ON THE EXISTENCE OF THE SELF MAP v92ON THE SMITH-TODA COMPLEX V (1) AT THE PRIME 3 MARK BEHRENS AND SATYA PEMMARAJU Abstract.Let V (1) be the Smith-Toda complex at the prime 3. We prove that there exists a map v92: 144V (1) ! V (1) that is a K(2) equivalenc* *e. This map is used to construct various v2-periodic infinite families in the 3-* *primary stable homotopy groups of spheres. Contents 1. Introduction and statement of results 1 2. The Adams spectral sequence of eo2 ^ V (1) 4 3. Calculation of the Adams E2 term ExtA*(F3, H*(V (1))) 11 4. The splitting of D(V (1)) ^ V (1) 16 5. The product rule 17 6. Selected AHSS calculations of ß*(V (1)) 21 7. The order of the fi1 action on V (1) 25 8. Proof that v92is a permanent cycle 26 9. Proof that v92extends over V (1) 32 References 40 1.Introduction and statement of results Let V (0) denote the mod 3 Moore spectrum. Let V (1) be the Smith-Toda complex obtained by taking the cofiber of the self map v1 : 4V (0) ! V (0) whi* *ch induces multiplication by v1 in K(1)-homology. This is an example of a type 2 complex. The periodicity theorem of Hopkins and Smith [10] states that there exists a v2-self map v : N V (1) ! V (1) which is a K(2)-equivalence. The purp* *ose of this paper is to provide a minimal such v2-self map. The main theorem of this paper is stated below. Theorem 1.1. There exists a self-map v92: 144V (1) ! V (1) whose effect on K(2) homology is multiplication by v92. The strategy of proving the theorem is straightforward and computational. We first prove that the element v92in the Adams spectral sequence (ASS) for comput* *ing ß*(V (1)) is a permanent cycle. We then prove that this map extends over V (1). We use the ASS instead of the Adams-Novikov spectral sequence (ANSS) because ____________ Date: March 19, 2003. 2000 Mathematics Subject Classification. Primary 55Q51; Secondary 55Q45, 55T* *15. 1 2 MARK BEHRENS AND SATYA PEMMARAJU v92is in Adams filtration 9, whereas it has Adams-Novikov filtration 0. Therefo* *re there are less potential targets for a differential supported by v92in the ASS * *than in the ANSS. The ASS E2-term is also easier to compute. Our method of showing that v92is a permanent cycle in the ASS is to consider all of the elements of the ASS which could be targets of differentials supporte* *d by v92which would not be detected in the ASS for eo2 ^ V (1). We then will show all of these potential targets either support non-trivial differentials, or are* * killed on earlier pages of the spectral sequence. This requires knowledge of the ASS of eo2 ^ V (1), as well as the E2 term of the ASS converging to ß*(V (1)). The spectrum eo2 is a connective cover of the spectrum EO2 discussed in [5]. This spectrum should be regarded as a chromatic level 2 analog to the spectrum * *bo. The spectrum eo2 is a ring spectrum, and thus there is a Hurewicz homomorphism h : V (1) ! eo2 ^ V (1). For our proof of Theorem 1.1, we need to know what the ASS for eo2 ^ V (1) looks like, and what the effect of the Hurewicz homomorphism is on Adams E2 terms, and this is accomplished in Section 2. The reader who like to avoid a digressio* *n on eo2-theory is invited to skip Section 2 and simply refer to Figure 2.2 and Prop* *osi- tion 2.5 for the relevant information. All of the methods in this section deriv* *e from unpublished work of Hopkins, Mahowald, and Miller. In retrospect, the resolution we use to compute Ext(eo2^ V (1)) should be compared to that of Ravenel [16, ch. 7]. In Section 3 we compute the E2-term of the ASS for V (1) through the 144-stem (the degree of v92). We rely on Tangora's computer generated tables [22] of H*(* *P*) where P* is the polynomial part of the dual Steenrod algebra. The periodic lamb* *da algebra [7] allows us to compute the E2 term H*(A*==E[ø0, ø1]) via a Bockstein spectral sequence (BSS). In some instances Christian Nassau's computer generated Ext tables were of welcome assistance. Differentials in the ASS are computed by using the Hurewicz image in the homo- topy of the spectrum eo2. In addition, we will prove a modified `Leibnetz rule'* * for differentials in the ASS for V(1). This product rule is our main tool for calcu* *lating difficult differentials in the ASS for V (1). The product rule is presented in* * Sec- tion 5. It is a generalization of a formula for Adams d2's that was communicated to us by Brayton Gray. Such technology is essential because V (1) is not a ring spectrum, so its ASS is not a spectral sequence of algebras. However, the S-mod* *ule structure of V (1) does make the ASS a spectral sequence of modules over the ASS for computing ß*(S), and this is used occasionally to propagate differentials. We make heavy use of the computation of the 3-primary stable stems through the 108 stem presented in [16]. We use these computations as input for the Atiy* *ah- Hirzebruch spectral sequence (AHSS) to make selective computations of ß*(V (1)) in certain ranges. These computations are described in Section 6 Section 8 is devoted to proving that v92is a permanent cycle in the ASS for V* * (1), and therefore detects an element of ß*(V (1)). It now remains to extend v92over V (1). This is simplified by a certain split* *ting of the complex D(V (1)) ^ V (1), where D(V (1)) is the Spanier-Whitehead dual. This splitting is the subject of Section 4. The splitting is also needed to pro* *ve the product rule. Using the attaching map structure of one of the wedge summands, the obstruc- tion to the extension of v92over V (1) is identified as an element of ß*(V (1))* *. In ON THE EXISTENCE OF v92 3 showing that this obstruction is zero, it is helpful to know what power fik1has* * the property that fik1: 10kV (1) ! V (1) is null. In Section 7, we show that the m* *ap is null precisely when k 5. In Section 9, we proceed by considering all of the elements in the ASS that m* *ight survive to the obstruction to extending v92over V (1), and show that they all e* *ither support differentials, or are the targets of differentials. Thus the map v92ext* *ends to a self map of V (1), completing our proof of Theorem 1.1. We will now indicate the construction of some v2-periodic elements of the sta* *ble stems which arise from the self-map v92. The authors learned of these construct* *ions from Katsumi Shimomura (compare with [18]). Theorem 1.1 allows us to deduce that certain elements of the ANSS for the sphere must be permanent cycles. In particular, we have the following consequence. Corollary 1.2. The elements fiiare permanent cycles in the ANSS for i 0, 1, 2* *, 5, 6 (mod 9). Proof.For dimensional reasons, v2 is a permanent cycle of the ANSS which detects a map v2 : S16 ! V (1). By [15], or Remark 8.2, the element v52in the ANSS is a permanent cycle which detects a map v52: S80 ! V (1). We denote the Spanier-Whitehead dual of v2 by v*2: 16DV (1) = 10V (1) ! S0. Let : V (1) ! S6 be projection onto the top cell. The elements fiiin Corollar* *y 1.2 are constructed by the following compositions. 9t fi9t: S144t-6,! 144t-6V (1) v2--! -6V (1) -!S0 9t fi9t+1: S144t+10v2-! 144t-6V (1) v2--! -6V (1) -!S0 9t v* fi9t+2: S144t+26v2-! 144t+10V (1) v2--! 10V (1) -2!S0 5 v9t fi9t+5: S144t+74v2-! 144t-6V (1) -2-! -6V (1) -!S0 5 v9t v* fi9t+6: S144t+90v2-! 144t+10V (1) -2-! 10V (1) -2!S0 It should be the case that the elements fi9t+3exist, but we are unable to ded* *uce this from the existence of our self map on V (1). Oka [15] indicates that if t* *he complex M(3, v31) has a v92self map then the elements fi9t+3exist. Shimomura's computations of ß*(L2V (0)) [20] demonstrate that the elements fi9t+3are present in ß*(L2(S0)). Shimomura [19] proves that fiicannot be a permanent cycle for i 4, 7, 8 (mod 9). In [18], many relations amongst the fii's are investigated co* *ntingent on the existence of the self-map constructed in this paper. In particular, Shim* *omura proves that if the elements fiifi41are permanent cycles for i 1 (mod 9), the* *n they are non-trivial. These elements are permanent cycles by Corollary 1.2. They sho* *uld be regarded as the substitutes for the the Adams-Novikov elements fi9t+4, which fail to exist. 4 MARK BEHRENS AND SATYA PEMMARAJU Some remarks as to how this paper came to be written are in order. The main result of this paper was the subject of the second author's dissertation comple* *ted at Northwestern University under the direction of Mark Mahowald. The first author required the result for his dissertation work at the University of Chicago unde* *r the direction of J. Peter May. Certain errors and gaps in the original work needed * *to be corrected. In the original thesis, the second author's main technique for obtai* *ning differentials in the ASS was to lift differentials from the ANSS using a techni* *cal lemma called the `ladder lemma'. We were unable to make the proof of this lemma rigorous, and so the product rule (Theorem 5.1) is used instead for many of the differentials. The authors would like to thank Mark Mahowald for his constant encouragement and assistance in this project. We also thank Katsumi Shimomura, for his useful correspondence concerning the construction of the fii's from the self map. The first author would also like to thank his advisor, J. Peter May, for many useful conversations related to this project. We are also appreciative of many useful comments made by the referee, including pointing out a substantially simpler pr* *oof of Lemma 8.12. Conventions. Throughout this paper we shall always be working in the stable homotopy category localized at the prime 3, and all homology will be with F3 coefficients. We shall use the following abbreviations. ASS: Adams spectral sequence ANSS: Adams-Novikov spectral sequence BSS: Bockstein spectral sequence AHSS: Atiyah-Hirzebruch spectral sequences The dual Steenrod algebra will be denoted by A*. If X is a spectrum, we will of* *ten use the notation Ext(X) to represent ExtA*(F3, H*(X)), the E2 term of the ASS for computing ß*(X). We will denote the Er term of this ASS by Er(X). We shall use the notation .