ON BP 2 -COOPERATIONS

Transcription

1 ON BP -COOPERATIONS D. CULVER Contents. Introduction Conventions Acknowledgements. The Adams spectral sequence for BP BP.. The (co)module structure of (A E()) 4.. An E()-module splitting of (A E()) 6.. The Bockstein spectral sequence for Q 9.4. Topological splitting 5. Calculations 7.. Brown-Gitler (co)modules 7.. Exact sequences.. Inductive calculations 5.4. Low degree computations 4 References 4. Introduction The purpose of this paper is to give a description of the algebra of cooperations for the second truncated Brown-Peterson spectrum, denoted by BP, at the prime. At chromatic height, the cooperations algebra BP was computed by Adams in []. When the prime is, the spectrum BP is connective complex K-theory, denoted by bu, and when the prime is odd, BP is the Adams summand of connective complex K-theory. In his calculation, Adams observed that the E -page of the Adams spectral sequence for BP BP has a non-canonical direct sum decomposition Ext E() (F, (A E()) ) V C

2 D. CULVER where the subspace V is concentrated in Adams filtration and C is v -torsion free. Adams also gave a complete description of C in terms of Adams covers of Ext E() (F, F ). The interest in studying BP -cooperations originates in Mark Mahowald s work on the Adams spectral sequence based on connective real K-theory, bo. Armed with his calculation of bo bo, Mark Mahowald was proved the -primary v -telescope conjecture in [6]. With Wolfgang Lellmann, he was able to compute the bobased Adams spectral sequence for the sphere, and showed that it collapses in a large range (cf. [7]). These calcuations have been extended in []. At chromatic height, the role of bo is played by tmf and the role of bu is played by the spectrum tmf (), in that it is a form of BP (cf. [5]). Partial calculations of tmf tmf has been achieved in [4]. The goal of this work is to compute the cooperations algebra for tmf (). This is motivated by the fact that one can descend from tmf () to tmf through the Bousfield-Kan spectral sequence of the cosimplicial resolution tm f tm f () (tm f () tm f ). Since the spectrum tmf () is a form of BP, for the purposes of calculations, we can replace tmf () by BP. Consequently, a natural choice for computing the cooperations algebra is the Adams spectral sequence Ext A (F, H (BP BP ; F )) = BP BP Z. There are two main contributions of this paper. The first is a structural result regarding the algebra BP BP. In particular, we will show there is a direct sum decomposition into a vector space V concentrated in Adams filtration, and a v -torsion free component. The second is an inductive calculation of BP BP. This inductive calculation is similar to the one produced in [4]. Moreover, this decomposition of BP BP implies that the methods developed in [] to calculate the bo-ass can be applied to the BP - ASS. One of our goals is to prove an analogous splitting for tmf tmf and compute the tmf-ass. Conventions. We will let A denote the mod Steenrod algebra and A its dual. We will let ζ k denote the conjugate of the the generator ξ k in the dual Steenrod algebra A. Given a Hopf algebra B

3 ON BP -COOPERATIONS and a comodule M over B, we will often abbreviate Ext B (F, M) to Ext B (M). Homology and cohomology are implicitly with mod coefficients. All spectra are implicitly -complete. Acknowledgements. I would like to thank Mark Behrens for suggesting this project and for innumerable helpful discussions along the way. I would also like to thank Prasit Bhattacharya and Nicolas Ricka for several helpful conversations. Finally, I thank Dylan Wilson and Paul VanKoughnett for pointing out an error in my original approach to proving Theorem... The Adams spectral sequence for BP BP In this section, we will prove BP BP decomposes into a v - torsion and v -torsion free component. This will be accomplished through the Adams spectral sequence Ext A (F, H (BP BP )) = BP BP In particular, we will begin by determining the structure of the E - page. Recall that the mod homology of BP is given by H BP (A E()) where E() denotes the subhopf algebra of the Steenrod algebra A generated by the first three Milnor primitives Q, Q, Q (cf. []): E() := E(Q, Q, Q ). As a subalgebra of the dual Steenrod algebra, the homology of BP is explicitly given as (A E()) = F [ζ, ζ, ζ, ζ 4, ζ 5,...], see [] for this calculation. By the Künneth theorem, we have H (BP BP ) H BP H BP and hence, via a change of rings, we find that the E -term of this spectral sequence is Here, the dual of E() is given by Ext E() (F, (A E()) ). E() E(ζ, ζ, ζ ).

4 4 D. CULVER The E() -comodule structure of (A E()) uniquely determines, and is uniquely determined by, a corresponding E()-module structure (given by the dual action of E()). Thus we may rewrite the E -page as Ext E() (F, (A E()) ) = Ext E() (F, (A E()) ) where the right hand side corresponds to Ext of modules. In order to calculate the Adams spectral sequence, ASS, we need to calculate this Ext group of modules over E(). Recall that the Adams spectral sequence for BP takes the form Ext E() (F, F ) = π BP and that the E -term is isomorphic to F [v, v, v ]. The main theorem of this section concerns the structure of Ext E() (F, (A E()) ) as a module over this three variable polynomial algebra. Theorem.. The E -page of the ASS for BP BP admits a decomposition as modules over F [v, v, v ] as C V where C is v -torsion free and is concentrated in even (t s)-degree, and V is concentrated in Adams filtration. Corollary.. The ASS for BP BP collapses at E. Proof. We will show how to prove this from the theorem. Suppose we have a d r -differential in the ASS for BP BP, d r x = y Since V is concentrated in Ext, it follows that y cannot be an element of V, as y necessarily has Adams filtration at least r. Since C is concentrated in even (t s)-degree, it also follows that x cannot be an element of C. Thus, we have x V and y C. Since BP BP is a BP -module, the differentials in the ASS are linear over Ext E() (F ). So multiplying by v on the differential gives = d r v x = v y. As y C, that v y = implies y =. And so there are no differentials in this spectral sequence... The (co)module structure of (A E()). We will now describe the structure of (A E()) as a module over E(), which is necessary in order to compute Ext. To do this, we will use first describe

5 ON BP -COOPERATIONS 5 the E()-comodule structure. Recall that the coproduct on the dual Steenrod algebra A is given by ψ : A A A ; ζ n ζ i ζn i i. i n This formula shows that, when restricted to (A E()), the coproduct satisfies ψ : (A E()) A (A E()), making (A E()) into a A -comodule algebra. Modding out by (A E()) on the left gives a map α : (A E()) E() (A E()). Applying the formula for the coproduct on A, we find that and α : ζ n ζ n + ζ ζ n + ζ ζ 4 n + ζ ζ 8 n, n > α : ζ n ζ n, n =,,,. Given a locally finite E() -comodule M, we may define a E()- module structure on M via the following formula: if α(x) is given by i x i x i then define Q n x := i= Q n, x i x Thus Q i ζ n = ζn i i+. Since the Q i are primitive elements in A, we see that Q i is a derivation on (A E()), and so we have completely determined the structure of (A E()) as a module over E(). Though the module structure on (A E()) is rather simple, (A E()) is very large, making calculating the Ext groups difficult. The following subsections will develop means of breaking up (A E()) into simpler pieces. This will rely heavily on the Margolis homology of (A E()). We will now briefly review Margolis homology. Let P be a module over an exterior algebra E(x) on one generator. Then the following i. x x x P P P is a chain complex since x squares to zero. The homology of this chain complex is called the Margolis homology of P, and is denoted by M (P; x). Modules P over the algebra E() will have three different

