NILPOTENCE IN THE STEENROD ALGEBRA
Kenneth G. Monks
University of Scranton
August 1991
I. Introduction and Notation
While all of the relations in the Steenrod algebra, A, can be deduced in pri*
*nciple from
the Adem relations, in practice, it is extremely difficult to determine whether*
* a given poly-
nomial of elements in A is zero for all but the most elementary cases. In his o*
*riginal paper
[Mi] Milnor states "It would be interesting to discover a complete set of relat*
*ions between
the given generators of A ". In particular Milnor shows that every positive di*
*mensional
element of A is nilpotent. Thus it would be desirable to find a simple closed*
* form for
nilpotence relations in A.
Let x 2 A . We say that x has nilpotence k, if xk = 0 and xk-1 6= 0 (take x*
*0 = 1).
In this case we write Nil(x) = k. In this paper we investigate Nil(x) for seve*
*ral infinite
families of Milnor basis elements of A at the prime 2.
The paper is organized as follows. First, an infinite family of subalgebras *
*and isomor-
phisms between them are constructed. The isomorphisms are used to produce infi*
*nite
families of elements having the same nilpotence. Next, we compute strong upper*
* and
lower bounds for the nilpotence of Milnor basis elements in these subalgebras. *
*Comparing
these bounds and extending to the families produced via the isomorphisms shows *
*that
Sq(2m (2k - 1) + 1) has nilpotence k + 2. Finally a strong lower bound for the *
*nilpotence
of Ptsis computed for all s; t 2 N. The main results are stated and discussed i*
*n Sections
II and III. Detailed proofs are presented in Section IV.
II. Nilpotence in an Odd Subalgebra of A
There is a doubling isomorphism (see Section IV) which implies that
Nil(Sq(2r1; : :;:2rm )) Nil(Sq(r1; : :;:rm ))
for every Milnor basis element in A . Thus it is natural to begin by asking wh*
*at the
nilpotence of Sq(r1; : :;:rm ) is when some or all of the ri are odd.
We begin by describing a family of isomorphic subalgebras Ok A and a famil*
*y of
isomorphisms between them.
_____________
1980 Mathematics Subject Classification (1985 Revision). Primary 55S10, 55S0*
*5; Secondary 57T05.
Typeset by AM S-*
*TEX
1
2 KENNETH G. MONKS
Definition 2.1. Let k 2 N. Let Ok be the Z2-subspace of A whose basis is the s*
*et of
Milnor elements
k+1 k+1
Bk = Sq (r1; : :;:rm ) | ri -1 (mod 2 ) fori < m; and rm 1 (mod 2 :)
We will write O = O0. Thus we have the vector subspace inclusions
A O = O0 O1 O2 : :::
Notice O is just the subspace of A generated by the Milnor basis elements Sq(r1*
*; : :;:rm )
with ri odd for all 1 i m .
Theorem 2.2. O kis a sub-algebra of A for all k 2 N.
Ok is not a Hopf subalgebra, but we do not require this for our purposes.
Definition 2.3. Let : O ! O be the Z2-linear map given by
(Sq(r1; : :;:rm )) = Sq(2r1 + 1; 2r2 + 1; : :;:2rm-1 + 1; 2rm - 1)
on elements of the basis.
For example, (Sq(5) + Sq(3; 1; 3))= Sq(9) + Sq(7; 3; 5).
Theorem 2.4. is an algebra monomorphism.
If we let (0)be the identity map on O , and (k)= O (k-1)for k > 1 then (k)i*
*s also
an monomorphism for every k. It is a routine calculation to check that
k k k k k *
* k
(k)(Sq(r1; : :;:rm )) = Sq 2 r1 + (2 - 1); : :;:2 rm-1 + (2 - 1); 2 rm - (*
*2 - 1)
(2:*
*5)
Using (2.5) it is elementary to see that (O k) = Ok+1 and thus that the rest*
*riction of
to Ok yields an isomorphism k between Ok and Ok+1. Hence for any x 2 O we have
Nil(x) = Nil((k)(x)) for all k 2 N. Thus
Corollary 2.6. Let Sq(r1; : :;:rm ) 2 O. Then
k k k k k k
Nil(Sq(r1; : :;:rm=))NilSq 2 r1 + (2 - 1); : :;:2 rm-1 + (2 - 1); 2 rm - (2 *
* - 1)
for all k 2 N.
