SPHERICAL CLASSES AND THE ALGEBRAIC TRANSFER

Transcription

1 TRNSCTIONS OF THE MERICN MTHEMTICL SOCIETY Volume 349, Number 10, October 1997, Pages S ( SPHERICL CLSSES ND THE LGEBRIC TRNSFER NGUYÊ N H. V. HU NG bstract. We study a wea form of the classical conjecture which predicts that there are no spherical classes in Q 0 S 0 except the elements of Hopf invariant one and those of Kervaire invariant one. The wea conjecture is obtained by restricting the Hurewicz homomorphism to the homotopy classes which are detected by the algebraic transfer. Let P = F 2 [x 1,...,x ]with x i = 1. The general linear group GL = GL(, F 2 and the (mod 2 Steenrod algebra act on P in the usual manner. We prove that the wea conjecture is equivalent to the following one: The canonical homomorphism j : F 2 (P GL (F 2 P GL induced by the identity map on P is zero in positive dimensions for >2. In other words, every Dicson invariant (i.e. element of P GL of positive dimension belongs to + P for >2, where + denotes the augmentation ideal of. This conjecture is proved for = 3 in two different ways. One of these two ways is to study the squaring operation Sq 0 on P (F 2 P, the range of j,and GL to show it commuting through j with Kameo s Sq0 on F 2 P (P, the GL domain of j. We compute explicitly the action of Sq0 on P (F 2 P for GL Introduction The paper deals with the spherical classes in Q 0 S 0, i.e. the elements belonging to the image of the Hurewicz homomorphism H : π (S s 0 = π (Q 0 S 0 H (Q 0 S 0. Here and throughout the paper, the coefficient ring for homology and cohomology is always F 2, the field of 2 elements. We are interested in the following classical conjecture. Conjecture 1.1. (conjecture on spherical classes. There are no spherical classes in Q 0 S 0, except the elements of Hopf invariant one and those of Kervaire invariant one. (See Curtis [9] and Wellington [21] for a discussion. Let V be an elementary abelian 2-group of ran. It is also viewed as a - dimensional vector space over F 2. So, the general linear group GL = GL(, F 2 Received by the editors pril 7, Mathematics Subject Classification. Primary 55P47, 55Q45, 55S10, 55T15. Key words and phrases. Spherical classes, loop spaces, dams spectral sequences, Steenrod algebra, invariant theory, Dicson algebra, algebraic transfer. The research was supported in part by the DGU through the CRM (Barcelona c 1997 merican Mathematical Society

2 3894 NGUYÊ N H. V. HU NG acts on V and therefore on H (BV in the usual way. Let D be the Dicson algebra of variables, i.e. the algebra of invariants D := H (BV GL = F2 [x 1,...,x ] GL, where P = F 2 [x 1,...,x ] is the polynomial algebra on generators x 1,...,x, each of dimension 1. s the action of the (mod 2 Steenrod algebra and that of GL on P commute with each other, D is an algebra over. One way to attac Conjecture 1.1 is to study the Lannes Zarati homomorphism ϕ : Ext,+i (F 2, F 2 (F 2 D i, which is compatible with the Hurewicz homomorphism (see [12], [13, p.46]. The domain of ϕ is the E 2 -term of the dams spectral sequence converging to π s(s0 = π (Q 0 S 0. Furthermore, according to Madsen s theorem [15] which asserts that D is dual to the coalgebra of Dyer Lashof operations of length, the range of ϕ is a submodule of H (Q 0 S 0. By compatibility of ϕ and the Hurewicz homomorphism we mean ϕ is a lifting of the latter from the E -level to the E 2 -level. Let h r denote the dams element in Ext 1,2r (F 2,F 2. Lannes and Zarati proved in [13] that ϕ 1 is an isomorphism with {ϕ 1 (h r r 0} forming a basis of the dual of F 2 D 1 and ϕ 2 is surjective with {ϕ 2 (h 2 r r 0} forming a basis of the dual of F 2 D 2. Recall that, from dams [1], the only elements of Hopf invariant one are represented by h 1,h 2,h 3 of the stems i =2 r 1=1,3,7, respectively. Moreover, by Browder [5], the only dimensions where an element of Kervaire invariant one would occur are 2(2 r 1, for r>0, and it really occurs at this dimension if and only if h 2 r is a permanent cycle in the dams spectral sequence for the spheres. Therefore, Conjecture 1.1 is a consequence of the following: Conjecture 1.2. ϕ = 0 in any positive stem i for >2. It is well nown that the Ext group has intensively been studied, but remains very mysterious. In order to avoid the shortage of our nowledge of the Ext group, we want to restrict ϕ to a certain subgroup of Ext which (1 is large enough and worthwhile to pursue and (2 could be handled more easily than the Ext itself. To this end, we combine the above data with Singer s algebraic transfer. Singer defined in [20] the algebraic transfer Tr : F 2 PH i (BV Ext,+i (F 2, F 2, GL where PH (BV denotes the submodule consisting of all -annihilated elements in H (BV. It is shown to be an isomorphism for 2 by Singer [20] and for = 3 by Boardman [4]. Singer also proved that it is an isomorphism for =4in a range of internal degrees. But he showed it is not an isomorphism for =5. However, he conjectures that Tr is a monomorphism for any. Our main idea is to study the restriction of ϕ to the image of Tr. Conjecture 1.3. (wea conjecture on spherical classes. ϕ Tr : F 2 PH (BV P(F 2 H (BV := (F 2 D GL GL is zero in positive dimensions for >2.