6 6 D. CULVER Margolis homology groups, namely the ones arising from restricting P to a module over the exterior algebras E(Q ), E(Q ), E(Q ). The following theorem demonstrates the importance of the Margolis homology groups. Theorem. (Margolis, cf [], [8]). Let P be a module over E where E is an exterior algebra on a (possibly countably infinite) set of generators x, x,... so that their degrees satisfy < x < x <. If P is bounded below, then P is free if and only if all of its Margolis homology groups vanish. Recall that two modules P and Q are stably equivalent if there are free modules F and F such that there is an isomorphism P F Q F. Corollary.4. If f : P Q is a map of bounded below E-modules, then f is a stable equivalence if and only if f induces an isomorphism in all Margolis homology groups. For later subsections, we will now record the Margolis homology of BP. Proposition.5. The Margolis homology of BP is given by H (BP ; Q ) = F [ξ, ξ ] H (BP ; Q ) = F [ξ ] E(ξ i i ) H (BP ; Q ) = F [ξ i i ] (ξ 8 i i ) Proof. This is an easy generalization of the proof given in [] for the case of bu... An E()-module splitting of (A E()). Definition.6. Let m A be a monomial, say it is m = ζ i ζ i ζ i. Define the length of m to be the number of odd exponents in m: l(m) := #{k i k ()}. If x is a sum of monomials all of length l, then we say x has length l and write l(x) = l. The following lemma states that the action by a Milnor primitive on (A E()) decreases length by exactly.

7 ON BP -COOPERATIONS 7 Lemma.7. Given m A a monomial and i =,,, we have l(q i m) = l(m). Proof. This follows from the formula for the action of Q i on ζ k and the fact that Q i acts via derivations. From Adams calculation of bu bu we expect there to be infinitely many torsion classes concentrated in Adams filtration. The purpose in introducing the notion of length is to locate a large amount of the torsion inside Ext E() ((A E()) ). Recall that the group Ext E() (F, M) is the group of primitive elements in the comodule M. Translating this into the language of modules, this corresponds to elements y (A E()) for which Q y, Q y, and Q y are all zero. One source, then, for torsion elements in Ext are the bottom cells of free submodules of (A E()). Suppose that x (A E()) generates a free E()-module. This is equivalent to the element Q Q Q x being nonzero. From Lemma.7, this implies that l(x) is at least. Motivated by this observation, we define S to be the E()-submodule of (A E()) generated by monomials of length at least, S := E(){m (A E()) l(m) }. Proposition.8. The Margolis homology of S is trivial. Proof. Let x S with Q i x =. If l(x) = then we can write x = Q Q Q y for some y of length, in which case x represents in H (S; Q i ). If l(x), then by.5, it follows that there is a y (A E()) with Q i y = x. So l(y), and so y S. Thus x represents zero in H (S; Q i ). So the only interesting case is when l(x) =. For concreteness, suppose Q x =. Then there is a y (A E()) with Q y = x. If Q y =, then since l(y) =, there is a z (A E()) with Q z = y and l(z) =. Then x = Q Q z, which shows that x represents zero in H (S; Q ). This is similarly true if Q y =. So we will assume that Q y = and Q y =. Observe that if l(x) =, then there are x, x, x of length with x = Q Q x + Q Q x + Q Q x

8 8 D. CULVER and so Q x = if and only if Q Q Q x =. So we may assume that x is of the form x = Q Q x for some x with l(x ) =. This implies Q x = Q x =. So assume that Q y = and Q y =. We will modify y to produce an element y with Q y = x and Q y =. Define y := y and x := Q y. Since Q x =, we see Q x = Q Q y = Q x = Thus, there is a y with Q y = x. Note that l(y ) = and note that y = y. We now ask if Q y is zero. If it is, then we stop, otherwise we take x := Q y and note again that Q x =. Thus we find y with Q y = x. We continue this procedure, producing elements y, y,... of length two, and we stop once we reach n with Q y n =. Such an n will occur eventually because y i = y i and (A E()) is a connective algebra. Let the procedure stops at n. Then we have produced y, y,..., y n and x,..., x n with Q y i = x i, Q y i = x i. In other words, we have produced a module with generators y,..., y n. Since Q y n = and l(y n ) =, there is a z n with Q z n = y n. So define y n := Q z n + y n. Then Q y n = x n and Q (y n ) = Q Q z n + Q y n = Q y n + Q y n = x n + x n = Thus we can find z n with Q z n = y n. Define y n := y n + Q z n. Then Q y n = x n and Q y n =. Keep performing this procedure to produce an element y with Q y = x and Q y =. Then we can find z with Q z = y and so we can define y := Q z + y. Then Q y = x and Q y = Q y + Q y = x + x = Thus there is z with Q z = y. Since l(z ) =, we have z S and so x = Q Q z, which shows x represents zero in H (S; Q ). Thus we have shown H (S; Q ) =. A similar proof for H (S; Q ) works with Q replacing Q and Q replacing Q. However, the proof that H (S; Q ) = requires some adjustment. The problem is that if we were to perform an analogous procedure, say with Q and Q, then the y i s would not decrease degree, rather y i = y i + 5

9 ON BP -COOPERATIONS 9 To rectify this, we may without loss of generality restrict to the case when x belongs to some particular weight as defined in [6] (we review this concept later in section.). Since the action by Q, Q, Q preserves the weight, all the y i will have the same weight as x. The subcomodule M (k) (A E()) of weight k is finite dimensional. So the procedure we have described will terminate at a finite stage if we restrict to x M (k) for some k. Corollary.9. The submodule S is a free E()-module. Proof. As S is bounded below, Theorem. implies that S is free. Recall the following important theorem. Theorem. ([8], pg. 45). Let B be a finite Hopf algebra over a field k, and let M be a module over B. Then the following are equivalent: M is free, M is projective, M is injective. In particular, we conclude that S is an injective E()-module. Considering the short exact sequence of E()-modules S (A E()) Q. Since S is injective, there is a splitting (A E()) S Q. Thus we get a corresponding decomposition for Ext. Our goal is to show that that Ext E() (Q) is v -torsion free... The Bockstein spectral sequence for Q. We begin with an overview of how to construct the Bockstein spectral sequence. Consider the short exact sequence of modules. Q (E() E()) Q Σ 7 Q. Recall that (E() E()) is an exterior Hopf algebra E(ζ ) with ζ primitive. The first map comes from including Q into Q. The second map is the projection onto ζ Q. Applying Ext E() gives