=to indicate two quantities are equal up to multiplication by* * a unit in F3. Finally, in Section 3 we give many elements in H*(P*) names which are derived from their May spectral sequence names. In all but one case, the sign of the element corresponding to a name coincides with the element whose Curtis algorit* *hm representative has a leading term with a positive sign. The one exception we ma* *ke is the element called k0 in bidegree (2, 20). We work under the sign convention that k0 is detected by the lambda algebra element -~4~3. The reason we make this exception is so that certain relations (Equation 3.1) are more uniform. 2. The Adams spectral sequence of eo2 ^ V (1) In this section we will define eo2 ^ V (1) and compute its ASS. The method of computing the E2 term of the ASS is to produce a finite complex Y (2) such that upon smashing it with eo2 ^ V (1) we get a wedge of k(2)'s. We know the ASS of this object, and we recover the E2 term of the ASS for ß*(eo2 ^ V (1)) by formi* *ng a periodic resolution of S0 out of copies of Y (2). There is nothing original i* *n this section. Most of the material here was first discovered by Hopkins, Mahowald, a* *nd Miller, but remains unpublished. ON THE EXISTENCE OF v92 5 Let E2 be the Hopkins-Miller spectrum at p = 3. It represents a Landweber exact cohomology theory whose coefficient ring is E2*= WF9[[u1]][u, u-1] where |u1| = 0 and |u| = -2. Here WF9 is the Witt ring with residue field F9. F* *ix a primitive 8throot of unity __!in F9. We will refer to it's Teichmüller lift i* *n WF9 also as !. The element __!satisfies the relation __!2+ __!+ 2 = 0 in F9. The spectrum E2 is a BP -ring spectrum, and the map : BP ! E2 has the following effect on coefficient rings. (v1) = u-2u1 (v2) = u-8 (vi)= 0, for i > 2 Let S2 be the Morava stabilizer group. It is the automorphism group of the Honda height 2 formal group law F2 over F9, and is contained in the non-commutative algebra WF9=(S2 = p, Sa = oe(a)S) as the multiplicative group of units. Here oe is the a lift of the Frobenius m* *ap. The Galois group Gal = Gal(F9=F3) acts on S2 by acting on WF9. It is cyclic of order 2 generated by the Frobenius automorphism oe. One may form the semi-direct product G2 = S2 o Gal. The spectrum E2 is an E1 ring spectrum, and the group G2 acts on E2 via E1 maps. The spectrum E2 and the A1 action of S2 are presented in [17]. There is a maximal finite subgroup G12< S2 of order 12 which is isomorphic to C3 o C4. It is generated by an element s of order 3, and an element t of order 4, given by * *the following formulas. s= -1_2(1 + !S) t= !2 These elements, as well as specific formulas for their action on E2*, are given* * in [5]. The subgroup G12 is not invariant under the Galois action, so following [* *6] we instead investigate a maximal finite subgroup G24 < G2 (of order 24) which contains G12 and fits into the following (non-split) short exact sequence. 1 ! G12! G24! Gal ! 1 The subgroup G24 is generated by the elements s, t, and _, where we define _ = !oe 2 G2. The spectrum EO2 is defined to be the homotopy fixed point spectrum EhG242. A complete computation of the homotopy of EO2, and its ANSS, is given in [6]. In [5], Goerss, Henn, and Mahowald compute the ANSS for EO2 *(V (1)), but their approach must be modified since they use G12 instead of G24. Their results may be conveniently summarized in Figure 2.1. In this figure, dots represent additi* *ve F3 generators, lines of length 3 represent multiplication by ff1, lines of leng* *th 7 6 MARK BEHRENS AND SATYA PEMMARAJU Figure 2.1. EO2 *(V (1)) is generated as a free module over F3[v29=2] on the above pattern. represent the Toda Bracket , and lines of length 10 represent mult* *iplica- tion by fi1. These products are given by the S-module structure. The homotopy is periodic with periodicity generator v29=2 of degree 72 on the displayed pattern. We want an Adams spectral sequence, but unfortunately the v2-periodicity in EO2 makes its homology trivial. We therefore need to take a connective cover. There is a nice connective cover of EO2 called eo2 which has been constructed by Hopkins, Mahowald, and others. Since the details of this construction are qu* *ite involved, we will instead define eo2^V (1) to be the connective cover of EO2^V * *(1). Since there is a gap in the homotopy of EO2 ^ V (1) between the 56 stem and the 72 stem (and hence by the periodicity of EO2's homotopy groups there is a gap between the -16 stem and 0 stem), taking the connective cover removes the periodic copies of the homotopy in negative dimensions. We remark that the reason that we cannot just define eo2 to be the connective cover of EO2 is that* * the there are infinitely many copies of BP <1> in the homotopy supported on negative periodicity generators whose unwanted homotopy eventually appears in positive degrees. Smashing with V (1) kills all of this troublesome v1-periodic homotopy. We will now produce a finite complex Y (2) which, when smashed with EO2, splits as a wedge of Morava K-theories. Let fl : S4 ! BO be a generator of ß4(BO) = Z. The map fl extends to a loop map fl1 : S5 ! BO. Let Ji(S4) ,! S5 be the ithfiltration of the James construction [11]. Then fl restricts to a* * map fli: Ji(S4) ! BO. Let Y (i) be the Thom spectrum (Ji(S4))fli. Then the homology of the ring spectrum Y (1) is given by H*(Y (1)) = F3[b2] where b2 has degree 4. The homology H*(Y (i)) is the additive subgroup generated by bk2for 0 k i. There are maps Y (i)^Y (j) ! Y (i+j) induced from the maps Ji(S4)xJj(S4) ! Ji+j(S4). The complex Y (1) is just the Thom spectrum (S4)fl, and it is a two c* *ell complex whose top cell is attached to the bottom cell by the attaching map which is the image of fl under the J-homomorphism. Therefore, Y (1) = S0 [ff1e4. It follows that the dual Steenrod operation P*1acts on H*(Y (1)) by the formula P*1(b2) = 1. ON THE EXISTENCE OF v92 7 Using the map Y (1) ^ Y (1) ! Y (2), we obtain the following formulas for the d* *ual action of the Steenrod algebra on H*(Y (2)). P*1(b2)= 1 P*1(b22)= 2b2 P*2(b22)= 1 In particular, we have the CW decomposition Y (2) = S0 [ff1e4 [2ff1e8. Our interest in Y (2) arises from the following proposition. Proposition 2.1. There is a splitting EO2 ^ V (1) ^ Y (2) ' K(2) _ 8K(2). Proof.One can easily compute ß*(EO2 ^ V (1) ^ Y (1)) from the AHSS arising from the cellular filtration of Y (1), but the associated graded arising from t* *his filtration gives too much ambiguity for our purposes. We therefore will use ins* *tead the homotopy fixed point spectral sequence H*(G24; ß*(E2 ^ V (1) ^ Y (1))) ) ß*(EO2 ^ V (1) ^ Y (1)) where ß*(E2 ^ V (1) ^ Y (1)) = F9[u, u-1, b2]. In [6], the action of G24 on E2*(V (1)) is given by the following formulas. s*(u)= u t*(u)= __!2u _*(u) = __!u The elements s, t 2 G12 correspond to the automorphisms s(x)= x +F2 u-2__!x3 = x + u-2__!x3 + O(x4) t(x)= __!2x of the Honda height 2 formal group F2 over F9[u, u-1], with 3-series [3]F2 = u-* *8x9. Under the canonical map of Thom spectra Y (1) ! MU, b2 maps to the element of the same name in (E2 ^ V (1))*(MU) = F9[u, u-1][b1, b2, b3, . .]. where the generators bi coincide with those of Adams in [1, II.4.5]. These bi c* *orre- spond to the coefficient of xi+1 in a strict map of formal groups, and as such,* * we have s*(b2)= b2 + u-2__! t*(b2)= b2 _*(b2)= b2. Therefore, the fixed points are given by ß*(E2 ^ V (1) ^ Y (1))G24= F3[(__!2u-4) 1, b32- __!2b2u-4] F9[u 1, b2]. Define a4 = __!2u-4 and a6 = b32-__!2b2u-4. In [5, 1.4], the E2 term of the hom* *otopy fixed point spectral sequence for EO2 ^ V (1) is computed to be H*(G24; ß*(E2 ^ V (1))) = F3[a41, fi] E[ff] 8 MARK BEHRENS AND SATYA PEMMARAJU where fi = . The cellular filtration of Y (1) gives an Atiyah-Hirz* *ebruch type spectral sequence that allows one to compute H*(G24; ß*(E2^ V (1) ^ Y (1))) from this. The d4's in this spectral sequence are multiplication by ff, and the* * d8's are given by the application of the Massey product . Thus we conclu* *de that ( G24,s = 0 Hs(G24; ß*(E2 ^ V (1) ^ Y (1))) = ß*(E2 ^ V (1) ^ Y (1)) 0, s > 0 and ß*(EO2 ^ V (1) ^ Y (1)) = F3[a41, a6]. The spectrum EO2 ^ Y (1) is a ring spectrum whose homotopy is concentrated in Adams-Novikov filtration 0, and since the obstruction for V (1) to be a ring sp* *ectrum lies in positive Adams-Novikov filtration (4.1), EO2 ^ V (1) ^ Y (1) is also a * *ring spectrum. Its homotopy is concentrated in even degrees, so it is complex-orient* *able [1]. The complex orientation ` : BP ! E2 ^ V (1) ^ Y (1) for which `*(vi) = 0 for i 6= 2 and `*(v2) = u-8 lifts to a complex orientation e`: BP ! EO2 ^ V (1) ^ Y (1). Here the effect on homotopy is given by e`*(vi) = 0 and e`*(v2) = -a24. There a* *re maps (for ffl = 0, 1) affl4aj6^OE S12j+8ffl^ BP -----! EO2 ^ V (1) ^ Y (1) that extend to maps _12j+8ffl: 12j+8fflK(2) ! EO2 ^ V (1) ^ Y (1). These maps give a splitting ` EO2 ^ V (1) ^ Y (1) = 12j K(2) _ 8K(2) . j 0 The composite EO2 ^ V (1) ^ Y (2) ! EO2 ^ V (1) ^ Y (1) ! K(2) _ 8K(2) (the second arrow is projection onto the first two wedge summands) is an equiva- lence. Corollary 2.2. There is a splitting eo2 ^ V (1) ^ Y (2) ' k(2) _ 8k(2). Proof.The spectrum k(2) _ 8k(2) is the connective cover of K(2) _ 8K(2). The Atiyah-Hirzebruch spectral sequence for (eo2 ^ V (1))*(Y (2)) is easily compute* *d, and one finds ß*(eo2 ^ V (1) ^ Y (2)) = F3[a4]. Therefore, eo2 ^ V (1) ^ Y (2) is the connective cover of EO2 ^ V (1) ^ Y (2). * *The previous proposition and the uniqueness of the connective cover combine to give this corollary. Remark 2.3. Hopkins and Miller, in [9], prove the following stronger result. eo2 ^ Y (2) ' BP <2>^3_ 8BP <2>^3 ON THE EXISTENCE OF v92 9 We will now construct a resolution of the sphere spectrum out of suspensions * *of Y (2). There are cofiber sequences S0 ! Y (2) ! 