NUMERICAL INVARIANTS OF PHANTOM MAPS C. A. MCGIBBON AND JEFFREY STROM Abstract Two numerical homotopy invariants of phantom maps, the Gray index G(f) and the essential category weight E(f), are studied. The possible values of these invariants are determined. In certain cases bounds on these values are given in terms of ra- tional homotopy data. Examples are provided showing that the Gray index can take any positive finite value. For certain cases it is shown that every essential phantom f : X ! Y has finite Gray index. However it is also shown that there exist spaces, e:g: CP 1, which are the domains of essential phantoms with infinite index. The same type of analysis is carried out on the essential category weight of a phantom map. If the loop space of X is homotopy equivalent to a finite complex, then every phantom f : X ! Y has E(f) = 1. However, in certain other cases it is shown that E(f) is strictly less than the rational Lusternik-Schnirelmann category of the domain. A homotopy classification of phantoms f : K(Z; n) ! Sm is given along with the values of E(f). The invariants G and E provide decreasing filtrations on the set of homotopy classes of phantoms from X to Y . A third filtration on this set is introduced for certain special targets. When the rational cohomology of the domain X is finitely generated, this filtration enables one to reduce the search for essential phantoms (into finite type targets) to a finite list of spheres. By a phantom map we mean a pointed map f from a CW -complex X to another space Y with the property that the restriction of f to the n-skeleton, f|Xn, is nullhomotopic for each natural number n. In this paper we study certain numerical invariants associated with the homotopy class of such a map, namely the Gray index and essential category weight. These invariants give rise to decreasing filtrations on the set Ph(X; Y ) of pointed homotopy classes of phantom maps from X to Y . ____________ Date: January 25, 2000. 1 2 C. A. MCGIBBON AND JEFFREY STROM 1. The Gray Index The first invariant is one we call the Gray index of a phantom map f and we denote it by G(f). It was first defined by Brayton Gray in his thesis [3]. Since f is a phantom map, one has a factorization of the following sort for each natural number. f X FF_______________//Y< denotes the 2n-connective cover of the sphere S2n. This connective cover has the rational homotopy type of S4n-1. 6 C. A. MCGIBBON AND JEFFREY STROM Example E Essential phantoms S2n<2n> ! S4n exist when n 2 and all of them have essential category weight at least 2. Uncountably many of them have E(f) = 2. For the moment let G = S2n<2n> . In the proof of Example E we show that there are no essential phantoms from G to a finite type target, so in this sense G resembles the finite dimensional groups in Example C. This motivates the last open question mentioned at the end of this paper. 3. A Filtration on Phantom maps into Spheres The case where the target of a phantom map is a sphere might at first glance seem bizarre to those used to thinking of Sn as a source. However, it is known that for phantom maps out of finite type domains, the spheres provide good test cases as targets. The following result of McGibbon and Roitberg [10 ] makes this precise and motivates a closer look at phantom maps into spheres. Theorem 5 The following are equivalent for any finite type domain X: i) Ph(X; Y ) = 0 for every finite type target Y . ii)Ph(X; Sn+1 ) = 0 for every n. iii)There exists a map from X to a bouquet of spheres which induces an isomorphism in rational homology. It is reasonable to ask if, for a given X, it is really necessary to check all spheres in part ii). A quick answer is no, in general it is only necessary to check those dimensions n + 1 where Hn (X; Q) 6= 0. One goal in this section is to improve on this answer-to reduce the search to those n for which the module of indecomposables QHn (X; Q) is nonzero. In doing this, we make use of the simple but useful filtration