10 D. CULVER an exact couple Ext E() (Σ 7 Q) Ext E() (Q) Ext E() (Q) Note that Ext s,t E() (Σ 7 Q) = Ext s,t 7 E() (Q) which shows that the connecting homomorphism is of bidegree (, 7). This is the correct degree for to be multiplication by v. In order to set up the BSS, we will want to show that the connecting map is indeed multiplication by v. Proposition.. For the short exact sequence of E() -comodules F (E() E()) F {ζ }, the connecting homomorphism in Ext E() induces multiplication by v. Proof. This short exact sequence of comodules induces a short exact sequence of cobar complexes C E() (F ) C E() ((E() E()) ) C E() (F {ζ }) To calculate the boundary map, let z = [a i a si ]ζ i be a cycle in the cobar complex for F {ζ }. This means that dz = d ([a i a si ] ζ ) i = i = s j= A lift of z to C E() ((E() E()) ) is [a i ψ(a ji ) a si ] ζ z = [a i a si ]ζ i

11 In the cobar complex for (E() E()), d z = d ([a i a si ]ζ ) i = i s j= ON BP -COOPERATIONS [a i ψ(a ji ) a si ]ζ + [a i a si ζ ]. Since z was a cycle, the first term is zero. So d z = [a i a si ζ ]. i Since ζ represents v in the cobar complex, and multiplication is induced by concatenation, it follows that the boundary map is indeed multiplication by v. The upshot of this proposition is that we have the following distinguished triangle in the derived category D E() of E() -comodules, Σ 7 v F [ ] F (E() E()) Σ 7 F. Tensoring with Q gives a distinguished triangle Σ 7 v Q[ ] Q (E() E()) Q Σ 7 Q. This allows us to consider the unrolled exact couple i Ext s,t E() (Q) v Exts,t 7 v v (Q) Ext s,t 4 (Q) E() E() Ext s,t E() (Q) (Q) Ext s,t 4 (Q) E() E() Exts,t 7 E s,t, E s,t 7, E s,t 4, This results in the Bockstein spectral sequence, which is of the form E = Ext E() (Q) F [v ] = Ext E() (Q). This spectral sequence is trigraded, where E s,t,r = Ext s,t E() (Q){vr },

12 D. CULVER and in this spectral sequence, E s,t,r converges to Ext s+r,t+7r (Q). The E() d k -differential has trigrading d k : E s,t,r k E s k,t 7k,r+k k and so in Adams indexing, all the differentials look like d -differentials. The utility of this spectral sequence is to prove that Ext E() (Q) is v -torsion free. Towards this end, consider the short exact sequence S Q Q where S is the E()-submodule of Q generated by length monomials (or rather images of monomials). Proposition.. The submodule S has trivial Q and Q -Margolis homology. Thus S is a free E()-submodule. Proof. Suppose that x S is such that Q i x =. So x represents an element in M (S, Q i ). We are tasked with showing that x represents the zero element. If l(x) =, then as x S there must be a y Q with l(y) = such that Q Q y = x. So x is zero in M (S ; Q i ). So suppose that l(x) =. Since the Margolis homology of Q is isomorphic to that of (A E()), and since the latter has Margolis homology concentrated in length zero, it follows that there is a y Q with Q i y = x. Since l(x) =, then l(y) =, and so y S. Thus showing that x is zero in the Margolis homology group M (S ; Q i ). If l(x) = and Q i x =, then x = in Q. This shows that the Margolis homology groups of S are both zero, and so by Theorem., the module S must be free over E(). Corollary.. There is a splitting of Q as an E()-module and thus we get a splitting Q S Q Ext E() (Q) Ext E() (S ) Ext E() (Q). To prove that Ext E() (Q) is v -torsion free, we will show that the v -BSS collapses at E. Since all the differentials look like d - differentials in Adams indexing, this will follow if we can prove Ext E() (Q) is concentrated in even (t s)-degree. Corollary.4. The groups Ext E() (S ) are concentrated in even (t s)- degree.

13 ON BP -COOPERATIONS Proof. Observe that if x (A E()) is a length monomial, then x = mζ i ζ j for some monomial m of length and i = j. Observe that every length monomial is in even degree. The degree of ζ k is k. So the degree of ζ i ζ j is ζ i ζ j = i + j which is even. Thus length elements of (A E()) are concentrated in even degree, and this remains true when projecting to Q. If x Q generates a free E()-module M, then the unique nonzero element in Ext E() (M) lives in degree x 4. Thus if x is in even degree, it determines an element in Ext, (S E() ) in even degree. Combining all these observations shows that Ext E() (S ) is concentrated in even degree. Proposition.5. The Ext-groups of Q are concentrated in even (t s)- degree. Proof. Since S is free, the Q and Q -Margolis homology groups of Q and Q are isomorphic. Consequently, the Margolis homology groups of Q are concentrated in length, and so in even degree. Observe that for x Q to generate a free E()-module, that the length of x must be. So Q has no free summands. By the classification theorem of modules over E() (cf. [], pg. 45), it follows that Q is a direct sum of finite lightning flash modules, a consequence of which is that Ext E() (Q) is a direct sum of Adams covers of Ext E() (F ). The Adams cover associated to a lightning flash in Q is uniquely determined by the two unique nonzero elements of its Margolis homology. In the case of Q, the Margolis homology is in even (t s)-degree, and this forces the Adams covers to be in even (t s)- degree. Corollary.6. The BSS for Q collapses at E. v -torsion free. Thus Ext E() (Q) is We can now prove Theorem.. Proof of Theorem.. We have shown that there is a splitting of E()- modules (A E()) S Q

14 4 D. CULVER and so applying Ext E() gives a decomposition Ext E() ((A E()) ) Ext E() (S) Ext E() (Q). We have shown that S is free, and so Ext E() (S) is concentrated in Ext. We have also just shown that Ext E() (Q) is v -torsion free. So we define V := Ext E() (S), C := Ext E() (Q). The previous two propositions show that C is concentrated in even (t s)-degree. This completes the proof of Theorem.. Corollary.7. The Adams spectral sequence for BP BP collapses at E. Remark.8. Note that the v -BSS for Q has many hidden extensions. Remark.9. A consequence of the discussion thus far is that the Ext E() (Q) is generated as a module over F [v, v, v ] by elements in Ext, E(). Remark.. One could attempt to generalize the above arguments to the spectra BP n BP m when m n. In this case, the homology of BP n is (A E(n)). One would like to show that Ext E(n) ((A E(m)) ) splits into a v n -torsion summand concentrated in Adams filtration, and a v n -torsion free summand. Towards this end, one could define S(m, n) (A E(m)) to be the E(n)-submodule generated by monomials of length at least n +, in analogy with S above. If it could be shown that S(m, n) is a free module over E(n), then we would have a splitting of E(n)- modules (A E(m)) = S(m, n) Q(m, n). An inductive argument with the v n -BSS would then allow one to show that Ext E(n) (Q(m, n)) is v n -torsion free. In the case n = m =, S(, ) coincides with the submodule S defined above. The problem is that the arguments presented hitherto to show that M (S; Q i ) = do not seem to generalize to other values of m and n.