3 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3895 In other words, there are no spherical classes in Q 0 S 0, except the elements of Hopf invariant one and those of Kervaire invariant one, which can be detected by the algebraic transfer. natural question is: How can one express ϕ Tr in the framewor of invariant theory alone, and without using the mysterious Ext group? Let j : F 2 (P GL (F 2 P GL be the natural homomorphism induced by the identity map on P.Wehave Theorem 2.1. ϕ Tr is dual to j, or equivalently, j = Tr ϕ. By this theorem, Conjecture 1.3 is equivalent to Conjecture 1.4. j = 0 in positive dimensions for >2. This seems to be a surprise, because by an elementary argument involving taing averages, one can see that if H GL is a subgroup of odd order then the similar homomorphism j H : F 2 (P H (F 2 P H is an isomorphism. Furthermore, j 1 is iso and j 2 is mono. Obviously, j = 0 if and only if the composite D = P GL proj F 2 (P GL j (F 2 P GL F 2 P is zero. So, Conjecture 1.4 can equivalently be stated in the following form. Conjecture 1.5. Let D +, + denote the augmentation ideals in D and, respectively. Then D + + P for any >2. The domain and range of j both are still mysterious. nyhow, they seem easier to handle than the Ext group. They both are well-nown for = 1,2. Furthermore, on the one hand, (F 2 P GL is computed for = 3 by Kameo [11], lghamdi Crabb Hubbuc [3] and Boardman [4]. On the other hand, F 2 (P GL is determined by Hu ng Peterson [18] for =3and4. Let F 2 (P GL := 2 (P 0F GL and(f 2 P GL := (F 2 P GL. They 0 are equipped with canonical coalgebra structures. We get Proposition 3.1. j = j : F 2 (P GL (F 2 P GL is a homomorphism of coalgebras. Let Sq 0 : PH (BV PH (BV be Kameo s squaring operation that commutes with the Steenrod operation Sq 0 : Ext,t (F 2, F 2 Ext,2t (F 2, F 2 through the algebraic transfer Tr (see [11], [3], [4], [17]. Note that Sq 0 is completely different from the identity map. We prove Proposition 4.2. There exists a homomorphism Sq 0 : P(F 2 H (BV P (F 2 H (BV, GL GL which commutes with Kameo s Sq 0 through the homomorphism j. These two propositions lead us to two different proofs of the following theorem.

4 3896 NGUYÊ N H. V. HU NG Theorem 3.2. j =0in positive dimensions for =3. In other words, there is no spherical class in Q 0 S 0 which is detected by the triple algebraic transfer. We compute explicitly the action of Sq 0 on P (F 2 H (BV for =3and4 GL in Propositions 5.2 and 5.4. The paper contains six sections and is organized as follows. Section 2 is to prove Theorem 2.1. In Section 3, we assemble the j for 0 to get a homomorphism of coalgebras j = j. By means of this property of j we give there a proof of Theorem 3.2. Section 4 deals with the existence of the squaring operation Sq 0 on P (F 2 H (BV that leads us to an alternative proof GL for Theorem 3.2. This proof helps to explain the problem. In Section 5, we compute explicitly the action of Sq 0 on P (F 2 H (BV for 4. Finally, in Section 6 GL we state a conjecture on the Dicson algebra that concerns spherical classes. cnowledgments I express my warmest thans to Manuel Castellet and all my colleagues at the CRM (Barcelona for their hospitality and for providing me with a wonderful woring atmosphere and conditions. I am grateful to Jean Lannes and Fran Peterson for helpful discussions on the subject. Especially, I am indebted to Fran Peterson for his constant encouragement and for carefully reading my entire manuscript, maing several comments that have led to many improvements. 2. Expressing ϕ Tr in the framewor of invariant theory First, let us recall how to define the homomorphism j. We have the commutative diagram F 2 P GL j F 2 P, (P GL where the vertical arrows are the canonical projections, and j is induced by the inclusion P GL P. Obviously, p(p GL P GL.So, j factors through (F 2 P GL to give rise to j : F 2 (P GL j (1 p (F 2 P (F 2 P GL, Y = 1 for any polynomial Y D = P GL. The goal of this section is to prove the following theorem. Theorem 2.1. j = Tr ϕ. Now we prepare some data in order to prove the theorem at the end of this section. Y,

