ELEMENTS OF LARGE ORDER IN ss*(SU(n)) DONALD M. DAVIS 1.Discussion of main theorem If p is a prime number, then the p-exponent of a topological space X, denoted expp(X), is defined to be the largest e such that some homotopy group of X cont* *ains an element of order pe. Our main theorem is a lower bound for expp(SU(n)) if p * *is an odd prime. Theorem 1.1. If p is an odd prime, then hn + 2p - 3i hn + p2 - p - 1i expp(SU(n)) n - 1 + __________ + _____________ : p2 p3 In other words, we show that, for some i, ssi(SU(n)) contains an element of ord* *er pe 2-p-1 with e n - 1 - [n+2p-3_p2] + [n+p_____p3]. Computing homotopy groups is a notoriously difficult problem. The groups ssi(* *Sn) are known only for relatively small values of i - n, and quickly become quite c* *ompli- cated, but the celebrated exponent theorem of [7] showed expp(S2n+1) = n for al* *l odd primes p. Groups ssi(SU(n)) have been computed for small values of i - n by many topologists, with all known results compiled in [12]. It was not until the work* * of the author in [9] that information about all homotopy groups of all SU(n) was obtai* *ned. There it was shown that (1.2) expp(SU(n)) n - 1: Theorem 1.1 of this paper improves upon that result. In this introductory sect* *ion, we discuss the extent to which Theorem 1.1 might be nearly optimal. Theorem 1.1 is proved by computing certain v1-periodic homotopy groups of SU(* *n). We will review the basic properties of v1-periodic homotopy groups in Section 2* *. Here ___________ 1991 Mathematics Subject Classification. 55T15. Key words and phrases. p-exponents, v1-periodic homotopy groups, unitary grou* *ps, unstable Novikov spectral sequence. 1 2 D. DAVIS we just point out that for each prime p there is a v1-periodic homotopy theory,* * and the prime is not usually written explicitly in the notation. The v1-periodic ho* *motopy groups yield lower bounds for p-exponents since each v1-periodic homotopy group* * of SU(n) occurs as a direct summand of some actual homotopy group of SU(n). It is also important to note from [9] that v-11ss2k(SU(n)) is cyclic, and |v-11ss2k-1* *(SU(n))| = |v-11ss2k(SU(n))|. It is possible that our methods might allow us to compute, for each integer n* *, the exact exponent of the order of the largest v1-periodic group of SU(n). We deno* *te this number by expp(v-11SU(n)). As observed above, we have (1.3) expp(SU(n)) expp(v-11SU(n)): This number expp(v-11SU(n)) will depend intimately on the precise form of n. In Theorem 1.1 we have sacrificed optimality for simplicity. It seems likely that * *Theorem 1.1 is quite close to the exact value of expp(v-11SU(n)). A refined version of Theorem 1.1 is stated as Theorem 1.4 below, the proof of* * which will occupy most of our effort. Throughout the paper, ( ) denotes the exponent* * of p in an integer. Theorem 1.4. Given n and an odd prime p, define integers N, i, j, and k by (1.5) n = N + 1 + (p - 1)i with 1 N p - 1 and (1.6) i = jp + k with 1 k p: For 0 t < i, define S0t= |{s : t s < i; s N mod p}| and S00t= |{s : t < s < i; s kp + N mod p2}|: If N = k, define S000t= |{s : t < s < i; s jp2 + Np + N mod p3}| ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 3 and 8 <1 ift Np + N mod p2 ffit= : 0 otherwise. Let 8 k: We define n to be p-clean if for all D ffl (i - j - D) < N - k + pD; p-cleaner if for all D ffl (1.8) (i - j - D) < N - k + ffl + (p - 1)D - ((D - ffl)!) + Sj+ffl- Sj+D; and p-cleanest if N = k and for all D 1 (i - j - D) < (p - 1)D - ((D - 1)!) + Sj+1- Sj+D: Let r = (i - j). Then expp(v-11SU(n)) 8 >>>n + Sj+ffl if n is not p-clean >>< n - 1 + ((i - j) . .(.i - j - r)) + Sj+r+1if N = k and n is p-cleaner (1.9>>>n)+ ((i - j) . .(.i - j - r)) + S if N = k and n is p-cleanest >>: j+r+1 n - 1 + Sj+ffl otherwise. All the notation of Theorem 1.4 will pervade the paper. The variants on n bei* *ng p- clean have to do with the avoidance of associated numbers having large p-expone* *nts. Note also that p-cleanest implies p cleaner, and p-cleaner implies p-clean. The bridge from Theorem 1.4 to Theorem 1.1 is provided by the following resul* *t, which will be proved at the end of Section 2. Theorem 1.10. Assume the notation of Theorem 1.4. Then (1.11) S0j+ffl [(n + 2p - 3)=p2] - 1 4 D. DAVIS with equality occurring if and only if k = N and j N or N + 1 mod p, while (1.12) S00j+ffl [(n + p2 - p - 1)=p3] - 1 with equality occurring if and only if k = N and j kp + N or kp + N - 1 mod p* *2. Now we provide the easy steps across the bridge from Theorem 1.4 to Theorem 1* *.1. Proof of Theorem 1.1.We use (1.3), (1.9), and Theorem 1.10. If k 6= N, we have 2-p-1 expp(SU(n)) n - 1 + S0j+ffl+ S00j+ffl n - 1 + [n+2p-3_p2] + [n+p_____p3* *]: We now consider the case in which N = k. If i - j - 1 2p3, then S000j+1 2, whi* *ch makes up for the 2 that we might possibly be missing if equality holds in both * *(1.11) and (1.12). If i - j - 1 < 2p3, then either n is p-cleanest or else p = 3 and i* * - j = 10, 19, 28, 37, or 46. If i - j - 1 < 2p3 and n is p-cleanest, then either equalit* *y holds in at most one of (1.11) or (1.12), in which case the desired result follows, o* *r else j kp + N mod p2. In this case, (i - j) 2, which implies the desired result. Finally, if p = 3 and i - j = 10, 19, 28, 36, or 46, one can check that equalit* *y holds in neither (1.11) nor (1.12), and so again the result follows. || In [9], an explicit, but not very tractable, formula for expp(v-11SU(n)) was * *given when p is odd. This was (1.13) expp(v-11SU(n)) = max {ep(k; n) : k n}; where iXj i j j (1.14) ep(k; n) = min{ (-1)i jiik : n j k}: i=1 In [9], following work of [8], the Little Fermat Theorem was applied to (1.13) * *to deduce (1.2), a precursor of Theorem 1.1. The formula given in (1.13) and (1.14) has the positive feature of being writ* *able on one line, but the negative feature of not offering much insight into the act* *ual values. These cyclic groups Z=pep(k;n)were first determined by Bendersky in [2* *] as the 1-line groups of the unstable Novikov spectral sequence (UNSS) for SU(n). I* *n [9], the relationship of the UNSS with v1-periodic homotopy groups was first noted. * *The calculation in [2] was by downward induction on n. In [11], the UNSS for SU(n) * *was ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 5 computed by increasing induction on n, with complete and tractable results obta* *ined, provided n p2-p+1. A comparison of the results of these two computations yields results in number theory, proved via algebraic topology. The methods of this pa* *per are an extension of those of [11]. We could compute most of the v1-periodic gro* *ups of SU(n), but displaying them in a comprehensible form becomes a serious bookkeepi* *ng problem. So instead we concentrate our attention on the largest groups. As discussed in [9], we can use a computer to compute the RHS of (1.13) for s* *mall values of n. This uses the periodicity in k of (1.14) proved in [8]. In the f* *ollowing table, we compare the computer result for expp(v-11SU(n)) with our estimates of Theorems 1.4 and 1.1 when p = 3 and n 45. The few cases in which Theorem 1.4 fails to be sharp are due to lack of control of coefficients in Lemma 3.6; a hu* *ge amount of analysis would be required in order to sharpen this more finely. The cases i* *n which Theorem 1.4 is stronger than Theorem 1.1 are due to approximations made in the proof and application of Theorem 1.10; they are due to our desire for a tractab* *le formula. We remark that the computer value when n = 34 is 1 greater than that presented in [9]; our theoretical results here led to the discovery of this cor* *rection. 