junior Associate at the Abdus-Salam ICTP P H Cho’n: Department of Mathematics, College of Science, Cantho University, 3/2 St., Ninh Kieu, Cantho, Vietnam e-mail: phchon@ctu.edu.vn; phchon.ctu@gmail.com Current address: P H Cho’n: Department of Mathematics and Application, Saigon University, 273 An Duong Vuong, District 5, Ho Chi Minh city, Vietnam L M Hà (B): Department of Mathematics-Mechanics-Informatics, Vietnam National University, Hanoi, 334 Nguyen Trai St., Thanh Xuan, Hanoi, Vietnam e-mail: minhha@vnu.edu.vn Mathematics Subject Classification (2000): Primary 55P47, 55Q45, 55S10, 55T15 DOI: 10.1007/s00229-011-0487-0 142 P H Cho’n, L M Hà be the subspace of H∗ (BVs ) consisting of all elements that are annihilated by all positive degree Steenrod squares, then there is an induced action of G L s on P H∗ (BVs ), and we have a map (F2 , F2 ), ϕs∗ : [P H∗ (BVs )]G L s → Ext s,s+∗ A from the coinvariant elements of P H∗ (BVs ) to the cohomology of the Steenrod algebra Moreover, the “total” dual algebraic transfer ϕ ∗ = ⊕ϕs∗ is an algebra homomorphism Calculations by Singer [26] for s ≤ and by Boardman [3] for s = showed that ϕs∗ is an isomorphism This shows that the algebraic transfer is highly nontrivial, and an interesting (F2 , F2 ) that are detected by the dual algebraic problem is to determine elements of Exts,s+∗ A transfer In rank 4, it was already known to Singer that ϕ4∗ is an isomorphism in a range It is from these preliminary calculation that he conjectured that ϕs∗ is a monomorphism for all s ≥ In [5], Bruner, Hà and Hu’ng showed that the entire family of elements {gi : i ≥ 1} is not in the image of the transfer, thus refuting a question of Minami concerning the so-called new doomsday conjecture [21] Here we are using the standard notation of elements in the cohomology of the Steenrod algebra as was used in [29,15,4] One of the main results of this paper is the proof that all elements in the family pi are in the image of the rank algebraic transfer Combining the results of Hu’ng [11], Hà [10] and Nam [23], we obtain a complete picture of the behaviour of the rank dual transfer It should be noted that in [12], Hu’ng and Qu`ynh claimed to have a proof that the family { pi : i ≥ 0} is detected by the dual algebraic tranfer, but the details have not appeared Our method is completely different Very little information is known when s ≥ Singer gave an explicit element in (F2 , F2 ), namely Ph , that is not detected by the dual algebraic transfer Another Ext 5,5+∗ A element, commonly denoted as Ph is also not detected (See Qu`ynh [25]) Using the lambda algebra, we are able to prove in [7,8] several non-detection results in even higher rank For example, h Ph as well as h Ph are not in the image of ϕ6∗ ; h 21 Ph is not in the image of ϕ7∗ Often, these results are available because it is possible to compute the domain of the algebraic transfer in the given bidegree Besides the lambda algebra, a relatively efficient tool to compute the cohomology of the Steerod algebra is the May spectral sequence, which passes from the cohomology of E A, the associated graded of the Steenrod algebra, to E Ext ∗,∗ A (F2 , F2 ), an algebra associated to the cohomology of A Much of the known information about the cohomology of the Steenrod algebra has been obtained by using this technique In this article, we construct a representation of the algebraic transfer ϕs in the May spectral sequence and apply this description to study the algebraic transfer Using this method, we recover, with much less computation, results in [25], [7], [8] and [12] Moreover, our method can also be applied, as illustrated in the case of the elements h n0 i, ≤ n ≤ and h n0 j, ≤ n ≤ 2, to degrees where computation of the domain of the algebraic transfer seems out of reach at the moment This article is the detailed version of the note with the same title [6] Organization of the paper The first two sections are preliminaries In Sect 2, we recall basic facts about May spectral sequence for an arbitrary A-module M In Sect 3, we present the algebraic transfer Detailed information about the representation of the algebraic transfer in the bar construction and the version of the algebraic transfer in the E -term of May spectral sequence are also given The structure of E ∞ p,− p,∗ (Ps ) is presented in Sect Three final sections contains the main results of this article May spectral sequence and algebraic transfer 143 May spectral sequence In this section, we recall May’s results on the construction of his spectral sequence The original papers [17] and [18] are our main references May’s chain complex to compute the cohomology of E A was also reworked in the framework of Priddy’s theory of Koszul resolutions [24] 2.1 The associated graded algebra E A The Steenrod algebra is filtered by powers of its augmentation ideal A¯ by setting: F p A = ¯ − p if p < Let E A = ⊕ p,q E 0p,q A where E 0p,q A = A if p ≥ and F p A = ( A) (F p A/F p−1 A) p+q be the associated graded algebra It is clearly a primitively generated Hopf algebra, and according to a theorem due to Milnor and Moore [20], E A is isomorphic to the universal enveloping algebra of its restricted Lie algebra of primitive elements, which j in this case, are the Milnor generators Pk ([19]) The following result from May’s thesis remains unpublished, but well-known Theorem 2.1 (May [17]) E A is a primitively generated Hopf algebra It is isomorphic