5 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3897 First we setch Lannes Zarati s wor [13] on the derived functors of the destabilization. Let D be the destabilization functor, which sends an -module M to the unstable -module D(M =M/B(M, where B(M is the submodule of M generated by all Sq i u with u M, i> u. Dis a right exact functor. Let D be its -th derived functor for 0. Suppose M 1,M 2 are -modules. Lannes and Zarati defined in [13, 2] a homomorphism : Ext r (M 1,M 2 D s (M 1 D s r (M 2, (f,z f z, as follows. Let F (M i be a free resolution of M i, i =1,2. class f Ext r (M 1,M 2 can be represented by a chain map F : F (M 1 F r (M 2 of homological degree r. We write f =[F]. If z =[Z] is represented by Z F (M 1, then by definition f z =[F(Z]. Let M be an -module. We set r = s =, M 1 =Σ M,M 2 =P M,where as before P = F 2 [x 1,...,x ], and get the homomorphism : Ext (Σ M,P M D (Σ M P M. Now we need to define the Singer element e (M Ext (Σ M,P M (see Singer [20, p. 498]. Let P 1 be the submodule of F 2 [x, x 1 ] spanned by all powers x i with i 1, where x =1. The-module structure on F 2 [x, x 1 ] extends that of P 1 = F 2 [x] (see dams [2], Wilerson [22]. The inclusion P 1 P 1 gives rise to a short exact sequence of -modules: 0 P 1 P 1 Σ 1 F 2 0. Denote by e 1 the corresponding element in Ext 1 (Σ 1 F 2,P 1. Definition 2.2. (Singer [20]. (i e = e 1 e } {{ 1 Ext } (Σ F 2,P. times (ii e (M =e M Ext (Σ M,P M, for M an -module. Here we also denote by M the identity map of M. The cap-product with e (M gives rise to the homomorphism e (M : D (Σ M D 0 (P M P M, e (M(z = e (M z. s F 2 is an unstable -module, the following theorem is a special case (but would be the most important case of the main result in [13]. Theorem 2.3 (Lannes Zarati [13]. Let D P be the Dicson algebra of variables. Then e (ΣF 2 :D (Σ 1 F 2 ΣD is an isomorphism of internal degree 0. Next, we explain in detail the definition of the Lannes Zarati homomorphism ϕ : Ext (Σ F 2, F 2 (F 2 D, which is compatible with the Hurewicz map (see [12], [13]. Let N be an -module. By definition of the functor D, we have a natural homomorphism: D(N F 2 N. Suppose F (N is a free resolution of N. Then the above natural homomorphism induces a commutative diagram

6 3898 NGUYÊ N H. V. HU NG DF (N DF 1 (N i i 1 F 2 F (N F 2 F 1 (N. Here the horizontal arrows are induced from the differential in F (N, and i [Z] =[1 Z] for Z F (N. Passing to homology, we get a homomorphism i : F 2 D (N Tor (F 2,N, 1 [Z] [1 Z]. Taing N =Σ 1 F 2, we obtain a homomorphism i : F 2 D (Σ 1 F 2 Tor (F 2,Σ 1 F 2. Note that the suspension Σ : F 2 D F 2 ΣD and the desuspension Σ 1 : Tor (F 2,Σ 1 = F 2 Tor (F 2,Σ F 2 are isomorphisms of internal degree 1 and ( 1, respectively. This leads us to Definition 2.4. (Lannes Zarati [13]. The homomorphism ϕ of internal degree 0 is the dual of ϕ =Σ 1 i (1 e (ΣF 2 Σ:F 2 D Tor (F 2,Σ F 2. Now we recall the definition of the algebraic transfer. Consider the cap-product Ext r (Σ F 2,P Tors (F 2,Σ F 2 Tors r (F 2,P, (e, z e z. Taing r = s = and e = e as in Definition 2.2, we obtain the homomorphism Tr : Tor (F 2,Σ F 2 Tor0 (F 2,P F 2 P, Tr [1 Z] = e [Z]=[1 E(Z] 1 E(Z, for Z F (Σ F 2, where e =[E] is represented by a chain map E : F (Σ F 2 F (P. Singer proved in [20] that e is GL -invariant, hence Im(Tr (F 2 P GL. This gives rise to a homomorphism, which is also denoted by Tr, Tr : Tor (F 2,Σ F 2 (F 2 P GL. Definition 2.5. (Singer [20]. The -th algebraic transfer Tr : F 2 PH (BV GL Ext,+ (F 2, F 2 is the homomorphism dual to Tr. We have finished the preparation of the needed data.

7 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3899 Proof of Theorem 2.1. Note that the usual isomorphism Ext (Σ F 2,P = Ext (Σ 1 F 2, ΣP sends e (F 2 toe (ΣF 2 =e (F 2 ΣF 2. Moreover, if e (F 2 =[E] is represented by a chain map E : F (Σ F 2 F (P thene (ΣF 2 =[E Σ ] is represented by the induced chain map E Σ : F (Σ 1 F 2 F (ΣP, which is defined by E Σ =ΣEΣ 1. By Theorem 2.3, e (ΣF 2 is an isomorphism. So, for any Y D,thereexists a representative ( of e 1 (ΣF 2ΣY, which is denoted by E 1 Σ ΣY F (Σ 1 F 2, such that E Σ E 1 Σ ΣY =ΣY. The cap-product with e (ΣF 2 =[E Σ ] induces the homomorphism Tr : Tor (F 2,Σ 1 F 2 Tor0 (F 2,ΣP F 2 ΣP, Tr [1 Z] = e (ΣF 2 [Z] =[1 E Σ (Z] 1 E Σ (Z, for Z F (Σ 1 F 2. It is easy to chec that Tr =Σ 1 Tr Σ. Moreover, set ϕ =Σϕ Σ 1 =i (1 e 1 (ΣF 2 : F 2 ΣD Tor (F 2,Σ 1 F 2. Obviously, Tr ϕ =Σ 1 Tr ϕ Σ. Now, for any Y D,wehave Tr ϕ (1 Y =Σ 1 Tr ϕ Σ(1 Y =Σ 1 Tr ϕ (1 ΣY By definition of j, we also have j (1 =Σ 1 Tr [ ] 1 E 1 Σ ΣY =Σ (1 1 E Σ (E 1 Σ ΣY =1 Σ 1 (ΣY =1 Y. Y =1 Y.The theorem is proved. 3. The homomorphism of coalgebras j = j The canonical isomorphism V = Vl V m,for=l+m, induces the usual inclusion GL GL l GL m and the usual diagonal : P P l P m. Therefore, it induces two homomorphisms ( ( (P GL D : F 2 P :(F 2 P GL F 2 (P GL l l (F 2 F 2 (Pm GLm, P l GL l (F 2 P m GLm. Here and in what follows, means the tensor product over F 2, except when otherwise specified.