6 D. DAVIS -1 | ||_nN____i__k__j_||expp(v1_SU(n))_Thm_1.4___Thm_1.1_|_ | 3 2 0 0 0 | 2 2 2 | || || || || 4 1 1 1 0 || 4 4 3 || || 5 2 1 1 0 || 4 4 4 || || 6 1 2 2 0 || 6 6 6 || || 7 2 2 2 0 || 7 7 7 || || 8 1 3 3 0 || 8 8 8 || || 9 2 3 3 0 || 10 10 9 || ||10 1 4 1 1 || 11 11 10 || ||11 2 4 1 1 || 12 12 11 || ||12 1 5 2 1 || 12 12 12 || ||13 2 5 2 1 || 14 14 13 || ||14 1 6 3 1 || 14 14 14 || ||15 2 6 3 1 || 16 16 16 || ||16 1 7 1 2 || 18 18 17 || ||17 2 7 1 2 || 19 19 18 || ||18 1 8 2 2 || 20 20 19 || ||19 2 8 2 2 || 21 21 20 || ||20 1 9 3 2 || 22 22 21 || ||21 2 9 3 2 || 22 22 22 || ||22 1 10 1 3 || 25 25 24 || ||23 2 10 1 3 || 26 25 25 || ||24 1 11 2 3 || 28 27 27 || ||25 2 11 2 3 || 29 29 28 || ||26 1 12 3 3 || 30 29 29 || ||27 2 12 3 3 || 31 30 30 || ||28 1 13 1 4 || 32 32 31 || ||29 2 13 1 4 || 34 33 32 || ||30 1 14 2 4 || 34 34 33 || ||31 2 14 2 4 || 34 34 34 || ||32 1 15 3 4 || 35 35 35 || ||33 2 15 3 4 || 37 37 37 || ||34 1 16 1 5 || 38 38 38 || ||35 2 16 1 5 || 39 39 39 || ||36 1 17 2 5 || 41 41 40 || ||37 2 17 2 5 || 42 42 41 || ||38 1 18 3 5 || 43 43 42 || ||39 2 18 3 5 || 43 43 43 || ||40 1 19 1 6 || 45 45 44 || ||41 2 19 1 6 || 45 45 45 || ||42 1 20 2 6 || 47 47 47 || ||43 2 20 2 6 || 50 50 48 || ||44 1 21 3 6 || 51 50 49 || |_45__2__21__3_6_|______52___________51________50____| ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 7 An optimistic conjecture would be that the elements of largest order in ss*(S* *U(n))(p) are v1-periodic, i.e., that expp(SU(n)) = expp(v-11SU(n)): The justification for such a conjecture would be its validity for spheres and t* *he fact that v1-periodic homotopy groups are related to K-theory, as are the spaces SU(* *n). The only cases in which the precise value of expp(SU(n)) has been established a* *re the following. Proposition 1.15. If p is an odd prime, then 8 n, then (2.10) ffm=e -svm-11h1 mod p: (3) If n (m) + 1 and 1 j n, then d(pm-(m)-1-j hm1)2n+1 has order pj in E2;2n+1+qm2(S2n+1). It equals vm-j-11h1 hj1mod S2j-1. (4) If (m) + 1 n and 1 j (m) + 1, then d(pm-n-j hm1)2n+1 has order pj in E2;2n+1+qm2(S2n+1). It equals vm-n-j+(m)1h1hn+j-(m)-11mod S2n+2j-2(m)-3. (5) The homomorphism 2 : E2;2n-1+qm2(S2n-1) ! E2;2n+1+qm2(S2n+1) is injective if n (m) + 1 and is multiplication by p otherwise. The precise form of the classes on the 2-line in (3) and (4) differs here fro* *m previous treatments such as [1] and [11], which were always written mod boundaries or cl* *asses which desuspend. The classes are easily seen to be precise by the method employ* *ed in the proof of Lemma 4.6 below. Our final preliminary is to introduce the spaces which are the factors in the* * decom- position of [14] of the p-localization of SU(n) and its quotients as a product * *of p - 1 spaces. Proposition 2.11. ([14]) There are spaces Xij(N) for 1 N p - 1 and 0 j i satisfying (1) Xjj(N) = S2N+1+jq; (2) there is a fibration Xi-1j(N) ! Xij(N) ! S2N+1+iq; 14 D. DAVIS (3) H*(Xij(N)) is an exterior algebra with a single generator of each dimensi* *on 2N + 1 + kq for j k i; p-1Y (4) SU(n)=SU(m) '(p) Xd(n;N)d(m;N)+1(N), where d(n; N) = [n-1-N_p-1]. N=1 The UNSS of Xij(N) has E2-term Exts;tU(P (BP*(Xij(N)))), where P (BP*(Xij(N))) has BP*-basis {x2N+jq+1; x2N+(j+1)q+1; : :;:x2N+iq+1}, with coaction induced fr* *om Lemma 2.5. Theorem 2.4 applies to the spaces Xij(N), and, because of the sparse* *ness results for spheres given in Theorem 2.8, the following is immediate. Proposition 2.12. If 1 N p - 1 and s = 1 or 2, then v-11ss2N+1+qm-s(SU(n)) v-11ss2N+1+qm-s(Xi0(N)); if N + 1 + (p - 1)i n < N + 1 + (p - 1)(i + 1): Combining this with Theorem 2.4, we see that it suffices to show that if n = N + 1 + (p - 1)i, there is an integer m such that p(|E1;2N+1+qm2(Xi0(N))|) is a* *t least as large as (1.9). We close this section with the proofs of Proposition 2.6 and Theorem 1.10. Proof of Proposition 2.6.The first part was proved as [11, 2.4]. From that proo* *f, we have X F* X X pj hi= ( hk mj)i+1bi: i i0 j;k0 Here bi 2 H2i(BP ), and by writing mj and bi on the right, we mean that we are multiplying by their conjugates, which becomes the right action of BP* on BP*BP after binomial coefficients are used to write these elements as elements of BP* H*BP . It turns out that bi= 0 unless i is a multiple of p - 1, but that is not* * relevant to the proof. P pj i+1 e1 * * e Since pj - 1_2|mj| = 1, each term in ( hk mj) is a multiple of some h1 . .* *h.rrm P 1 P 1 with m 2 H*BP and ei- _2|m| = i + 1. Then ei- _2|mbi| = 1. || ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 15 Proof of Theorem 1.10.We begin by showing that if N > k, then the two inequalit* *ies are true without the -1 on the RHS. For the first, hi - N - 1i hj - N - 1i hk - N - 1i hj - N - 1i S0j= _________ - __________= j + __________ - __________: p p p p The putative lower bound is [((k + 2)(p - 1) + p2j + N - pj)=p2], and so the de* *sired inequality is equivalent to hk - N - 1i h(k + 2)(p - 1) + N - pji hj - N - 1i (2.13) __________ ______________________ + __________: p p2 p Since 1 k < N p - 1, the LHS of (2.13) equals -1. Increasing j by p does not change either side of this inequality, and so it suffices to consider j between* * N + 1 and N + p, inclusive, so that [(j - N - 1)=p] = 0. The RHS of (2.13) satisfies h(k + 2)(p - 1) + N - p(N + 1)i h(k + 2 - N)(p - 1) - pi RHS ____________________________ = _____________________ < 0; p2 p2 since k + 2 - N 1. Thus LHS RHS, as desired. Next we prove that if N > k, (1.12) is true without the -1 on the RHS. We have hpj + k - 1 - kp - Ni hj - kp - Ni S00j= ___________________ - ___________; p2 p2 while the putative lower bound is [(N + 1 + (p - 1)(pj + k) + p2 - p)=p3]. Incr* *easing j by p2 increases both sides by p - 1. Thus we may assume that j = kp + N + for 0 < p2, so that [(j - kp - N)=p2] = 0. After k is subtracted from both sides,* * the desired inequality becomes, with D = N - k, h(D + )p - D - 1 i h(D + + 1)(p2 - p + 1) - 1i (2.14) _________________ __________________________ : p2 p3 Since 0 < D < p - 1, the LHS of (2.14) equals [(D + - 1)=p], and so letting s = D + reduces (2.14) to hs - 1i hs(p2 - p + 1)i (2.15) _____ ____________ p p3 with s > 0. For 1 s p, both sides of (2.15) are 0. Since increasing s by p increases the LHS of (2.15) by 1, and increases the RHS of (2.15) by no more th* *an 1, this establishes (2.15), and hence the desired result. 16 D. DAVIS Now assume N k. Then (1.11) becomes hk - N - 1i hj - Ni hN + 1 + (p - 1)(jp + k) + 2p -i3 j + __________ - ______ ______________________________ - 1: p p p2 We subtract j from both sides, and note that increasing j by p changes both sid* *es by the same amount, so that it suffices to consider N j < N + p, and the desir* *ed inequality is hk - N - 1i h(k - N + 2)(p - 1) - p(j - N)i __________ ____________________________ - 1: p p2 For 0 k - N < p and 0 j - N < p, this is always satisfied, with equality if a* *nd only if k - N = 0 and j - N = 0 or 1, as claimed. Similarly, if N k, then (1.12) becomes hpj + k - 1 - kp - Ni hj - kp - N + 1i hN + 1 + (p - 1)(pj + k + p)i ___________________ - ______________ __________________________ - 1; p2 p2 p3 which is unaffected when j is increased by p2. We let D = k - N and j = kp + N - 1 + ap + b for 0 a; b; D < p. The inequality simplifies to h(b - D - 1)p + D - 1i h(b - D)(p2 - p) - D - ap2 + 1i ____________________ ____________________________ - 1; p2 p3 which is always satisfied, with equality if and only if D = 0, a = 0, and b = 0* * or 1. || 3. The cellular spectral sequence In this section, we sketch the proof of Theorem 1.4 when n is p-cleaner and N* * 6= k by showing, modulo two proofs postponed until Section 4, that for suitably chos* *en m (3.1) p(|E1;2N+1+qm2(Xi0(N))|) n - 1 + Sj+ffl: The relations (1.5) and (1.6) which define N, i, j, and k will be used througho* *ut the rest of the paper. Also ffl will always be as in (1.7). We will usually t* *hink of N and m as being fixed, so that they may be omitted from some notation. The expression Es2(-) will always mean Es;2N+1+mq2(-), and we let Xi`= Xi`(N). Note that X``= S2N+1+q`. We can organize the computation of Es2(Xi0) as a cellular spectral sequence (* *CSS) with Es;`1= Es2(X``) for 1 s 2 and 0 ` i, and dr : E1;`r! E2;`-rrinduced ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 17 @ 2 * *`-r by pulling the class in E12(X``) back to E12(X``-r+1) and then applying -! E2(X* *`-r). We will usually omit the subscripts of differentials. This subscript just equa* *ls the difference of the second superscripts. It serves no major purpose, although we * *should keep in mind that short differentials take place before longer ones. Our desire* *d Es2(Xi0) is filtered with subquotients Es;`1, 0 ` i. Each group E1;t2(Xi0) is cyclic, * *since, if nonzero, it is isomorphic to E1;t2(SU(n)), which is cyclic by [2]. Thus our de* *sired p-exponent is given by the sum Xi (|E12(Xi0)|) = (|E1;`1|): `=0 Note that for s = 1 or 2 Es;`1= Z=pmin(N+(p-1)`; (m-`)+1): This is a consequence of Theorem 2.8(1). Perhaps "cellular" is not the best na* *me for this spectral sequence, since each new sphere doubles the number of cells o* *f the CW -complex SU(n), but we hope that the name suggests that it is measuring the effect of building SU(n) one sphere at a time. We are computing the CSS only in certain degrees that yield large groups. In * *[11] the entire CSS was computed for SU(n) with n p2 - p + 1, but to do this for arbitrary n would cause too many cases to have to be considered. We will always choose m so that (m - i) is large, although not arbitrarily large. The largest * *groups E1;2N+1+qm2(Xi0) are obtained when m is chosen in this way. There are variation* *s in the exact form of the CSS depending upon relationships among j, k, and N. In th* *is section, we will establish (3.1) under