to the universal enveloping algebra of the restricted Lie algebra of its primitive elements j {Pk | j ≥ 0, k ≥ 1} Moreover, k P ji , P k = δi,k+ P j+ for i ≥ k; j (2) ξ(Pk ) = 0, where ξ is the restriction map (of its restricted Lie algebra structure) (1) j Here, δi,k+ is the usual Kronecker delta The set of primitive elements R = {Pk | j ≥ 0, k ≥ 1} is equipped with a natural total order, given by P ji < P k if i < k or i = k, j < An element θ ∈ F p A but θ ∈ F p−1 A is said to have weight − p The following result determines the weight of a given Milnor generator Sq(R), where R = (r1 , r2 ) Theorem 2.2 (May [17]) The weight of a Milnor generator Sq(R), where R = (r1 , r2 , ), is w(R) = i, j iai j where ri = j j j is the binary expansion of ri In fact, May has showed that Sq(R) is the sum of (P ji )ai j and terms of strictly greater weight, so they can be identified in the associated graded E A In particular, the weight of P ji is just its subscript j In the language of Priddy’s theory of Koszul resolutions [24], Thej orem 2.1 implies that E A is a Koszul algebra with Koszul generators {Pk | j ≥ 0, k ≥ 1} and quadratic relations: P ji P k = P k P ji if i = k + , i− P ji P i− + P i− P ji + P j+ = 0, P ji P ji = The following theorem is first proved in May’s thesis, but see also [24] for a modern treatment Theorem 2.3 (May [17], Priddy [24]) H ∗ (E A) is the homology of the complex R, where R is a polynomial algebra over F2 generated by {Ri, j |i ≥ 0, j ≥ 1} of degree 2i (2 j − 1), and with the differential is given by j−1 δ(Ri, j ) = Ri,k Ri+k, j−k k=1 Cup product in H ∗ (E A) correspond to products of representative cycles in R 144 P H Cho’n, L M Hà It is more convenient for our purposes to work with the homology version The dual complex, denoted as X¯ in [18], is an algebra with divided powers on the generators P ji In fact, X¯ is embedded in the bar construction for E A by sending γn (P i ) to j {P ji |P ji | |P ji } (n factors) and the product in X¯ corresponds to the shuffle product (see [2], pp 40) Under this embedding, the differential for X¯ is exactly the differential of the bar construction This embedding technique was successfully exploited by Tangora ([29], Chap 5) to compute of the cohomology of the mod Steenrod algebra, up to a certain range 2.2 May spectral sequence Let M be a left A-module of finite type and bounded-below M admits a filtration, induced from that of A, given by F p M = F p A · M It is clear that F p M = A · M = M if p ≥ 0, and p F p M = The associated graded 0 module E M = p,q E p,q M where E p,q M = (F p M/F p−1 M) p+q is a bigraded mod0 ule over the associated graded algebra E A Let B(A; M) be the usual bar construction with the induced filtration: F p B(A; M) = F p1 A¯ ⊗ · · · ⊗ F pn A¯ ⊗ F p0 M n where the sum is taken over all (n + 1)-tuples ( p0 , , pn ) such that n + i=0 pi ≤ p This filtration respects the differential, and in the resulting spectral sequence, we have E 1p,q,t (M) = F p B p+q (A; M) F p−1 B p+q (A; M) t Here p is the filtration degree, p+q is the homological degree and t is the internal degree The differential d of the spectral sequence is the connecting homomorphism of the homology of the following short exact sequence: 0→ F p−1 B(A; M) F p B(A; M) F p B(A; M) → → → F p−2 B(A; M) F p−2 B(A; M) F p−1 B(A; M) On the other hand, E 1p,q,t (M) is isomorphic to B p+q (E A; E M)−q,q+t as F2 -trigraded vector space Under this identification, d is exactly the canonical differential of the bar construction B∗ (E A; E M) Hence E 2p,q,t (M) ∼ = H p+q (E A; E M)−q,q+t and we can summarize the situation in the following theorem Theorem 2.4 ([18]) Let M be an A-module of finite type and bounded-below There exists a third-quadrant spectral sequence converging to H∗ (A; M), whose E -term is E 2p,q,t (M) = H p+q (E A; E M)−q,q+t The differentials d r : E rp,q,t (M) −→ E rp−r,q+r −1,t (M) are F2 -linear maps May spectral sequence and algebraic transfer 145 The algebraic transfer Let Vs be an s-dimensional F2 -vector space The mod cohomology of BVs is a polynomial algebra which we will write as Ps = F2 [x1 , , xs ] where xi are in degree / π∗ (S ) admits an algebraic analogue at The geometric stable transfer π∗ (BVs )+ the E level of the Adams spectral sequence: A ϕs : Tor s,s+t (F2 , F2 ) / Tor A (F2 , Ps ) ∼ = (F2 ⊗ A Ps )t 0,t This map was constructed by W Singer in [26] and further investigation (see [3,5,10– 12,21,23]) shows that it is highly nontrivial In this section, we will refine this algebraic gadget by constructing a map between May spectral sequences: E r ψs : E rp,q,t (F2 ) / E rp,q−s,t−s (Ps ), which “converges” to the algebraic transfer More precisely, E ∞ ψs in homological degree p + q = s coincides with the induced map of the algebraic transfer in the corresponding associated graded modules 3.1 The algebraic transfer and the bar construction There are several ways to describe the algebraic transfer [3,8,14,23,26] Each has its own advantages and disadvantages We choose to follow the presentation in [23] because we need an explicit lift of the algebraic transfer on the bar construction Let Pˆ1 be the unique A-module extension of P1 = F2 [x1 ] obtained by formally adding a generator x1−1 of degree −1 and requiring that Sq n (x1−1 ) = x1n−1 Let u : A → Pˆ1 be the uniquely determined A-map that sends an operator θ to θ (x1−1 ), and let ψ1 be its restriction ¯ Clearly, ψ1 maps onto P1 Let ψs : A¯ ⊗s → Ps be defined by to the augmentation ideal A the following recursive formula: θs (xs−1 )θs (ψs−1 ({θs−1 | |θ1 })), ψs ({θs | |θ1 }) = (3.1) |θs |>0 where we use the “bar notation” for elements of A¯ ⊗s and standard notation for the coproduct (θ ) = θ ⊗ θ It is sometimes more convenient to use another form of Eq (3.1): ψs ({θs | |θ1 }) = θs (xs−1 ψs−1 ({θs−1 | |θ1 })) + xs−1 θs ψs−1 ({θs−1 | |θ1 }) (3.2) In [23], it is shown that ψs is a chain-level representation of the algebraic transfer Theorem 3.1 (Nam [23], Section 5.1) For each s ≥ 1, ψs is a chain-level representation of the algebraic transfer A A (F2 , F2 ) → Tor 0,t (F2 , Ps ) ∼ ϕs : Tor s,s+t = (F2 ⊗ A Ps )t Singer [26] showed that the image of ϕs actually lies in the G L s -invariant subspace of F2 ⊗ A Ps / B∗−s (A; Ps ) between We extend ψs to a chain homomorphism ψ˜ s : B∗ (A; F2 ) the bar constructions Let ψ˜ s ({θn | |θ1 }) = {θn | |θs+1 } ⊗ ψs ({θs | |θ1 }) (3.3) 146 P H Cho’n, L M Hà Proposition 3.2 The map ψ˜ s is a chain homomorphism Proof We have ψ˜ s (∂({θn | |θ1 })) = n−1 {θn | |θi+1 θi | |θs+1 }ψs ({θs | |θ1 }) i=s+1 s + (3.4) {θn | |θs+2 }ψs ({θs+1 | |θ j+1 θ j | |θ1 }) j=1 Since, ψs ({θs+1 | |θ j+1 θ j | |θ1 }) −1 = θs+1 (xs−1 θ j+1 θ j (x −1 j θ1 (x ) ) ) −1 −1 = θs+1 (xs−1 θ j+1 (x −1 j θ j (θ j−1 (x j−1 θ1 (x ) ))) ) −1 + θs+1 (xs−1 θ j+1 (θ j (x −1 j θ1 (x ) )) ), where the action of θi (x −1 y) is understood as the right hand side of (3.1) Then the second term on the right hand side of (3.4) equals {θn | |θs+2 }θs+1 (ψs ({θs | |θ1 })) Therefore, ψ˜ s (∂({θn | |θ1 })) = ∂(ψ˜ s ({θn | |θ1 })) The proof is complete Our next result shows that the chain map ψ˜ s just constructed respects the May filtration Proposition 3.3 For each s ≥ 1, ψ˜ s restricts to chain map: F p ψ˜ s : F p B∗ (A; F2 ) → F p B∗−s (A; Ps ) Thus, ψ˜ s induces a map between May spectral sequences: E r ψs : E rp,q,t (F2 ) → E rp,q−s,t−s (Ps ), r ≥ Proof Since the coproduct preserves May filtration, the general case follows at once, if we can prove the theorem for the case s = 1, ψ˜ : B∗ (A; F2 ) → B∗−1 (A; P1 ) We need to prove that if i i k ¯ Sq Sq ∈ F−k+1 B1 (A, F2 ) = F−k+1 A, then, Sq Sq (x1−1 ) ∈ F−k+1 B0 (A, P1 ) = F−k+1 P1 ik i1 But, Sq Sq (x1−1 ) = Sq Sq ik i1 i1 i k−1 ik (x12 −1 ), ¯ The assertion follows ∈ F−k+1 A and x12 −1 ∈ F0 P1 while Sq Sq The second statement follows immediately from the former ik i i k−1 May spectral sequence and algebraic transfer 147 3.2 A description of E ψs Proposition 3.3 allows us to use the May spectral sequence to study properties of the chainlevel representation ψs of the s-th algebraic transfer We are going to give an explicit formula for the maps in E page: 0 E A E A E ψs : Tor u,v (F2 , F2 ) → Tor u−s,v−s (F2 , E Ps ) Our construction is similar to Singer’s original construction of the algebraic transfer in [26] Therefore, we will give only a sketch construction Consider the short exact sequence of A-modules / P1 ι1 / Pˆ1 π1 / F2 / 0, where ι1 is the inclusion and π1 is the obvious quotient map Note that π1 has degree There is a corresponding short exact sequence of E A-modules: / E P1 E ι1 / E Pˆ1 E π1 / F2 / 0, and this in turn induces a short exact sequence of the bar constructions: −→ B∗ (E A; E P1 ) −→ B∗ (E A; E Pˆ1 ) −→ B∗ (E A; F2 ) −→ Tensoring with E M, where M is any A-module of finite type and bounded-below, we obtain −→ B∗ (E M; E P1 ) −→ B∗ (E M; E Pˆ1 ) −→ B∗ (E M; F2 ) −→ The connecting homomorphism of this exact sequence is 0 E A E A E ψ1 (M) : Tor s,s+∗ (E M, F2 ) → Tor s−1,s−1+∗ (E M, E P1 ) More generally, we obtain 0 E A E A E (ψ1 × Pk−1 )(M) : Tor s,s+∗ (E M, E Pk−1 ) → Tor s−1,s−1+∗ (E M, E Pk ) Splicing these maps when ≤ k ≤ s together, we obtain 0 E A E A (E M, F2 ) → Tor 0,∗ (E M, E Ps ) E ψs (M) : Tor s,s+∗ When M = F2 , E ψs = E ψs (F2 ) is the E -level of the algebraic transfer in the May spectral sequence The following is the main theorem of this section Proposition 3.4 The E page of the algebraic transfer, E ψs , is induced by the chain-level map E ψs (M) : E M ⊗ (E A)⊗s → E M ⊗ E Ps , given inductively as follows θs (E ψs−1 (M)(m{θs−1 | |θ1 }))θs (xs−1 ) E ψs (M)(m{θs | |θ1 }) = |θs |>0 148 P H Cho’n, L M Hà Proof It is sufficient to prove that E (ψ1 × Ps−1 )(M) is given by i s−1 )→ m{θs }(x1i xs−1 s−1 m ⊗ θs (x1i xs−1 )θs (xs−1 ) i |θs |>0 i s−1 ) ∈ B(E M, E Ps−1 ), Indeed, for any cycle x = m{θs }(x1i xs−1 i i s−1 s−1 + m ⊗ θs (x1i xs−1 ) = ∂(x) = θs (m) ⊗ x1i xs−1 Then pre-image of x under E πs is s−1 −1 x = m{θs }(x1i xs−1 xs ) i Therefore, s−1 −1 s−1 xs + m ⊗ θs (x1i xs−1 )xs−1 ∂(x ) = θs (m) ⊗ x1i xs−1 i i s−1 m ⊗ θs (x1i xs−1 )θs (xs−1 ) i + |θs |>0 = |θs |>0 (3.5) i s−1 m ⊗ θs (x1i xs−1 )θs (xs−1 ) The proof is complete 0 Example 3.5 For s = 1, the chain level of the transfer, E ψ1 : E − p,q A → E − p,q−1 P1 , for p ≥ 0, q > p, sends 2i (2 p+1 −1)−1 i i } → Pp+1 (x1−1 ) = x1 {Pp+1 i i+ p−1 = Sq Sq The non-trivial cycles in E A are {P1i } ∈ E 0,2i i+ p −1 x12 ∈ E− p,q−1 P1 i P = P , and E ψ1 ({P1i }) = x12 −1 ∈ E 0,∗ 1 i Since the A-generators of P1 are exactly those of the form x12 −1 , i ≥ 0, thus E ψ1 is an isomorphism Example 3.6 For s = 2, we have 2i (2 j1 −1)−1 2i (2 j2 −1)−1 x2 {P ji |P ji } → E ψ1 ({P ji })P ji (x2−1 ) = x1 1 (3.6) The nontrivial cycles in (E A)⊗2 are {P1i |P1 } + {P1 |P1i }( j > i + 1) and {P ji |P ji } for j > i j j (see [29, page 49]) Their corresponding images in E P2 are i j j