8 3900 NGUYÊ N H. V. HU NG Set F 2 D = F 2 (P GL := 2 (P 0F GL, (F 2 P GL := (F 2 P GL. 0 It is easy to see that F 2 (P GL and(f 2 P GL are endowed with the structure of a cocommutative coalgebra by D and P, respectively. The coalgebra structure of (F 2 P GL was first given by Singer [20]. Proposition 3.1. j = j : F 2 (P GL (F 2 P GL is a homomorphism of coalgebras. Proof. This follows immediately from the commutative diagram F 2 D j (F2 P GL (F 2 D D l (F 2 D m j l j m (F2 P P l GL l (F 2 P m GLm. Remar. ccording to Singer [20], Tr = Tr is a homomorphism of coalgebras. One can see that ϕ = ϕ is also a homomorphism of coalgebras. Then, so is j = Tr ϕ. This is an alternative proof for Proposition 3.1. Now let F 2 (BV := GLPH ( ( F 2 PH (BV = (F 2 P GL, 0 GL 0 ( P (F 2 (BV := GLH P (F 2 H (BV = 0 GL 0 F 2 (P GL. Passing to the dual, we obtain the homomorphism of algebras j : F 2 PH (BV P(F 2 H (BV. GL GL s an application of j, we give here a proof for Conjecture 1.4 with =3. Theorem 3.2. j 3 : F 2 (P GL3 3 (F 2 P 3 GL3 is zero in positive dimensions. Proof. We equivalently show that j3 : F 2 PH (BV 3 P(F 2 H (BV 3 GL 3 GL 3 is a trivial homomorphism in positive dimensions. F 2 PH (BV 3 is described by Kameo [11], lghamdi Crabb Hubbuc [3] GL 3 and Boardman [4] as follows. F 2 PH (BV 1 has a basis consisting of h r, r GL 1 0, where h r is of dimension 2 r 1 and is sent by the isomorphism Tr 1 to the dams element, denoted also by h r, in Ext 1,2r (F 2,F 2. ccording to [11], [3],

9 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3901 [4], F 2 PH (BV 3 has a basis consisting of some products of the form h r h s h t, GL 3 where r, s, t are non-negative integers (but not all such appear, and some elements c i (i 0 with dim(c i =2 i+3 +2 i+1 +2 i 3. We will show in Lemma 3.3 that any decomposable element in P (F 2 H (BV 3 GL 3 is zero. Then, since j is a homomorphism of algebras, j3 sends any element of the form h r h s h t to zero. On the other hand, by Hu ng Peterson [18], F 2 D 3 is concentrated in the dimensions 2 s+2 4(s 0 and 2 r+2 +2 s+1 3(r>s>0. Obviously, these dimensions are different from dim(c i for any i. Thenj3 also sends c i to zero. To complete the proof of the theorem, we need to show the following lemma. Lemma 3.3. Let D = F 2 D. Then the diagonal D : D 3 D 1 D 2 D 2 D 1 is zero in positive dimensions. Proof. Let us recall some informations on the Dicson algebra D. Dicson proved in [10] that D = F2 [Q 1,Q 2,...,Q 0 ], a polynomial algebra on generators, with Q s =2 2 s. Note that Q s depends on, and when necessary, will be denoted Q,s. n inductive definition of Q,s is given by Definition 3.4. Q,s = Q 2 1,s 1 + v Q 1,s, where, by convention, Q, =1,Q,s =0fors<0and v = (λ 1 x 1 + +λ 1 x 1 +x. λ i F 2 Dicson showed in [10] that 1 v = Q 1,s x 2s. Now we turn bac to the lemma. Since D is symmetric, we need only to show that the diagonal s=0 :D 3 D 2 D 1 is zero in positive dimensions. For abbreviation, we denote x 1,x 2,x 3 by x, y, z, respectively, Q i = Q 3,i (x, y, z for i =0,1,2, q i = Q 2,i (x, y fori=0,1. s is well nown, F 2 D 1 has the basis {z 2s 1 s 0}, andf 2 D 2 has the basis {q 2s 1 1 s 0}. By Hu ng Peterson [18], F 2 D 3 has the basis {Q 2s 1 2 (s 0,Q 2r 2 s 1 2 Q 2s 1 1 Q 0 (r>s>0}. For 3, every monomial in Q 0,...,Q 1 which does not belong to the given basisiszeroinf 2 D. Note that the analogous statement is not true for 4 (see [18].