15 ON BP -COOPERATIONS 5.4. Topological splitting. In the previous subsection, we established a decomposition π (BP BP ) = V C where V is the F -vector space of v -torsion elements and C is v - torsion free. In this section, we will establish that this splitting of homotopy groups in fact lifts to the stable homotopy category. That is, we will show that there is an equivalence of spectra BP BP HV C with HV the Eilenberg-MacLane spectrum with π (HV) = V and π C = C. Let X denote BP BP. We will establish this spectrum level splitting by showing there is a map X HV and defining C to be the fibre. This will produce a fibre sequence C X HV. We will show that there is a section to the map X HV. In the previous section, we established a splitting (A E()) = S Q with S a free E()-module. Dualizing gives a decomposition (.) A E() = S Q and S is free as an E()-module. In applying the change of rings theorem for an A-module M, one has to use the sheering isomorphism A E() F M A E() M. where the left hand side is endowed with the diagonal action. In the case of H X, we have the isomorphism A E() A E() A E() (A E()). Coupled with the decomposition., we see that as a module over A, the cohomology H X decomposes as H X (A E() S ) (A E() Q ).

16 6 D. CULVER As S is free as an E()-module, the first factor is free as an A- module. Let us denote this free factor by F. Note that H (HV) is precisely F. The idea is to show that the maps F H X F in the splitting of H X lift to maps of spectra via the Adams spectral sequence. Consider the Adams spectral sequence Ext A (F, H X) = [X, HV]. Since F is free as an A-module, the E -page is concentrated in Adams filtration, and so it collapses at E. Note that Ext A (F, H X) = hom A (F, H X), so the inclusion of F into H X determines a map of spectra X HV. For the map in the other direction, we shall use the Adams spectral sequence again, Ext A (H X, F) = [HV, X]. For this spectral sequence, we can apply the change-of-rings isomorphism on the E -term, Ext A (H X, F) Ext E() (A E(), F). By Theorem 4.4 in [9], A is free over E(). Since S is also locally finite, it follows that F = A E() S is locally finite. Thus F is a locally finite free E()-module. If {b α } is an E()-basis, then F = α since F is locally finite. Thus Ext s, E() ((A E()), F) E(){b α } = E(){b α } α α Ext s, E() ((A E()), E(){b α }). Since E() is self-injective, it follows that each component group on the right-hand side is zero when s >. Thus the Ext s groups are trivial for s >. So the E -page of the ASS is concentrated in Adams filtration, and hence collapses. Therefore we have the desired map of spectra. Thus we get the section of the cofibre sequence C X HV

17 and hence the desired splitting ON BP -COOPERATIONS 7 X C HV. Remark.. If we could prove the analogous splitting for Ext E(n) ((A E(m)) ) when m n, then the above argument could be used to show that BP n BP m splits as a spectrum into an analogous wedge C(n, m) HV(n, m).. Calculations In this section we develop techniques to provide an inductive calculation of Ext E() ((A E()) )... Brown-Gitler (co)modules. The majority of this and the following three sections are adapt the techniques of [] to our setting. Let E(n) denote the sub-hopf algebra of the Steenrod algebra which is generated by the first n + Milnor primitives, Q,..., Q n. Let E(n) denote the dual of this algebra. This will be a quotient of the dual Steenrod algebra and it is given by E(n) = E(ζ,..., ζ n+ ) where the ζ i s are the images of ζ i in A. In E(n), these elements are primitive, and so E(n) is a self-dual Hopf algebra. Following [6], we define a weight filtration on A which induces a filtration on the A -subcomodule (A E(n)) = A E(n) F F [ζ,..., ζ n+, ζ n+,...]. We define the weight of the generators ζ k by and extend multiplicatively by wt(ζ k ) := k wt(xy) := wt(x) + wt(y). The Brown-Gitler comodule N i (j) is the subspace of (A E(i)) spanned by elements of weight less than or equal to j. From the coproduct formula for the dual Steenrod algebra (.) ψ(ζ k ) = ζ i ζ i j i+j=k

18 8 D. CULVER we see that N i (j) is an A -subcomodule of (A E(i)). The algebra (A E(i)) can also be regarded as a comodule over E(i), in fact it is a comodule algebra. For ease of notation, we shall write For i =, we shall write and for i = we write BP i j := N i (j). bu j := N (j), HZ j := N (j) As in [], we can define a map of ungraded rings ϕ i : (A E(i)) (A E(i )) which is defined on generators by ϕ i : ζ l k {ζ l k k > k = and extended multiplicatively. So, for example, ϕ (ζ 4 ζ 5 ) = ζ ζ 4. Lemma.. The map ϕ i is a map of ungraded E(i) -comodules. Proof. Since (A E(i)) is generated by {ζ,..., ζ i+, ζ i+,...}, it is enough to check that ϕ i commutes with coaction on these generators. This follows immediately from the coproduct formula (.) and the fact that E(i) is exterior. Let M i (j) denote the subspace of (A E(i)) spanned by the monomials of weight exactly j. Observe that the coaction on (A E(i)) (as a E(i) -comodule) preserves the weight. Thus the subspaces M i (j) are E(i) -subcomodules. In particular we have shown Proposition.. There is a splitting of E(i) -comodules (A E(i)) j M i (j) Lemma.4. For i >, the map ϕ i maps the subspace M i (j) isomorphically onto N i ( j/ ).