i j j {P1i |P1 } + {P1 |P1i } → x12 −1 x22 −1 + x12 −1 x22 −1 , 2i (2 j −1)−1 2i (2 j −1)−1 x2 {P ji |P ji } → x1 By induction we obtain the following result which is extremely useful for computation in later sections Corollary 3.7 2i (2 j1 −1)−1 i E ψs ({P j s | |P ji }) = x1 s 2i s (2 js −1)−1 xs Proof The results follows immediately from the fact that Pst s are primitive in E A May spectral sequence and algebraic transfer 149 Two hit problems The algebraic transfer is closely related to an important problem in algebraic topology called the hit problem Call a polynomial hit in Ps if it is a linear combination of elements in the images of positive Steenrod squares The quotient of Ps by this subspace is exactly the range of the algebraic transfer F2 ⊗ Ps The hit problem asks for a construction of a F2 -basis for the space of “nonhit” elements F2 ⊗ Ps For a comprehensive survey of the hit problem and related questions in modular representation theory, we recommend Wood [33] and the upcoming book [31] by Walker and Wood The hit problem in rank s ≤ is relatively straightforward, but it appears to be very difficult in general In fact, the case s = was completely analyzed just recently in a 200 page preprint by Sum [28] Very little information is available for higher rank, except at a certain “generic degrees” [9,23] From the point of view of the algebraic transfer, the elements in generic degrees corresponds to the Adams subalgebra of Ext∗,∗ A (F2 , F2 ) generated by the elements denoted as h i , i = 0, 1, Our goal is to find more “exotic” elements in the cohomology of the Steenrod algebra that are also detected by the algebraic transfer In the May spectral sequence for Ps in homological degree 0, we have the edge homomorphism E 2p,− p,t (Ps ) ∼ = H0 (E A, E Ps ) p,− p+t = (F2 ⊗ E A (E Ps )) p,− p+t E∞ p,− p,t (Ps ), where the targets E ∞ p,− p,t (Ps ), when p varies, are associated graded components of (F2 ⊗ t Ps ) Thus, the hit problem for E Ps , considered as a module over the restricted Lie algebra E A whose structure is completely known, should be helpful as a first step toward understanding the general structure of F2 ⊗ Ps The second hit problem should be simpler in principle But there are several questions that remain unsolved, even in the rank case In [30], Vakil described a recursive algorithm to compute the filtration degree of a given element of E P1 , but no closed formula was obtained We plan to investigate this second hit problem, and its deeper relationship with the original hit problem elsewhere Given a homogeneous polynomial f ∈ Ps We denote by E( f ), E r ( f ) and [ f ] the corresponding classes of f in E (Ps ) = E Ps , E r (Ps ) and F2 ⊗ A Ps respectively Note that E( f ) is uniquely determined by those monomials of highest filtration degree, which we will call the essential part of f , and denote as ess( f ) For example, ess(x17 x213 x313 + x19 x211 x313 + x18 x212 x313 ) = x17 x213 x313 , because x17 x213 x313 is in filtration −4 while the latter two monomials are in filtrations −5 and −9 respectively Lemma 4.1 Let f ∈ Ps be a homogeneous polynomial If E( f ) is a nontrivial permanent cycle in the May spectral sequence for Ps , then ess( f ) is non-hit in Ps Proof It is clear that E ( f ) = E (ess( f )) and thus E ∞ ( f ) = E ∞ (ess( f )) Suppose on the contrary that ess( f ) is hit in Ps Write ess( f ) = i Sq ti f i It follows that there exists r > such that E r ( f ) = in E r Example 4.2 Consider g = x y + x y ∈ P3 in degree Since g = Sq (x y ) + (P2 ) is Sq (x y ), we have that [g] = in F2 ⊗ A Ps On the other hand, E(g) ∈ E −1,9 nontrivial However, when passing to the E terms, we have g = Sq (x y ) + x y , 150 P H Cho’n, L M Hà 0 P in P2 , so E(g) = Sq E (x y ) ∈ E −1,3 A · E 0,6 and x y ∈ F−4 P2 is trivial in E −1,9 E (Ps ) Thus E(g) is trivial in E (P2 ) One may wonder whether the converse is true, that is if f is a homogeneous polynomial such that E( f ) is trivial in E (Ps ), is it necessarily true that f is hit in Ps ? The following example implies that this is not the case Example 4.3 Let m = x17 x213 x313 ∈ P3 , it is not difficult to check that m is nonhit in P3 On the other hand, m = Sq (x17 x211 x313 ) + x19 x211 x313 + x18 x212 x313 + x17 x212 x314 + x18 x211 x314 , where x19 x211 x313 ∈ F−5 P3 and the last three monomials are in lower filtrations Therefore, P3 E(m) = Sq E(x17 x211 x313 ) ∈ E −4,37 So E(m) = in E (Ps ) The point of Example 4.3 is that in the class [m], one can choose a different representative which is in lower filtration than m Our last example shows that if a homogeneous polynomial is hit (that is, it is trivial in F2 ⊗ A Ps ), then we may have to wait for a while before this element is killed in the spectral sequence Example 4.4 Consider n = x1 x22 x32 + x12 x2 x32 + x12 x22 x3 = Sq (x1 x2 x3 ), so n is hit in P3 It can be checked by direct inspection that E(n) is nontrivial in E (P3 ) because it is (P3 ) The element n does not survive to E because the above formula E A-nonhit in E −2,7 shows that n = d (n ), where n = Sq ⊗ (x1 x2 x3 ) ∈ E 0,1,5 (P3 ) The following is the main result of this section Proposition 4.5 Let f ∈ Ps be a homogeneous polynomial in filtration degree p Then f is a nontrivial permanent cycle in the May spectral sequence for Ps if and only if ess( f ) is non-hit in Ps and there does not exist any non-hit polynomial g ∈ Fq Ps , with q < p, such that ess( f ) − g is hit