described in the next result. Theorem 6 Fix n 1 and let Y be a finite type space with the same (n + 1)-type and the same rational type as K(Z; n + 1). Then there is a decreasing filtration, with nth term zero, on the group Ph(X; Y ). The filtration is induced by the cup product length filtration {Fk} on the group Hn (X; Z) [X; K(Z; n)] [Y; Y (n)]: Phantom maps f : X ! Y in filtration k have have essential category weight E(f) k. This filtration is natural with respect to maps between NUMERICAL INVARIANTS OF PHANTOM MAPS 7 domains X ! X0 and maps between targets whose homotopy groups satisfy the above hypothesis. Of course this result is applicable when Y is an odd sphere or the loops on an odd sphere. However, it does not apply to even spheres which are, in a sense, more complicated. This prompted the question- could the even spheres in Theorem 5 be replaced by the loops on odd spheres? The next result shows that the answer is no. Example F Let X = 2S5. Then Ph(X; Sn) = * and Ph(X; Sn) = * for all odd n. However, there exist essential phantoms X ! S4. On the positive side, we found the following unexpected stable result, which says that the even spheres can be ignored if you are willing to suspend the domain once. Proposition 7 Assume X has finite type. Then Ph(X; S2n+1) = Ph(X; S2n+1) = * for all n if and only if Ph(X; Y ) = * for all finite type targets Y . In particular Y could be the iterated loop space of another space, and so the triviality of phantoms from X to odd spheres and loops on odd spheres implies the triviality of phantoms from kX to any finite type target for each k 1. Of course, the filtration described in Theorem 6 might be trivial in certain cases where the cup product length filtration is not. Indeed, the whole group Ph(X; Y ) might be trivial in such cases (for example, if X = Sn x Sm ). However the following result shows this filtration is well behaved in the sense that the existence of nontrivial phantoms ultimately depends on the indecomposable terms. Theorem 8 Let X be a finite type domain. If Ph(X; Sn+1 ) = * for all n such that QHn (X; Q) 6= 0, then Ph(X; Y ) = * for all finite type targets Y . 4. Proofs The proofs use basic facts about phantom maps and the towers that determine them, see [1], Chapters 9 and 11, or [6]. In particular for any two spaces X and Y there are bijections of pointed sets Ph(X; Y ) lim1[X; Y (q)] lim1[Xn; Y ]: Proof of Proposition 1 The proof of part a) amounts to showing that the map Ph(X; Y ) Ph(X=X1;_Y ) induced by the quotient map X ! X=X1 is surjective. Let Y denote the universal cover of Y . 8 C. A. MCGIBBON AND JEFFREY STROM __ Zabrodsky shows in Section 2_of [16 ] that the covering ss : Y ! Y induces a surjection Ph(X; Y) ! Ph(X; Y ). __ __ We claim there is another surjection Ph(X=X1; Y) ! Ph(X; Y). It is here we need the two homotopy groups to be finitely generated. Apply __(q) the functor [ ; Y ] to the cofiber sequence X1 ! X ! X=X1 to obtain the exact sequence __(q) ____//_ __(q) ____//_ __(q) [X=X1; Y ] [X; Y ] [X1; Y ] which we abbreviate Aq ! Bq ! Cq; write A = {Aq} and similarly for B and C. We want to show that the map of towers A ! B induces a surjection on lim1. Since the fundamental group of X is finitely generated we may as- sume that X1 is a finite bouquet of, say, k circles. The tower of groups __(q) Cq = [X1; Y ] is then constant; each nonzero term in it is isomorphic to the direct sum of k copies of ss2Y and each nontrivial structure map __(q) is an isomorphism. We claim that the image of Bq = [X; Y ] in this constant tower stabilizes_(becomes constant) at some finite stage q. To see this note that Y F x T where T is a product of circles and ss1F __(q) is finite. Since the torus T is a retract of each Y , it follows that Bq ! Cq is surjective modulo torsion. And since the torsion subgroup of Cq is finite, it follows that the images of Bq cannot become ever smaller as q increases; instead they become constant. Let C0 denote this constant subtower of images. Then