19 ON BP -COOPERATIONS 9 Proof. Given a monomial in M i (j), it can be written as ζ lx where x is a monomial which is a product of ζk i for k. In this case, the weight of x is j l. Observe that ϕ i (ζ l x) = ϕ i(x) and the weight of ϕ i (x) is j l. Write x as Then the weight of x is x = ζ i ζ i ζ i 4 4. wt(x) = i + 4i + 8i n i n +. Since i >, it follows that i is even, and hence wt(x) is divisible by 4. It follows that j l is divisible ( ) by 4, whence j l is divisible by. So ϕ i (x) j l belongs to M i. This shows that ϕ i maps the subspace spanned by monomials of the form ζ l ) x isomorphically. Letting l vary, we see that the image ( j l onto the subspace M i of ϕ i restricted to M i (j) maps isomorphically onto N i ( j/ ). Remark.5. The inverse to the isomorphism in the previous lemma is given by ϕ i : N i ( j/ ) M i (j); ζ i ζ i ζ a ζi ζ i where a = j wt(ζ i ζ i ). Corollary.6. There is an isomorphism of graded E(i) -comodules M i (j) Σ j N i ( j/ ). Corollary.7. There is an isomorphism of E(i) -comodules M i (j) Σ M i (j + ) which is given by multiplication by ζ. In light of Corollary.6, we will always make the identification in the rest of this paper. M (j) Σ 4j N (j)

20 D. CULVER.. Exact sequences. Inspired by [], we construct exact sequences relating the Brown-Gitler comodules N (j) and M (j). Recall that Consider the F -linear map (E() E()) E(ζ ). τ : (A E()) (A E()) (E() E()) defined on the monomial basis by ζ i ζi ζi +ɛ ζ i 4 4 ζ i ζi ζi ζi 4 4 ζ ɛ. Note this is not a map of E() -comodules. For example, in (A E()), there is the coaction α(ζ ) = ζ + ζ ζ + ζ ζ 4 + ζ whereas in (A E()) (E() E()), we have α( ζ ) = ζ + ζ. However, we do have that τ is an isomorphism of F -vector spaces. Following [], we will put a decreasing filtration on (A E()). Define F j (A E()) := τ M (k) (E() E()) k j which gives the decreasing filtration (A E()) = F (A E()) F (A E()) F (A E()). The following observation will be useful in later arguments, Observation.8. Let x be a monomial in (A E()). If x F j ((A E()) ), then wt(x) is bounded below by j. If the power of ζ in x is odd, then its weight is bounded below by j + 4. The coproduct formula (.) shows that this is a filtration by E() -subcomodules. Passing to filtration quotients gives a map on the associated graded comodule algebra (.9) τ : E (A E()) (A E()) (E() E()) which is in fact a map of E() -comodules. For example, in the example of α(ζ ) above, the terms which prevented the map τ from being a comodule map were the terms ζ ζ and ζ ζ 4. Note that ζ F, whereas ζ and ζ4 are both in F. In general, the coproduct

21 ON BP -COOPERATIONS formula shows that the coaction of an element x of (A E()) is the same as the coaction on τ(x) modulo elements of higher filtration. Proposition.. The map (.9) is an isomorphism of E() -comodules. Proof. To check that this is a map of E() -comodules, it is enough to check that it is a map of E()-modules. So let x be a monomial in (A E()), say x = ζ i ζi ζi +ɛ ζ i 4 4. Then as Q i acts via a derivation, we have that (.) Q i x = ζ i ζi Q i(ζ i +ɛ )ζ i ζ i ζi ζ i j j (Q iζ i j j )ζi j+ j+. j> When ɛ =, the action by Q i commutes with the action of E() on (A E()) (E() E()), since the action by Q i preserves the weight. Suppose then that ɛ =. Then Q i τ(x) = Q i (ζ i ζi ζi ζi 4 4 ζ ). Since Q i is a derivation, and since the target of τ is endowed with the diagonal E()-action, we can rewrite this as (.) Q i (τ(x)) = ζ i ζi ζi ζi 4 4 Q i ζ + ζ i ζi ζi ζ i j j (Q iζ i j j )ζi j+ j+ ζ. j> Note that τ carries the sum in (.) to the sum in (.). When i =,, then Q i ζ = in (E() E()), and so the first term in (.) vanishes. In (A E()), Q i ζ =, but the first term in (.) does land in strictly higher filtration than x, and so the first term in (.) vanishes in the associated graded E (A E()). When i =, then in both E (A E()) and (E() E()), Q ζ =. In this case, τ carries the first term in (.) to the first term in (.). This shows that τ is a comodule map in the associated graded comodule. To see that this map is an isomorphism, just note that τ is an F - isomorphism and induces a linear isomorphism on the associated graded. So τ is an isomorphism of comodules. Define quotients Q j (A E()) := (A E()) /F j+ (A E()),

22 D. CULVER then this is an E() -comodule and it inherits a filtration from (A E()). The map τ induces an isomorphism of F -vector spaces τ : Q j (A E()) BP j (E() E()) which induces an isomorphism of associated graded E() -comodules, τ : E Q j (A E()) BP j (E() E()). Lemma.. There is a short exact sequence of E() -comodules (.4) Σ 4j bu j bu bu j+ Q j (A E()) Proof. Observe there is a commutative diagram BP j E(ζ ) (A E()) E(ζ ) (A E()) τ. bu j+ This factorization arises because if x BP j, then the image of x ζ ɛ in (A E()) has weight bounded by wt(τ (x ζ ɛ )) 4j + 4 = 4j +, and hence must lie in bu j+. The projection of (A E()) onto Q j (A E()) restricts to give a surjection ρ : bu j+ Q j (A E()). Note that the kernel of ρ is the intersection of F j (A E()) with bu j+. So let x be a nonzero element of this intersection. We may suppose that x is in fact a monomial, say it is the monomial ζ i ζi ζi +ɛ ζ i 4 4. Since x is an element of bu j+, its weight is bounded above by 4j +. Since x F j (A E()), Observation.8 implies that wt(x) is at least 4j. Observation.8 also implies that if ɛ =, then the weight of x is bounded below by 4j + 4, which is a contradiction. Thus ɛ =. All of this implies that the kernel of ρ is ker ρ = M (j) bu where bu is the subcomodule F {, ζ } of (A E()). By.6, we get the desired short exact sequence. Lemma.5. There is an exact sequence of E() -comodules (.6) Σ 4j bu j bu j Q j (A E()) Σ 4j+5 bu j

23 ON BP -COOPERATIONS Proof. As an F -vector space, the image of bu j in (A E()) E(ζ ) under τ is τ(bu j ) = (BP j E(ζ )) (M (j ) F {}) (M (j) F {}) which gives the following exact sequence on the level of F -vector spaces ϕ ψ M (j) bu j Q j ω (A E()) M (j ) F {ζ }. The map ψ is defined by the following commutative diagram bu j ψ Q j (A E()) = τ ψ bu j BP j E(ζ ) where ψ is defined on monomials by ψ : ζ i ζi ζi +ɛ ζ i 4 4 ζ i ζi ζi ζi 4 4 ζ ɛ. We need to check that the maps are maps of E() -comodules. Let us start with ω. Note the commutative diagram τ ker π ker π τ Q j (A E()) BP j E(ζ ) ω π = M (j ) F {ζ } M (j ) F {ζ } where π is the projection. So in order to check that ω is a comodule map, we just need to check that τ ker π is a subcomodule of Q j (A E()). That ω is a comodule map follows then follows from the following lemma. Lemma.7. τ ker π is a subcomodule of Q j (A E()).