Proof Suppose f is trivial in E r (Ps ) Then E r −1 ( f ) = E r −1 (ess( f )) is in the image of the differential d r −1 Therefore, there exist θr −1,i ∈ A¯ and fr −1,i ∈ Ps such that E r −1 (ess( f )) = E r −1 (θr −1,i fr −1,i ) ∈ E r −1 (Ps ) i In the E r −2 -term, we have E r −2 (ess( f )) = E r −2 (θr −1,i fr −1,i ) + fr −2 ∈ E r −2 (Ps ), i where fr −2 is in the image of d r −2 Thus, there exist θr −2,i ∈ A¯ and fr −2,i ∈ Ps such that E r −2 (θr −2,i fr −2,i ) ∈ E r −2 (Ps ), fr −2 = i and we can write E r −2 (ess( f )) = E r −2 (θki f ki ) ∈ E r −2 (Ps ) i,r −2≤k≤r −1 May spectral sequence and algebraic transfer 151 Repeating this process, in the E -term, we obtain E (ess( f )) = r −1 E (θki f ki ) ∈ E (Ps ) i,k=1 Note that E (Ps ) = E (Ps ), we finally have r −1 ess( f ) = θki f ki + g ∈ Ps , i,k=1 where g is some polynomial in Fq Ps , with q < p Thus, g is hit in Ps iff so is ess( f ) Conversely, if there exists g ∈ Fq Ps with q < p such that ess( f ) − g is hit in Ps then, in E rp,∗,∗ Ps , E r ( f ) = E r (ess( f ) − g) Since ess( f ) − g is hit in Ps , ess( f ) − g is a trivial permanent cycle and so is f Companion to Proposition 4.5 is the following key result, due to Wood [32], which will be used frequently in our applications Theorem 4.6 (Wood [32]) Let f be a monomial of degree d in Ps with exactly r odd exponents and suppose r < μ((d − r )/2) Then f is hit In the above theorem, μ(d) is the least number k for which it is possible to write d as a sum k (2 i − 1) of k numbers d = i=1 First application: a non-detection result In this section, we use the representation of the algebraic transfer in the E -term of the May spectral sequence to investigate its image The first author Cho’n’s thesis (in preparation, 2010) contains a complete analysis of the algebraic transfer in rank up to Notably, we recover all previously known results, obtained by various different methods Here is our first main result Theorem 5.1 The following elements in the cohomology of the Steenrod algebra (F2 , F2 ); (1) h Ph ∈ Ext 6,16 A (F2 , F2 ); (2) h 20 Ph ∈ Ext 7,18 A (F2 , F2 ), ≤ n ≤ 5; (3) h n0 i ∈ Ext 7+n,30+n A 152 P H Cho’n, L M Hà (F2 , F2 ), ≤ n ≤ 2, (4) h n0 j ∈ Ext 7+n,33+n A are not detected by the algebraic transfer Note that according to Bruner [4], h 60 i = h 30 j = Thus the last two results are best possible We would like to emphasize that the dimension of the above elements go far beyond the current computational knowledge of the hit problem (F2 , F2 ) and Ph ∈ Ext 5,16 (F2 , F2 ) are not in Corollary 5.2 ([26,25]) Ph ∈ Ext5,14 A A the image of the dual algebraic transfer The claims of this corollary are not new-they are due to Singer [26] and Qu`ynh [25] respectively We believe that our proof is much less computational Proof (Proof Of Corollary 5.2) It is well-known that the subalgebra generated by the h i s is in the image of the dual of the algebraic transfer Thus, if Ph or Ph were detected, then h Ph and h 20 Ph would be as well, since the dual of the algebraic transfer is an algebra homomorphism This contradicts Theorem 5.1 We are now ready to prove Theorem 5.1 The proof will be divided into parts, corresponding to the four nontrivial elements in the cohomology of the Steenrod algebra under consideration The general strategy is as follows Suppose an element in x ∈ Tor s,s+∗ A (F2 , F2 ) has nontrivial image y, via the transfer ϕs in F2 ⊗ A Ps , we choose a representative x¯ for this element in the E (F2 ) term of the May spectral sequence Because of naturality, its image y¯ in E (Ps ) must be a nontrivial permanent cycle and represents y We then show that this is not possible using knowledge of the A-module structure of Ps and a combination of degree arguments as well as computation in the May spectral sequence 5.1 h Ph is not detected According to Tangora [29], a representation for h Ph in the E -term of the May spectral R Its dual, therefore, contains sequence is R1,1 0,2 (F2 ), X = {P11 |P11 } ∗ {P20 |P20 |P20 |P20 } ∈ E −4,10,16 where ∗ denotes the shuffle product (see [2, pp 40]) Corollary 3.7 allows us to find the image of X under E ψ6 : (E ψ6 )(X ) = x1 x2 x32 x42 x52 x62 + all its permutations = Sq (x1 x2 x3 x4 x5 x6 ) Therefore, (E ψ6 )(X ) ∈ E (P6 ) is hit in P6 In the bar construction, we must have (h Ph )∗ = X + x, where x ∈ F p B(F2 ) with p < −4 Thus, if h Ph is detected, then ψ6 ((h Ph )∗ ) = E ψ6 (X ) + y, where y ∈ F p P6 with p < −4, is nonhit in P6 On the other hand, by direct computation, we see that there are only two possible monomials (or their permutations) in F−5 P6 , namely x14 x24 x32 x40 x50 x60 and x14 x22 x32 x42 x50 x60 But these two monomials are clearly hit in P6 May spectral sequence and algebraic transfer 153 5.2 h 20 Ph is not detected The argument is similar to the first case, so we will only give a sketch proof Again, from [29], we know that a representation for h 20 Ph : ∗ h 20 Ph = {P10 }2 ∗ {P12 } ∗ {P20 }4 + {P10 }3 ∗ {P11 } ∗ {P20 }2 ∗ {P30 }, where {a}n denotes {a|a| |a} (n factors) Its image under E ψ7 is x10 x20 x33 x42 x52 x62 x72 + x10 x20 x30 x41 x52 x62 x76 + all their permutations, and again one can easily verify that this image is hit in P7 The argument now follows a similar line as in the first case, noting that there does not exist any non-hit polynomial in F p P7 with p < −4 in degree 11 5.3 h n0 i is not detected It is sufficient to provide a proof for n = According to Tangora ([29]), h 50 i is represented R R (R in the E -term of the May spectral sequence by x = R0,1 0,2 0,3 1,1 R0,3 + R1,2 R0,2 ) By direct inspection, we see that the dual of h 50 i has a representation in the E -term of the May spectral sequence as x ∗ = {P20 }2 ∗ {P30 }3 ∗ {P11 } ∗ {P10 }6 + {P20 }3 ∗ {P30 }2 ∗ {P21 } ∗ {P10 }6 The image of this element under E ψ12 is x x + x · · · x x x x x x x + all their permutations x10 · · · x60 x71 x82 x92 x10 11 12 10 11 12 It can easily be checked, using Wood’s result ([32]), that this image is hit in P12 Therefore, ψ12 ((h 50 i)∗ ) = E ψ12 (x ∗ ) + X, for some X ∈ F p P12 in filtration p < −8 and degree 23 Using Wood’s result again, we obtain that X is hit in P12 Thus, ψ12 ((h 50 i)∗ ) is hit in P12 5.4 h n0 j is not detected We need only to consider the case n = According to Tangora [29], a representation of R R R We compute its dual h 20 j in the E -term of the May spectral sequence is R0,1 0,2 0,3 1,2 in the bar construction, the result is {P10 }3 ∗ {P20 }2 ∗ {P30 }2 ∗ {P21 }2 + {P10 }3 ∗ {P20 } ∗ {P21 } ∗ {P30 }3 ∗ {P11 } + x where x = {P10 }3 ∗ {P20 } ∗ {P30 }3 ∗ {P10 |P30 } ∈ F−9 B(A; F2 ) The image under E ψ9 of the first two summands is x42 x52 x65 x75 x86 x96 + x4 x52 x65 x76 x86 x96 + all their permutations which are all hit in P9 by using Wood’s test for the case r = 2, d = 26 Let X ∈ F−9 P9 be the image of x We claim that X also consists of monomials in which there are at most 154 P H Cho’n, L M Hà odd exponents, and therefore X is also hit Indeed, looking at Corollary 3.7, we see that an odd exponent is obtained whenever there appears a P ji with i > Since x only contains monomials {a1 | |a9 }, where is primitive in A, and the coproducts of P21 and P11 , in A, are (P21 ) = ⊗ P21 + P20 ⊗ P20 + P21 ⊗ 1, (P11 ) = ⊗ P11 + P10 ⊗ P10 + P11 ⊗ Second application: p0 is in the image of the transfer In this sections, we show that our method can also be used to detect elements in the image of the algebraic transfer This completes the proof of a conjecture in [11] which describe the behaviour of the algebraic transfer in rank The following is our second main result (F2 , F2 ) is in the image of the fourth algebraic Theorem 6.1 The element p0 ∈ Ext 4,37 A transfer This result is announced [12], but the details have not appeared All previous methods of computation such as [26,23,10,11] have failed to work with p0 As an immediate corollary of the above theorem, the entire family of elements { pi } belongs to the image of the dual algebraic transfer i (F2 , F2 ), i ≥ is in the image Corollary 6.2 Every element in the family pi ∈ Ext 4,37·2 A of the algebraic transfer Proof it is well known that the cohomology of the Steenrod algebra (in fact, any cocommutative Hopf algebra) admits an action of a Steenrod algebra in which Sq is an independent operation, not equal to the identity ([16]) In the family, { pi }, one has Sq ( pn ) = pn+1 On the other hand, it follows from work of Boardman [3] (see also Minami [21]) that there exists a similar Sq operation, first constructed by Kameko [13], that commutes with the classical Sq on the E xt group Therefore, if p0 belongs to the image of the dual algebraic transfer, so does the entire family { pi } so that its Proof of Theorem 6.1 According to Tangora, p0 is represented by R0,1 R3,1 R1,3 dual p0∗ in the E -term of May spectral sequence contains p¯ = {P13 } ∗ {P10 } ∗ {P31 |P31 } + {P13 |P13 } ∗ {P21 } ∗ {P40 } In fact, by direct inspection, a cycle representing p0∗ can be chosen as p0∗ = p¯0 +x + y +z +t, where x = {P12 |P12 } ∗ {P21 } ∗ {P40 } ∈ F−5 B(A; F2 ); y = {P20 |Sq(6)} ∗ {P31 |P31 } ∈ F−6 B(A; F2 ); z = {P40 |P30 } ∗ {P10 } ∗ {P31 } + {Sq(7)|P40 } ∗ {P10 } ∗ {P31 } + {P30 |Sq(4, 1)} ∗ {P13 } ∗ {P40 } + {P30 |Sq(7)} ∗ {P13 } ∗ {P40 } + {Sq(0, 3)|P30 } ∗ {P21 } ∗ {P40 } + {P30 |P20 |P22 } ∗ {P40 } + {P30 |Sq(3)|P22 } ∗ {P40 } + {P12 |P20 |P40 } ∗ {P40 } + {P20 |Sq(5)|P40 } ∗ {P31 } + {P20 |P12 |P40 |P40 } + {Sq(6)|P40 |P40 } ∗ {P10 } ∈ F−7 B(A; F2 ); t = {P10 |Sq(6, 3)} ∗ {P21 } ∗ {P40 } + {P10 |Sq(5, 3)|P30 } ∗ {P40 } + {P30 |P30 |Sq(2, 2)} ∗ {P40 } ∈ F−9 B(A; F2 ) May spectral sequence and algebraic transfer 155 We want to compute the image of p0∗ under the chain level map ψ4 To simplify notation, we write (a, b, c, d) for the monomial x1a x2b x3c x4d ∈ P4 , and make use of a ∗ operation which is similar to the shuffle product For example, (a, b, c) ∗ (d) = (a, b, c, d) + (a, b, d, c) + (a, d, b, c) + (d, a, b, c) Note that we only care about the image modulo hit elements The computation is greatly simplified because of the following observation Lemma 6.3 Let f be a monomial of degree 33 in P4 (i) If f has exactly one odd exponent, then f is hit (ii) If f ∈ F−9 P4 , then f is hit Proof Since μ( 33−1 ) = 2, the first statement follows immediately from Wood’s criterion (Theorem 4.6) So we need only be concerned with monomials which have exactly odd exponents For the second statement, there is simply no monomial with exactly odd exponents in degree 33 having filtration degree at most −8 It follows from the above lemma that the image of t is hit in P4 Modulo hit elements, the image of p0∗ is ψ4 ( p0∗ ) = X + X (12) + X (132) + X (1432) + Y, where X = x10 x27 x313 x413 + x10 x213 x37 x413 + x10 x213 x313 x47 + x10 x213 x317 x43 + x10 x217 x313 x43 + x10 x217 x33 x413 ; Y = (7, 7) ∗ (5) ∗ (14) + (16, 5, 7) ∗ (5) + (18, 3, 7) ∗ (5) + (20, 1, 7) ∗ (5) + (11, 3, 14) ∗ (5) + (11, 3) ∗ (5) ∗ (14) + (5, 2) ∗ (13, 13) + (17, 1, 2) ∗ (13) + (14, 9, 3, 7) + (9, 14, 3, 7) + (9, 3, 14, 7) + (7, 14, 3, 9) + (7, 9, 3, 14) + (14, 7, 3, 9) + (9, 3, 7, 14) + (9, 7, 3, 14) + (20, 1, 5, 7) + (16, 9, 1, 7) + (9, 16, 1, 7) + (5, 16, 9, 3) + (9, 5, 16, 3) + (18, 3, 9, 3) + (9, 3, 18, 3) + (9, 5, 14, 5) + (9, 5, 5, 14) + (5, 9, 5, 14) (12), (132), (1432) are elements of the symmetric group S4 permuting the four variables of P4 The notation X (12) means it is obtained from X by applying the permutation (12) Since X = Sq (x27 x313 x47 ) + Sq (x27 x313 x49 + x29 x313 x47 + x213 x313 x43 ) + x213 x37 x413 + x217 x33 x413 (mod A¯ P4 ) = Sq (x213 x37 x411 + x217 x33 x411 ) + Sq (x213 x35 x411 ) (mod A¯ P4 ), we see that all four summands X, X (12), X (132) and X (1432) are hit in P4 We can also simplify Y Observe that 156 P H Cho’n, L M Hà (7, 7) ∗ (5) ∗ (14) = Sq ((7, 7) ∗ (3) ∗ 14)) +(7) ∗ (9) ∗ (3) ∗ (14) + (7, 7) ∗ (3) ∗ (16) + others; (7, 7) ∗ (3) ∗ (16) = Sq ((7, 7) ∗ (3) ∗ (8)) + Sq ((11, 11) ∗ (3) ∗ (4)) +(13, 13) ∗ (3) ∗ (4) + others; (7) ∗ (9) ∗ (3) ∗ (14) = Sq ((7) ∗ (5) ∗ (3) ∗ (14)) + Sq ((7) ∗ (5, 5) ∗ (14)) +(11) ∗ (5) ∗ (3) ∗ (14) + (7) ∗ (5) ∗ (3) ∗ (18) + others; (7) ∗ (5) ∗ (3) ∗ (18) = Sq ((7) ∗ (5) ∗ (3) ∗ (10)) + Sq ((7) ∗ (9) ∗ (3) ∗ (10)) +Sq ((11) ∗ (5, 5) ∗ (10)) + (13) ∗ (5) ∗ (3) ∗ (10) + others, where “others” means monomials that are either hit or in filtration degree strictly less than −5 This shows that Y = (3, 5) ∗ (11) ∗ (14) + Y2 (modulo hit), where Y2 is in F p P4 with p < −5 Thus, E(Y ) = E((3, 5) ∗ (11) ∗ (14)) ∈ E −5,5,33 (P4 ) Lemma 6.4 Y1 = (3, 5) ∗ (11) ∗ (14) is a nontrivial permanent cycle in the May spectral sequence for P4 The proof of this lemma is elementary but rather technical, so we postpone it to Sect Assume that Lemma 6.4 is proved, then ψ4 ( p0∗ ) is non-hit in P4 Thus, p0 is in the image of the fourth algebraic transfer The proof is complete Proof of lemma 6.4 In this final section, we prove Lemma 6.4 Our strategy is to prove that Y1 represents a nontrivial permanent cycle in the May spectral sequence for P4 Proposition 7.1 Y1 is nonhit in E P4 , considered as an E A-module In other words, it is a nonzero element in term E = H0 (E A, E P4 ) Proof We prove this by contradiction Assume that there exist A1 , B1 , C1 , D1 , E ∈ , such that E −4,4,∗ Y1 = P10 (A1 ) + P11 (B1 ) + P12 (C1 ) + P13 (D1 ) + P14 (E ) (7.1) Such an expression is called a hit presentation of Y1 Here, we are using a simple fact that can be verified directly from the relations in Theorem 2.1 that the P1i generate E A We also assume that this presentation is reduced, that is, one can not extract a smaller hit presentation of Y1 It is convenient to visualize the process in terms of “lighting flash” For example, everytime we choose a monomial in A1 to hit a certain monomial, the operation P10 potentially can create extra monomials, which we again need to hit by some other monomials The process will stop once we find that besides the monomials in Y1 , there are no other extra monomials The pair of a monomial and an arrow chosen must not be repeated A forced presentation means one in which the choice of the first monomial dictates unique choices of the remaining monomials We begin with some simple observations Lemma 7.2 If A1 , B1 , C1 , D1 and E , provide a hit presentation of Y1 , then C1 does not contain any permutations of (3, 12, 7, 7) and B1 does not contain any permutations of (3, 10, 7, 11) May spectral sequence and algebraic transfer 157 Proof If C1 contained (3, 12, 7, 7), then the right hand side of (7.1) must contain (3, 16, 7, 7), but there is no other way to hit (3, 16, 7, 7) to eliminate it from Y1 Note that while expression P10 (3, 15, 7, 7) contains (3, 16, 7, 7) in A-module P4 , it is trivial in E A-module E P4 because of the filtration discrepancy This also explain why the hit problem for E Ps should be simpler The statement for B1 follows easily since P11 (3, 10, 7, 11) contains (3, 12, 7, 11) and the only other way to hit (3, 12, 7, 11) is with P12 (3, 12, 7, 7), which is not possible Lemma 7.3 Y1 = (5, 3) ∗ (11) ∗ (14) in E −5,5,33 Proof In fact, we have the following identity: (3, 5, 11, 14) + (3, 5, 14, 11) = (5, 3, 11, 14) + (5, 3, 14, 11) + P10 [(3, 5, 11, 13) + (3, 5, 13, 11) + (5, 3, 11, 13) + (5, 3, 13, 11) (7.2) + (5, 5, 11, 11)] + P11 [(3, 6, 11, 11) + (6, 3, 11, 11)] Now consider the following list of pairs of elements in degree 33 • • • • a = (3, 5, 11, 14) + (3, 5, 14, 11), a = (5, 3, 11, 14) + (5, 3, 14, 11); b = (5, 6, 11, 11), b = (6, 5, 11, 11); c = (9, 6, 7, 11) + (6, 9, 7, 11), c = (9, 6, 11, 7) + (6, 9, 11, 7); d = (10, 5, 7, 11) + (5, 10, 7, 11), d = (10, 5, 11, 7) + (5, 10, 11, 7) Lemma 7.4 The above elements induces equivalent classes in E Proof Because of symmetry and Lemma 7.3, we need only to show that a, b, c and d are equal in E This follows from the following explicit formula a = P10 [(3, 5, 11, 13) + (3, 5, 13, 11)] + P11 (3, 6, 11, 11) + P12 (3, 4, 11, 11) + b (7.3) b = P10 [(5, 9, 7, 11)] + P12 (5, 6, 7, 11) + c (7.4) c = P10 [(5, 9, 7, 11) + (9, 5, 7, 11)] + d (7.5) The proof of Proposition 7.1 follows from the following key observation Lemma 7.5 In any partial hit presentation of a, there appears an even number of elements in the list above (including a) It follows from this lemma that all these elements are “inseparable” in the sense that everytime we try