the hypothesis that N 6= k and n is p-cl* *eaner, relegating two proofs to Section 4. Then in Sections 5 and 6 we will discuss t* *he modifications required when these hypotheses are not satisfied. The following result, which will be proved in Section 4, is essential to our * *bounds on the size of the v1-periodic homotopy groups. The reader should keep in mind * *that we are only dealing with groups of total degree 2N + 1 + mq with (m - i) large. Here we do not need to make any assumptions about n, N, and k. Proposition 3.2. All differentials E1;b! E2;aare 0 if b < i and a j + ffl. 18 D. DAVIS This says that there are no differentials in a broad band in the CSS. Indeed, t* *he only possible nonzero differentials either emanate from the top group, E1;i, or else* * hit into the bottom band, E2;twith t < j + ffl. Recall j = [i-1_p]. On the other hand, it will usually be the case that all groups E2;a1with a < * *j + ffl are killed by differentials. There is an exception to this when N = k and a = j* *, but aside from that, the cases in which this does not happen are so rare and so dif* *ficult to describe that we shall not attempt to describe them explicitly in this paper. W* *e are making a small sacrifice in optimality here for a big gain in simplicity of sta* *tement. In Section 4 we will describe the general pattern of differentials by which the* * groups E2;a1with a < j + ffl are killed. In Example 6.15, we give an example in which * *a group E2;a1with a < j is nonzero. Now we can see how (3.1) is obtained when N 6= k and n is p-cleaner. As just * *noted, all of the groups E2;awith a < j + ffl are usually killed by differentials from* * E1;`with ` < i, while by Proposition 3.2 no other groups E2;aare killed by such differen* *tials. We wish to choose m so that _e:= (m - i) + 1 is as large as possible with the property that the generator ff(m-i)=_eof E12(Xii) pulls back to a class z 2 E12* *(Xij+ffl). This implies that differentials E1;i! E2;aare 0 if a j + ffl. Later in this se* *ction we shall determine this _e. P 1;` P 1;` Then `i (|E1 |) = A - B, where A = `i (|E1 |) and B is the sum of the exponents of the orders of the groups killed by differentials. Since (m - i) is* * large, (m - `) = (i - `) for 0 ` < i, and hence, for d > 0, we have (|E1;i-d1|) = min(1 + (d); N + (p - 1)(i - d)). We may assume without loss of generality that* * this will always equal 1 + (d), for in the rare case in which N + (p - 1)(i - d) < 1* * + (d), both A and B will be modified in the same way. Thus A = _e+ i + (i!), and j-1+fflX j-1+fflX B (|E2;`1|) = (1 + (i - `)) = j + ffl + (i!) - ((i - j - ffl)!): `=0 `=0 We obtain Xi (3.3) (|E1;`1|) = A - B _e+ i - j - ffl + ((i - j - ffl)!): `=0 In order to explain how we obtain an estimate of the threshold value _eof (m - i) + 1, we introduce some notation and conventions for working in the unstable * *cobar complex. Most of these conventions were employed in [11]. ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 19 Convention 3.4. (1)We are dealing with UC*(Xi0(N)), where 1 N p-1, in total degree 2N + 1 + qm, with (m - i) large. Also, i = jp + k with 1 k * * p. We write Xbafor subquotients of Xi0(N). (2) We write y` for the generator x2N+q`+1 of BP*(Xi0(N)). (3) We write h for h1 and v for v1. (4) We often omit powers of v on the left side of terms of the cobar complex.* * In particular, hey` means vm-e-`hey`. This can be done because we are in fix* *ed degree 2N + 1 + qm. P P s (5) If c`y` and d`y` are classes in UC (X), we write X X c`y` d`y` mod L if, for each `, c`- d` is defined on a lower sphere than is d`. The "L" s* *tands for "lower sphere." This will usually be used in the context of d` being* * the "leading term" of c`, i.e., the term which desuspends least far. Note tha* *t the dimension of the lower sphere varies from term to term in this sum. The following definition will be extremely important. Definition 3.5.The excess exc(fl) of an element fl of s is defined to be the s* *mallest n such that fl2n+1 is an element of UCs(M). This means that if fl = fl1 . . .fls, then for 1 i s, fli (fli+1 . . .fls2* *n+1) must satisfy 2.3. In Lemma 4.2, we give a formula for excess of certain monomi* *als when s = 2. One of our main results for pulling back is given in the following lemma, whi* *ch will be proved in Section 4. Lemma 3.6. Assume that N 6= k, ` j + ffl, and (3.7) (i - `)(p - 1) - ((i - ` - 1)!) + S` < ei i(p - 2) + ` + N - ((i - `)!) + S`: Let z be a cycle in UC1(Xi`+1) satisfying Xi (3.8) z = (cthet+ Lt)yt t=`+1 20 D. DAVIS with (3.9) et= ei- (i - t)(p - 2) + ((i - t)!) - St; ct 2 Z(p), and exc(Lt) < et. Assume also that if ` + 1 kp + N mod p2 and c`+26 0 mod p, then c`+1 c`+2 mod p. Then z pulls back to a cycle z1 in UC1(X* *i`) satisfying z1 = z + (c`he`+ L`)y`, where e` satisfies (3:9) and exc(L`) < e`. M* *oreover, if ` kp + N mod p2 and c`+16 0 mod p, then c` c`+1 mod p in z1. Note that this result is self-perpetuating as long as ei satisfies (3.7). The* * require- ment that n be p-cleaner is needed in order that the first inequality in (3.7) * *is satisfied for all ` j + ffl. The upper bound on ei in (3.7) guarantees that z1 satisfies* * the un- stable condition. Now we can easily obtain our estimate for _eof (3.3). We wish to choose ei so* * that ej+ffl, defined by (3.9), satisfies ej+ffl= N + (j + ffl)(p - 1), since this is* * the largest value for which hej+fflyj+fflsatisfies the unstable condition of 2.3. Thus by Lemma 3* *.6 there is a cycle z in UC1(Xij+ffl) satisfying Xi z = (cthet+ Lt)yt t=j+ffl with (3.10) ei= N + (j + ffl)(p - 1) + (i - j - ffl)(p - 2) - ((i - j - ffl)!) + Sj+* *ffl The RHS of (3.10) is our value for _e. Substituting this into (3.3) yields (3.1* *) when N 6= k and n is p-cleaner, modulo the proofs of 3.2 and 3.6. These proofs will * *form the main content of Section 4. 4. Proofs of claims about the CSS In this section we establish the claims about the CSS made in Section 3, hence establishing (3.1) when N 6= k and n is p-cleaner. That is, we prove Propositio* *n 3.2 and Lemma 3.6. We also discuss the pattern of differentials which usually kill* * the groups E2;a1when a < j + ffl. We begin with some more general background regarding the unstable cobar com- plex. These general results do not assume any of the special notation and conve* *ntions of Section 3. The following basic formulas were used or proved in [6]. ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 21 Lemma 4.1. (1) v1 = ph1 + j(v1) and (h1) = h1 1 + 1 h1; P p+1-i i i (2) v2 = ph2+ (1 - pp-1)hp1v1+ j(v2) - (p + 1)vp1h1+ aiv1 p h1, where ai* *2 Z. (3) d(vn1) = j(vn1) - vn1and d(vahbvc) = (j(va) - va) hbvc - va (hb)vc - vahb (j(vc) - vc) where (hb) = (hb) - hb 1 - 1 hb. The first part of this lemma will be used very frequently in the context of rep* *lacing ph1 by v1- j(v1). For example, it is used in establishing the following formula* * about excess, which was defined in 3.5. Lemma 4.2. If a b and a d, then i j i j exc(pahb vchdve) = max b - (p - 1)(c + d); d - min a; |b - (p - 1)c - pd| - * *(p - 1)e: Proof.The effect of ve on the excess is clear, and so we will assume e = 0. Us* *ing Lemma 4.1(1), we can write the given monomial as (v - jv)ahb-a vchd = vahb-a vchd + . . .hb-a va+chd; which has excess i * * j max max (b - a - (p - 1)(c + d); d); max (b - a - (p - 1)(a + c + d); * *d) = max (b - a - (p - 1)(c + d); d): On the other hand, it can also be written as hb vc(v - jv)ahd-a = hb va+chd-a + . . .hb vchd-ava; which has excess max (max (b - (p - 1)(c + d); d - a); max(b - (p - 1)(c + d); d - pa)) = max (b - (p - 1)(c + d); d - a): The excess of the given monomial is the minimum of the excesses of these two wa* *ys of writing it. Letting B = b - (p - 1)(c + d), this is min(max (B - a; d); max(B; d - a)) = max (B; d) - min(|B - d|; a); as claimed. || 22 D. DAVIS We will need the following fact, which follows from Lemma 4.1(2). Lemma 4.3. In E12(S3) hp1v1 vp1h1 mod S1. The following result, proved in [11, 2.12], will be very useful. Lemma 4.4. hn+p-11 h12n+1 -vp-11h1 hn12n+1 mod S2n-1 if n > 1. The following lemma, proved in [11, 2.13], will also be used many times. It i* *s the one place where the number of v1's on the left is important. Lemma 4.5. d(v`1hn+11) -(` + n + 1)v`1h1 hn1 mod S2n-1: A related result about the unstable cobar complex for spheres is given in the* * next lemma. Lemma 4.6. Let = (oe), and let z = fflvoe-e-1he h + L 2 UC2;2n+1+qoe(S2n+1) be a cycle with ffl 2 Z(p)and exc(L) < e - p + 1 n - . Then z = d(ufflvoe-(e+-p+2)he+-p+2 + L0); where u is a unit in Z(p), and exc(L0) < e + - p + 2. The same conclusion hold* *s for z = fflvoe-e+p-2h he-p+1+ L. Proof.If ffl 0 mod p, then z is defined on