4Y (1) Y (1) ! Y (2) ! S8 where the first maps are the evident inclusions. Splicing these together gives * *the following 2-periodic resolution of S0. S0 oo____ 3Y (1)oo_____S10 oo_____ 13Y (1)oo_____S20oo_____ . . . | | | | | | | | | | fflffl| fflffl| fflffl| fflffl| fflffl| Y (2) 3Y (2) 10Y (2) 13Y (2) 20Y (2) The homology long exact sequences associated to this resolution break up into s* *hort exact sequences as a result of the following lemma. Lemma 2.4. The maps H*(eo2 ^ V (1)) ! H*(eo2 ^ V (1) ^ Y (2)) H*(eo2 ^ V (1) ^ Y (1)) ! H*(eo2 ^ V (1) ^ Y (2)) are injective. Proof.The natural map of Thom spectra Y (1) ! MU makes MU a Y (1)-ring spectrum, and therefore the Eilenberg-MacLane spectrum HF3 = H is a Y (1)-ring spectrum. Thus there is a retraction 1^j ~ H _____//__________________________77__________________________* *_________________________________H/^/Y_(2)H/^/Y_(1)H __________________________________________________________* *_________________________________________________ ______________________________________________________* *______________________________________________________________@ _____________________________________________* *________________________________________________________ Id and we may conclude that H*(eo2^ V (1)) ! H*(eo2^ V (1) ^ Y (2)) is an inclusio* *n. The complex Y (2) is, up to suspension, Spanier-Whitehead self-dual, and so we also have that the projection map H*(eo2 ^ V (1) ^ Y (2)) ! H*( 8eo2 ^ V (1)) is surjective, hence, the previous map in the cofiber sequence H*(eo2 ^ V (1) ^ Y (1)) ! H*(eo2 ^ V (1) ^ Y (2)) must be injective. We may therefore apply ExtA*(F3, H*(eo2 ^ V (1) ^ -)) to this resolution, and get long exact sequences, hence a spectral sequence. Our spectral sequence takes the form (for ffl = 0, 1) (2.1)E2k+ffl,s,t1= Exts,t(eo2^V (1)^ 10k+3fflY (2)) ) Exts+2k+ffl,t+2k+ffl(eo2^* *V (1)) Applying Corollary 2.2, and using the known computation Ext(k(2)) = F3[v2] where |v2| = (1, 17), we may express the E1 term of 2.1 by E*,*,*1= F3[v2, fi] E[ff, a]. The tridegrees of these elements are |v2| = (0, 1, 17), |a| = (0, 0, 8), |fi| =* * (2, 0, 10), and |ff| = (1, 0, 3). The only possible differentials are d1(fffiivj2a) .=fii+1vj2 10 MARK BEHRENS AND SATYA PEMMARAJU Figure 2.2. The ASS for ß*(eo2 ^ V (1)) is a free module over F3[v92] on the displayed pattern. but this d1 arises from the composite Y (2) ! S8 ! 8Y (2). The element vj2a 2 Ext(eo2^V (1)^Y (2)) is born on the zero cell of Y (2), and so must map to zero* * when projected onto the 8-cell of Y (2). Therefore, the spectral sequence 2.1 collap* *ses at E1, and we are left with a computation of the E2-term of the ASS for ß*(eo2^V (* *1)). The differentials in the ASS are easily inferred from the differentials in th* *e ANSS computed in [5]. Figure 2.2 displays the complete ASS chart. In this chart, F* *3- generators are represented by dots, ff multiplication is displayed with solid l* *ines, and ON THE EXISTENCE OF v92 11 the Massey product is displayed with dotted lines. Solid lines of n* *egative slope represent Adams differentials. The x-axis represents the t - s degree, an* *d the y-axis represents the homological degree s. We remark that the ASS for eo2^ V (1) is additively identical to the ANSS. The only difference is that v2 has Adams filtration 1, whereas it has Adams-Novikov filtration 0. We finish this section with a computation of the effect of the eo2 Hurewicz homomorphism on the ASS. We will use the resolution of S0 by the ring spectrum Y (1). Mahowald, in [12], investigates a geometric Thom isomorphism Y (1) ^ Y (1) ' Y (1) ^ ( S5+) under which we may make the identification ß*(eo2 ^ V (1) ^ Y (1) ^ Y (1)) = F3[a4, a6, r] Here r has degree 4. We may regard F3[a4, a6, r] as being contained in F3[u-1, * *b2] F3[b2] where a4 and a6 are contained in the first factor as described earlier, * *and the element r corresponds to 1 b2. The main result of [12] states that the right * *unit of the associated Hopf algebroid (F3[a4, a6], F3[a4, a6, r]) is given by the following formulas where b2 maps to b2 1 + 1 b2. ß*(eo2 ^ V (1) ^ Y (1))e//_ß*(eo2 ^ V (1) ^ Y (1) ^ Y (1)) || || F3[a4, a6] F3[a4, a6, r] e(a4) = (__!2u-4)= __!2u-4 1 = a4 e (a6) = (b32- __!2b2u-4)= (b32- __!2u-4b2) 1 - __!2u-4 b2 + 1 b32 = a6 - a4r + r3 The Hurewicz image of h0 is represented by r, and the Hurewicz image of h1 is represented by r3. The d1 supported by a6 identifies the Hurewicz image of h1 w* *ith h0a4 = ff . a. This observation may be used to prove the following proposition. Proposition 2.5. The Hurewicz homomorphism h : Ext(V (1)) ! Ext(eo2 ^ V (1)) is described by h(h0)= ff h(b0)= fi h(v2)= v2 h(h1)= ff1a h(g0)= fia 3.Calculation of the Adams E2 term ExtA*(F3, H*(V (1))) The ASS for computing ß*(V (1)) has as its E2 term ExtA*(F3, H*(V (1))) As a comodule over the dual Steenrod algebra, we have H*(V (1)) = E[ø0, ø1]. 12 MARK BEHRENS AND SATYA PEMMARAJU Figure 3.1. ExtP*(F3, F3), from Tangora's tables. ON THE EXISTENCE OF v92 13 Figure 3.2. The BSS ExtA*==E[ø0,ø1,ø2](F3, F3) P [v2] ) ExtA*==E[ø0,ø1](F3, F* *3). 14 MARK BEHRENS AND SATYA PEMMARAJU Figure 3.3. The E2 term ExtA*==E[ø0,ø1](F3.F3) of the ASS for computing ß*(V (1)). ON THE EXISTENCE OF v92 15 This is a subalgebra of the Steenrod algebra, but V (1) is not a ring spectrum,* * so this algebra structure is not the consequence of a geometric multiplication. Fo* *r the purposes of computing Ext, though, we may use the algebra structure. A change of rings isomorphism identifies the E2 term of the ASS as the cohomology of a Hopf algebra. ExtA*(F3, E[ø0, ø1]) ~=ExtA*==E[ø0,ø1](F3, F3) = H*(A*==E[ø0, ø1]). We may identify A*==E[ø0, ø1] = P [,1, ,2, . .]. E[ø2, ø3, . .].. The cohomology of this Hopf algebra is the cohomology of the subalgebra of the periodic lambda algebra (see [7] for a description of the periodic lambda algeb* *ra) given by __ __ (1)= <~i, vj : i 0, j 2> . We only need the E2 term of the Adams spectral sequence through the dimension_of v92, which is 144. Since the dimension of v4 is 160, the cohomology of (2)coin* *cides with H*(P*) P [v3] in the range we are interested in. Figure 3.1 displays H*(P*). It was produced from Tangora's Curtis tables in [* *22]. The y axis is the homological degree s, and the x-axis is t-s, where t is the i* *nternal degree. Solid lines of different slopes indicate multiplication by h0 and h1. D* *otted lines indicate the Massey product <-, h0, h0>. The element b0 = , s* *o b0 multiplication can be read off as composites of h0 multiplication and applicati* *on of the above Massey product. Rectangles indicate that a generator supports a polynomial algebra on b0, that is, all multiples of b0 on the generator are non* *-zero in the calculated range, but they have not been written down in an effort to ma* *ke the chart less cluttered. The Massey product representatives are the ones produ* *ced by using the full tags produced by the Curtis algorithm. In Figure 3.1, certain generators have been given names. In our summary of conventions at the end of Section 1, we indicated that the signs of these eleme* *nts will be chosen (with the exception of k0) so that the leading term of the correspond* *ing element of the lambda algebra has coefficient +1. The following table summarizes this choice of signs for some of the low dimensional generators, by comparing o* *ur name, the lambda algebra name, and a Massey product representation. _________________________________________ _Generator_|Lambda_Name_|_Massey_Product__ _____b0____|_~2~1_+_~1~2__|__ _____g0____|____~2~3_____|___ _____k0____|___-~4~3_____|___ _____b1____|_~6~3_+_~3~6__|__ There are certain relations which may be read off of Figure 3.1 up to sign. We indicate the proper sign of some of the low dimensional relations. By choosing * *the sign of k0 as we have, these relations look more uniform. h1b0= h0g0 (3.1) h1g0 = h0k0 h1k0 = h0b1 16 MARK BEHRENS AND SATYA PEMMARAJU Figure 3.2 is a chart of the BSS ExtA*==E[ø0,ø1,ø2](F3, F3) P [v2] ) ExtA*==E[ø0,ø1](F3, F3) It is straightforward to calculate the differentials of this spectral sequence * *com-_ pletely by explicitly finding the differentials in the periodic lambda algebra * * (1) and then finding the representatives in the Curtis table. In Figure 3.2, the E1-term consists of H*(A*==E[ø0, ø1, ø2]) P [v2] which is isomorphic to H*(P*) P [v2, v3] in our range of computation. It is implicit * *in the chart that every generator supports a P [v2], but all v2 multiples are omit* *ted unless they are targets of differentials, or otherwise contribute to hidden ext* *ensions. When v2 multiples are displayed, they are represented by dash-dot lines. Hidden extensions are represented by dashed lines, and differentials are represented by negatively sloped solid lines. We now have computed the E2 term of the ASS. It is displayed in Figure 3.3. Unlike in Figure 3.2, in this chart v2 is not implicit unless specifically indi* *cated. Small solid dots on the chart represent, like all of the previous charts, F3-ba* *sis elements. Small circles represent polynomial algebras on v2. Otherwise v2 multi- plication is represented explicitly by dash-doted lines. It should be noted tha* *t if an element is represented by a circle, it does not mean that that element suppo* *rts infinitely many non-trivial multiplications by v2. It just means that throughou* *t the indicated range, all multiplications by v2 are non-trivial. 