we have an exact sequence of towers A _____//B_____//C0____//_0 Since C0 is a constant tower, lim1C0 = *, and it follows, using the six term lim - lim1 sequence, that_the induced_map lim1A ! lim1B is surjective. Thus Ph(X=X1; Y) ! Ph(X; Y) is surjective. The two surjections we have constructed fit into a commutative square __ __ Ph(X=X1; Y) _____//Ph(X; Y) | | | | fflffl| fflffl| Ph(X=X1; Y ) _____//Ph(X; Y ) which clearly shows that Ph(X=X1; Y ) ! Ph(X; Y ) is surjective, and completes the proof of part a). NUMERICAL INVARIANTS OF PHANTOM MAPS 9 The proof of part b) is easier. The conditions on X or on Y imply that the quotient map X ! X=Xn induces a surjection of towers [X=Xn; Y (q)] _____//[X; Y (q)] which yields a surjection Ph(X=Xn; Y ) ! Ph(X; Y ). Proof of Theorem 2 For each space X, we will construct a map : X ! W which is universal with the property of being a phantom map with infinite Gray index. We will then show that this map must be essential under the given hypotheses. A phantom map f : X ! Y has infinite Gray index if and only if it factors as in the diagram f X _________//KKYOO | KKKOEnKK | | KKK | fflffl|n K%%| X [ CXn _____//Wn for each n, where n : X [ CXn ! Wn is the universal phantom map [4] out of X [ CXn. Fix X, and let W be the pushout of the diagram consisting of all of the maps X ! Wn. In other words, W is the space formed from the disjoint union of X and each of the Wn by identifying OEn(x) with x for each x 2 X. Let : X ! W (X) denote the natural inclusion. It is easy to see that is a phantom map with infinite Gray index, and that any phantom f : X ! Y with infinite Gray index out of X must factor through . It follows that a space X is the domain for an essential phantom map with infinite Gray index if and only if the map : X ! W is essential. Before we prove Theorem 2, we need a lemma. Lemma 2.1 Let X be of finite type, and let u 2 Hn (X; G). Suppose _ f : X ! Kff W and u = f*(v)Wfor some v 2 Hn ( Kff; G). Then there is aWfinite subwedge ff2IKffsuch that (j O f)*(i*(v)) = u, where i : ff2IKffand j is its left inverse. Proof of Lemma 2.1 Since X is of finite type XmW is compact, so f(Xm ) must be contained in a finite subwedge ff2IKff. It is easy to verify the claim for this subwedge. 10 C. A. MCGIBBON AND JEFFREY STROM Since H*(X; Z=p) is not locally finite as an Ap module, there is a class u 2 H*(X; Z=p) and cohomology classes 1; 2; : : :such that n . .1.(u) 6= 0 for each n. Choose such a u. Extend the map X ! W to a cofiber sequence _____// ____//_ __j__// X W W=X X: If is nullhomotopic, then the map j must have a section. We will show that no map X ! W=X can have u in its image, and hence, the map j cannot have a section. Since W is formed from the Wn by gluing along X, we have the following cofibration W X _____//W_____// nWn=X: By Lemma 2.1, if u is in the image of a map induced by W X _____//n Wn=X then we can restrict to a finite subwedge, so it suffices to show that u cannot be in the image of a map induced by fW X _____//n2I Wn=X: W where I is a finite index set. Assume there is a v 2 H*( n2I Wn=X; G) such that u = f*(v). From the diagram X ___________//X_________//* | | | | | | fflffl| fflffl| W fflffl| X [ CXn ______//_Wn______//_Kff | | | | | | fflffl| fflffl| W fflffl| Xn _______//_Wn=X_____// Kff in which all the rows and columns are cofibrations, we find that there are cofibration sequences W Xn ____//_Wn=X _____// Kff in which each Kffis a finite complex. Wedging these together, we have the following cofibration sequence W W W W n2IXn ____//_n2IWn=X _____// n2I Kff: NUMERICAL INVARIANTS OF PHANTOM MAPS 11 Now let N be an upper bound for I and consider N . .1.(u)W= f*(N . .1.(v)). Since N is larger than the dimension of n2IXn, we see that N . .1.(v) is in the image of a map induced by W W W X ____//_n2IWn=X _____// n2I Kff: by Lemma 2.1, we can restrict to a finite subwedge, which means that N . .1.