24 4 D. CULVER Proof. First note that ker π = (BP j E(ζ )) (M (j ) F {}). Consider a monomial x = ζ i ζi ζi +ɛ ζ i 4 4 in τ ker π. So that τ(x) = ζ i ζi ζi ζi 4 4 }{{} define y to be this ζ ɛ In particular, the weight of y is bounded above by 4j, and when the weight of y is 4j, we must have ɛ =. Moreover x = yζ ɛ. Applying any of the Milnor primitives Q i to x gives Q i x = Q i (y)ζ ɛ + yq iζ ɛ. Applying τ to this expression gives (.8) τ(q i x) = Q i y ζ ɛ + (yq iζ ɛ ). It needs to be checked that both terms are in the kernel of π. First suppose that ɛ =. Then x = y is an element of BP j, and in this case the action of the Milnor primitives preserves the weight, and so τ(q i y) = (Q i y) is an element of ker π. Suppose then that ɛ =, in which case wt(y) 4j 4. First consider the case when wt(y) < 4j 4. Then wt(yq i ζ ) < 4j for i =,,. This implies that (yq i ζ ) BP j F {}. Since the action by the Q i s preserves the weight of y, we find then that both terms on the right hand side of (.8) lie in ker π. We are thus left with checking the case when wt(y) = 4j 4. When i = or i =, then yq i ζ has weight 4j, and so is zero in Q j (A E()). Thus, for i =,, τ Q i x = (Q i y)ζ Q i y ζ, and since the weight of Q i y is still 4j 4, this is an element of ker π. Finally, we have which is mapped under τ to Q (x) = Q (yζ ) = Q (y)ζ + y Q y ζ + y. The both terms lie in BP j E(ζ ), and so belong to ker π. This completes the proof that τ ker π is a E() -subcomodule.

25 ON BP -COOPERATIONS 5 It is clear that the map ϕ is a E() -comodule map. Indeed, we can regard M (j) as a subspace of bu j. If x M (j) is a monomial, then an odd power of ζ cannot occur in x. So the action of Q, Q, Q on x preserves the weight, and consequently lies in M (j). Thus M (j) is a subcomodule of bu j. Finally, we check that ψ is a map of comodules. Let K denote the cokernel of ϕ. Then we have an induced morphism of short exact sequences of E() -comodules ϕ ψ M (j) F {} bu j K ψ M (j) E(ζ ) Q j (A E()) Q j (A E()) Since ψ is the composite ψ ψ, we may conclude that ψ is a comodule map. Remark.9. The quotients Q j (A E()) have finite filtrations projected from the filtration on (A E()). Applying Ext E() to this filtration produces a spectral sequence E = Ext E() (E Q j (A E()) ) = Ext E() (Q j (A E()) ). Since E Q j (A E()) is isomorphic to BP j (E() E()) as an E() -comodule, we have that the E -page is Ext E() (BP j ) which we know consists of v -torsion elements on the -line and a v - torsion free component concentrated in even (t s)-degree. The spectral sequence is a linear over Ext E() (F ), which implies that this spectral sequence collapses. Consequently, for the purposes of the inductive calculations in the following section, we will regard Q j (A E()) as BP j (E() E())... Inductive calculations. In the last section, we produced the exact sequences of comodules (.4) and (.6). As in [4], we regard these as providing spectral sequences converging to Ext E() (bu j+ ) and Ext E() (bu j ) respectively. Following [4], we write Mi [k i ] = M to denote the existence of a spectral sequence s k Ext i,t+k i (M E() i ) = Ext s,t (M). E()

26 6 D. CULVER We shall abbreviate M i [] by M i. Below, we shall always be identifying bu j with M (j) via the isomorphism ϕ : M (j) Σ 4j bu j. The exact sequence (.4) gives a spectral sequence (.) ) Σ 8j+4 Q j (A E()) (Σ j+4 bu j bu = Σ 8j+4 bu j+ and (.6) gives a spectral sequence (.) Σ j bu j Q j (A E()) Σ j+5 bu j [] = Σ 8j bu j In the exact sequences, there were the comodules Q j (A E()), and in Remark.9, it was pointed out that Ext E() (Q j (A E()) ) Ext E() ((E() E()) BP j ) Ext E() (BP j ). In order to carry out the computation, we must then calculate Ext E() of BP j. Lemma.. For any j, there are isomorphisms BP j E() M (k) E() Σ k bu k/ k j k j E() j k= k/ l= Σ k+l HZ l/ Proof. The first isomorphism just follows from Proposition.. From an application of Corollary.6 we obtain M (k) E() Σ k bu k/ which gives the second isomorphism. Applying Proposition. and Corollary.6 again, but with i = gives the third isomorphism. Example.. We have M (4) E() Σ 8 bu E() Σ 8 (HZ Σ HZ Σ 4 HZ ). From [], we have Ext E() (HZ k )/v -torsion Ext E() (F ) k α(k)

27 ON BP -COOPERATIONS 7 where Ext(M) n denotes the nth Adams cover of Ext(M). In [], this is a consequence of the fact that the Margolis homology groups M (HZ k ; Q ) and M (HZ k ; Q ) are both one dimensional. Observe that when v is inverted then (.4) v Ext E() (HZ k ) v Ext E() (F ) k α(k) = F [v ±, v ] F M (HZ k ; Q ) Example.5. Below, we illustrate this isomorphism for HZ. 4 6 In the chart, blue indicates elements obtained by inverting v. Note that as a module over F [v ±, v ] that this Ext-group is generated by the element in (, ). Since v is inverted, this module is also generated by the element in (, ), which is precisely the nonzero element of M (HZ ; Q ). As a first step to determining Ext E() (BP j ), we will compute v -inverted Ext groups first. This will allow us to locate the starting points of the Adams covers within the integral Ext groups. Proposition.6 ([4]). We have the isomorphism v Ext E() (BP j ) F [v ±, v ]{ζ i ζi i + i j} Proof. Given an A-comodule M, there is an isomorphism v Ext A (M) v Ext A() (M). This is an algebraic analogue of Serre s theorem that rational stable homotopy is the same as rational homology. Consider the following

28 8 D. CULVER sequence of isomorphisms v Ext E() (BP j ) v Ext A ((A E()) BP j ) v Ext A() ((A E()) BP j ) v Ext A() ((A E()) ) F [v ± ] v Ext A() (BP j ) where the last line follows from the Künneth formula. Calculating v Ext A() is extremely simple, it is the free F [v ±]-module generated by the Q -Margolis homology of M, v Ext A() (M) = F [v ± ] F M (M; Q ). From [], the Q -Margolis homology of (A E()) is M ((A E()) ; Q ) = F [ζ ]. Since the action by Q preserves the weight on (A E()), there is an associated weight filtration on which implies that M (BP ; Q ) = F [ζ, ζ ] M (BP j ; Q ) = F {ζ i ζi i + i j}. In the v -inverted Adams spectral sequence v Ext A() ((A E()) ) = HQ bu = Q [v ] ζ is detecting v. Thus we get the desired isomorphism. Remark.7. Each of the monomials in v Ext E() (BP j ) determines an Adams cover of Ext E() (F ). Here is an algorithm for determining the the Adams covers associated to a monomial ζ i ζi : () If i 4, then the next element in Adams cover is ζ i ζi 4 ζ, and there is the relation v ζ i ζi = v ζ i ζi 4 ζ. If i 4 < 4, then the process terminates. () If i 4 4, then repeat the previous step. Continue this until the exponent of ζ is or. () Perform the previous steps on ζ until the exponent on ζ is or. Then continue onto ζ 4, ζ 5,... and so on until the process terminates.