In particular, if n is odd then Nil(Sq(n))= Nil Sq(2kn - (2k - 1))for all k *
*2 N. For
example, since Nil(Sq(7))= 4, every element of the family
Sq(7); Sq(13); Sq(25); Sq(49); Sq(97); Sq(193); : : :
also has nilpotence 4.
NILPOTENCE IN THE STEENROD ALGEBRA 3
Theorem 2.4 reduces the problem of computing the nilpotence of elements of O*
* to that
of finding the nilpotence of elements in O - O1. For the case m = 1 this says *
*that the
nilpotence of the Milnor elements Sq(n) with n 1 (mod 4) is completely determ*
*ined
by the nilpotence of the elements Sq(n) with n -1 (mod 4). We begin to attack*
* this
question by obtaining a strong upper bound.
Theorem 2.7. Let Sq(r1; : :;:rm ) 2 O. Then
n o
Nil(Sq(r1; : :;:rm )) min k | rm < 2(k-1)m+1 - 1 :
Corollary 2.8. If n is odd then Nil(Sq(n)) min k | n < 2k - 1.
For example Sq(15; 31)4 = 0 since 31 < 2(4-1)2+1- 1 = 127. As a possible app*
*lication,
notice that Sq(r1; : :;:rm )2 = 0 whenever rm < 2m+1 -1 and Sq(r1; : :;:rm ) 2 *
*O. Elements
whose square is zero have been useful in the past in developing Ptshomology the*
*ory.
It should be noted that this upper bound appears to be quite good. Computer *
*calcu-
lations show that we actually have equality in Corollary 2.8 for every n -1 (m*
*od 4)
less than 143 with the exception of n = 67 and n = 131 (note also that these ex*
*ceptions
eliminate the possibility that one might actually be able to prove equality in *
*all cases).
We now obtain a lower bound on nilpotence for certain of these elements.
Theorem 2.9. Let n be odd. Then
k
Nil(Sq(n)) > max k | n -1 (mod 2 ) :
Combining all of the previous results yields
Theorem 2.10. Nil Sq 2m (2k - 1) + 1 = k + 2 for all m; k 1.
Notice for m = 1 this implies Nil(Sq(2k - 1)) = k + 1.
As an illustration of the theorem consider that 524281 = 23(216 - 1) + 1. T*
*hen by
Theorem 2.10 we have immediately that Sq(524281)17 6= 0 and Sq(524281)18 = 0, w*
*hich
would be a truly monumental computation by usual means.
Table 2.11 gives a comparison between the nilpotence bounds obtained in this*
* section
and the actual values of Nil(Sq(n)) for odd n less than 64. In the table the va*
*lues labeled
Nil are the actual values of Nil(Sq(n)) obtained from computer calculations. Th*
*e values
labeled High are the upper bounds for Nil(Sq(n)) obtained from Corollary 2.6 an*
*d Corollary
2.8. Similarly, the values labeled Low are the lower bounds obtained from Corol*
*lary 2.6
and Theorem 2.9. Finally, the values labeled Gap are just the difference betwee*
*n the upper
and lower bounds. Thus the nilpotence is completely determined whenever the gap*
* is zero.
This occurs at the values of n given in Theorem 2.10.
4 KENNETH G. MONKS
Table 2.11: Comparison of Nilpotence Bounds with Computed Values
n Nil High Low Gap n Nil High Low Gap
1 2 2 2 0 33 3 3 3 0
3 3 3 3 0 35 6 6 3 3
5 3 3 3 0 37 5 5 3 2
7 4 4 4 0 39 6 6 4 2
9 3 3 3 0 41 4 4 3 1
11 4 4 3 1 43 6 6 3 3
13 4 4 4 0 45 5 5 4 1
15 5 5 5 0 47 6 6 5 1
17 3 3 3 0 49 4 4 4 0
19 5 5 3 2 51 6 6 3 3
21 4 4 3 1 53 5 5 3 2
23 5 5 4 1 55 6 6 4 2
25 4 4 4 0 57 5 5 5 0
27 5 5 3 2 59 6 6 3 3
29 5 5 5 0 61 6 6 6 0
31 6 6 6 0 63 7 7 7 0
III. Nilpotence of Pts
Let Pts= Sq(r1; : :;:rt) where ri= 0 for all i < t and rt = 2s. There is an *
*old conjecture
which has been growing in notoriety ([Da], [Conf]) which says Nil(Sq(2s))= 2s +*
* 2 for all
k (or equivalently, Nil(P1s) = 2s + 2). One naturally might ask what the corre*
*sponding
conjecture would be for Nil(Pts) for any t. Some sample calculation leads one i*
*mmediately
to the following. Let brc denote the greatest integer less than or equal to th*
*e rational
number r.