(u) is in the image of a map to a finite dimensional space, which is clearly a contradiction. Proof of Proposition 3 This proof uses a different description of the Gray index-one that requires X and Y to be nilpotent and of finite type. Let r : X ! Xo denote the rationalization map. By Theorem 5.1 of [6] the phantom maps X ! Y are precisely those maps which factor through the rationalization of X and hence Ph(X; Y ) = r*[Xo; Y ]. The set [Xo; Y ] contains no essential phantom maps since Xo is a rational space (and hence the universal phantom map out of it is trivial, [4]). Therefore for each essential map OE : Xo ! Y there is a largest natural number fl = fl(OE) such that the restriction OE to (Xn)o is null homotopic (or equivalently that OE extends over (X=Xn)o) for n fl. It follows that if f : X ! Y is an essential phantom map then its Gray index can be defined as G(f) = lub{fl(OE) | r*(OE) = f}: This least upper bound will obviously be finite if the set over which it is taken is finite. This happens when Map*(X; ^Y) is weakly con- tractible because the function r* is then a bijection between Ph(X; Y ) and [Xo; Y ], by Theorem 5.4 of [6]. The second assertion has already been discussed in the paragraph following Example B. We give a different proof here. We need to take a closer look at [Xo; Y ]. By Theorem 5.2, ibid, there is a bijection of pointed sets Y [Xo; Y ] Hk(X; ssk+1(Y ) R) k where R denotes a rational vector space whose cardinality equals that of the real numbers. Assume that Hk(X; ssk+1(Y ) Q) = 0 for k N. It is possible to construct a subcomplex K X such that XN-1 K XN and the natural map Hk(X; Q) _____//Hk(K; Q) 12 C. A. MCGIBBON AND JEFFREY STROM is an isomorphism for k N. If f : X ! Y is a phantom map with G(f) N, then f factors through a phantom map f0 : X=K ! Y . Since Y [(X=K)o; Y ] Hk(X=K; ssk+1(Y ) R) = 0; k this shows that f0 ' *, so f ' *. Corollary 3.1 Suppose that X, Y and Y 0are finite type nilpotent spaces such that the rationalization map r : X ! Xo induces bijections Ph(X; Y ) [Xo; Y ] and Ph(X; Y 0) [Xo; Y 0]. If g : Y ! Y 0is a map which induces a monomorphism on rational homotopy groups, then g* : Ph(X; Y ) ! Ph(X; Y 0) is also one-to-one. Moreover G(f) = G(g O f) for each phantom f : X ! Y . Proof Since f is a phantom map, we can write f = OE O r for some OE : Xo ! Y . Suppose G(f) = n, so OE|(Xn)o' *, but OE|(Xn+1)o6' *. There is a cofibration j (Xn+1=Xn)o _i___//(X=Xn)o ____//_(X=Xn+1)o: Since f does not factor through j, f Oi 6' *. Since g induces a monomor- phism on rational homotopy groups and (Xn+1=Xn)o is a wedge of ra- tional spheres, (gOf)Oi 6' *, and so gOf does not extend to (X=Xn+1)o. Proof of Theorem 4 Let f : X ! Y be a phantom map, and suppose cat(Xo) = n. To show that E(f) < n, we need to show that f O j 6' * in the diagram j f Bn(X) ______//X_____//Y?? "" |rn| |r|"""" fflffl|jo fflffl|f"" Bn(X)o ____//_Xo Since cat(Xo) = n the inclusion jo has a right inverse s; this a basic fact due to Ganea. It follows that j*ois injective, and so f O jo 6' *. Now f O j ' r*n(f O jo), so we will be done once we show that r*nis bijective. To this end it suffices to show that Map*(W; bY) is weakly contractible by Theorem 5.4 of [6]. To simplify notation, let H = X. The proof that Map(H; bY) = * implies that Map*(Bn(H); bY) = * goes back to Zabrodsky, [16 ]. The weak contractibility of Map*(H?. .?.H; bY) follows from the first assumption and the exponential rule. One then uses the principal H-fibration H _____//H ? . .?.H_____//Bn(H) NUMERICAL INVARIANTS OF PHANTOM MAPS 13 and the Zabrodsky lemma (5.5 of [6]) to obtain the result for the base space Bn(H). Finally, suppose that cat(Xo) = 1. Let f : X ! Y be an essential phantom and refer again to the diagram used in this proof. The ex- tension f : Xo ! Y is not a phantom: it restricts nontrivially to some finite dimensional skeleton of Xo. Since j : Bn(X) ! X is at least an (n - 1)-equivalence, it follows that f O jo is essential for n sufficiently large. Since r*nis still bijective, this implies that E(f) is bounded above by such n. This