10 3902 NGUYÊ N H. V. HU NG Using the above inductive definitions of Q,s and v,weget Q 0 =q 2 0z+q 0 q 1 z 2 +q 0 z 4, or (Q 0 =q0 2 z+q 0q 1 z 2 +q 0 z 4. This implies easily that every term in (Q 2r 2 s 1 2 Q 2s 1 1 Q 0 is divisible by q 0,so it equals zero in F 2 D 2 as shown above. In other words, or Similarly, (Q 2r 2 s 1 2 Q 2s 1 1 Q 0 =0. Q 2 = q v 3 = q q 0z + q 1 z 2 + z 4, (Q 2 =q1 2 1+q 0 z+q 1 z 2 +1 z 4, (Q 2s 1 2 =(q1 1+q 2 0 z+q 1 z 2 +1 z 4 2s 1. By the same argument as above, we need only to consider terms in (Q 2s 1 2 which are not divisible by q 0. Such a term is some product of powers of q1 2 1, q 1 z 2, 1 z 4. If it contains a positive power of z then this power is even and it equals zero in F 2 D 1. Otherwise, it should be q 2(2s Obviously, q 2(2s 1 1 equals zero in F 2 D 2.So, (Q 2s 1 2 =0fors>0. In summary, = 0 in positive dimensions. The lemma is proved. Then, so is Theorem 3.2. s Tr 3 is an isomorphism (see Boardman [4], we have an immediate consequence. Corollary 3.5. ϕ 3 : Ext 3,3+i i. (F 2, F 2 (F 2 D 3 i is zero in every positive stem 4. The squaring operation: the existence Liulevicius was perhaps the first person who noted in [14] that there are squaring operations Sq i : Ext,t (F 2, F 2 Ext +i,2t (F 2, F 2, which share most of the properties with Sq i on cohomology of spaces. In particular, Sq i (α =0ifi>, Sq (α =α 2 for α Ext,t (F 2, F 2, and the Cartan formula holds for the Sq i s. However, Sq 0 is not the identity. In fact, Sq 0 : Ext,t (F 2, F 2 Ext,2t (F 2, F 2, [b 1... b ] [b b2 ], in terms of the cobar resolution (see May [16]. Recall that H (BV is a divided power algebra H (BV =Γ(a 1,...,a generated by a 1,...,a, each of degree 1, where a i is dual to x i H 1 (BV. Here and in what follows, the duality is taen with respect to the basis of H (BV consisting of all monomials in x 1,...,x.

11 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3903 Let γ t be the t-th divided power in H (BV andforanya H (BV let a (t =γ t (a. So a (t i is the element dual to x t i. One has a (2r i a (2r i =0, and a (t i = a (2r 1 i a (2rm i if t =2 r rm,0 r 1 < <r m. In [11] Kameo defined a GL -homomorphism where a (i1 1 a (i Sq 0 : PH (BV PH (BV, a (i1 1 a (i a (2i1+1 1 a (2i +1, is dual to x i1 1 xi. (See also [3]. Crabb and Hubbuc gave in [8] a definition of Sq 0 that does not depend on the chosen basis of H (BV as follows. The element a(v =a 1 a is nothing but the image of the generator of Λ (V under the (sew symmetrization map Λ (V H (BV =Γ (V =(V V } {{ } Σ. times Let F : H (BV H (BV be the Frobenius homomorphism defined by F (x = x 2 for any x, and let c : H (BV H (BV be the degree-halving dual homomorphism. It is obviously a surjective ring homomorphism. Then Sq 0 can be defined by Sq 0 (c(y = a(v y. Since y er c if and only if a(v y =0,Sq 0 is a monomorphism of GL -modules. Further, it is easy to see that csq 2i+1 =0,cSq 2i = Sqi c.sosq0 maps PH (BV to itself. Using a result of Carlisle and Wood [6] on the boundedness conjecture, Crabb and Hubbuc also noted in [8] that for any d, thereexistst 0 such that Sq 0 : PH 2t d+(2 t 1(BV PH 2 t+1 d+(2 t+1 1(BV is an isomorphism for every t t 0. Kameo s Sq 0 is shown to commute with Sq 0 on Ext (F 2,F 2 through the algebraic transfer Tr by Boardman [4] for = 3 and by Minami [17] for general. One denotes also by Sq 0 the operation Sq 0 : F 2 PH (BV F 2 PH (BV GL GL induced by Kameo s Sq 0. It preserves the product. Further, for = 3, it satisfies (see Boardman [4]. Sq 0 (h r h s h t =h r+1 h s+1 h t+1, Sq 0 (c i =c i+1 Lemma 4.1. Sq 2r+1 Sq 0 =0,Sq 2rSq0 = Sq 0 Sq r. Proof. We need a formal notation. Namely, for a H 1 (BV, set (a (t [2] = a (2t. In general, (a (t [2] γ 2 (γ t (a = ( 2t 1 t a (2t (see Cartan [7].