S2e-2p+1. Since, by 2.8(1,5) or [1* *, 3.3], the iterated suspension E2;2e-2p+1+oeq2(S2e-2p+1) ! E2;2e-2p+2+3+oeq2(S2e-2p+2+3) is the 0-homomorphism, z is a boundary on S2e-2p+2+3, as desired. If ffl 6 0 mod p, we claim that, with E = e + - p + 2, (4.7) poe-Ehoe = (v - jv)oe-EhE = voe-EhE + L0 ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 23 with exc(L0) < E, while, if oe = sp , (4.8) d(poe-Ehoe)= spoe-E+ h hoe-1+ L1 = sh (v - jv)oe-E+ hE--1 + L1 (4.9) = s(v - ph)oe-E+ h hE--1 + L2 (4.10) = svoe-E+ h hE--1 + L3 (4.11) = svoe-e-1he h + L4; where exc(Li) < E - - 1 for 1 i 4. Then, if us 1 mod p, d(uffl(voe-EhE + L0)) = fflvoe-e-1he h + ufflL4 = z + (ufflL4 - L) with ufflL4 - L a cycle of excess less than e - p + 1, and so, as in the first * *part of the lemma, ufflL4 - L = d(L00) with exc(L00) < E, completing the proof, modulo verification of the numbered equations above. At (4.7), we merely use 4.1(1), and at (4.8) we just need oe-1Xij d(hoe) = oejhj hoe-j: j=1 The term with j = 1 has the largest excess. At (4.9), we have expanded (v -jv)o* *e-E+ and incorporated all terms except the first into L2, since j(v)hE--1 = hE--1 v* * has lower excess than hE--1 . The term voe-E+ is then moved to the left of the , wh* *ere it becomes (jv)oe-E+ , which we expand, using 4.1(1). In (4.10), we note that w* *hen (4.9) is expanded, all terms except the first have lower excess, since, if j 1, (ph)jh hE--1 = (v - jv)j-1h2 (v - jv)hE--2 : Finally, at (4.11) we use Lemma 4.4 and the definition of E. The last sentence of the lemma follows from the above analysis and Lemma 4.4. || Another result of the same sort is the following. Lemma 4.12. If x 2 UC2;2n+1+oeq(S2n+1) is an unstable cycle such that {x} has order pj in E2;2n+1+oeq2(S2n+1), then x = d(uvoe-n-jhn+j + L), with u a unit in* * Z(p) and exc(L) < n + j. Moreover, if e < j, then exc(peL) < n + j - e. 24 D. DAVIS Proof.By Theorem 2.8(4), there is a unit u in Z(p)such that x - ud(poe-n-jhoe) = d(L1) with exc(L1) n. We have poe-n-jhoe= (v - jv)oe-n-jhn+j = voe-n-jhn+j + L2; with exc(L2) < n + j. The result follows, with L = uL2 + L1. The last claim of * *the lemma follows from the explicit form of L2; we can say nothing about multiplica* *tion by pe decreasing the excess of L1. || Corollary 4.13. If x 2 UC2;2n+1+oeq(S2n+1) is an unstable cycle with exc(x) e, then x = d(uvoe-tht+ L) for some t e + (oe) + 1, u a unit, and exc(L) < t. Proof.The class {x} is defined in UC2;2e+1+oeq(S2e+1), where it has order pj fo* *r some j (oe) + 1. The result now follows from Lemma 4.12. || The following lemma will be very useful. Lemma 4.14. The boundary @ : E12(Xtt) ! E22(Xt-10) satisfies X @((cthet+ Lt)yt) = (cthet+ Lt) Hy`; ` ` + 1) (4.17) exc(Xt arh(r)v(r)) max {et- (t - `)(p - 1); t - `}: We have et- (t - `)(p - 1)= ei- (i - t)(p - 2) + ((i - t)!) - St- (t - `)(p - 1) ei- (i - `)(p - 2) + ((i - ` - 1)!) - St- (t - `) ei- (i - `)(p - 2) + ((i - ` - 1)!) - S`- 1: The last inequality is true because we assume ` + 1 6 kp + N mod p2. We also * *have (4.18) t - ` i - ` ei- (i - `)(p - 2) + ((i - ` - 1)!) - S`- 1; with the latter inequality implied by (3.7). Next we verify the claim about excess when ` + 1 kp + N mod p2. Now we have S0`+1= S0`+2+ 1 and S00`= S00`+1+ 1, which cause the inequalities proved above * *in the case ` + 1 6 kp + N to fail for exc((N + `(p - 1))c`+1he`+1 h) if c`+1is a unit* *, and for exc(cthet arh(r)v(r)) if t = ` + 2, and c`+2 and ar are units. The special hypo* *thesis in the lemma says that we are computing @(uh^e-p+1y`+1+ uh^ey`+2); with ^e= ei- (i - ` - 2)(p - 2) + ((i - ` - 2)!) - S`+2 ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 27 and u a unit in Z(p). We write ` + 1 = Kp + N, with K k mod p, and, working mod terms of excess less than ^e- 2(p - 1), find that @(uh^e-p+1y`+1+ uh^ey`+2) u((Kp + N - 1)(p - 1) + N) h^e-p+1 h i j! +h^e (Kp+N-1)(p-1)+N-1_2h2 - hv yKp+N-1 i j (4.19) u h^e-p+1 h + p(N-1-K+Kp)_2h^e h2 - h^e (vh - ph2) yKp+N-1 : In the first , we have argued as in the proof of Lemma 4.14, and in the second we have used Lemma 4.1(1). In (4.19), the two terms with ph^e h2 can, by Lemma 4.1(1), be written as h^e-1 h2, which has excess ^e- 1 - 2(p - 1), and we can u* *se Lemma 4.4 to write h^e vh as (vph^e-p+1+ L) h with exc(L) < ^e- p + 1. The fir* *st term cancels the initial h^e-p+1 h, and exc(L h) ^e- 1 - 2(p - 1). Thus (4.19) has excess ^e- 2(p - 1) - 1 = ei- (i - ` - 2)(p - 2) + ((i - ` - 2)!) - S`+2- 2(p - 2) - 3 ei- (i - `)(p - 2) + ((i - ` - 1)!) - (S`- 2) - 3; as claimed in (4.16). It is our lack of knowledge of the precise excess here th* *at is the problem in the examples in the table in Section 1 where Theorem 1.4 does not ag* *ree with the actual value of exp3(v-11SU(n)). Next we establish the technical condition in the last sentence of Lemma 3.6 w* *hen ` kp + N mod p2. Excess considerations imply that the only term which we need to consider is c`+1he`+1y`+1, with c`+1 a unit in Z(p). We will show that when * *this is pulled back to UC1(X`+1`), the new term added is c`+1he`+1-(p-1)y`, mod L. 28 D. DAVIS We have, mod L, @(c`+1he`+1y`+1)= (N + `(p - 1))c`+1he`+1 hy` (N - k)pc`+1he`+1 hy` (k - N)pc`+1h he`+1-p+1y` (k - N)c`+1h he`+1-py` (4.20) = N-k_m-`c`+1d((he`+1-p+1+ L0)y`) with exc(L0) < e`+1- p + 1. Here the first "=" follows from the method of Lemma 4.14, the first "" follows from the assumption that ` kp + N mod p2, the next "" uses Lemma 4.1(1) to replace ph by v - j(v), the next "" uses Lemma 4.4, while the final "" uses Lemma 4.5 to control the coefficient and either Lemma 4* *.6 or Lemma 4.12 for the L0-part. Since m k and ` N mod p, the fraction in the last line of (4.20) is -1 mod p, and so, as claimed in the last line of Lemma * *3.8, there is a cycle z1 satisfying z1 c`+1he`+1y`+1+ c`+1he`+1-p+1y` mod L: This completes the proof of Lemma 3.6. || Next we prove Proposition 3.2. Proof of Proposition 3.2.We will show that there is a cycle Xb (4.21) (u`he`+ L`)y` `=j+ffl with exc(L`) < e`, u` a unit in Z(p), e` (p - 1)` + N, and (ubheb+ Lb)yb a gen* *erator of E12(Xbb). The numbers e` will be less than or equal to the numbers a` define* *d in the following lemma. Note that the conclusion of the lemma implies that the cyc* *le (4.21) satisfies the unstable condition. Lemma 4.22. Suppose i, j, k, N, and ffl are as in Theorem 1.4, and b < i. Fo* *r ` going from b down to j + ffl, define numbers M` and a` iteratively by Mb = 0, a* *nd a` = M`+ (i - `) + 1 ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 29 with ii j j (4.23) M` = max {t - ` - `(p-1)+Nt-`; at- (t - `)(p - 1) : ` < t b}: Then a` (p - 1)` + N for j + ffl ` b. ii j j Proof.We first verify that the lemma is true when M` = t - ` - `(p-1)+Nt-`. T* *he desired inequality in this case is ii j j t - ` - `(p-1)+Nt-`+ (i - `) + 1 `(p - 1) + N: Using Lemma 4.25, this inequality will follow from i i jj t + (i - t) `p+N-ti-t+ 1 `p + N; which is implied by Lemma 4.26, since logp(ff) + 1 ff. Now we illustrate the proof in the general case by assuming that for ` < `1 <* * `2 < t, we have a` = a`1- (`1 - `)(p - 1) + (i - `) + 1; a`1 = a`2- (`2 - `1)(p - 1) + (i - `1) + 1; ii jj a`2 = t - `2 - `2(p-1)+Nt-`2+ (i - `2) + 1: The desired inequality is then easily simplified to the following. ii jj (4.24) t + (i - `) + (i - `1) + (i - `2) + 3 `2p + N + `2(p-1)+Nt-`2: We write the final binomial coefficient as (`2(p-1)+N)...(t-`2+1)_(`2p+N-t)!, a* *nd observe that (i - `), (i - `1), and (i - `2) are all factors of the numerator. Since `2p + N (` + 2)p + N (j + ffl + 2)p + N i + 2p > t + 2p; the only way that (4.24) might fail is if one of (i - `), (i - `1), or (i - `2)* * is large. If more than one of these is large, then (`2 - `) is very large, and this will * *cause `2p + N - t to totally dominate all the (-)'s. We assume that it is (i - `1) which is large, with the other cases being hand* *led similarly. Similarly to Lemma 4.25, we have ii jj i j! (i - `1) - `2(p-1)+Nt-`2 (i - t + `2 - `1) i`2p+N-t-t+`2-`1; 30 D. DAVIS and so, using also Lemma 4.26, the desired inequality, (4.24), follows from (i - `) + (i - `2) + [logp(`2p + N - t)] + 3 `2p + N - t; which is clear from the preceding discussion. Indeed, it it true that `2p + N - t p(i-`)+ p(i-`2): A final case deserving mention is the case ` = b. Here we require (i - b) + 1 (p - 1)b + N. For b j + ffl, this is implied by the fact that (x) < x. || Now we complete the proof of Proposition 3.2. We are trying to establish exis* *tence of the cycle (4.21). Assume that we have a cycle Xb z`+1= (uthet+ Lt)yt t=`+1 with exc(Lt) < et at, where at is as defined in Lemma 4.22. Then @(z`+1) is evaluated as in Lemma 4.14. By Lemma 4.2 and the last part of Lemma 4.12, exc(@(z`+1)) M`, where M`is as in (4.23). Then since a` = M`+(i-`)+1, Lemma 4.12 implies that this @(z`+1) can be written as d(u`he`+ L) with exc(L) < e` * *a`. Letting z` = z`+1- u`he`- L extends the induction. || The following two combinatorial lemmas were used above. Lemma 4.25. Assume that ` < t < i < `p + N. Then ii j j i i jj (i - `) - `(p-1)+Nt-` (i - t) `p+N-ti-t: Equality holds if p(i-`)> max {`p + N - i; t - `}. Proof.Note that t - ` + 1 i - ` `(p - 1) + N. Thus ii j j (i - `) - `(p-1)+Nt-` ! (`p - t + N)! = __________________________________________________ (`(p - 1) + N) . .