completes the proof of Theorem 4. Proof of Theorem 6 We again use the identification Ph(X; Y ) = lim1[X; Y (q)]. Write Gq = [X; Y (q)] and let Hq denote the image of Gq in Gn for q n and let Hq = 0 for smaller values of q. Notice that Gn = [X; (Y )(n)] Hn (X; Z) since Y has the same (n + 1)-type as K(Z; n + 1). If we let K be the kernel of the surjection of towers G ! H, we obtain a short exact sequence of towers 0 _____//K_____//G____//_H____//_0: The homotopy groups of Y are finite in dimensions above n because Y has the same rational type as K(Z; n + 1). It follows that the kernels Kq are finite for q n; hence lim1K = 0. From the six term lim- lim1 sequence applied to the short exact sequence of towers just displayed, it follows that lim1G lim1H. The filtration {Fk} on the groupTGn then induces a filtration on the tower H; the kth stage being H Fk. Having identified Ph(X; Y ) with lim1H we define the kth stage of the filtration on Ph(X; Y ) to be the image " lim1(H Fk) ! lim1H: It is easy to check that the identifications used here are natural with respect to the maps stated in the hypothesis. It is well known that cup products of length k vanish in the co- homology of a space of L-S category less than k. Since the category of Bk(X) is at most k, the claim that phantoms in filtration k have E(f) k then follows from Theorem 10 in [13 ] and naturality. Proof of Proposition 7 To show that Ph(X; Y ) = * for all finite type Y , it suffices to show that all phantom maps from X to spheres are trivial. Since Ph(X; Y ) Ph(X; Y ) it follows from the hypoth- esis that all phantoms from X to odd-dimensional spheres are trivial. In the even dimensional case there exist maps S2n-1 x S4n-1 ! S2n which are rational equivalences in general (and homotopy equivalences 14 C. A. MCGIBBON AND JEFFREY STROM in the Hopf invariant one cases). By Theorem 7.2 of [6] these maps induce surjections Ph(X; S2n-1 x S4n-1) ! Ph(X; S2n). The hy- pothesis on X forces the first group to be trivial and the result follows. Now assume that Ph(X; Y ) = * for all finite type Y . We have to show that Ph(X; S2n-1) = * for all n. There are maps S2n-1 ____//_S2n ____//_S2n-1 whose composite has degree 2. This composite, being a rational equiv- alence, induces a surjection of Ph(X; S2n-1) to itself. This surjec- tion must also be the trivial endomorphism, because Ph(X; S2n) Ph(X; S2n) = *. This forces Ph(X; S2n-1) = *. Proof of Theorem 8 The hypothesis is that Ph(X; Sn+1 ) = * whenever QHn (X; Q) 6= 0. Assume then that uff2 Hn (X; Z) is a class whose rational image projects to a nonzero indecomposable. We claim that there is a map gff: X ! Sn+1 such that some nonzero multiple of ufflies in the image of g*ff. Assume for the moment that this claim is true. Choose a minimal set of integral classes {uff} whose images span the graded vector space QH*(X; Q). Let Y g : X ! Snff+1 ff be given by gffin the ffth coordinate. This map induces a surjection in rational cohomology since it is evidently surjective on the module of rational indecomposables. Since the product has finite type, Theorem 2 of [10 ] shows that iY j g* : Ph Snff+1; Y ! Ph(X; Y ) is surjective for every finite type target Y . Now we are done, because itQis proved in [4] that there are no essential phantom maps out of Snff+1. It remains to prove the claim made in the first paragraph. Let n = nff. By hypothesis Ph(X; Sn+1 ) = * and so, since X also has finite type, it follows that the tower {[X; (Sn+1 )(k)]} is Mittag-Leffler by Theorem 2 of [8]. By Lemma 3.2, ibid, it follows that the image of [X; Sn+1 ] in [X; (Sn+1 )(k)] has finite index for each k. Taking k = n, we see that the image of [X; Sn+1 ] ! Hn (X; Z) has finite index, which proves the claim and completes the proof. NUMERICAL INVARIANTS OF PHANTOM MAPS 15 Proof of Example A To establish the existence of essential phan- tom maps CP 1 ! S2n+1, one could use Gray's proof for n = 1 (it adapts easily for larger n) or one could compute, as in [6], with the aid