Conjecture 3.1. Nil(Pts) = 2bs=tc + 2 for all s 0; t 1.
Our main result regarding this conjecture is
Theorem 3.2. Nil(Pts) 2bs=tc + 2 for all s 0; t 1.
This theorem generalizes an original result of Davis [Da], who first proved *
*this theorem
for the special case t = 1.
It is well known that the conjecture is true if bs=tc = 0, i.e. if s < t. We*
* can also prove
the conjecture for bs=tc = 1.
Theorem 3.3. If bs=tc = 1 then Nil(Pts) = 4.
The conjecture has been verified by computer calculation for all s; t such t*
*hat s + t < 16
and s - 2t < 4 as well as several other cases. For the case t = 1 the conjectu*
*re was
originally verified by Davis for s 5 [Da]. A summary of the calculation is giv*
*en in Table
3.4. It is interesting to note that for many of the 56,627 Milnor basis element*
*s, x, which
are a summand of Sq(64)13, the product Sq(64) . x is nonzero, and yet the sum o*
*f all such
products is still zero.
The Theorems in this section were first presented in the author's Ph.D. thes*
*is [Mo].
NILPOTENCE IN THE STEENROD ALGEBRA 5
Table 3.4: Powers of Sq(2s)
Number of Milnor Basis Elements Which are a Summand of Sq(2s)k
k s = 0 s = 1 s = 2 s = 3 s = 4 s = 5 s = 6
1 1 1 1 1 1 1 1
2 0 1 1 2 3 5 8
3 1 3 5 13 32 84
4 0 2 8 27 131 629
5 2 15 72 473 3,632
6 0 7 93 876 11,454
7 8 153 2,222 37,128
8 0 59 2,070 61,447
9 69 3,297 136,738
10 0 1,093 100,618
11 1,313 158,089
12 0 46,325
13 56,627
14 0
IV. Proof of Results
We begin by recalling some results from [Mi] to which we will need to refer *
*in the proofs
that follow. The mod 2 Steenrod algebra is a graded Z2-vector space with basis *
*all formal
symbols Sq(r1; r2; : :):where ri 0 and ri> 0 for finitely many i. As usual, it *
*is convenient
to write Sq(r1; : :;:rm ) for Sq(r1; : :;:rm ; 0; 0; : :):when rm 6= 0. Let R =*
* (r1; : :;:rm ). It
will also be convenient to write Sq(R) for the Milnor basis element Sq(r1; : :;*
*:rm ).
The product is given by
X
Sq(r1; r2; : :):. Sq(s1; s2; : :):= Sq(t1; t2; : :):
X
where the summation is taken over all matrices X = (xij) satisfying:
X
xij= sj (4:*
*1)
i
X
2jxij= ri (4:*
*2)
j
Y
(xh0; xh-1 1; : :;:x0h) 1 (mod 2) (4:*
*3)
h
where (n1; : :;:nm ) is the multinomial coefficient (n1 + . .+.nm )!=n1! . .n.m*
*!. We will
say such a matrix X is Sq(r1; r2; : :):Sq(s1; s2; : :):-allowable. Each such al*
*lowable matrix
yields a summand Sq(t1; t2; : :):given by
X
th = xij (4:*
*4)
i+j=h
6 KENNETH G. MONKS
In this case we will say that X is the matrix associated with Sq(t1; t2; : :):(*
*for the product
Sq(r1; r2; : :):. Sq(s1; s2; : :):). The value of x00 is never used and may be *
*assumed to be
zero.
When evaluating the multinomial coefficients in (4.3) it is well known (e.g.*
* [Ma]) that
(n1; : :;:nmP) is odd if and only if the ni have disjoint binary expansions. Mo*
*re formally,
let n = jffj(n)2j be the binary expansion of an integer n. Then
Lemma 4.5. (n1; : :;:nm ) is odd if and only if for each k < 1, ffk(ni) = 1 for*
* at most
one i.
In particular, if (n1; : :;:nm ) is odd then at most one of the ni is odd. W*
*e will make
frequent use of this fact.