(.i - ` + 1) . (i - ` - 1) . .(.t - ` + 1) i j = __(`p-t+N)!__(i-t-1)!(`+N-i)!(`p+N-i)!_(`p-`+N)...(i-`+1)(i-t-1* *)!_(i-`-1)...(t-`+1) i i jj ii j j ii jj = (i - t) `p+N-ti-t- `(p-1)+Ni-`- i-`-1t-`: ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 31 ii j j The condition p(i-`)> max {`p + N - i; t - `} guarantees that `(p-1)+Ni-`= 0 * *and ii jj i-`-1t-`= 0 by the standard result relating the mod p value of binomial coeff* *icients to the p-ary expansions of the numbers. || i j Lemma 4.26. If 0 < i b, then p(i bi) logp(b). Proof.Assume pj b < pj+1. We show i b . .(.b - i +j1) ______________ j: (i - 1)! We have max {p(k) : b - i < k b} j. The i - 1 numbers p(k) other than the one which achieves the maximum are less than or equal to corresponding numbers (1); : :;:(i - 1) in the denominator. || We close this section by discussing the pattern of differentials which usuall* *y kill all elements in E2;a1for a < j + ffl. As discussed in Section 3, these differential* *s are not essential to Theorem 1.1. That theorem assumed that all such elements are kille* *d. If some are not, then our theorem becomes weaker than it might be; however, the au* *thor believes that the theorem as stated is the correct compromise between tractabil* *ity and optimality. The main differentials are (4.27) dap+N-a : E1;ap+Nap+N-a! E2;aap+N-ais usually surjective. The cases in which this differential is not surjective occur when there is a va* *lue of t slightly greater than a for which (i - t) is very large. In such a case, there * *can be a shorter nonzero differential E1;ap+N! E2;t. If this happens, there will usual* *ly be a nonzero differential from E1;ap+N- to E2;afor some small positive integer . The surjectivity of the differential in (4.27) requires that if |E2;a1| > |E1* *;ap+N1|, then there are shorter differentials which kill some of the multiples of the generat* *or of E2;a1. This will indeed be the case, with most of the killing done by E1;ap+N- for sm* *all positive values of . 32 D. DAVIS We illustrate with the case p = 5, i = 55 + 2, and N = 4. For 0 t 55 + 1, we have 8 >>>1 if t 6 2 mod 5 >>> 2 >>>2 if t 2 mod 5 but t 6 2 mod 5 >< 2 3 3 if t 2 mod 5 but t 6 2 mod 5 (|E1;t1|) = (|E2;t1|) = > 3 4 >>>4 if t 2 mod 5 but t 6 2 mod 5 >>> 4 >>>5 if t = ff5 + 2 with 1 ff 4 :6 if t = 2. We are not concerned here with E2;i1. If 0 a < 54 and a 6= 1, the pattern of differentials into E2;ais as follows.* * If (|E2;a1|) > 2, then d4a+2 : E1;5a+2! E2;ahits all multiples of p2. If (|E2;a1|* *) > 1, then a nonzero d4a+3 from E1;5a+3hits p times the generator of E2;a1. Finally, * *d4a+4: E1;5a+44a+4! E2;a4a+4is surjective for all a satisfying 0 a < 54 and a 6= 1. The large group E2;2causes the unusual behavior when a = 1. The group E2;2is hit by E1;9, E1;11, E1;12 Z=52, E1;13, and E1;14, each differential, starting w* *ith the shortest, hitting the element of order p in the remaining group E2;ar. (The dif* *ferential from E1;12also hits elements of order p2.) Finally E1;8! E2;1is surjective. So * *in the notation of (4.27), a large group when t = 2 blocked the expected nonzero diffe* *rential E1;9! E2;1. Since the claims surrounding (4.27) are not necessary to our main theorem, we need not provide complete justification for them. We just outline the backgroun* *d for the basic differential. We use (2.10) to represent the generator of E1;ap+N1, * *mod p, by hyap+N. Under favorable conditions, this pulls back to a cycle z in UC1(Xap+* *Na+1) satisfying, mod L, i j @(z) a(p-1)+Nap+N-ah hap+N-aya 2 UC2(Xaa): This class is a generator of E22(Xaa). Here we have used (2.7), 2.6, and 2.5. 5.Results when N = k and n is p-cleaner In this section, we establish the second and third cases of (1.9) by proving * *that if N = k and n is p-cleaner, then Xi (|E1;`1|) n - 1 + ((i - j) . .(.i - j - r)) + Sj+r+1; `=0 ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 33 and the estimate can be improved by 1 if n is p-cleanest. We continue to employ freely all the notation of Theorem 1.4. We still have Proposition 3.2 and that, for all practical purposes, E2;t1= 0 * *for all t < j. Most of this section will be devoted to establishing the following resul* *t. Theorem 5.1. Assume the notation of Theorem 1.4, with N = k. Assume that for all D 1 (5.2) (i - j - D) (p - 1)D - ((D - 1)!) + Sj+1- Sj+D: Let r = (i - j). Either (1) for some m with (5.3) (m - i) + 1 = N + j + r + 1 + i(p - 2) - ((i - j - r - 1)!) + Sj+r+1; the differentials E1;i! E2;j+tare 0 for t > r, while for 0 t r, they ha* *ve image of order p, and, if equality never occurs in (5:2), then at most r* * of them are nonzero; or (2) if (m - i) + 1 = N + j + r + 2 + i(p - 2) - ((i - j - r - 1)!) + Sj+r+1; then the differentials E1;i! E2;j+tare 0 for t > r, while for 0 t r, th* *ey have image of order p; or (3) r = 0 and if (m - i) + 1 = N + j + 2 + i(p - 2) - ((i - j - 1)!) + Sj+1; then, for t 0, the differential E1;i! E2;j+tis possibly nonzero only for* * t = 2 or 3 (but not both) and, if nonzero, its image has order p. Thus, with R denoting the RHS of (5:3), we have, similarly to (3:3), Xi i j (|E1;t1|) R + i + (i!) - j + (i . .(.i - j + 1)) + r + 1 t=0 (5.4) = n - 1 + ((i - j) . .(.i - j - r)) + Sj+r+1; and if equality never occurs in (5:2), then the RHS of (5:4) can be increased b* *y 1. This completes the proof of the second and third cases of (1.9). In order to prove Theorem 5.1, we will need the following analogue of Lemma 3* *.6. 34 D. DAVIS Lemma 5.5. Assume the notation of Theorem 1.4 with N = k, and that (5.6) (i - `)(p - 1) - ((i - ` - 1)!) + S` < ei i(p - 2) + ` + N - ((i - `)!) + S`: Let z be a cycle in UC1(Xi`+1) satisfying Xi z = (cthet+ Lt)yt t=`+1 with (5.7) et= ei- (i - t)(p - 2) + ((i - t)!) - St; ct 2 Z(p), ` j, and exc(Lt) < et. Assume also that if j 6 N mod p and ` + 1 jp2 + Np + N mod p3 and c`+2 6 0 mod p, then c`+1 c`+2 mod p. Then z pulls back to a cycle z1 in UC1(Xi`) satisfying z1 = z + (c`he`+ L`)y`, where e` sati* *sfies (5:7) and exc(L`) < e`. Moreover, if j 6 N mod p and ` jp2 + Np + N mod p3 and c`+16 0 mod p, then c` c`+1 mod p in z1. As in Lemma 3.6, the last sentence is needed to make this lemma self-perpetua* *ting. Proof.When `; ` + 1 6 Np + N mod p2, the proof is the same as that of Lemma 3* *.6 in Section 4. We will analyze two steps in pulling back (he + L)y`+1 when ` = ffp2 + Np + N; a condition which will be assumed through the remainder of this proof. Let b = (i - `). We will show that (he + L)y`+1 pulls back, mod L, to 8 ` + 1 cannot mess up the pulling back c* *laimed in (5.8), but this can be accomplished as in the proof of (4.17) in 3.6, using * *the LHS of (5.6). The part about c 1 in (5.8) when ff j 6 N translates to the last sente* *nce of Lemma 5.5. This specific information is required to obtain the exponent of h* * on y`-1in this case, which is the content of the assumption after (5.7). The RHS o* *f (5.6) is required so that z1 satisfies the unstable condition. Now we prove (5.8). We consider first the case j 6 N mod p. Then `(p-1)+N = p2(ffp + N - ff) and i - ` = p(j - N - ffp), and so @((he + L)y`+1) = p2(ffp + N - ff)(he + L) hy` = -p2(ffp + N - ff)(h he-p+1+ L1)y` ffp + N - ff e-p+2 = p___________d(h + L2)y` j - N - ffp ffp + N - ff e-p+1 0 = ___________d((h + L )y`); j - N - ffp where in each case the L has excess less than that of the accompanying term. He* *re we have used Lemma 4.4 at the second step, Lemma 4.5 at the third step, and Lemma 4.12 (applied to p(hhe-p+1+L1)2(e-p)+1) at the last step to know that the p red* *uces the excess of L2. Thus we can pull back to a cycle ffp + N - ffe-p+1 0 z0= (he + L)y`+1+ (___________h + L )y`: ffp + N - j (Note that the fraction here is 1 mod p if ff j mod p, and is a unit if ff 6 N 36 D. DAVIS mod p.) Now we have, omitting a factor (` - 1)(p - 1) + N, which is 1 mod p, ! 