of Miller's proof of the Sullivan conjecture, that Ph(CP 1; S2n+1) [CPo1; S2n+1] H2n(CP 1; ss2n+1(S2n+1) R) where again R is a rational vector space with the cardinality of the real numbers. Thus essential phantoms CP 1 ! S2n+1 exist. By Proposi- tion 1 they have Gray index at least 2n - 1. Now if X = CP 1=CP m where m n and Y = S2n+1 it is easy to see that [Xo; Y ] = * and hence Ph(X; Y ) = *. Thus 2n-1 is also an upper bound on the finite values of the Gray index in this case. To check that the suspension operator here raises the Gray index by exactly one, consider first the adjoint of the double suspension. Note that E2 : Y ! 22Y is a rational equivalence when Y is an odd sphere. Hence by Corollary 3.1, E2(f) has the same index as f. The adjoint correspondence E2(f) 7! 2f clearly raises the Gray index by 2. It follows that G(f) 2n. Since the inequality in the other direction is obvious, the proof follows. Proof of Example B By the Hilton-Milnor theorem (S2_S2) ffSnff where the set {nff} = {2; 3; 4; :::}: It follows that for each odd n 3 there is a map fn : Sn ! S2 _ S2 which induces a monomorphism on integral (and hence on rational) homotopy groups. Hence Example B follows from Example A and Corollary 3.1. Proof of Example D The first part deals with phantoms into odd dimensional spheres. Consider the cofiber sequence Bk _____//K_____//Ck where K = K(Z; 2n) and Bk = Bk(K). As in the proof of Theorem 4, we will once again exploit the equality [Xo; Y ] [X; Y ] = Ph(X; Y ); in other words, every map is phantom when X is any one of the terms in this cofiber sequence and Y is a finite complex. The same is then true of Ck, and this forces the induced sequence of phantom maps sets to be exact. The first map in the cofiber sequence induces an epimorphism in rational cohomology with kernel . Thus, Ck is (2n(k + 1) - 1)- connected, Ph(Ck; S2nq+1) = * for q k, which shows that the map Ph(K; S2nq+1) ____//_Ph(Bk; S2nq+1) 16 C. A. MCGIBBON AND JEFFREY STROM is a bijection for k q and is trivial for k < q. Hence, the category weight of each essential phantom K ! S2nq+1 is exactly q. The second part deals with phantoms into even dimensional spheres. Since the rational cohomology of the domain is concentrated in even degrees, the phantoms in this case must factor through a Whitehead product for degree reasons. So each phantom in part ii) has the form K(Z; 2a) _____//S4k-1____//_S2k Thus, 2ab = 4k - 2 for some b, which means a and b must both be odd, and 2k = ab + 1. The given formula comes from taking a = 2n - 1 and b = 2q + 1. The essential category weight of these phantoms can be calculated just as in part i). Proof of Example E The existence of essential phantom maps S2n<2n> ! S4n was shown in Example 6.1 of [9]. Let G = (S2n<2n> ). Note that there is a map h : G ! S4n-1, which is a rational equiv- alence. This map h can be taken to be the lift of the classical Hopf invariant to (2n - 1)-connected covers. Since S4n-1 is homotopy equivalent to a bouquet of spheres, it follows that Ph(G; Y ) = *, for all finite type targets Y , by Theorem 5 iii). Thus all phantoms f : BG ! S4n have E(f) 2. To obtain the phantoms with E(f) = 2, we will study the "projective plane" B3G. We will show that the natural map B3G ! BG ' S2n<2n> induces an epimorphism Ph(S2n<2n> ; S4n) ! Ph(B3G; S4n) and that the latter set is nontrivial. To obtain the surjection recall that Ph(X; Y ) = r*[Xo; Y ] in the nilpotent finite type case. For X = S2n<2n> and Y = S4n it was shown that r* is a bijection in [9]. So it suffices to show that the induced map [BGo; S4n] ! [B3Go; S4n] is surjective. The fiber of the inclusion B3G ! BG is the 3-fold join G ? G ? G which is rationally (12n - 5)-connected. Thus B3G ! BG induces an isomorphism in rational homology in degrees 4n - 1 and 8n - 2 and so [(B3G)o; S4n] [(BG)o; S4n] H4n-1(S4n-1; ss4n(S4n) R): To obtain Ph(B3G; S4n) 6= * it suffices to show that tower Tq = [B3G; (S4n)(q)]; q = 1; 2; 3; : : :is not Mittag-Leffler. Consider the