12 3904 NGUYÊ N H. V. HU NG We start with a simple remar. Let x H 1 (BV ; then Sq r (x s = ( s r x s+r.letadenote the dual element of x. Then, by dualizing, ( t r Sq r (a(t = a (t r. r s a consequence, Sq 2r+1 (a (2t+1 =0and ( 2t 2r Sq 2r (a(2t = Let α = a (i1 1 a (i 2r a (2t 2r = ( t r r. By the Cartan formula, we have ( a (t r [2] = ( Sq r a(t [2]. Sq Sq r 0 (α=sq (a r (2i1+1 1 a (2i +1 = Sq r1 (a(2i1+1 1 Sq r (a(2i +1. r 1+ +r =r The term corresponding to (r 1,...,r equals 0 if at least one of r 1,...,r is odd. Hence Sq 2r+1 Sq 0 (α = 0. Furthermore, Sq 2r Sq0 (α= Sq 2r1 (a (2i1+1 1 Sq 2r (a (2i +1 = = The lemma is proved. r 1+ +r =r r 1+ +r =r r 1+ +r =r { =Sq 0 Sq r (α. Sq 2r1 } (a (2i1 1 Sq 2r (a (2i a 1 a { } [2] [2] Sq r1 (a (i1 1 { Sq r (a (i } a1 a Proposition 4.2. For every positive integer, there exists a homomorphism Sq 0 : P(F 2 GL H (BV P (F 2 GL H (BV that sends an element of degree n to an element of degree 2n + and maes the following diagram commutative: F 2 GL PH (BV j P(F 2 GL H (BV F 2 PH (BV GL Sq 0 Sq 0 j P(F 2 H (BV. GL Proof. Since Sq 0 : H (BV H (BV isagl -homomorphism, we can define Sq 0 D =1 GL Sq 0 and get a commutative diagram

13 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3905 H (BV F 2 GL H (BV Sq 0 Sq 0 D H (BV F 2 GL H (BV, where the horizontal arrows are the canonical projections. Next, we show that SqD 0 sends the primitive part to itself. In other words, suppose α H (BV satisfies Sq (1 r α=1 Sq α r =0 GL GL for any r>0; we want to show that Sq r (Sq0 (1 α = 0 GL for any r>0. By definition of Sq 0 and Lemma 4.1, we have for every r>0 Sq (Sq r 0 (1 α = 1 Sq Sq r 0 (α GL GL { 1 Sq 0 (Sq r/2 (α, r even, = GL 0, r odd, { Sq 0 Sq r/2 (1 α, r even, = GL 0, r odd, =0. Therefore, the above commutative diagram gives rise to a commutative diagram PH (BV j P(F 2 GL H (BV Sq 0 Sq 0 D PH (BV j P(F 2 GL H (BV. By definition of j, the homomorphism j factors through F 2 GL PH (BV and the previous diagram induces the commutative diagram stated in the proposition, in which Sq 0 D is re-denoted by Sq0 for short. The proposition is proved. s an application of Proposition 4.2, we give an alternative proof of Theorem 3.2. By Kameo [11], lghamdi et al. [3] and Boardman [4], F 2 PH (BV 3 hasa GL 3 basis consisting of some products of the form h r h s h t and certain elements c i (i 0 with Sq 0 (c i =c i+1 for any i 0. By Lemma 3.3, j3 vanishes on any product h rh s h t. Maing use of Proposition 4.2, one has j3(c i =j3(sq 0 i (c 0 =(Sq 0 i (j3(c 0.

14 3906 NGUYÊ N H. V. HU NG One needs only to show that j3 (c 0 = 0. Recall that dim(c 0 =8. Theonlyelement of dimension 8 in D 3 is Q 2 2. Obviously, Q 2 2 = Sq 4 Q 2. So P(F 2 H (BV 3 8 = GL 3 (F 2 D 3 8 = 0. Therefore, j 3 (c 0 = 0. Theorem 3.2 is proved. 5. The squaring operation: an explicit formula for 4 Let d (i 1,...,i 0 be the dual element of Q i 1 1 Qi0 0 D, where the duality is taen with respect to the basis of D consisting of all monomials in the Dicson invariants Q 1,...,Q 0. It is well-nown that P (F 2 H (BV 1 = Span {d (2 s 1 s 0}, GL 1 P (F 2 H (BV 2 = Span {d (2s 1,0 s 0}. GL 2 By means of the definition of Sq 0 one can easily show that Sq 0 (d (2s 1 =d (2 s+1 1, Sq 0 (d (2s 1,0 =d (2 s+1 1,0. In this section we compute Sq 0 explicitly on P (F 2 H (BV for =3and4. GL Theorem 5.1 (Hu ng Peterson [18]. PD3 := P (F 2 H (BV 3 has a basis consisting GL 3 of d (2s 1,0,0, s 0, d (2r 2 s 1,2 s 1,1, r > s > 0. They are of dimensions 2 s+2 4 and 2 r+2 +2 s+1 3, respectively. Remar. It is easy to chec that PD3 has at most one non-zero element of any dimension. Proposition 5.2. Sq 0 : PD 3 PD 3 is given by Sq 0 (d (2 s 1,0,0 =0, Sq 0 (d (2 r 2 s 1,2 s 1,1 =d (2 r+1 2 s+1 1,2 s+1 1,1. Proof. For brevity, we denote x 1,x 2,x 3 by x, y, z and a 1,a 2,a 3 by a, b, c, respectively. The first part of the proposition is an immediate consequence of dimensional information. To prove the second part we start by recalling that, from Definition 3.4, we have Q 3,0 = Q 0 = x 4 y 2 z 1 + (symmetrized. Suppose m, n are non-negative integers. Let x α y β z γ be the biggest monomial in Q m 2 Q n 1 with respect to the lexicographic order on (α, β, γ. We claim that x α+4 y β+2 z γ+1 appears exactly one time in Q m 2 Qn 1 Q 0, or equivalently (a Q m 2 Q n 1 Q 0 = x α+4 y β+2 z γ+1 +(otherterms.