4. The splitting of D(V (1)) ^ V (1) The complex V (1) may be visualized with the following cell diagram. 0 1 _________5____6_________________________________* *______________________________________________________________@ _________________________________________________* *_____________________ O ___O________ O ___O Here the uncurved lines represent the Bockstein fi (attaching map .3) and the curved line represents the Steenrod operation P 1(attaching map ff1). The top V (0) is attached to the bottom V (0) by v1, but this is not explicitly indicat* *ed in the cell diagram. Let D(V (1)) ' -6V (1) be the Spanier-Whitehead dual of V (1* *). In this section we will decompose D(V (1)) ^ V (1) into irreducible subcomplexe* *s. Since V (1) is self dual, we will have also provided a splitting of V (1) ^ V (* *1). Define finite complexes Y1 and Y2 as follows. i fi j Y1= cofiber -1V (1) -! 4V (0) -!1 -6V (1) i ff j Y2= cofiber -2V (1) -!1 -5V (1) Here is projection onto the top V (0). Figure 4.1 displays cell diagrams of t* *hese complexes. We will prove the following Proposition 4.1. There is a splitting D(V (1)) ^ V (1) ' Y1 _ Y2. Proof.Since D(V (1)) ^ V (1) ' -6V (1) ^ V (1), it suffices to split the latte* *r. Consider the twist map ø : V (1) ^ V (1) ! V (1) ^ V (1) ON THE EXISTENCE OF v92 17 Figure 4.1. The irreducible subcomplexes of D(V (1)) ^ V (1) The map ø decomposes V (1) ^ V (1) into +1 and -1 eigenspaces T+ and T- . We claim that these are 6Y1 and 6Y2, respectively. More precisely, the self-maps (1 + ø)=2 and (1 - ø)=2 are idempotents on V (1) ^ V (1), and thus give a split* *ting V (1) ^ V (1) ' T+ _ T- . The structure of T- is straightforward from the action of the Steenrod algebra. To prove the existence of the fi1 attaching map in T+ , we will use the seconda* *ry cohomology operation OE corresponding to the Adem relation P 2P 1= 0. This secondary operation detects fi1. In [23], Thomas proves the Cartan formula OE(xy) = OE(x)y + xOE(y) + (P 1fi(x))(fiP 1fi(y)) + (fiP 1fi(x))(P 1fi(y)* *). Let ei denote the generator of H*(V (1)) in dimension i. Evaluating OE on e0 ^ * *e0, we get OE(e0 ^ e0) = 0 ^ e0 + e0 ^ 0 + e5 ^ e6 + e6 ^ e5 = e5 ^ e6 + e6 ^ e5. Similarly, we see OE(e1 ^ e0 + e0 ^ e1) = -e6 ^ e6. 5. The product rule The statement of the product rule will require_some notation related to Adams resolutions which we will give presently. Let H be the fiber of the unit j : S0* * ! H, where H is the Eilenberg-MacLane spectrum_HF3. The standard Adams reso- lution is defined by letting Ws = H (s)and then defining Ws,rto be the cofiber __(s)__(s+r) __(s) __(s) H =H . In particular, Ws,1 = Ws = H and Ws,1= H ^ H . For a 18 MARK BEHRENS AND SATYA PEMMARAJU spectrum X the resolution may be written as X oo_______W1 ^ X oo____ W2 ^ X oo____. . . | | | | | | fflffl| |fflffl fflffl| W0,1^ X W1,1^ X W2,1^ X The ASS for ß*(Ws,r^ X) is a truncated version of the ASS for ß*(X); it is the spectral sequence obtained by only including the Ei,j1terms for s i < s + r, * *and omitting differentials supported on Ei,jrfor i < s. There are several important maps relating the spectra Ws,r. fs,r,k: Ws+r,k! Ws,r+k gs,r,k: Ws,r+k! Ws,r @s,r,k: Ws,r! Ws+r,k ~s1,s2,r: Ws1,r^ Ws2,r! Ws1+s2,r ~ is induced by the product on E. The remaining maps are compatible in all of t* *he ways one might expect them to be, and the sequence Ws+r,kf-!Ws,r+kg-!Ws,r@-! Ws+r,k is a cofiber sequence. It is easy to see that an element x 2 ß*(Ws,1^ X) persists to the Er term of * *the ASS if and only if it lifts to an element ~x2 ß*(Ws,r^ X). In fact, we have Es,tr= _________Im_{ßt-s(Ws,r^_X)_!_ßt-s(Ws,1^_X)}_________@ Im {ßt-s+1(Ws-r,r^ X) -!ßt-s(Ws ^ X) ! ßt-s(Ws,1^ X)} and dr(x) is computed as @(~x) 2 ß*(Ws+r,1^ X). Define, for xi in ß*(V (1)), ___ |x |__ __ F (x1, x2) = __x1. x2+ (-1) 1x1. x22 ß*(V (0)) * * ___ where __xiis the image of xi in ß*(V (0)) under the projection V (1) ! 5V (0),* * xiis the image of __xiin ß*(S) under the projection V (0) ! S1. If xi2 ß*(Wsi,r^ V (* *1)), then F (x1, x2) will be regarded as an element of ß*(Ws1+s2,r^ V (0)). Theorem 5.1 (Product Rule). Suppose xi2 E1(V (1)) persist to the Er-term of the ASS and fi1. F (x1, x2) = 0, thought of an element of ß*(Ws+2,r-2^ V (1)) (* *the product is induced from the V (0)-module structure of V (1)). Then it follows t* *hat fi1 . F (x1, x2) 2 ß*(Ws+2,r-1^ V (1)) lifts to an element G(x1, x2) in ß*(Ws+r* *,1^ V (1)) and we have the following formula for dr(x1x2). dr(x1x2) = (drx1) . x2 + (-1)|x1|x1 . (drx2) - G(x1, x2) Example 5.2. We will use the product rule to compute d3(v22). The element_v2 is* * a permanent cycle for dimensional reasons. We have __v2= ffi1= h1 and __v2= fi1 =* * b0. Here we have given the ASS names of these elements. Therefore, G(v2, v2) = (h1b0 + b0h1)b0 = -h1b20 so the product rule says that d3(v22) = d3(v2)v2 + v2d3(v2) - G(v2, v2) = h1b20. ON THE EXISTENCE OF v92 19 One can also use the Hurewicz homomorphism V (1) ! eo2 ^ V (1) to get this formula. Proof of product rule.Let X be the cofiber of fi1 : 10V (0) ! V (1). The close* *st substitute for a product on V (1) is the map ~ : V (1) ^ V (1) ! X formed by projecting onto the wedge summand 6Y1 (Proposition 4.1) and collaps- ing out the bottom two cells of the top V (1). If yi are elements of ß*(V (1)),* * then the image of y1 ^ y2 2 ß*(V (1) ^ V (1)) under the composition V (1) ^ V (1) ! X ! 11V (0) is F (y1, y2). We shall need various filtered forms of X. Define Xs,r= Ws,r^ X, and define eXs,rand eeXs,rto be the following cofibers. 10Ws,r^ V (0) fi1-!Ws+2,r-2^ V (1) ! eXs,r 10Ws,r^ V (0) fi1-!Ws+2,r-1^ V (1) ! eeXs,r Note that the maps fi1 above may be chosen to raise the s-index by 2 because th* *ey have Adams filtration 2. Then we have the following cofiber sequences (by Verdi* *er's axiom). eXs,r! Xs,r! Ws,2^ V (1) Xee e s,r! Xs,r! Ws+r,1^ V (1) Xes,r+1! eXs,r! Xs+r,1 Xes,r+1! eeXs,r! 12Ws+r,1^ V (0) Since the Adams filtration of fi1 is greater than 0, there is an equivalence * *H^X ' H ^ V (1) _ H ^ 11V (0), thus Xs,1splits in a similar manner. We need a splitt* *ing map that behaves well with respect to the other maps floating about. Consider t* *he splitting j induced on the cofibers below (the rows are cofiber sequences). Xes,r+1__________//eeXs,r_______//_ 12Ws+r,1^ V (0) _____________________________* *___________________ || | _____________________________* *___________________ || | j ____ ________________________* *__________ || fflffl| fflffl_______________________* *_______ Xe ___________//eX_______________// Xs+r,1 _Id____________________* *________________ s,r+1 s,r ______________________ | | ________________________* *_________________ | | _________________________* *_________ fflffl| ff--_________________________* *_________lffl| 12Ws+r,1^ V (0)_____ 12Ws+r,1^ V (0) 20 MARK BEHRENS AND SATYA PEMMARAJU We consider j to be a nice splitting, because the projection it induces onto * *the other wedge summand fills in the following triad of cofiber sequences. (5.1) eXs,r+1________//_eeXs,r______// 12Ws+r,1^ V (0) ___ || | _____ || | j ____ || fflffl| fflffl___ eXs,r+1________//_eXs,r___________// Xs+r,1 @1 ____ | @2| _____ | | ____ fflffl| fflffl| fflffl_ * _______// Ws+r,1^ V (1)_____ Ws+r,1^ V (1) Let exi2 ß*(Wsi,r) be lifts of xi2 ß*(Wsi,1). Let s = s1+ s2. It is useful to* * keep in mind the following diagram. Ws,1^AVA(1)oo__g_______Ws,r^_V_(1)______@______// Ws+r,1^]V](1)__ ____________________ ____________________* *___________ __|'|___________________'||_____________________ '||___________________* *____________ __fflffl|________________fflffl|_ fflffl|______________* *_____ Xs,1oo_______g_________Xs,r___________@________//_OXs+r,1OOOOO | | | |~| |~| ~| | | W ^ W| ^ V (1)(2) Ws1,1^ Ws2,1^ V (1)(2)gWs1,r^oWs2,r^oV_(1)(2)@//__s1+r,1 Ws2,rs * * (2) ^ 1,rWs2+r,1^ V (1) The element x1 ^ x2 2 ß*(Ws1,1^ Ws2,1^ V (1)(2)) lifts to fx1^ fx22 ß*(Ws1,r^ Ws2,r^ V (1)(2)). We then have @(fx1^ fx2) = dr(x1) ^ x2 + (-1)|x1|x1 ^ dr(x2). The element x1 . x2 is equal to O ~(x1 ^ x2) 2 ß*(Ws,1^ V (1)). We want to compute dr(x1 . x2), which means we first need to lift x1 . x2 to ß*(Ws,r^ V (1* *)). Now ~(fx1^ fx2) is a lift of ~(x1^ x2), but this element will not lift to a lif* *t of x1. x2 without a little modification. The following sequence is exact. ß*(Xes,r) ! ß*-11(Ws,r^ V (0)) .fi1--!