projection lim T ! T4n-1. We will show that the image of this homo- morphism has infinite index in H4n-1(B3G; Z) and then invoke Lemma 3.2 of [8], as we did in the proof of Theorem 8. Since B3G has finite type it is enough to show that there are no essential maps g : B3G ! S4n which are nontrivial in H4n-1( ; Q). NUMERICAL INVARIANTS OF PHANTOM MAPS 17 Suppose that g is such a map. There is a cofibration _H___// _p3_//_ G ? G G B3G in which p3 induces an isomorphism in H4n-1( ; Q). It follows that the map f = g O i : G ! S4n is rationally a 4n-equivalence. Further- more, the composite __H__// __f__// G?G G S4n is nullhomotopic. It follows that the adjoint f O eHof f O H is also nullhomotopic, and so the composite ixi q V eH f S x S _____//G x G_____//G G ____//_G ____//_2S4n: is nullhomotopic. Here i : S ! G denotesVthe (2n-1)-connective cover of E : S2n-1 ! S2n and the map GxG ! G G is the usual quotient map. >From this diagram we get a composite map OE : S x S ! 2S4n which is evidently nullhomotopic. Fix a prime p, and let Lp denote the Neisendorfer localization functor ([9],[2]). We will derive a contradiction by showing that Lp(OE) is an essential map. To prove this, we first decompose the map OE into more manageable pieces. Recall that the adjoint of the Hopf construction has the form eHq = E - Ess1 - Ess2 where : GxG ! G is the loop multiplication on G, and ss1 and ss2 are the two projections. The last two terms on the right are "subtracted" using the loop multiplication on G. Applying the loop map f gives f(Heq) = f(E) - f(Ess1) - f(Ess2): Up to p-completion, the functor Lp is an inverse to the connective cover functor in many important cases; in particular Lp(i : S ! G) is the p-completion of E : S2n-1 ! S2n. This localization functor also takes loop maps to loop maps (e.g. [2], [9]). Since the map Lp(fEss1(i x i)) factors through the p-completion of a map S2n-1 ! 2S4n it is evidently null homotopic. The same goes for the term involving Ess2. It follows that the Neisendorfer localization of the map OE is homotopic to the p-completion of the composite ixi 2n 2n ef S2n-1 x S2n-1 ____//_S x S ____//_S2n ____//_2S4n: 18 C. A. MCGIBBON AND JEFFREY STROM We will show that this completed composite induces a nontrivial map on H4n-2( ; Z=pr) for large enough r. The key point is that if f : X ! Y induces a nontrivial map in H4n-2( ; Z=pr), then so does the p-completed map. Since the original map f : G ! S4n was assumed to induce a nontrivial map in H4n-1( ; Q), the adjoint induces a nontrivial map in H4n-2( ; Q) (this uses the fact that, rationally, G is a sphere). It follows that the whole uncompleted composite is nontrivial on H4n-2( ; Q), and so is nontrivial on H4n-2( ; Z=pr) for r large enough. Thus, OE cannot be null-homotopic as our initial assumption implied. In other words, no rationally essential map G ! S4n extends to B3G, and we conclude that the tower Tq is not Mittag-Leffler. This completes the proof for Example E. Proof of Example F The existence of essential phantoms from 2S5 ! S4 is proved in [6], Proposition 8.2. Since 2S5 has the rational homotopy type of S3, phantom maps into spheres of dimension n 6= 4 will necessarily be trivial. The evaluation map 22S5 ! S5 is a rational equivalence and so, by Theorem 7.1c of [6], all phantoms from 2S5 into finite type targets are trivial. 5. Some open questions Question 1 What is the Gray index of the universal phantom map out of RP 1 ? Question 2 Does every essential phantom X ! Y have finite Gray index when both spaces are nilpotent and of finite type? Question 3 Suppose that X is a space such that every phantom from X to a sphere has infinite essential category weight. Does it follow that every phantom from X to a finite complex (or a finite-type target) also has infinite essential category weight? Question 4 Does there exist a space whose universal phantom map has essential category weight which is finite, but larger than 1? 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