15 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3907 Indeed, suppose to the contrary that it appears more than once in Q m 2 Qn 1 Q 0.That means there exists a monomial x α y β z γ in Q m 2 Qn 1, which is different from xα y β z γ, and a permutation σ on the set {4, 2, 1} such that x α+4 y β+2 z γ+1 = x α +σ(4 y β +σ(2 z γ +σ(1. Since α +4 = α +σ(4 and 4 σ(4, this implies α α. Combining this with the fact that (α, β, γ is the biggest monomial in Q m 2 Q n 1 with respect to the lexicographic order on (α, β, γ, one gets α = α and σ(4 = 4. Similarly, β = β, γ = γ and σ is the identity permutation. This contradiction proves (a, or equivalently (b d (m,n,1 =1 a (α+4 b (β+2 c (γ+1 +(otherterms. Here and throughout the proof, means the tensor product over GL 3. By definition of the squaring operation (c Sq 0 (1 a (α+4 b (β+2 c (γ+1 =1 a (2α+9 b (2β+5 c (2γ+3. Now a direct computation using Definition 3.4 shows that Q 2 Q 1 Q 0 = x 12 y 4 z + x 10 y 6 z + x 10 y 5 z 2 + x 10 y 4 z 3 + x 9 y 6 z 2 + x 9 y 5 z 3 + x 8 y 6 z 3 + x 8 y 5 z 4 + (symmetrized. Note that x 9 y 5 z 3 and its symmetrized terms are the only terms of the form x odd y odd z odd in Q 2 Q 1 Q 0. On the other hand, Q 2m 2 Q2n 1 = x 2α y 2β z 2γ +(otherterms, where x 2α y 2β z 2γ is the biggest monomial in this polynomial with respect to the lexicographic order on (2α, 2β,2γ. Focusing on monomials of the form x odd y odd z odd and using the same argument as in the proof of (a, we have Q 2m+1 2 Q 2n+1 1 Q 0 = x 2α+9 y 2β+5 z 2γ+3 +(otherterms. This is equivalent to (d 1 a (2α+9 b (2β+5 c (2γ+3 = d (2m+1,2n+1,1 +(otherterms. Combining (b, (c and (d, we get Sq 0 (d (m,n,1 =d (2m+1,2n+1,1 +(otherterms. pplying this for (m, n, 1 = (2 r 2 s 1, 2 s 1, 1, we obtain Sq 0 (d (2r 2 s 1,2 s 1,1 =d (2 r+1 2 s+1 1,2 s+1 1,1 +(otherterms. In addition, Sq 0 maps PD3 to itself (by Proposition 4.2 and PD 3 consists of at most one non-zero element of any dimension. So the proposition is proved. Theorem 5.3 (Hu ng Peterson [18]. PD4 := P (F 2 H (BV 4 has a basis consisting of GL 4 d (2s 1,0,0,0, s 0, d (2 r 2 s 1,2 s 1,1,0, r > s > 0, d (2t 2 r 1,2 r 2 s 1,2 s 1,2, t>r>s>1, d (2 r 2 s+1 2 s 1,2 s 1,2 s 1,2, r > s+1>2. They are of dimensions 2 s+3 8, 2 r+3 +2 s+2 6, 2 t+3 +2 r+2 +2 s+1 4 and 2 r+3 +2 s+1 4, respectively. Remar. PD4,aswellasPD 3, has at most one non-zero element of any dimension.