2+Np-ffp-1)2 ffp+N-ff e-p+1 0 @(z0) = (he + L) (p(ffp________2h - hv) + (_______ffp+N-jh + L ) h * * y`-1 = (fihe-1 h2 + __j-ff_ffp+N-jhe-p+1 h + L00)y`-1: Here we have used, as usual, hv = vh - ph2 and hpv vph, and then combined the (he-1 h2)-terms and the (he-p+1 h)-terms. If ff 6 j, then the leading term is a unit times he-p+1 h, and so the claim follows from Lemma 4.6, while if ff j, t* *hen all terms have excess e - 2p + 1, and so the claim follows from Corollary 4.13. Now assume j N mod p, so that b := (i - `) 2. As before, @((he + L)y`+1) = (he + L) p2(ffp + N - ff)hy`: If ff 6 N mod p, this is d((uhe-p+b+ L0)y`) by Lemma 4.12. We need here that exc(p2L) < e - 2, which will be true if L was obtained as in Lemma 4.12, using * *also that, in the notation of 4.12, (oe) 2. Then @((he + L)y`+1- (uhe-p+b+ L0)y`) is dominated by the term @(uhe-p+by`), which is easily obtained. If, on the other hand, ff N mod p, then all terms T in @((he + L)y`+1) sat* *isfy exc(T ) e - 2 - p. Here we need again that L has been obtained as in Lemma 4.12, and we only need to decrease the excess of L by 2. This @((he + L)y`+1) * *is d((che-p+b+ L0)y`) with c 0 mod p, and from here we proceed similarly to the previous paragraph. || Proof of Theorem 5.1.We choose ei equal to the RHS of (5.3), and use Lemma 5.5 to deduce that heiyi pulls back to a cycle (5.9) zj+r+1 heiyi+ . .+.cj+r+1h(j+r+1)(p-1)+Nyj+r+1 mod L with cj+r+1 2 Z(p). The first part of (5.6) for ` j + r + 1 is implied by (5.2* *), and the second part is automatic. First we consider the case r = 0 and cj+1 a unit. Then, mod L @(cj+1h(j+1)(p-1)+Nyj+1) uh(j+1)(p-1)+N hyj -uh hj(p-1)+Nyj with u a unit. However, this can be cancelled if equality never occurs in (5.2)* *. We refer back to Theorem 2.8(2), which says that the initial term ff(m-i)(p-1)=eiy* *i (which ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 37 we have been representing as heiyi mod L) is congruent mod p to -sv(m-i)(p-1)-* *1hyi, where (m - i)(p - 1) spei-1 mod pei: This yieldsia contributionj-shijhi-jyj to @(zj+1). Note that the binomial coeff* *icient here is j(p-1)+Ni-j= j(p-1)+Nj(p-1)+N= 1, and, because i - j = j(p - 1) + N, * *the class of h hi-jyj generates E22(Sjq+2N+1). We obtain @(zj+1) = (-s - u)h hj(p-1)+Nyj: We can choose m so that s -u mod p. Here it is important to note that the value of the unit u does not depend on this s. In our pulling back, we use (2.9* *) to view the initial term as -v(m-i)(p-1)-eiheiyi, independent of s. The assumption* * that equality never occurs in (5.2) implies that the LHS of (5.6) can be made 1 stro* *nger, so that, in @(z) which is relevant at each stage of pulling back, the term ff * *hi-` will have excess strictly less than the bound being used for the excess of term* *s such as che`+1 h. This will prevent the value of s from contributing to the coeffici* *ent u. Since E2;j1 Z=p in the case under consideration, we obtain that if m is chosen * *so that s -u mod p, then di-j= 0 : E1;i! E2;j, so that the last clause of Theorem 5.1* *(1) is satisfied. If equality does occur in (5.2), then we cannot assert that this * *cancellation can be achieved, but Theorem 5.1(1) is still satisfied. We had not discussed th* *is kind of cancellation when N 6= k and n is p-cleaner because h hi-j-fflyj+fflis a bo* *undary mod L in these cases. Next we consider the case where r = 0 and cj+1 0 mod p. Then the last term of (5.9), cj+1h(j+1)(p-1)+Nyj+1, will still be unstable if the exponent of h is* * increased by 1. It will usually be the case that the whole sum (5.9) will still be unstab* *le if each term is multiplied by h (and divided by v to keep the degree unchanged). To see* * this, we note that in (5.7) et+1- et p - 1, with equality if and only if t Np + N mod p2 and N 6 j mod p (5.10) or t jp2 + Np + N - 1 mod p3: If neither of the conditions in (5.10) is satisfied when t = j + 1, then (5.9) * *remains unstable after being multiplied by h, since all terms et with t > j + 1 satisfy* * et < t(p - 1) + N. In this case, for any m with (m - i) + 1 exactly 1 greater than t* *he 38 D. DAVIS value stated in (5.3), d : E1;i! E2;jwill be possibly nonzero due to the -sh h* *i-j term discussed in the two preceding paragraphs, and Theorem 5.1(2) is satisfied* * in this case. Since we do not assert cancellation here, we do not need the assumpt* *ion of no equality in (5.2). If, however, j + 1 Np + N mod p2 with N 6 j mod p, then it is possible th* *at the penultimate term of (5.9) is uh(j+2)(p-1)+Nyj+2 with u a unit. (This would * *be the case where ` = j + 1, c 0 mod p, and b = 1 in (5.8).) All preceding terms wil* *l be of the form cthetytwith et< t(p - 1) + N. If eiis 1 greater than the value on t* *he RHS of (5.3), heiyi will pull back to a cycle w in UC1(Xij+3) with exponents of h o* *n the leading terms 1 greater than those of (5.9), and there may be a nonzero differe* *ntial E1;i! E2;j+2due to @(h(j+3)(p-1)+Nyj+3) u0h(j+3)(p-1)+N hyj+2: Then pw pulls back to a cycle whose leading terms equal those of (5.9), and the differential into E2;jis 0. Indeed, the component from yj+1 is 0 due to cj+1 * *0, while the component from yi is 0 since w has been multiplied by p. Thus we obta* *in Theorem 5.1(3) with t = 2. When j + 1 jp2 + Np + N - 1 mod p3, the situation is somewhat similar. If j N, then in (5.8) with ` - 1 = j + 1, we have b > 1 and c is not a unit, and * *we deduce that all terms in (5.9) remain unstable after the exponents of h are inc* *reased by 1, and so Theorem 5.1(2) is satisfied. If j 6 N, then b = 1 and c is a unit * *in (5.8). In this case, the last three terms of (5.9) could be uh(j+3)(p-1)+Nyj+3+ u0h(j+2)(p-1)+Nyj+2+ ch(j+1)(p-1)+Nyj+1 with u and u0units, but c 0 mod p, and et < t(p - 1) + N if t > j + 3. Hence,* * if ei is 1 greater than the value on the RHS of (5.3), heiyi will pull back to a c* *ycle w in UC1(Xij+4) with exponents of h on the leading terms 1 greater than those of (5.* *9), and there may be a nonzero differential E1;i! E2;j+3due to @(h(j+4)(p-1)+Nyj+4) u00h(j+4)(p-1)+N hyj+3: Then pw pulls back to a cycle whose leading terms equal those of (5.9), and, as* * before, the differential into E2;jis 0. Thus we obtain Theorem 5.1(3) with t = 3. ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 39 Assume next that 0 < r < p and cj+r+1 is a unit. Then, mod L, @(zj+r+1) uh h(j+r)(p-1)+Nyj+r; and so di-j-r: E1;i1= E1;ii-j-r! E2;j+ri-j-r E2;j+r1 Z=p is surjective. Then by Lemma 4.6 @(pzj+r+1) = d((u0h(j+r)(p-1)+N+ Lj+r)yj+r); and so pzj+r+1 pulls back to a cycle wj+r satisfying wj+r pzj+r+1- u0h(j+r)(p-1)+Nyj+r mod L: This continues, yielding, for 1 t r, surjective differentials di-j-t: E1;ii-j-t! E2;j+ti-j-t= E2;j+t1 Z=p and classes wj+t pwj+t+1- uh(j+t)(p-1)+Nyj+t mod L: Finally, @(wj+1) contains significant contributions from its top (prff(m-i)(p-1* *)=eiyi) and bottom (uh(j+1)(p-1)+Nyj+1) classes. Similarly to the case above with r = 0 and* * cj+1 a unit, the top hits -prsh hi-jyj if (m - i)(p - 1) spei-1 mod pei, while the bottom hits u(j(p - 1) + N)h(j+1)(p-1)+N hyj -u(i - j)h hj(p-1)+Nyj: Since (i - j) = r and i - j = j(p - 1) + N, we find that if m is chosen so that* * s has a certain mod p value, then the two contributions cancel in E2;p Z=pr+1, and so di-j= 0. Hence here we have Theorem 5.1(1). Again we need the assumption of no equality in (5.2) in order to assert that the coefficients obtained in pulli* *ng back do not depend on s. If r p, then in the procedure of the previous paragraph, there will be at le* *ast one value of t for which j + t N mod ip. This willjhave two effects: (a) |E2;* *j+t1| = p1+(i-j-t) p2, and (b) the coefficient (j+t)(p-1)+N1which occurs in @(wj+t+1) * *is a multiple of p. Since i - j - t = j(p - 1) + N - t and (j + t)(p - 1) + N have t* *he same 40 D. DAVIS p-exponent for the values of t being considered, these effects will effectively* * cancel. Indeed, if (j + t)(p - 1) + N = up , then, mod L, i j @(wj+t+1) (j+t)(p-1)+N1h(j+t+1)(p-1)+N hyj+t -up h h(j+t)(p-1)+Nyj+t has order p in E2;j+t1. Thus d : E1;ii-j-t! E2;j+ti-j-thas image of order p, a* *nd pwj+t+1 extends as in the previous paragraph. If (cj+r+1) = d > 0, then the first d of the differentials described in the t* *wo previous paragraphs will usually be 0, and then the pattern described there wil* *l begin. For example, if d = 2, then p2h(j+r+1)(p-1)+Nyj+r+1 pulls back to a class congr* *uent, mod L and ignoring unit coefficients, to z p2h(j+r+1)(p-1)+Nyj+r+1+ ph(j+r)(p-1)+Nyj+r+ h(j+r-1)(p-1)+Nyj+r-1; and @(z) h h(j+r-2)(p-1)+Nyj+r-26= 0 2 E2;j+r-2: It could happen that pdh(j+r+1)(p-1)+Nyj+r+1 is preceded in (5.9) by a term suc* *h as uh(j+r+2)(p-1)+Nyj+r+2, which will dominate in the calculation of @(zj+r+1). T* *hat could make it so that these first differentials are not zero, but we still obta* *in the pattern and results of the previous two paragraphs. || 6. Results when n is not p-cleaner In this section, we prove the remaining cases of (1.9). That is, we prove Theorem 6.1. If n is not p-clean, then Xi (|E1;`1|) n + Sj+ffl; `=0 while if n is p-clean but not p-cleaner, then Xi (|E1;`1|) n - 1 + Sj+ffl: `=0 The following result gives the description of the CSS (in the degree specifie* *d in Convention 3.4(1)) in this case. ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 41 Lemma 6.2. Assume n is not p-cleaner. Let D be the largest integer satisfying* * ffl D < i - j - ffl such that (6.3) (i - j - D) N - k + ffl + D(p - 1) - ((D - ffl)!) + Sj+ffl- Sj+D; and then choose the smallest 0 such that (6.4) + (!) - (Sj+D- - Sj+D) N - k + Dp - (i - j - D): Such a exists satisfying D. For j +ffl ` j +D -, the differential di-`is possibly nonzero. (Here and in* * the rest of this lemma, all differentials emanate from E1;i.) However, if ` N mod* * p, then either di-`is not surjective or else di-`-ffi= 0 for 1 ffi p. Similarly,* * if k 6= N and ` kp + N - 1 mod p2, then either di-`is not surjective or else di-`-ffi= * *0 for 1 ffi p + 1. In addition, if k = N and ` = ffp2 + Np + N, then either 8 <> p2 ifff N j mod p | coker(di-`)| : = p2 otherwise and di-`+1= 0 if ff j 6 N mod p, or else di-`-ffi= 0 for 1 ffi p. If () > 0 (or = 0), then (E2;j+D-1) > 1, and the order of the image of E1;i! E2;j+D- is less than or equal to pfl+1, where fl is the difference of th* *e two sides of (6:4). There will be no other differentials into groups E2;`with ` j + ffl, provide* *d ei = (m - i) + 1 satisfies (6.5) ei j + k + 1 - D(p - 1) + i(p - 2) - ((i - j - D - 1)!) + Sj+D: If = 0, then either there exists such an m for which the image of the differ* *ential into E2;j+D has order less than or equal to pfl, or else ei can be made 1 large* *r than the RHS of (6:5) and still have no additional differentials. 5 4 4 For example, if N = 5, k = 2, and j = (pp - 1 + p - 1)=(p - 1), then D = p * *and = 3. Before proving this lemma, we deduce Theorem 6.1 from it. 42 D. DAVIS Proof of Theorem 6.1.Lemma 6.2 shows that if n is not p-cleaner, then, analogou* *sly to (3.3), we have Xi (6.6) (|E1;`1|) = A - B `=0 with A = i + (i!) + RHS of (6.5) and B j + D - + (i . .(.i - j - D + + 1)) - (Sj+ffl- Sj+D- ) + fl + 1; where fl is as in the lemma. The Sj+ffl- Sj+D- in B is due to the cases in Le* *mma 6.2 in which di-tmight not be surjective for j + ffl t < j + D - . Simplifying somewhat gives the first step of Xi (|E1;`1|) i(p - 1) + k - Dp + + Sj+D + Sj+ffl- Sj+D- - fl `=0 +((i - j - D + ) . .(.i - j - D)) = i(p - 1) + N + Sj+ffl = n - 1 + Sj+ffl; while substituting the definition of fl gives the second step, along with the o* *bservation that (i - j - D + t) = (t) for t 6= 0 in the range under consideration. Clearly = 0 satisfies (6.4) if and only if n is not p-clean. The final port* *ion of Lemma 6.2 says that in this case either B may be decreased by 1 or A may be increased by 1 in (6.6). || Proof of Lemma 6.2.It is immediate from (6.3) that inequality (6.4) is satisfie* *d for = D. This establishes the claim that the smallest such is less than or equal * *to D. Inequality (6.3) is the antithesis of (1.8), which was used in the proof of L* *emma 3.6 to guarantee that (3.7) could be satisfied. If D is chosen as in (6.3), then th* *e cycle z of (3.8) with ` = j + D yields a term T in @(z) of the form i j T = (j+D)(p-1)+Ni-j-Dhei hi-j-Dyj+D: ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 43 The binomial coefficient here is a unit, since (j + D)(p - 1) + N = (i - j - D) + (Dp + N - k); i L j so if (i - j - D) is a large number L, the coefficient is ffpf+dfpL, with d <<* * pL; in fact, d is not much larger than L. The excess of T is at least i - j - D, and, * *by the analysis of (4.18), this does not satisfy exc(T ) e with e as in (4.16). Thus * *Lemma 3.6 must be modified for subsequent pulling back. In this situation, eican be increased over its value in (3.10). Recall that t* *his value of e in (3.10) was obtained as the largest value that would allow pulling back * *to a class with ej j(p - 1) + N. In our new situation, we choose the largest ei tha* *t will allow pulling back to a class with ej+D+1 i - j - D + p - 1. This is appropria* *te since i j @(hi-j-D+p-1yj+D+1) = (j+D)(p-1)+N1hi-j-D+p-1 hyj+D i j (j+D)(p-1)+N1h hi-j-Dyj+D; whose excess is equal to or less than that of the term T , obtained independent* *ly of the value of ei. The same pullback analysis which yielded (3.10) now yields ei = i - j - D + p - 1 + (i - j - D - 1)(p - 2) - ((i - j - D - 1)!) + Sj+D (6.7)= j + k + 1 - D(p - 1) + i(p - 2) - ((i - j - D - 1)!) + Sj+D; which is (6.5). With ei chosen in this way, and et satisfying (3.9), we have a cycle Xi (6.8) z = (cthet+ Lt)yt t=j+D+1 with exc(Lt) < et. By the analysis in the proof of Lemma 3.6, together with the observations so far in this proof, all terms T in @(z) have exc(T ) i - j - D.* * If = 0, then, by (6.4), (6.9) i - j - D = (j + D)(p - 1) + k - Dp (j + D)(p - 1) + N - (i - j - D); and so {@(z)} is possibly nonzero in E22(S(j+D)(p-1)+N). Moreover, its order i* *s less than or equal to pfl+1, where fl is the difference in (6.4). 44 D. DAVIS In fact, we can do 1 better than this. If equality is obtained in (6.9),iand * *in (6.8)j we have cj+D+1 a unit, ej+D+1 = i - j - D + p - 1, and N 6= k (so that (j+D)(p* *-1)+N1 is a unit), then, for some m with (m - i) + 1 equal to (6.7), the two terms in i i j j (6.10@(z)) uffm-i=ei hi-j-D + (j+D)(p-1)+N1cj+D+1hi-j-D+p-1 h yj+D; will be cancelling elements of orderip in E2;j+D,jand so here we have fl = 0 and im(di-j-D) = 0. The coefficient u = (j+D)(p-1)+Ni-j-Dis a unit since the top p* *art equals i - j - D + N -ik + Dp, and (i - j - D) N - k + Dp, so the binomial coefficient d+dj is of the form cpcpd, which is a unit. If one of the hypotheses at the beginning of this paragraph fails, then we wi* *ll not get the cancellation and will still have the nonzero differential due to ffm-i=eihi* *-j-D, but now we can increase ei by 1 over its value in (6.7) and still have all terms un* *stable. The argument in this case ( = 0) continues as in the paragraph, soon to follow, which begins "If pR is the order." On the other hand, if > 0, then (6.9) does not hold, and so by Corollary 4.13 @(z) = d(uhe + L)yj+D with e i - j - D + (i - j - D) + 1 and exc(L) < e. Thus z pulls back to the cycle z0= z - (uhe + L)yj+D 2 UC(Xij+D). Similarly to the procedure in the proof of Lemma 3.6, this cycle z0 pulls bac* *k to X z0+ (cj+D-thej+D-t+ Lj+D-t)yj+D-t t1 with ej+D-t i - j - D + (i - j - D) + 1 - (p - 2)t +((i - j - D + 1) . .(.i - j - D + t)) - Sj+D-t + Sj+D and exc(Lj+D-t) < ej+D-t, as long as ej+D-t (j + D - t)(p - 1) + N. Using the fact that ((i - j - D + 1) . .(.i - j - D + t)) = (t!) since (i - j - D) is lar* *ge, this last inequality simplifies to t + (t!) + (i - j - D) < N - k + Dp + Sj+D-t - Sj+D: The smallest t for which this fails will be the of (6.4). One can verify that * *in this pulling back from j + D to j + D - , the factor on the right side of the cannot be the determining factor for excess, i.e., that the analogue of (4.18) is sati* *sfied. ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 45 Next we establish the claimed possible differentials into E2;j+D- when > 0.* * We have a cycle Xi (6.11) z"= (cthet+ Lt)yt t=j+D-+1 as described in the two previous paragraphs. The leading term in @("z) will be i j T = cj+D-+1 (j+D-)(p-1)+N1hej+D-+1 hyj+D- : The binomial coefficient will be a unit, since, if it were not, the LHS of (6.4* *) would be no larger for than it is for - 1. For the same reason, Sj+D- = Sj+D-+1 ,* * a fact which will be used later in this paragraph. If cj+D-+1 is a unit and ej+* *D-+1 has its maximum possible value of i - j - D + (i - j - D) + 1 - (p - 2)( - 1) + (( - 1)!) - Sj+D-+1 + Sj+D; then, using Lemma 4.4, T is a unit multiple of h hffyj+D- , where ff = (p - 1)j + k - D + (i - j - D) - (p - 2) + (( - 1)!) - Sj+D-+1 + Sj+D: By Theorem 2.8(4), the order of the subgroup of E22(Sq(j+D-)+2N+1 ) generated by h hffis ff - ((j + D - )(p - 1) + N) + (i - j - D + ) + 1 = fl + 1; where fl is, as in the lemma, the difference of the two sides of (6.4). This es* *tablishes the claim of the lemma that the order of the image of the differential into E2;* *j+D- is less than or equal to pfl+1. (If cj+D-+1 is not a unit or ej+D-+1 does n* *ot have its maximum value, then the order of the image of the differential will be less* * than pfl+1.) If pR is the order of the image of the differential from E1;ito E2;j+D- , and* * "zis as in (6.11), then @(pRz") is a cycle of excess (j + D - )(p - 1) + N - (j + D - )* * - 1, and so by Corollary 4.13, it is d(u0h(j+D-)(p-1)+N + L)yj+D- , where u0is a un* *it and L has lower excess. Thus there is a cycle zj+D- = pRz"+ (uh(j+D-)(p-1)+N + L)yj+D- 2 UC1(Xij+D- ): This begins an induction which we would like to use to give us classes z`in