ß*-1(Ws+2,r-2^ V (1)) Our assumption that fi1 . F (fx1, fx2) 2 ß*(Ws+2,r-2^ V (1)) is trivial implies* * that F (fx1, fx2) lifts to an element eF2 ß*(Xes,r). Let y be the image of eFin Xs,r* *, and define z = ~(fx1^ fx2) - y 2 ß*(Xs,r). We claim that (1) z lifts to ez2 ß*(Ws,r^ V (1)), and (2) ezis a lift of x1 . x* *2 2 ß*(Ws,1^ V (1)). With regard to claim (1), we need only check that the image of z in ß*( 11Ws,* *r^ V (0)) is zero. The image of both ~(fx1^fx2) and y in ß*( 11Ws,r^V (0)) is F (f* *x1, fx2), therefore the image of z, their difference, is zero. Claim (2) is established b* *y noting that the sequence ß*(Xes,r) ! ß*(Xs,r) ! ß*(Ws,2^ V (1)) ON THE EXISTENCE OF v92 21 is exact. Therefore the image of y in ß*(Ws,2^ V (1)) is zero, so its image O* * g(y) 2 ß*(Ws,1^ V (1)) is zero. So, we have g(ez) = O ' O g(ez) = O g(z) = O ~(x1 ^ x2) = x1 . x2 and claim (2) is established. We are left with identifying @(ez). We have @ez= O @(~(fx1^ fx2) - y) = dr(x1) . x2 + (-1)|x1|x1 . dr(x2) - O @(y) We must evaluate O @(y). In Diagram 5.1 the boundary maps @1 and @2 are displayed. There is a map of cofiber sequences relating @1 to @ in the commutat* *ive diagram displayed below. eXs,r________________//_Xs,r HH vv HHH vvv @1HHH$$H zzv@vvv Xs+r,1 Therefore, @y = @1eF. Furthermore, Diagram 5.1 reveals the relationship between @1 and @2. Thus we have O@1(Fe) = @2(Fe), and we just need an explicit descri* *ption of the latter. The map of cofiber sequences Ws+r,1^ V (1)_____//eeXs,r______//_eXs,r____@__//_ Ws+r,1^ V (1) | | 2 | f || |||| || f || fflffl| || fflffl| fflffl| Ws+2,r-1^ V (1)____//eeXs,r__// 11Ws,r^ V (0)fi1//_ Ws+2,r-1^ V (1) tells us that @2(Fe) is a lift of fi1 . F (fx1, fx2) to ß*(Ws+r,1^ V (1)) and a* *s such, deserves to be called G(x1, x2). This completes our verification of the formula. Remark 5.3. The theorem holds under a weaker assumption. The proof of the theorem does not require x1 and x2 to survive to Er, but only that @(x1) . x2 + (-1)|x1|x1.@(x2) have Adams filtration greater than or equal to s+r. We will ne* *ed this technical generalization for some of our applications of the product rule. 6. Selected AHSS calculations of ß*(V (1)) In our calculation of differentials in the ASS it is helpful to know some of * *the homotopy groups of V (1). The 3-component of the homotopy groups of spheres is known completely through the 108 stem. A table summarizing these elements may be found in [16, A3]. Thus one may write down the E1-term of the AHSS M E1s,t= ßt+s(S) ) ßt+s(V (1)) s-cells of V(1) in this range. The complex V (1) only has cells in dimensions 0, 1, 5, and 6. We shall denote an element in the E1 term by the notation fl[k] where fl 2 ß*(S) a* *nd k is the cell supporting it. The differentials are determined by the attaching * *maps, 22 MARK BEHRENS AND SATYA PEMMARAJU and are given by the following formulas. ( d1(fl[k]).=3fl[k - 1]k = 1, 6 0, otherwise ( d4(fl[k]).=ff1fl[1]k = 5 0 otherwise 8 ><[1]k = 6 d5(fl[k]).=>[0]k = 5 :0 otherwise While a complete determination of the AHSS through the 108 stem should be a relatively straightforward task, we restrict ourselves to a few vicinities wh* *ere we need the data. These partial charts are given on the next few pages, and are re* *ferred to in subsequent sections. All but two of the differentials are immediate. We do not know if the dot- ted differential (1) exists in (6.4) because we are unsure of whether or not fi* *322 . We will see in the proof of Lemma 9.7 that the differential (1)* * must exist. The only other differential which isn't clear is d5(x68[5]) in (6.3). We* * compute ff1<3, ff1, x68> .=3 6= 0. This is a hidden extension in the ANSS for ß*(S0) in the computations in [16]. The indeterminacy of <3, ff1, x68> is trivial, and the indeterminacy of is contained in ff1.ß72(S0) = 3.ß75(S0), so it doesn't enter into the above comput* *ation. We conclude that <3, ff1, x68> 6= 0, so it has no choice but to be a non-zero m* *ultiple of fi22fi21. Portions of the AHSS for ß*(V (1)) (6.1) Stem_55_ Stem_56_ Stem_57_ Stem_58_ ff14[0]eeL ff14[1]jjUUiifi22[5]U fi22[6] LLLL ttiiiiiUUUUU fi22ff1[0]LLLLfi22ff1[1] ff13[6] L fi51[5] ff13[5] fi2fi21ff1[6] fi51[6] (cont'd on next page) ON THE EXISTENCE OF v92 23 * * Portions of the AHSS for ß*(V (1)), @ (6.2) Stem_63_ S* *tem_64_ Stem_65_ Stem_66_ @ ff16[0]jjUUUU* *ff16[1]jjUU fi22fi1ff1[0]fi22fi1ff1[1]@ UUU * * UUUUUU jjUU@ fi22fi1[1] f* *f15=2[5].3ooff15=2[6]_ @ * * @ * * @ * * @ (6.3) Stem_68_ * *Stem_69_ Stem_70_ Stem_71_ @ x68[0] * * x68[1] fi22fi1ff1[5] ff18=3[0]j.3@ * * eeKKUU@ ff17[1]oo___* *ff16[6] KKK@ * * @ ff16[5] * * @ fi22fi1[6] * * @ 24 MARK BEHRENS AND SATYA PEMMARAJU Portions of the AHSS for ß*(V (1)), cont'd (6.4) Stem_77_ Stem_78_ Stem_79_ Stem_80_ fi22fi21[5] fi32[0] ff20[0]jjUUUUff20[1] ee________ UUUU ff18=3[6] fi22fi21[6](1)fi32[1]______ff19[5]_______ ____ eeKKKK fi5[5] KKKfi22fi21ff1[5]KK K fi22fi21ff1_ 3 [5] fi5[6] (6.5) Stem_87_ Stem_88_ Stem_89_ Stem_90_ ff22[0]kkWWWWW ff22[1] fl2ff1[5] fi6[0] WWWW eeKKKK fi6=2[1]kkVVVV ff21=2[5]KKKKfi5fi1[5] fi6=3ff1[5] VVVVV jjUUUKKUUU fi6=3[5] fi6=3[6] ff21=2[6] [5] [6] fl2ff1[6] fl2[6] fi5fi1[6] (6.6) Stem_97_ Stem_98_ Stem_99_ Stem_100_ x92[5] fi6ff1[5]oo_ ff25[0]oo__ fi2fi5[0] ] " ] " fi6=3fi1[5] fi6ff1_3[5] |[0]| ff25[1] || | ff23[6] x92[6] | fl2fi1ff1[5] X[1]| | [6]fi6=3fi1[6]|| fi5fi21[5] |ff24=2[5] | fl2fi1[6] XY||fi6ff1[6] fi6=3fi1ff1[5] |fi6ff1_ 3 [6] fl2fi1ff1[6] fi5fi21[6] ON THE EXISTENCE OF v92 25 7.The order of the fi1 action on V (1) In this section we will prove the following proposition. Proposition 7.1. The map fi51: 50V (1) ! V (1) induced from smashing the map fi51: S50 ! S0 with V (1) is null. Corollary 7.2. Regarding ß*(V (1)) as a module over ß*(S), we have the relation fi51. x = 0 for all x 2 ß*(V (1)). We remark that in ß*(S) we have the relation fi61= 0, and fi51is non-zero. The power of fi51in Proposition 7.1 is minimal, since in ß*(V (1)) the image of the* * element fi41under the inclusion of the bottom cell is non-trivial. Corollary 7.2 follows from Proposition 7.1 since the element fi51. x may be e* *x- pressed as the following composite. 5 S50+k x-! 50V (1) fi1-!V (1) We will first prove the following lemma. Lemma 7.3. The element fi51in ß50(V (1)) is trivial. Proof.There are no elements in the 50 stem of Adams filtration greater than b50. Therefore, it suffices to show that the element b50in the ASS for ß*(V (1)) is * *the target of a differential. In the ASS for ß*(eo2 ^ V (1)) there is a differential d6(v32h0) .=b50. Using the results of Proposition 2.5, we may conclude that if v32h0 supports no shorter differentials in the ASS for V (1), then it must kill b50. Upon investi* *gating the E2 term of the ASS for V (1), we see that there is no element in smaller Ad* *ams filtration that could be the target of a shorter differential. Proof of Proposition 7.1.We will demonstrate that the Spanier-Whitehead adjoint of fi51 fi51: S50 ! D(V (1)) ^ V (1) is null. Let X be the fiber of the composite V (1) ! 5V (0) fi1-! -5V (0) where the first arrow is projection onto the top V (0). By Proposition 4.1, X m* *ay be regarded as a subcomplex of Y1, which may in turn be regarded as a subcomplex of D(V (1)) ^ V (1). We wish to show that the composite 5 S50 fi1-!S0 ,! X ,! Y1 ,! D(V (1)) ^ V (1) is null. We will show that the shorter composite S50 ! Y1 is null. 26 MARK BEHRENS AND SATYA PEMMARAJU Consider the following diagram, whose two bottom rows are cofiber sequences. S50_; ____|;; _____fi51;; _____| ;; f _________fflffl|*;;;0 _____ S ;; _____ | ;; _____ | ;; ""______ fflffl| ;ÆÆ -6V (0)_________//_X________//_V (1) | | || | | || fflffl| fflffl| || -6V (1)_________//Y1________//V (1) In this diagram, the map S50 ! V (1) is null by Lemma 7.3. Therefore the lift f exists making the diagram commute. We will complete the proof of the proposition once we establish the following Claim. The image of the map ß56(V (0)) ! ß56(V (1)) is trivial. The claim follows easily from the AHSS for ß*(V (0)) and ß*(V (1)). A portion of the AHSS for ß*(V (0)) is displayed below. Stem_55_ Stem_56_ Stem_57_ ff14[0] ff14[1] fi22ff1[0] fi22ff1[1] There are no differentials, and ß56(V (0)) is of rank 2. We now consider the im* *age in V (1). The same portion of the AHSS for V (1) is displayed in (6.1), in whi* *ch the same generators of ß56(V (0)) have been killed by differentials, and the cl* *aim follows. 8.Proof that v92is a permanent cycle In this section we will prove that the element v92is a permanent cycle in the* * ASS for ß*(V (1)). We will let h : V (1) ! eo2 ^ V (1) be the Hurewicz homomorphism. We will first use the product rule (5.1) to determine d2(vi2). Lemma 8.1. There are the following differentials on vi2in the ASS. d*(v2)= 0 d3(v22)= h1b20 d2(v32)= -b0k0h1 d2(v42)= -b0k0h1v2 d2(v52)= -b0k0h1v22 d2(v62)= b0k0h1v32 d2(v72)= b0k0h1v42 d2(v82)= b0k0h1v52 d2(v92)= 0 Proof.These formulas are just obtained by iterated application of the product r* *ule. The differential d3(v22) is computed in this manner in Example 5.2. One then us* *es the following formulas in the ANSS, which are derived in [16, 5.1.20] __ vk2= ffik kh1vk-12 (mod v1) ___ `k' vk2= fik 2 vk-22k0 + kvk-12b0 (mod 3, v1) ON THE EXISTENCE OF v92 27 to inductively determine d*(vk+12) from d*(vk2). We should point out that this formula differs by a sign from the formula in [16] because the elements we are referring to as b0 and k0 are normalized differently. Remark 8.2. In [15], Oka demonstrates that v52is a permanent cycle in the ANSS for ß*(V (1)). Above, we have shown that it supports a d2 in the ASS for ß*(V (* *1)). In fact, there is a differential d2(v3b0g0) .