16 3908 NGUYÊ N H. V. HU NG Proposition 5.4. Sq 0 : PD4 PD 4 is given by Sq 0 (d (2s 1,0,0,0 =Sq 0 (d (2r 2 s 1,2 s 1,1,0 =0, Sq 0 (d (2 t 2 r 1,2 r 2 s 1,2 s 1,2 =d (2 t+1 2 r+1 1,2 r+1 2 s+1 1,2 s+1 1,2, Sq 0 (d (2 r 2 s+1 2 s 1,2 s 1,2 s 1,2 =d (2 r+1 2 s+2 2 s+1 1,2 s+1 1,2 s+1 1,2. Proof. We denote x 1,x 2,x 3,x 4 by x, y, z, t and a 1,a 2,a 3,a 4 by a, b, c, d, respectively, for brevity. The first part of the proposition is an immediate consequence of dimensional information. We claim that Q 0 = x 8 y 4 z 2 t + (symmetrized. It can be checed by a routine computation using Definition 3.4. Here we give an alternative argument. Indeed, the Dicson algebra D 4 = F2 [Q 3,Q 2,Q 1,Q 0 ] has exactly one non-zero element of dimension 15. To chec the equality we need only to show that the right hand side is GL 4 -invariant. Recall that GL 4 is generated by the symmetric group Σ 4 and the transformation x x + y, y y, z z, t t. So, it suffices to chec that the right hand side is invariant under this transformation. We leave it to the reader. Suppose m, n, p, q are non-negative integers with q>0. Let x α y β z γ t δ be the biggest monomial in Q m 3 Qn 2 Qp 1 Qq 1 0 with respect to the lexicographic order on (α, β, γ, δ. By the same argument as in the proof of Proposition 5.2 we have Q m 3 Q n 2 Q p 1 Qq 0 = xα+8 y β+4 z γ+2 t δ+1 +(otherterms. In other words, (a d (m,n,p,q =1 a (α+8 b (β+4 c (γ+2 d (δ+1 +(otherterms. Here and throughout this proof, denotes the tensor product over GL 4. By definition of the squaring operation (b Sq 0 (1 a (α+8 b (β+4 c (γ+2 t (δ+1 =1 a (2α+17 b (2β+9 c (2γ+5 d (2δ+3. Using the same method that we used to compute Q 0 above, we can show that Q 3 Q 2 = x s1 y s2 z s3 t s4, Q 1 = s 1+s 2+s 3+s 4=20 s i=0 or a power of 2 s 1+s 2+s 3+s 4=14 s i=0 or a power of 2 x s1 y s2 z s3 t s4. In particular, we have Q 3 Q 2 =(x 16 y 2 zt + symmetrized + (other terms, Q 1 =(x 8 y 4 zt + symmetrized + (other terms. Here, in both cases, any other term is of the form x even y even z even t even.so Q 3 Q 2 Q 1 =(x 17 y 9 z 5 t 3 + symmetrized + (other terms, where x 17 y 9 z 5 t 3 and its symmetrized terms are the only terms of the form x odd y odd z odd t odd in Q 3 Q 2 Q 1. On the other hand, Q 2m 3 Q 2n 2 Q 2p 1 Q2q 2 0 = x 2α y 2β z 2γ t 2δ +(otherterms, where x 2α y 2β z 2γ t 2δ is the biggest monomial in the polynomial with respect to the lexicographic order on (2α, 2β,2γ,2δ. gain, we focus on monomials of the form

17 SPHERICL CLSSES ND THE LGEBRIC TRNSFER 3909 x odd y odd z odd t odd and use the same argument as in the proof of Proposition 5.2 to get Q 2m+1 3 Q 2n+1 2 Q 2p+1 1 Q 2q 2 0 = x 2α+17 y 2β+9 z 2γ+5 t 2δ+3 +(otherterms, or equivalently (c 1 a (2α+17 b (2β+9 c (2γ+5 d (2δ+3 = d (2m+1,2n+1,2p+1,2q 2 +(otherterms. Combining (a, (b and (c, we get Sq 0 (d (m,n,p,q =d (2m+1,2n+1,2p+1,2q 2 +(otherterms. pply this for (m, n, p, q =(2 t 2 r 1,2 r 2 s 1,2 s 1,2 and (2 r 2 s+1 2 s 1, 2 s 1, 2 s 1, 2. Combining the resulting formulas and the facts that Sq 0 maps PD4 to itself (by Proposition 4.2 and that PD 4 has at most one non-zero element of any dimension, we obtain the last two formulas of the proposition. 6. Final remar Recall that D := F 2 D.Let D :D D l D m l+m= l,m>0 be the diagonal defined at the beginning of Section 3. Conjecture 6.1. (Hu ng Peterson [19]. The diagonal D is zero in positive dimensions for any >2. This conjecture is proved in Lemma 3.3 for = 3 and has been proved for 2 <<10 in [19]. It implies that j (respectively, ϕ vanishes on the decomposable elements in F 2 PH (BV with respect to the product given by Singer [20] GL and discussed in Section 3 (respectively, in Ext (F 2,F 2 with respect to the cup product for 2 <<10. Note added in proof. Conjecture 6.1 has been established by F. Peterson and the author in the final version of [19]. References 1. J. F. dams, On the non-existence of elements of Hopf invariant one, nn. Math. 72 (1960, MR 25: J. F. dams, Operations of the nth ind in K-theory, and what we don t now about RP, New Developments in Topology, G. Segal (ed., London Math. Soc. Lect. Note Series 11 (1974, 1 9. MR 49: M.. lghamdi, M. C. Crabb and J. R. Hubbuc, Representations of the homology of BV and the Steenrod algebra I, dams Memorial Symposium on lgebraic Topology 2, N. Ray and G. Waler (ed., London Math. Soc. Lect. Note Series 176 (1992, MR 94i: J. M. Boardman, Modular representations on the homology of powers of real projective space, lgebraic Topology: Oaxtepec 1991, M. C. Tangora (ed., Contemp. Math. 146 (1993, MR 95a: W. Browder, The Kervaire invariant of a framed manifold and its generalization, nn. Math. 90 (1969, MR 40: D. P. Carlisle and R. M. W. Wood, The boundedness conjecture for the action of the Steenrod algebra on polynomials, dams Memorial Symposium on lgebraic Topology 2, N. Ray and G. Waler (ed., Lond. Math. Soc. Lect. Note Series 176 (1992, MR 95f:55015

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