29 ON BP -COOPERATIONS 9 Observe that in the spectral sequences (.) and (.) that upon inverting v, all the terms in the E -page are in even degree. So these spectral sequences collapse after v -localization. We also know from our general structural results that there can be no differential originating from a torsion class and hitting a v -torsion free class. Thus we have Proposition.8. In the spectral sequences (.) and (.), the only nontrivial differentials must be between torsion classes. Consequently, the spectral sequences (.) collapse immediately. Proof. Since the v -torsion free component is concentrated in even (t s)-degree, there are no differentials between v -torsion free classes. Recall that we had the decomposition and that the BSS (A E()) = S Q Ext E() (Q) F [v ] = Ext E() (Q) collapses. The latter implies that Ext E() (Q) is generated by the elements in Ext, as a module over F [v, v, v ]. Thus Ext E() (bu j ) is generated as a module over F [v, v, v ] by elements in Ext,. E() Now consider one of the spectral sequences (.) or (.). Note that the differentials are linear over F [v, v, v ]. So if there were a differential d(x) = y where x is a torsion class and y is v - torsion free, then v y would be a permanent cycle in the spectral sequence. But this would contradict that Ext E() (bu j ) is generated over F [v, v, v ] by elements in Ext,. A similar argument shows that there cannot be a differential d(y) = x. A consequence of this is that in the inductive calculations, we can essentially ignore the torsion classes on the E -page. We will now carry out the inductive calculations. We begin with some remarks on the v -inverted calculations, starting with developing analogues of Lemmas 4.9 and 4. of [4]. Since the v -inverted versions of (.6) and (.4) collapse at E, we get summands (.9) v Ext E() (Σ 8k BP k ) v Ext E() (Σ 8k bu k ) (.) v Ext E() (Σ 8k+4 BP k ) v Ext E() (Σ 8k+4 bu k+ )

30 D. CULVER We will identify the generators of these summands. This is accomplished by contemplating the following portion of the 8k-fold suspension of the exact sequence (.6): Σ 8k bu k Σ 8k (E() E()) BP k. M (4k) Let ζ i ζi v Ext E() (BP k ), then in the diagram above we have ζ i ζi ζ i ζi ζ aζi ζi where a := 8k 4i 8i. Similarly, we could contemplate the diagrams below coming from (.4): Σ 8k+4 bu k+ Σ 8k+4 (E() E()) BP j M (4k + ) This results in Lemma.. The summands (.9) and (.) are generated as modules over F [v ±, v ] by elements ζ a ζi ζi with i + i j and a = 8j 4i 8i (resp. a = 8j + 4 4i 8i ). Next we need to determine the generators arising from the terms Σ 4j bu j in the case of (.6) and Σ 4j bu j bu in the case of (.4). Because of Proposition.8, we obtain summands (.) Ext E() (Σ j bu j )/v tors Ext E() (Σ 8j bu j ) (.) Ext E() (Σ j+4 bu j bu )/v o tors Ext E() (Σ 8j bu j+ )

31 ON BP -COOPERATIONS Proposition.4. Assume inductively that that the Ext E() (Σ 4j bu j ) has generators of the form {ζ i ζ i }. Then the summand of (.) has generators of the form {ζ i ζ i } and the summand (.) has generators {ζ i ζ i } {ζ 4, ζ }. The proof of this proposition follows by considering the diagrams Σ k bu k Σ 8k bu k Σ 4k M (k) M (4k) Σ k+4 bu k bu Σ 8k+4 bu k+ Σ 4k M (k) bu M (4k + ) For example, in the first diagram, given a monomial ζ i Ext E() (Σ 4k bu k ), we would obtain ζ i ζ i ζ a ζi ζ i ζ i in ζ i ζ i ζ i ζ i the right hand vertical arrow follows from the fact that a = 4k wt(ζ i ζ i ) = i. There remains two questions regarding these inductive calculations: What is the role of the summand Ext E() (Σ k+5 bu j ), and how does one determine the v -extensions in the spectral sequences on summands (.9) and (.)? The following lemma will be useful. Lemma.5. The composite τ M (j ) E(ζ ) (A E()) Q j (A E()) is a map of E() -comodules.

32 D. CULVER Proof. Denote the composite by χ. Consider an element y ζ ɛ in M (j ) E(ζ ). Then we need to check that Q i yζ ɛ = χ(q i(y ζ ɛ )). Note that Q i (y ζ ɛ) = (Q iy) ζ ɛ + y Q iζ ɛ and under χ this is χ(q i (y ζ ɛ )) = (Q iy)ζ ɛ + yq iζ ɛ. If i =,, then yq i ζ ɛ lies in Fj (A E()), and so is zero in Q j (A E()). Moreover, in M (j ) E(ζ ), Q i (y ζ ɛ ) = (Q iy) ζ ɛ. And so χ commutes with Q and Q. For the case of Q, if ɛ =, there is nothing to check. So let ɛ =. Then which is sent under χ to In Q j (A E()), one has Q (y ζ ) = (Q y) ζ + y χ(q (y ζ )) = (Q y)ζ + y. Q (yζ ) = (Q y)ζ + y as Q acts as a derivation and Q ζ =. Thus Q also commutes with χ. Corollary.6. We have the following diagram, where the rows are exact, in the category of E() -comodules, Σ 4j bu j bu j Q j (A E()) Σ 4j+5 bu j M (j ) M (j ) E(ζ ) M (j ) F {ζ } Corollary.7. Consider the summand v Ext E() (Σ j bu j ) v Ext E() (Σ 8j BP j ) v Ext E() (Σ 8j bu j ), generated over F [v ±, v ] by the generators ζ 4 ζi ζi (A E()) where i + i = j. In the summand (.8) v Ext E() (Σ j+5 bu j []) v Ext E() (Σ 8j bu j )