=b0k0h1v22 and v52 v3b0g0 is a permanent cycle in the ASS. This differential is establish* *ed in Lemma 9.5. We must eliminate the possibility that v92supports a dr for r > 2. We will ma* *ke a list of all elements in the ASS in the 143-stem of Adams filtration greater t* *han 11. It is given in the table below, with references to the lemma that takes car* *e of it, as well as the Adams filtration (AF). ________________________ _AF_|__Element__|Lemma___ __29__|_h0b140__|__8.3__ __23__|g0h0v22b90_|8.3__ __22__|b1h0v2b90_|_8.4__ 18 | h0b60v52 |8.3 7 | j1v2b08 | 8.5 _____|__v3h1b0___|_8.6__ 17 |k0h0v42b50 |8.9 6 _____|__j1k0b0___|8.10__ __13__|v3h0b40v32_|8.13_ 12 |g0h0b0v72 | 8.3 3 2 | j1g0v2b023|8.15 _____|v3k0h0v2b0_|8.14__ Lemma 8.3. If x 2 E2(V (1)) is an element of the ASS for V (1), and its Hurewicz image h(x) 2 E2(eo2^V (1)) is non-zero, then x cannot be the target of a differ* *ential supported by v92. Proof.We have h(dr(v92)) = drh(v92), but h(v92) 2 E2(eo2 ^ V (1)) is a permanent cycle, so dr(v92) must be in the Hurewicz kernel. Lemma 8.4. Suppose that y 2 E6(V (1)). Then h0b30y = 0 in E6(V (1)). Similarly, if z 2 E7(V (1)), then b60z = 0 in E7(V (1)). In particular, if x is an elemen* *t of E2(V (1)) and x = h0b30y or x = b60z for y or z as before, than x is not the ta* *rget of a non-trivial dr for r 6 or r 7, respectively. Proof.The element h0b30dies in E6(S0) (this is just the Toda differential d5b1 * *.= h0b30). Similarly, there is a differential d6(h0b0v32) .=b60giving the relation* * b60= 0 in E7(S0). Then use the S0-module structure of V (1). The following lemmas take care of the other possible targets. We work from hi* *gh- est to lowest Adams filtration to eliminate the possibility of intervening diff* *erentials as we go along. 28 MARK BEHRENS AND SATYA PEMMARAJU Lemma 8.5. In E5(V (1)) there is a non-trivial differential d5(j1v2b70) .=k0v2b100. Proof.In E5(S0), we have d5j1 .=k0b30. The element v2b70is a permanent cycle in the ASS for V (1), so the differential follows from the S-module structure of V (1). Lemma 8.6. In E4(V (1)) there is a non-trivial differential d4(v3h1b80) .=b90k20. Proof.Our AHSS calculations (6.3) prove that ß72(V (1)) = 0. Therefore, the ASS for ß*(V (1)) should have no non-trivial permanent cycles in the 72-stem. The E2 term contains k20b20, b30k0v2, and b40v22. The element b40v22supports a * *non- trivial d3, and d5j1v2 .=b30k0v2. The only possibility for eliminating k20b20i* *s for d4(v3h1b0) .=k20b20. Therefore d4(b80v3h1) .=k20b90. For Lemma 8.9 we need to know the differentials supported by vi2g0 for small * *i. These are given below. Lemma 8.7. We have the following Adams differentials on g0vi2in Er(V (1)). d4(g0v2)= b30h0 d3(g0v22)= b20k0h0 d2(g0v32)= -g0b0k0h1 d3(g0v42)= 0 d3(g0v62)= v42b20k0h0 Remark 8.8. The element g0v42is actually a permanent cycle, and this should be regarded as anomalous. The AHSS element which it corresponds to is [1], and this bracket is defined only because of the anomalous relation ff1fi5 = 0 in ß*(S). Proof.We will first explain how the term G(g0vi2, v2) and the term G(g0vi2,_v22* *) in the product rule is computed. The Adams-Novikov element which detects g0vi2is given by vi2b0h0 - ivi-12k0h0 (mod v1). (We will just work modulo (v1) since we will be mapping everything into_V_(1)__ for the product rule anyways.) The Adams-Novikov element which detects g0vi2is given by ` ' i vi-2k g + ivi-1b g (mod v ). 2 2 0 0 2 0 0 1 We recall from Lemma 8.1 the following formulas. __v __2 2= h1 v2= -h1v2 ___ ___ v2 = b0 v22= k0 - b0v2 Using the relations 3.1 and the relation v2b1h1 = 0, we may apply the product rule (5.1) iteratively to get the requisite differentials. Specifically, first * *apply the product rule to g0. v2, to get d4(v2g0), then apply the product rule to (g0v2) * *. v2 to get d3(g0v22). In E2(S0), g0 supports a d2, and v32supports a d2 in E2(V (1)), * *thus d2(v32g0) may be deduced from the S-module pairing of Adams spectral sequences. The problem is, we can no longer apply the product rule to v2 multiplication to* * get d3v42g0. However, we may instead apply the product rule to the product v22. v22* *g0 to get the formula for d3(v42g0) and similarly to v22. v42g0 to get the formula* * for d3(v62g0). ON THE EXISTENCE OF v92 29 In particular, we have the following lemma, which follows immediately. Lemma 8.9. In E3(V (1)) there is a differential d3(v62b30g0) = k0h0v42b50. Lemma 8.10. In E5(V (1)) there is a differential d5(v42h0b20j1) .=j1k0b60. Proof.We need to compute the differential supported by v42h0. Observe that the differential d6(v32h0) .=b50in the ASS for eo2^V (1) lifts to a differential in* * E6(V (1)). The element v22h0 must be a permanent cycle since there is nothing for it to ki* *ll. Hence, by the product rule, ____ _____ b50.=d6(v32h0) = -G(v22h0, v2) = -b0(h0v22b0 - h0v22h1). ____ _____ We conclude that the image of h0v22in ß*(V (1)) is fi31. The image of h0v22in ß*(V_(1))_has to be in Adams filtration greater than_or_equal_to that of the im* *age of h0v22, and so we may conclude that the image of h0v22is actually zero. Where* *as d3(v22) is non-zero, the differential (d3(v22))v22h0 is zero, and there are no * *permanent cycles in higher Adams filtration. Therefore (@(v22)) . v22h0 = 0, and we are * *in_a_ position to use the version of the product rule explained in Remark 5.3. Since * *v22 is detected by k0, we have __ _____ ___ ____ . d5(v42h0) = -G(v22, v22h0) = -b0(v22. v22h0+ v22. v22h0) = -b0(k0 . b30) = -k* *0b40. We then use the S-module structure of V (1), and the differential d5j1 .=k0b30to obtain d5(j1 . v42h0b20) = k0h0b50v42 j1k0b60. By Lemma 8.9, k0h0b50v42is the target of a d3. We will need a couple of lemmas to prove Lemmas 8.13 and 8.14. Lemma 8.11. In E4(V (1)) there is a differential d4(v3h0b0) .=b1b30. Proof.In the Adams spectral sequence for ß*(V (1)) in the vicinity of the 65 st* *em, we have the following elements and differentials. Stem_63_ Stem_64_ Stem_65_ Stem_66_ b60h0oo____b30g0v2_ b30h0v22``_ b50v2 ______ b20v22h1 b1b30ee___ b20h0k0v2__(1)_b40k0_____ _______ ee________________ b1b0h0v2oo_____v42 (2)__b20b1h1oo_____v32g0______________________* *_______ ____ ____________ v3h1 v3h0b0 b1v22 Here we know for dimensional reasons that b1, k0, v2, h0v22, and h0k0v2 are per- manent cycles, and so all b0 multiples of them are permanent cycles. We have the solid differentials from Lemmas 8.1 and 8.7. The only means by which the correct homotopy can be achieved is for one of the dotted differentials (1) to exist an* *d the dotted differential (2) to exist. The differential (2) is the desired different* *ial. 30 MARK BEHRENS AND SATYA PEMMARAJU Lemma 8.12. The element v32b1 is a permanent cycle. Proof.In [16] the element fi6=3is shown to exist in ß82(S0). It is detected mod* *ulo the ideal (3, v1) in the ANSS by the element v32b1. Thus, the image of fi6=3in ß82(V (1)) is detected in the ASS by the element v32b1, and this element must therefore be a permanent cycle. Lemma 8.13. In E2(V (1)) there is a non-trivial differential d2(v3h0b40v32) .=v2h0j1k0b40. Proof.In Lemma 8.11, we showed that d2(v3h0b0) = 0. In Lemma 8.1, we showed that d2(v32) = -b0k0h1 = -b0b1h0. __ ___ We also know v32, and hence v32, are trivial in E2 (they are in higher Adams fi* *l- tration). We will now use the product rule (5.1) to compute the differential on (v3h0b0) . v32. The term G(v3h0b0, v32) is trivial in E2 by the previous consi* *dera- tions. The following manipulations are made possible with the hidden extension h0 . (v3h0b0) .=b1g0v2. d2((v3h0b0) . v32)= (v3h0b0) . (b0b1h0) .=b2 1g0b0v2 .=v 2h0j1k0b0 Now multiply by b30. Lemma 8.14. In E5(V (1)) there is a non-trivial differential d5(v3k0h0v22b30) .=b1b60v32. Proof.In E2(V (1)), there is a Massey product (8.1) v32b1b302 . We are regarding the elements in the Massey product as detecting the following maps in homotopy. b1b30$ fi2fi31ff1[5] : 64V (0) ! V (1) v1 $ v1 : 4V (0) ! V (0) v21h2$ ffi3: 43V (0) ! V (0) This Massey product therefore detects the Toda bracket (8.2) . We saw in Lemma 8.11 that b30b1 = 0 in E5(V (1)). Therefore, the Toda bracket is detected in a higher Adams filtration modulo an indeterminacy contained in t* *he subgroup [ 69V (0), V (1)] O ffi3. We computed ß69(V (1)) and ß70(V (1)) using the AHSS (6.3). Both of these groups are rank 1, generated by x68[1] and fi22fi1ff1[5], respectively. So we * *have [ 69V (0), V (1)] is of rank 2, generated by elements fi22fi1ff1[5]1 and x68[1]* *0. Here, the subscripts 0, 1 are used to indicate what cell of the source V (0) the elem* *ent is ON THE EXISTENCE OF v92 31 born on. We will conclude that the indeterminacy is zero using the V (0)-module structure of V (1). (8.3) fi22fi1ff1[5]1 O ffi3= fi3 . fi22fi1ff1[5]0 = 0 (8.4) x68[1]0 O ffi3= (fi3[1] . x68[1])0 = 0 Equation 8.3 follows from the relation fi3fi1 = 0 in ß*(S). In Equation 8.4, we mean the product of fi3[1] 2 ß43(V (0)) and x68[1] 2 ß69(* *V (1)) under the module map ~ : V (0) ^ V (1) ! V (1). The following diagram commutes. ß43(V (0)) ß69(V (0))~//_ß112(V (0)) | | | | fflffl| fflffl| ß43(V (0)) ß69(V (1))~//_ß112(V (1)) Since x68[1] 2 ß69(V (1)) is born on the 1-cell, we may lift it to an element x* *68[1] 2 ß69(V (0)). Thus in order to show there is no indeterminacy it suffices to show that fi3[1] . x68[1] = 0 2 ß112(V (0)). At this point we remind the reader that by our descriptions of the elements fi3[1] and x68[1] as elements in ß*(V (1)),* * we mean that their images under the projection on to the top cell are fi3 and x68, respectively. These elements are not necessarily uniquely determined in ß*(V (0* *)), but any two representatives will differ by the image of an element of ß*(S) und* *er the inclusion of the bottom cell of V (0). Now ß69(S) = 0, so x68[1] is unique* *ly determined. However, ß43(S) is generated by ff11, so any two elements of ß43(V * *(0)) which project to fi3 on the top cell must differ by ff11[0] 2 ß43(V (0)). Unde* *r the product map V (0) ^ V (0) ! V (0), all homotopy carried by the smash product of the 1-cells is annihilated. We will necessarily have fi3[1] . x68[1] = 0 2 ß112* *(V (0)) for all possible representatives of fi3[1] if we can show ff11[0] . x68[1] = 0 2 ß112(V (0)). This is straightforward: in ß43(V (0)) we have ff11[0] = v101ff1[0], and ff1. x* *68= 0 2 ß*(S). Thus the Toda bracket 8.2 has no indeterminacy. We conclude that the Toda bracket 8.2 must be zero in ß*(V (1)) modulo higher Adams filtration. There are only three elements in the correct range to kill t* *he corresponding Massey product 8.1, and these elements are v32v3h0b0, fl2v22, and v3k0h0v22. In Lemma 8.13, we proved that v32v3h0b0 supported a non-trivial d2. The element fl2v22must support a non-trivial d3; this follows from the S-module pairing and the d3 supported by v22that was proved in Lemma 8.1. Thus we must have d5(v3k0h0v22) .