U* *C1(Xi`) for ` = j +D- down to j +ffl satisfying certain properties. The anticipated beh* *avior 46 D. DAVIS would be that differentials E1;i! E2;`would be surjective except in the followi* *ng cases: o If ` N mod p, then 8 >><= p if k 6= N or ` 6 kp + N mod p2 | coker(di-`)| >> p2 if k = N j and ` Np2 + Np + N mod p3 >: 2 = p otherwise; o If ` kp+N mod p2, then di-`+1is not surjective if k 6= N or ` jp2+Np+N mod p3 and j 6 N mod p. In each case, if a class z` supports a differential whose image has order ps, t* *hen psz` pulls back to some z`-1. Unfortunately, when we pass an ` which is congruent to kp + N mod p2, we may lose some control. Thus we cannot assert that differentials necessarily behave* * as we expect that they usually will, causing the lack of precision in Lemma 6.2. * *The following result will be adequate for our purposes; indeed it completes the pro* *of of Lemma 6.2 and hence of Theorem 1.1. See Remark 6.14 for more on how Proposition 6.12 translates into Lemma 6.2. || Proposition 6.12. In the notation of Lemma 6.2, assume ` < j + D - . Assume that z`+12 UC1(Xi`+1) is a cycle of the form Xi z`+1= (u`+1h(`+1)(p-1)+N+ L`+1)y`+1+ Ttyt t=`+2 with exc(Tt) t(p - 1) + N for ` + 2 t i. Here and throughout the statement and proof of this proposition, u is always a unit, and exc(L) is less than that* * of the accompanying power of h. Then (1) If ` j + ffl and ` 6 N mod p and ` 6 kp + N - 1 mod p2, then di-`: E1* *;i! E2;`is surjective, and p(i-`)+1z`+1 pulls back to a cycle z` = p(i-`)+1z`+1+ (u`h`(p-1)+N+ L`)y` 2 UC1(Xi`): (2) Assume ` - 1 j + ffl and ` = ap + N. Let c = (a(p - 1) + N), b = (i - `), and s = max (b - c; 0). Then | im(di-`)| = ps. If it is not the case th* *at ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 47 b = c < (jp - a + k), then di-`+1is surjective, and ps+1z`+1 pulls back t* *o a cycle z`-1 = ps+1z`+1+ (c`h`(p-1)+N+ L`)y` +(u`-1h(`-1)(p-1)+N+ L`-1)y`-12 UC1(Xi`-1): If b = c < (jp - a + k), then di-`+1= 0 is not surjective. If o is the sm* *allest positive integer with ` - o j + ffl such that di-`+o: E1;i! E2;`-ois non* *zero, then, if ps is the order of the image of this di-`+o, psz`+1 pulls back t* *o a cycle z`-o 2 UC1(Xi`-o) of the form Xo z`-o = psz`+1+ (cth(`-t)(p-1)+N+ L`-t)y`-t t=0 with ct2 Z(p)and co a unit. Proof.(1) Mod L, i j @(z`+1) u`+1 `(p-1)+N1h(`+1)(p-1)+N hy` u0h h`(p-1)+Ny` by Lemma 4.4. By Theorem 2.8, this generates E22(S`(p-1)+N). Thus the different* *ial is surjective. By Lemma 4.6, @(p(i-`)+1z`+1) can be written as d(uh`(p-1)+N+ L`* *)y`, and so the pulling back here follows as usual. The hypothesis ` 6 kp + N - 1 is necessary to insure that @(T`+2y`+2) is not significant. (2) The calculation here is similar to (4.19). Assume without loss of general* *ity that the unit attached to h(`+1)(p-1)+Ny`+1is 1. Write i - ` = upb and a(p - 1) + N * *= u0pc. Then, mod L, @z`+1 p(a(p - 1) + N)h(`+1)(p-1)+N hy` -u0pch h`(p-1)+N-1y`: 0 c-b `(p-1)+N If c b, then this is u_up d((h + L)y`), and so z`+1 pulls back to 0 c-b `(p-1)+N z` = z`+1+ (-u_up h + L)y`: 48 D. DAVIS Omitting a factor ((` - 1)(p - 1) + N) 1 mod p, we have i j @z` h(`+1)(p-1)+N (`-1)(p-1)+N-1_2h2 - hv ! 0 c-b `(p-1)+N -u_up h h y`-1: We note that the coefficient of h2 is a multiple of p, and replace hv by vh - p* *h2. Then we use hpv vph to obtain a term -h`(p-1)+N h. If c > b, this term -h`(p-1)+N h will be the leading term, giving the surjective differential into E2;`-1, and t* *he claimed pullback. If c = b, two leading terms combine to yield 0 `(p-1)+N (6.13) @z` (-1 - u_u)h hy`-1: 0 Since pb(u+u0) = jp-a+k, the coefficient (-1- u_u) is a unit if it is not the c* *ase that b < (jp - a + k), and when the coefficient is a unit, (6.13) has higher excess * *than the ph2-terms. Thus @z` generates E22(S(`-1)(p-1)+N), and the argument is finis* *hed as in the previous case. If b < (jp - a + k), then the coefficient in (6.13) is a * *multiple of p, and so this term is not a generator, and di-`+1is not surjective. We cont* *inue pulling z` back as far as possible, to z`-o+1 for some o 1. If im(di-`+o) has * *order ps with s > 0, then by Lemma 4.12 @(psz`-o+1) = d(uh(`-o)(p-1)+N+ L)y`-o; and the pullback is completed as usual. If b > c, @(z`+1) -u0pc+1h h`(p-1)+Ny` has order pb-cin E2;`, pb-cz`+1pulls* * back to 0 `(p-1)+N z` pb-cz`+1- u_uh y`; and @(z`) u00h`(p-1)+N hy`-1, which is a generator of E2;`-1 Z=p. The pullback of pz` is achieved as usual. || Remark 6.14. The translation from Proposition 6.12 to Lemma 6.2 is straightfor- ward. Assume ` = ap + N. If k 6= N, then b = 0 in 6.12, and so di-` = 0. If also a k, then c = 0 and (jp - a + k) > 0, so that di-`+1is not surjective. ELEMENTS OF LARGE ORDER IN ss*(SU(n)) 49 Now assume k = N. If a 6 N, then c = 0 and b > 0 so that (| coker(di-`)|) = (b + 1) - b = 1: Now assume also that a = ffp + N. If N 6 j, then b = 1 and c 1, so that | coker(di-`)| = |E2;`| = p2: If N j and ff 6 N, then b 2 and c = 1, and so (| coker(di-`)|) = (b + 1) - (b - 1) = 2: Similarly, if N j and ff N, then b 2, c 2, and | coker(di-`)| p3. Finally,* * if N 6 j and ff j, then b = c = 1 < (jp - a + k), and so di-`+1is not surjective. The scenario in Lemma 6.2 in which di-`-ffi= 0 for 1 ffi p or p + 1 is due * *to the possibility in Proposition 6.12 that di-(`+p)+ocould be 0 for o < p. We close with an example, promised earlier, in which there is a nonzero group* * E2;a1 with a < j. Recall that our results are predicated on these groups being 0. W* *hen these groups are nonzero, it just means that our Theorems 1.1 and 1.4 are a bit weaker than they might be. These cases were omitted because they are very rare * *and difficult to describe. Example 6.15. Let p = 3, i = 39+ 10, and N = 2. (Thus n = 2 . 39+ 23, j = 38+ * *3, and k = 1.) We claim that E2;71 Z=3. This was suggested by our theory, and was verified by computer calculation using (1.13). From our point of view, the thing that causes this anomaly is the large group E2;101 Z=310. It is killed by differentials from some of the groups E1;twith 23* * t 32. In the usual pattern, there would have been nonzero differentials E1;23! E* *2;7, E1;26! E2;9, and E1;29! E2;9. However, what actually happens is that there are shorter nonzero differentials Z=9 E1;2213! E2;913; E1;2214! E2;814; E1;20! E2;7 Z=9; Z=27 E1;19! E2* *;6: Groups whose order is not indicated are Z=3. Without going into much detail, we mention that the differential E1;20! E2;7is obtained by pulling hy20 back to z hy20+ . .+.h20y10+ h19y9 + h16y8: 50 D. DAVIS Here we have omitted unit coefficients. The terms in . . .have small enough pow* *ers of h as to be insignificant. The h20y10 comes from writing h h10y10 as d(h20y* *10). The leading term in @(z) is h19 hvy7 h17 hy7 h h15y7, which is the element of order 3 in E2;71. References 1. M. Bendersky, Unstable towers in the odd primary homotopy groups of spheres,* * Trans Amer Math Soc 276 (1985) 529-542. 2. ________, Some calculations in the unstable Adams-Novikov spectral sequence,* * Publ RIMS 16 (1980) 739-766. 3. ________, The v1-periodic unstable Novikov spectral sequence, Topology 31 (1* *992) 47-64. 4. M. Bendersky, E.B. Curtis, and H.R. Miller, The unstable Adams-Novikov spect* *ral sequence for generalized homology, Topology 17 (1978) 229-248. 5. M. Bendersky, D. M. Davis, and M. Mahowald, v1-periodic homotopy groups of S* *p(n), Pacific Journal of Math 170 (1995) 319-378. 6. M. Bendersky, D. M. Davis, and M. Mimura, v1-periodic homotopy groups of exc* *eptional Lie groups: torsion-free cases, Trans Amer Math Soc 333 (1992) 115-135. 7. F. R. Cohen, J. C. Moore, and J. A. Neisendorfer, The double suspension and * *exponents of the homotopy groups of spheres, Ann of Math 110 (1979) 549-565. 8. M. C. Crabb and K. Knapp, The Hurewicz map on stunted complex projective spa* *ces, Amer Jour Math 110 (1988) 783-809. 9. D. M. Davis, v1-periodic homotopy groups of SU(n) at odd primes, Proc London* * Math Society 43 (1991) 529-544. 10.D. M. Davis and M. Mahowald, Some remarks on v1-periodic homotopy groups, Pr* *oc Adams Symposium, London Math Soc Lecture Notes Series 176 (1992) 55-72. 11.D. M. Davis and H. Yang, Tractable formulas for v1-periodic homotopy groups * *of SU(n) when n p2- p + 1, to appear in Forum Math. 12.A. T. Lundell, Concise tables of James numbers and some homotopy of classica* *l Lie groups and associated homogeneous spaces, Springer Verlag Lecture Notes in Math 1509 (1* *992) 250-272. 13.B. Harris, On the homotopy groups of classical groups, Ann of Math 74 (1961)* * 407-413. 14.M. Mimura and H. Toda, Cohomology operations and the homotopy of compact Lie* * groups-I, Topology 9 (1970) 317-336. Lehigh University, Bethlehem, PA 18015 E-mail address: dmd1@lehigh.edu