33 ON BP -COOPERATIONS let x i for ( i j ) denote the generator of (.8) corresponding to ζ i. Then in the E -page of the spectral sequence (.), we have v ζ 4 ζi ζi = x i We will now discuss the v -extensions concerning the other generators in the summand in (.), v Ext E() (Σ 8j BP j ) v Ext E() (Σ 8j bu j ), and the generators of the summand v Ext E() (Σ 8j+4 BP j ) v Ext E() (Σ 8j+4 bu j+ ) in (.). That is the monomials of the form ζ a ζi ζi (A E()) where i + i j and a = 8j 4i 8i and a > 4 (resp. a = 8j + 4 4i 8i ). The notion of length discussed in section induces an increasing filtration of (A E()) by E() -comodules, G l (A E()) := E(){m (A E()) l(m) l}. Since the action by Q i lowers length by exactly one, it follows that the filtration quotients are trivial E() -comodules G l /G l = F {m l(m) = l}. Applying Ext E() gives a spectral sequence converging to Ext E() ((A E()) ). This spectral sequence is of the form E s,t,l = E (A E()) F [v, v, v ] = Ext E() ((A E()) ) and we call it the length spectral sequence. By examining the induced short exact sequences in cobar complexes, one easily derives that the d -differential is d (x) = v (Q x) + v (Q x) + v (Q x) which implies that Ext E() ((A E()) ) has the following class of relations, v (Q x) = v (Q x) + v (Q x). In particular, if m is a length monomial, then we have the following relations in Ext: (.9) v (ζ 8 m) = v (ζ 4 m) + v (ζ m).

34 4 D. CULVER This suggests that given a monomial m of length in we should have the relations Ext E() (Σ 8j+4ɛ bu j+ɛ ), (.4) v (m) = v (ζ 8 ζ m) + v (ζ 8 ζ4 m) Of course, under the hypothesis of Corollary.7, this does not make sense, but the following simple observation shows that this is a possibility for all other monomials in the summands v Ext E() (Σ 8j+4ɛ BP j ). Lemma.4. For the generators ζ a ζi ζi v Ext E() (Σ 8j BP j ) v Ext E() (Σ 8j bu j ) the exponent a is always divisible by 4 and a 4. For the generators ζ a ζi ζi v Ext E() (Σ 8j+4 BP j ) v Ext E() (Σ 8j+4 bu j+ ) the exponent a is always divisible by 4 and a 8. In the next several propositions, we prove that when m = ζ aζi ζi is a generator of v Ext E() (Σ 8j+4ɛ BP j ), then the other monomials occuring in (.4) are also generators of v Ext E() (Σ 8j+4ɛ bu j+ɛ ). The proofs are fairly direct; one simply breaks into several cases and makes sure certain inequalities hold. Proposition.4. Consider a monomial generator ζ a ζi ζi in v Ext E() (Σ 8j BP j ) and suppose that a > 8. Then the monomials are generators. ζ a 8 ζ i ζi +, ζ a 8 ζ i +4 ζ i v Ext E() (Σ 8j bu j ) Proof. Under the hypothesis of the proposition, we have i + i j and a = 8j 4i 8i Since the weights of the proposed monomials are still 8j, all that needs to be checked is that i + i + j

35 ON BP -COOPERATIONS 5 Since a = 8j 4i 8i > 8 we have j i i > and hence i + i < j. Therefore, i + i + < j which proves the proposition. Proposition.4. Consider a monomial generator ζ 8ζi ζi in the summand v Ext E() (Σ 8j BP j ). Then, the monomials ζ i ζi +, ζ i +4 ζ i are generators in the summand v Ext E() (Σ j bu j ). Proof. It needs to be checked that the monomials ζ i +4 ζ i and ζ i ζi + are generators of v Ext E() (Σ 4j bu j ). The proof will be broken up into several different cases. We will begin with the case in which i =. So assume that i =, we will consider two further sub-cases. Let k = j/. Suppose first that k = j. In this case, there is the exact sequence Σ k bu k Σ 8k bu j Σ 8k (E() E()) BP k Then we need to show the following () i + 4i + 4 = 8k = 4j () i + k = j. Since ζ 8 ζi ζi is a generator of the summand we know that v Ext E() (Σ 8j BP j ), 8 + 4i + 8i = 8j Σ 4k+5 bu k. which shows the first condition. Observe that this implies i is even. From this equality, we can write 8i + 8 = 8j 4i

36 6 D. CULVER which dividing by 8 and using the fact that i is even shows i + = j i j, showing the second condition. So consider the case when k = j. Then we have a sequence Σ k bu k bu j Σ 4j bu j (E() E()) BP k. We will consider the monomials ζ i +4 ζ i and ζ i ζi + separately. Consider first ζ i +4 ζ i. For this to be a generator of v Ext E() Σ 4j bu j, it would have to be a generator originating in the summand So we need to show () i i = 8k + 4 = 4j () i k = j v Ext E() (Σ 4j BP k ). The first condition follows as before. We can again write which dividing by 8 gives 8i = 8j 4i 8 i = j i j since i = and i is even. So consider now the monomial ζ i ζi +. There are two further sub-cases to consider for this monomial. Suppose first that i >. Then if this monomial is to be a generator of v Ext E() (Σ 4j bu j ), it would have to originate from the summand We thus need to check that () i + 4i + 4 = 8k + 4 = 4j () i + k = j. v Ext E() (Σ 4j BP k ). The first condition follows as before. For the second condition, note gives 8i = 8j 4i 8 i = j i.

37 ON BP -COOPERATIONS 7 Since we are assuming i >, then i / >, which implies i j as desired. So consider then the case when i =. In this case, for the monomial to be a generator of v Ext E() (Σ 4j bu j ), it would have to originate from the summand v Ext E() (Σ jk bu k bu ). Thus we need to check that ζ i + is a generator of v Ext E() (Σ 4k bu k ). For this to be true, it needs to be the case that Writing i + = k = j. 8i = 8j 4i 8 = 8j 6 and dividing by 8 shows that indeed i = j. This shows that ζ i + is indeed a generator of v Ext E() (Σ 4k bu k ). This completes the case when i =. So consider the final case when i =. Then we need to show that the monomials ζ 4ζj and ζ j are generators of v Ext E() (Σ 4j bu j ). Let k := j/. We again need to separate into the subcases when k = j and k = j. Suppose first that k = j. Then we have the sequence Σ k bu k Σ 8k bu j Σ 8k (E() E()) BP k Σ 4k+5 bu k. In this case, the monomial ζ 4ζj would have to be a generator originating from the summand v Ext E() (Σ 4j BP k ). In order to check that it is indeed a generator, it just needs to be observed that and that 4 + 4j 4 = 4j j = k. For the monomial ζ j to be a generator, it would have to be a generator originating from v Ext E() (Σ 4k bu k ), i.e. we need to check that ζ j is a generator for v Ext E() (Σ 4k bu k ). This is immediate since j = 4k.