=v32b1b30 and the lemma is proven after multiplying both sides by b30. Lemma 8.15. In E2(V (1)) there is a differential d2(j1g0v32b20) = j1g0k0b30h1 .=k1b60. Proof.This lemma follows immediately from the S-module pairing of Adams spec- tral sequences. We know b20j1 survives to E5(S) and we have computed d2(g0v32) in E2(V (1)) as part of Lemma 8.7. 32 MARK BEHRENS AND SATYA PEMMARAJU We have established that every possible target of a differential supported by* * v92 is either the target of a shorter differential or the source of a differential.* * We may conclude that v92is a permanent cycle. 9. Proof that v92extends over V (1) In this section we will prove that if v92: S144! V (1) is a map detected by t* *he element v922 E2(V (1)), then it extends over V (1) to a self-map ve92: 144V (1) ! V (1). Applying Spanier-Whitehead duality, this is equivalent to finding an extension * *cor- responding to the dotted arrow in the diagram below. D(V8(1))8^_V_(1) _______|_____ ________ | ______ fflffl| S144____v9____//V (1) 2 The inclusion of the wedge summand Y1 of D(V (1)) ^ V (1) (Proposition 4.1) has the property that the composite Y1 ,! D(V (1)) ^ V (1) ! V (1) is just projection onto the top V (1). It therefore suffices to extend v92over* * the complex Y1 as displayed below. _Y1::___ _____|_____ _______ | _____ fflffl| S144___v9__//V (1) 2 | |ffi fflffl| -5V (1) The vertical column forms a cofiber sequence where ffi is given as the composite V (1) -! 5V (0) fi1-! -5V (0) '-! -5V (1). Here, is projection onto the top V (0) and ' is inclusion of the bottom V (0)* *. Thus there is a solution to the extension problem if and only if the composite ffi O* * v92= 0. The map ffi is also given by the composite V (1)____//________'5V/99(0)/_f5Vi(1)1//_ -5V (1) _____________________________________________________________* *_______________________ __________________________________________________________* *______________________________________________________________@ ______________________________________________________* *______________________________________________________________@ ffi0 Here ffi0 is the geometric v1-Bockstein. In this section we will prove (9.1) fi1 . ffi0(v92) = 0 from which it follows that ffi O v92= 0, and thus v92extends over V (1). Since v92has Adams filtration 9, ffi0(v92) has Adams filtration 9. We may calculate ffi0 in Ext using the periodic lambda algebra. We have d(v92) = v91~27 ON THE EXISTENCE OF v92 33 thus we have __ __ __ (1)___*_//_(0)_'*_//_(1) v92Ø____//v81~27Ø__//0 and so ffi0*(v92) = 0 viewed as an element of E2(V (1)). We may conclude that t* *he Adams filtration of ffi0(v92) is 10. Our strategy to prove that fi1 . ffi0(v9* *2) = 0 is to make a list of all of the elements in E2(V (1)) in Adams filtration greater tha* *n or equal to 10 in the 139-stem. We then will prove that each of these elements is * *either not a permanent cycle, or is killed by a differential or at least has the prope* *rty that composition with fi1 is zero. A list of the elements, as well as the lemmas tha* *t deal with it, is given below.__________________________ _AF_|___Element___|__Lemma______ __26__|_h0v2b120__|____8.4_____ __25__|_k0h0b110__|____8.4_____ __20__|_g0h0v32b70__|__9.1_____ __19__|_b1h0v22b70__|__9.4_____ 15 | v62h0b40 | 9.7, 9.8 2 5 | j1v2b06 | 9.7, 9.9 _____|__v3h1v2b0___|_9.7,_9.9__ 14 | j1k0v2b40 | 9.7, 9.10 5 3 | v2k0h0b05 | 9.7, 9.11 _____|__v3k0h1b0___|9.7,_9.10,_9.11 __13__|b50|9.12____ __10__|_v3h0b20v42__|__9.13____ Lemma 9.1. If x is an element of Er(V (1)) whose Hurewicz image h(x) 2 Er(eo2^ V (1)) supports a non-trivial differential, then x cannot be a permanent cycle. Proof.This is obvious; h is a map of spectral sequences. We shall need the following lemma. Lemma 9.2. There is the following pattern of differentials in the ASS for ß*(V * *(1)) in the vicinity of the 68-stem. Stem_67_ Stem_68_ Stem_69_ v2b40h1ee__ b50g0 b50v2h0 ""____________eeKKKK_ b1b30h0oo_____b20v32_____b40h0k0________KKKK_______ _________ KK j1b0 ____b0v22k0_____b0h1v32 _____ v42h0 __v2k20______h0b1v22_ __ j1h1 g1h1 Here only one of the dotted differentials occurs. Remark 9.3. We will find later (see the proof of Lemma 9.7) that in fact j1h1 must be a permanent cycle, and thus the dotted differential supported on b0v22k0 must be the non-trivial one. 34 MARK BEHRENS AND SATYA PEMMARAJU Proof.We will first deduce the following portion of the ASS chart for ß*(V (1)). Stem_56_ Stem_57_ Stem_58_ Stem_59_ b40v2 b30h1v2ee__ b40g0 v2b40h0 `_______`____ee____________ b30k0jjUUUUUb1b20h0oo____b0v32_______k0b30h0_____________________ UUUU ___________(2)________ Uj1 (1) ___v22k0______ h1v32 ____ g1 The differential on j1 follows from the differential in E5(S). The differentia* *l on b0v32is a consequence of Lemma 8.1. Since the AHSS calculation (6.1) told us th* *at ß57(V (1)) = 0, we may conclude that one of the dotted differentials (1) exists. Also, we have shown that ß58(V (1)) has rank 1, hence something must kill b40g0. Both v2b40h0 and k0b30h0 are permanent cycles, so the only candidate to support* * the dotted differential (2) is h1v32. Note that this differential is present in the* * ASS for eo2 ^ V (1). Moving up to the vicinity of the 68-stem of the ASS, the two solid differenti* *als in the statement of the lemma are propagated by b0 multiplication. Our AHSS calculations (6.2) tell us that ß67(V (1)) = 0. Now v2k20must be a permanent cy* *cle since v2k0 is a permanent cycle for dimensional reasons. Since v2b40h1 must van* *ish, one of the dotted differentials must occur. We have computed ß68(V (1)) to be of rank 2, so there can be no more differentials originating from the 69 stem. Lemma 9.4. The element b1h0v22b702 E6(V (1)) must be zero. Proof.In Lemma 9.2 we showed that v22b1h0 2 Er(V (1)) is a permanent cycle. The result then follows from the fact that in E7(S) there is a relation b60= 0. We are now in a position to prove the differential promised in Remark 8.2. We will need this differential later in this paper. Lemma 9.5. In E2(V (1)), there is a non-trivial differential d2(v3g0) .=k0h1v22= b1h0v22. Proof.Our AHSS computations (6.3) show that ß69(V (1)) has rank 1. In the proof of Lemma 9.2, we computed all of the Adams differentials in E*(V (1)) supported in the 69 stem. That data is used to compute the following portion of the ASS f* *or ß*(V (1)). Stem_68_ Stem_69_ Stem_70_ b50g0eeK b50v2h0 b70 KKK eeKKKK b20v32 KKKKKb40h0k0oo____b20g0v22KKKKKKK b0v22k0 b0h1v32 b20v2b1 v2k20 h0b1v22jj___h0b0j1_____ ___________ j1h1 g1h1 v3g0_ The differential supported by b20g0v22is a consequence of Lemma 8.7. The differ* *ential supported by b20v2b1 is a consequence of the Toda differential on b1 in E5(S) a* *nd ON THE EXISTENCE OF v92 35 the S-module structure of V (1). One more element in the 69 stem must be the target of an Adams differential, and the only possibility is the dotted differe* *ntial. This is the differential we wanted. There is an elaborate pattern of activity in Adams filtration 13-15 in the 13* *9- stem. We would like to understand which elements in this range of filtrations are permanent cycles, and which aren't, and this is accomplished in Lemma 9.7. As a consequence, certain linear combinations of the possible obstructions to t* *he extension of v92will be eliminated. First we need the following differential. Lemma 9.6. In E3(V (1)), there is a non-trivial differential d3(v3h1v2) .=b0j1h1. Proof.In Lemma 8.6, we showed that v3h1 supports a non-trivial d4 and thus is a d3 cycle. We now apply the product formula (5.1) to deduce the differential on v3h1 . v2. Computer assisted lambda algebra calculations yield the formula ____ v3h1 = -g1 2 E2(V (0)). _____ We wish to compute v3h1, which is obtained by computing the Bockstein on g1. Now g1 is a d1-cycle when it is considered as an element of E1(S0), so we need to compute the Adams differential of g1 in E*(S0) to get this Bockstein. We use the main theorem of Bruner ([3, VI.1.1]) to understand the relationship between Steenrod operations in Ext and differentials in the ASS. Using Bruner's formula, we may compute (9.2) d2(g1) = d2(P 0(g0)) = v0 . fiP 0(g0) .=v0j1. Here we should point out that our indexing of the Steenrod operations is differ* *ent from that of Bruner's, but conforms to the perhaps more common indexing as given in [13]. The element v0 detects the degree p map on spheres. The Steenrod operation fiP 0(g0) .=j1 is computed using the May spectral sequence for H*(P*). Specifically, The element g0 is detected by by h2,0h1,0. On the May E2-term, we compute (using the Cartan formula) fiP 0(h2,0h1,0) = fiP 0(h2,0) . P 0(h1,0) - P 0(h2,0) . fiP 0(h1,0) = b2,0h1,1* *- h2,1b1,0 and (b2,0h1,1- h2,1b1,0) detects j1._____ We conclude from Equation 9.2 that v3h1 .