to find a hit presentation for one element by adding arrows to the process, we always create extra terms which is exactly another element of the list We will provide a detailed argument for b The argument for other elements are similar and will be omitted Clearly, there are four ways to hit b: P10 (5, 5, 11, 11); P12 (5, 6, 7, 11); P12 (5, 6, 11, 7); P11 (3, 6, 11, 11) The first choice ends up with b For the second choice, note that b = P12 (5, 6, 7, 11) + (9, 6, 7, 11) + (5, 10, 7, 11) For the extra terms, if we use either P10 (9, 5, 7, 11) or P10 (5, 9, 7, 11), we arrive at either c or d The only other way to hit them is with P12 [(9, 6, 7, 7) + (5, 10, 7, 7)] (we cannot use 158 P H Cho’n, L M Hà P11 (3, 10, 7, 11) because of Lemma 7.2.), but then this will give us either c or d The third choice is similar to the second, by symmetry Finally, P11 (3, 6, 11, 11) = P12 (3, 4, 11, 11) + (3, 6, 13, 11) + (3, 6, 11, 13) Now we can hit (3, 6, 13, 11) with either P01 (3, 5, 13, 11) or P12 (3, 6, 7, 13) So there are four cases: • • P10 [(3, 5, 11, 13) + (3, 5, 13, 11)] or P12 [(3, 6, 7, 13) + (3, 6, 13, 7)]; P01 (3, 5, 11, 13) + P12 (3, 6, 13, 7) or P01 (3, 5, 13, 11) + P12 (3, 6, 7, 13) The first choice yields extra term a The second yields (3, 10, 7, 13) + (3, 10, 13, 7) for which we have only one possible continuation by P10 [(3, 9, 7, 13) + (3, 9, 13, 7)] + P12 [(3, 5, 14, 7) + (3, 5, 7, 14)] (7.6) and this again ends with a In the third case, the extra term is (3, 5, 11, 14) + (3, 10, 13, 7) and again for the latter monomial (3, 10, 13, 7), there is only one continuation with P01 (3, 9, 13, 7)+ P12 (3, 5, 14, 7) and we obtain (3, 5, 14, 11) Proposition 7.6 [Y1 ] is a nontrivial cycle in the May spectral sequence E r (P4 ) Proof Suppose there exists x ∈ E −3,4,33 such that d (x) = [Y1 ] Write B for the bar 0 construction B(E A, E P4 ) and let Z rp = {z|z ∈ F p B, ∂z ∈ F p−r B} Since ∂(θ1 θ2 ⊗ m) = ∂(θ1 ⊗ θ2 m) we can assume that x = such that P1i (m i ) ∈ F−5 B Thus, we can write i i P1 ⊗ m i where m i ∈ F−3 B , Y1 = P10 (m ) + P11 (m ) + P12 (m ) + P13 (m ) + P14 (m ) ∈ E −5,5,33 and by direct inspection, we see that there are no choices for m , while the m i , i > might be a combination (of permutations) of monomials in the collection (Ci ) below (C1 ) = {(1, 1, 14, 15), (1, 7, 8, 15), (4, 5, 7, 15)} (C2 ) = {(1, 3, 10, 15), (2, 3, 9, 15), (3, 3, 8, 15)} (C3 ) = {(1, 2, 3, 19), (3, 7, 7, 8), (5, 6, 7, 7)} (C4 ) = {(1, 1, 1, 14), (1, 1, 2, 13), (1, 1, 4, 11), (1, 1, 7, 8), (1, 3, 3, 10), (1, 4, 5, 7), (2, 3, 3, 9), (3, 3, 3, 8), (3, 3, 5, 6)} It is straightforward to verify that P14 (m) ∈ F−6 (E P4 ) for all m ∈ (C4 ) Also, direct computation shows that P11 (m ) and P12 (m ) are all E A-hit in E P4 , for example P11 (1, 1, 14, 15) = (2, 2, 14, 15) + (1, 1, 16, 15) = P10 (1, 2, 14, 15) + P12 (1, 1, 12, 15) for all i = 0, 1, 2, For C , we have P (1, 2, 3, 19) ∈ Thus P1i (m i ) are all trivial in E −5 F−6 There are two monomials left, namely (3, 7, 7, 8) and (5, 6, 7, 7) On the other hand, P13 (3, 7, 7, 8) = (3, 8, 11, 11) + (3, 16, 7, 7) = P12 (3, 4, 11, 11) + (3, 16, 7, 7), May spectral sequence and algebraic transfer 159 Similarly, we have so P13 (3, 7, 7, 8) = (3, 16, 7, 7) in E −5 P13 (5, 6, 7, 7) = (9, 10, 7, 7) + (5, 6, 11, 11) Indeed, In fact, we claim that P13 (3, 8, 7, 7) = P13 (5, 6, 7, 7) in E −5,5,33 P13 ((3, 8, 7, 7) + (5, 6, 7, 7)) =P13 ((3, 6, 9, 7) + (3, 6, 7, 9)) + P12 ((3, 12, 7, 7) + (5, 10, 7, 7) + (3, 6, 13, 7) + (3, 6, 7, 13) + (5, 6, 7, 11) + (5, 6, 11, 7) + (3, 4, 11, 11)) + P11 ((3, 10, 11, 7) + (3, 10, 7, 11)) So P13 (m ) is a sum of permutations of (9, 10, 7, 7) + (5, 6, 11, 11), and we need to show that it can never be equal to Y1 , which is the sum of permutations of (5, 6, 11, 11), where always precede The proof now uses similar argument as in the proof of Lemma 7.3, noting that elements in the “orbit” of (9, 10, 7, 7) are • • • • • (10,9,7,7), (5,10,7,11) + (5,10,11,7), (6,9,7,11) + (6,9,11,7), (10,5,7,11) + (10,5,11,7), (9,6,7,11) + (6,9,7,11) The fact that Y1 is nontrivial This completes the proof that Y1 is non trivial in E −5,5,33 r in E −5,5,33 , r = 4, is proved by using similar argument In fact, there are very few choices of monomials in degree 33 that can jump filtration, and the problem becomes much simpler than that at E Finally, we would like to remark that R Bruner has independently used computer software to calculate the space F2 ⊗ A P4 in degree 33 He found a unique G L -invariant element, and has also confirmed by Bruner (2009 personal communication) that our hand-computed element Y is the same as his (though of different presentations in F2 ⊗ A P4 ) Acknowledgments The authors would like to thank R R Bruner, N H V Hu’ng, W H Lin and J P May for many enlightening e-mail exchanges and discussions We are grateful to the referee for helpful comments and corrections This article is partially supported by a Grant from the Vietnam National Foundation for Science and Technology Development (NAFOSTED) and VNU Grant QG-10-02 References [1] Adams, J.F.: On structure and applications of the Steenrod algebra Comment Math Helv 32, 180–214 (1958) [2] Adams, J.F.: 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thesis... (x12 −1 ), ¯ The assertion follows ∈ F−k+1 A and x12 −1 ∈ F0 P1 while Sq Sq The second statement follows immediately from the former ik i i k−1 May spectral sequence and algebraic transfer 147