=j1._ We then apply the product formula, keeping in mind that __v2= h1 and __v2= b0, and get d3(v3h1 . v2) = -b0( j1h1 - g1b0) .=b0j1h1. 36 MARK BEHRENS AND SATYA PEMMARAJU Lemma 9.7. We have the following differentials in the 139 stem of the ASS for V (1) in Adams filtrations 13-15. d3(j1v22b50).=b70h1j1 d3(v3h1b60v2).=b70h1j1 d3(v62h0b40).=b70h1j1 or v62h0 is a permanent cycle d5(v52k0h0b30).=b70k20v2 d5(j1k0v2b40).=b70k20v2 d5(v3k0b50h1).=b70k20v2 An F3-basis of permanent cycles in this range of Adams filtration is listed bel* *ow. v62h0b40+ a1(j1v22b50) + a2(v3h1b60v2) j1v22b50 v3h1b60v2 v52k0h0b30 v3k0b50h1 j1k0v2b40 v3k0b50h1 b50 Here a1 and a2 are elements of F3. We are unable to determine the signs or the coefficients ai. A diagram of this portion of the ASS chart is given below for * *the reader's convenience. AF_ Stem_138_ Stem_139_ oo```````` 19 k20v2b70oo____oo```````` ________________________________________________________________________* *_____________________________________________________UTØØED|%$@ 18 b70h1j1llXX Ø|ØØØ ________________________________________________________________________* *_____________________________________________________XXXXXiiSS@ 15 NNNSSSSSSSSv62h0b40NNNØØ||ØØ NNØ|ØØØ NNNNNj1v22b50ØØ|Ø Ø|ØØ Ø|Øv3h1b60v2Ø ________________________________________________________________________* *______________________________________________________!|ØØØ 14 Ø@Av52k0h0b30||Ø Ø PQØj1k0v2b40| v3k0b50h1 ________________________________________________________________________* *_____________________________________________________ 13 b50 Proof.The method of proof is to divide these elements by a maximal power of b0, and then multiply by b0 successively until all of the elements in question * *are present in the E2 term. We begin with the vicinity of the 79 stem. In our AHSS calculations (6.4), we computed ß78(V (1)) and ß79(V (1)) modulo one differenti* *al which we were unable to determine (this is the dotted differential labeled (1) * *in (6.4)). We conclude that either ß78(V (1)) has rank 2 and ß79(V (1)) has rank 1* *, or ß78(V (1)) has rank 1 and ß79(V (1)) is trivial. We will see shortly however th* *at the latter is the case, i.e. that the dotted differential (1) must exist. We displa* *y below ON THE EXISTENCE OF v92 37 the ASS in the same range. Stem_77_ Stem_78_ Stem_79_ Stem_80_ v2b50h1eeK b60g0 v2b60h0 b80 KKK eeKKKK eeKKKK b1b40h0oo____v32b30_KKKKKk0b50h0oo____b30g0v22KKKKKKKKKKKKK b20j1 v22k0b20 v32b20h1 b30v2b1 v42h0b0 b0v2k20 b1v22b0h0jjUUUj1b20h0 eeKKKUUUK v32k0h0jj___j1h1b0oo_____v3h1v2____KKKb0v3g0K ____________ KKK (2) v3k0_ v52 By comparing with the ASS chart in the vicinity of the 68-stem in the proof of Lemma 9.2, we see that the elements b0v2k20, v2b60h0, k0b50h0, and b1v22b0h0 ar* *e per- manent cycles, and the elements v32b30and v32b20h1 support the indicated differ* *entials. The differential on v3h1v2 was the subject of Lemma 9.6. In Lemma 9.2, we were unable to determine whether v2b40h1 was killed by b0v22k0 or j1h1. Since j1h1b0* * is the target of a differential, this ambiguity is now resolved: d4(v22k0b20) =.v2* *b50h1. The differential supported by b30g0v22was established in Lemma 8.7. The differe* *n- tial supported by b30v2b1 follows from the Toda differential on b1 in E*(S0). T* *he differentials supported by b0v3g0 and v52were proven in Lemmas 9.5 and 8.1, re- spectively. There is nothing remaining in stem 79, so we conclude that ß79(V (1* *)) is trivial. Therefore the dotted differential (1) exists in the AHSS chart (6.4* *). The AHSS chart now tells us that ß78(V (1)) is of rank 1, and the only way for the * *ASS to produce the same answer is for there to exist the dotted differential (2) si* *nce neither b20j1 nor v42h0b0 are permanent cycles. We now multiply everything by b0 and move into the vicinity of the 89 stem. The ASS chart is displayed below. Stem_87_ Stem_88_ Stem_89_ Stem_90_ b60h1v2eeK b70g0 v2b70h0 b90 KKK ffLLLLL ffLLLL b1b50h0oo_____v32b40KKKKKKk0b60h0oo_____v22b40g0LLLLLLLLL L LL j1b30 v22k0b30 v32h1b30 b1v2b40 v42b20h0 b20v2k20 b1v22b20h0jjVVj1b30h0V aaCCC ffLLVVVVLL b0h0v32k0jjUUUb20h1j1oo_____v22j1jjVVLLLLb20v3g0LCCCC UUU VVVVVCCC LL v3h2 v3k0b0 CCb0v2v3h1CC v52b0 C h1v3k0 v42k0 v22g1 v3h2h0 All of the indicated differentials follow from our computations near the 79 stem except for the one supported by h1v3k0. For that we consider the image under 38 MARK BEHRENS AND SATYA PEMMARAJU projection_onto_the top cell of V (1). We saw in the proof of Lemma 9.6 that v3h1=.j1. We have __________ _______ . . _______ d(v3h1k0)= d(v3h1k0) = d(j1k0) = k20b30= b20v2k20. This can only happen if d5(v3h1k0) = k20b20v2. Our AHSS calculations (6.5) tell* * us that ß89(V (1)) is of rank 2, so there can be no more differentials originating* * from the 90 stem. We now multiply once more by b0 and arrive in the crucial region around the 99 stem. Our AHSS computations (6.6) tell us that the rank of ß98(V (1)) is 3 and * *the rank of ß99(V (1)) is 4. We turn now to the ASS. Stem_97_ Stem_98_ Stem_99_ Stem_100_ b70v2h1eeK b80g0ffM v2b80h0 b100 KKK MMMM ffNNNN b1b60h0oo_____v32b50KKKKKKMk0b70h0oo______b50g0v22MMMMNNNNN M NN b40j1 b40v22k0 v32b40h1 b50v2b1 oo`````oo```` v42b30h0 b30k20v2 v22b1b30h0kkVVVVb40h0j1 MULTØØØØ ffNNVVVVNN v32b20h0k0jjUUb30h1j1oo____v2v3b20h1UkkVNNNb30v3g0NNØØØØ eeKK UUU ffMV(4)VVMMMMØØØØ NN v22v3h0b0(3)K b20v3k0 MMMMØØh0v62ØØ v52b20 K ØMMØØØ v2fl2 g0v52 ØØØb0v22j1ØØ v42k0b0 Ø_!ØØ b0v3h2 v42b1 ØØ v2j1k0 v32k20 01 v2k1 b0h1v3k0 v22h1j1 h0v2fl2 Aside from the differentials supported by (possibly) g0v52, h0v62, and v2j1k0, * *all of the differentials displayed follow from our calculations near the 89 stem. If w* *e had only the differentials arising from b0 multiplication on elements in the vicini* *ty of the 89 stem, we would have created groups of the correct rank in the 98 and 99 stem* * as predicted from the AHSS. Therefore any additional differentials must preserve t* *he rank of the E1 term. We easily see that d5(v2j1k0) = b30k20v2 from the differe* *ntial on j1 and the S-module structure of V (1). The problem is that h0v62could support a d4 killing b30k20v2, and this would make both v2j1k0 and b0h1v3k0 into permanent cycles. We claim that this cannot happen. For suppose that d4(h0v62) .=b30k20v2. Then there is no linear combinat* *ion of elements containing h0v62which is a permanent cycle. Our AHSS calculation indicates there is some element in ß99(V (1)) such that its image in ß93(S) und* *er projection onto the top cell of V (1) is fi6ff1. The only element which can acc* *ount for this is h0v62. Therefore, if h0v62is not a permanent cycle, it must support* * a d3 killing b30h1j1. This possibility is indicated by the dashed differential (4). Also, we cannot eliminate the possibility that d3(g0v52) .=v32b20h0k0, since * *g0v52is in the same Adams filtration as b20v3k0. This possibility is indicated with the* * dashed differential (3). Upon taking the pattern of differentials supported by the 99 * *stem ON THE EXISTENCE OF v92 39 and multiplying by b40, we get the promised pattern of differentials supported * *in the 139 stem. Lemma 9.8. Choose the correct coefficients ai2 F3 so that v62h0b40+ a1(j1v22b50* *) + a2(v3h1b60v2) is a permanent cycle. The composite of any element that this perm* *a- nent cycle detects with fi1 must be null. Proof.The element v62h0 + a1(j1v22b0) + a2(v3h1b20v2) is a permanent cycle by Lemma 9.7. Now apply Corollary 7.2. Comparing with the elements present in the 149 stem of E2(V (1)) we see that there is no possibility of a hidden fi1 e* *xten- sion. Lemma 9.9. Choose the correct sign so that j1v22b50 v3h1b60v2 is a permanent cycle. This element must be the target of a differential. Proof.The element j1v22 v3h1b0v2 is a permanent cycle (Lemma 9.7). Now apply Corollary 7.2. Lemma 9.10. Choose the correct sign so that j1k0v2b40 v3k0b50h1 is a permanent cycle. The composite of any element that this permanent cycle detects with fi1 must be null. Proof.Again, apply Corollary 7.2. Comparing with the elements present in the 149 stem of E2(V (1)) we see that there is no possibility of a hidden fi1 extension. Lemma 9.11. Choose the correct sign so that v52k0h0b30 v3k0b50h1 is a permanent cycle. This element must be the target of a differential. Proof.In Lemma 9.5, it was established that v52 v3g0b0 was a permanent cycle, therefore the element h0 . (v52 v3g0b0) = h0v52 v3h1b20 is a permanent cycle. Now d5(j1) .=k0b30in the ASS for ß*(S). Using the S-module structure of V (1), we deduce that there is a differential d5(j1 . (h0v52 v3h1b20)) .=v52k0h0b30 v3k0b50h1. Note that v52k0h0b30 v3k0b50h1 might be the target of a shorter differential, i* *n which case the conclusion of the lemma is still satisfied. Lemma 9.12. The element b50must be the target of a differential. Proof.This follows immediately from Corollary 7.2 and the fact that is a permanent cycle (Lemma 9.7). Lemma 9.13. In E2(V (1)) there is a non-trivial differential d2(v3h0b20v42) .=v22h0j1k0b20. Proof.In Lemma 8.11, we showed that d2(v3h0b0) = 0. In Lemma 8.1, we showed that d2(v42) = -b0k0h1v2 = -b0b1h0v2. Computer assisted lambda algebra computations reveal that in E2, we have ______ v3h0b0= -h0j1 _______ v3h0b0= 0 40 MARK BEHRENS AND SATYA PEMMARAJU __ ___ We also have v42= h1v32and v42= 0 in E2. We will now use the product rule (5.1) to compute the differential on (v3h0b0) . v42. The term ______ ___ _______ __ G(v3h0b0, v42) = v3h0b0. v42- v3h0b0. v42 ___ _______ is trivial in E2 because v42and v3h0b0are trivial in E2. 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