Proof of Theorem 2.2. It suffices to show that x . y 2 O for all x; y 2 B0. We *
*will prove a
slightly stronger result which we will need later, namely
Lemma 4.6. Let Sq(r1; : :;:rm ); Sq(s1; : :;:sn) 2 O. If Sq(t1; : :;:tp) is a *
*summand of
Sq(r1; : :;:rm )Sq(s1; : :;:sn) then Sq(t1; : :;:tp) 2 O and p = m + n.
Proof. Let X = (xij) be the matrix associated with Sq(t1; : :;:tp). Sq(r1; : :*
*;:rm ) 2 O
implies that ri is odd for each 1 i m . Thus xi0is odd for each 1 i m by (*
*4.2).
Combining this with (4.3) shows xijis even whenever i + j m, and j > 0. Let d *
*< n
and assume that xmj is odd for j d and xij is even whenever i + j m + d, j > *
*0,
and i < m. Then Sq(s1; : :;:sn) 2 O implies sd+1 is odd and thus xm d+1 is odd *
*by (4.1).
Once again invoking (4.3) shows xijis even whenever i + j = m + d + 1, and j > *
*d + 1.
Thus by finite induction on d we have shown xij is odd if and only if j = 0, i *
* m or
i = m, j n. Applying (4.4) shows Sq(t1; : :;:tp) 2 O. Further tp = xmn is odd,*
* therefore
p = m + n.
Proof of Theorem 2.4. It is easy to see from the definition that is injective.*
* Let R =
(r1; : :;:rm ), S = (s1; : :;:sn), and T = (t1; : :;:tm+n ). To show that is a*
* homomorphism
we will prove that Sq(T ) is a summand of the product Sq(R)Sq(S) if and only if*
* (Sq(T ))
is a summand of (Sq(R))(Sq(S)) for every Sq(R); Sq(S) 2 O . Let Xb = (bxij) be*
* a
(Sq(R))(Sq(S))-allowable matrix. As shown in the proof of Lemma 4.6, bxijis od*
*d if
and only if j = 0, i m or i = m, j n. Thus there exist nonnegative integers x*
*ijsuch
that 8
><2xij+ 1 if j = 0, i m or i = m, j < n
bxij= > 2xmn - 1 if i = m and j = n (4:*
*7)
: 2x
ij otherwise.
Given such an allowable matrix Xb we can define the matrix X = (xij). On the o*
*ther
hand, if we are given a Sq(R)Sq(S)-allowable matrix, X = (xij), we can define a*
* matrix
Xb = (bxij) by (4.7). We now wish to show that Xb is (Sq(R))(Sq(S))-allowable i*
*f and
only if X is Sq(R)Sq(S)-allowable. We must verify that each of the conditions (*
*4.1),(4.2),
and (4.3) hold for X if and only if they hold for bX.
NILPOTENCE IN THE STEENROD ALGEBRA 7
ae
-1 if j = n
Let 1 j n and define fflj = . Then (Sq(S)) = Sq(bs1; : :*
*;:bsn)
1 otherwise
where bsj= 2sj + fflj. Thus checking (4.1) we have
Xm m-1X !
bxij= bsj() 2xij + (2xmj + fflj)= 2sj + fflj
i=0 i=0
Xm
() xij= sj
i=0
Again letting (Sq(R)) = Sq(br1; : :;:brm) we have bri= 2ri + 1 for 1 i < m *
*and
brm= 2rm - 1. Verification for (4.2) breaks up into two cases. If 1 i < m then
0 1
Xn Xn
2jbxij= bri() @ 2j2xijA + (2xi0+ 1)= 2ri+ 1
j=0 j=1
Xn
() 2jxij= ri
j=0
But if i = m we have
0 1
Xn n-1X
2jbxmj= brm () @ 2j(2xmj + 1)A+ 2n (2xmn - 1)= 2rm - 1
j=0 j=0
Xn n-1X
() 2 2jxmj + 2j - 2n = 2rm - 1
j=0 j=0
Xn
() 2 2jxmj + (2n - 1) - 2n = 2rm - 1
j=0
Xn
() 2jxmj = rm
j=0
Verification of (4.3) follows easily from the observation that for any multi*
*nomial coeffi-
cient (a1; : :;:ah) we have
(a1; : :;:ah) (2a1 + fl1; : :;:2ah + flh) (mod 2)
where fli = 1 for at most one 1 i h and is zero otherwise. This follows imme*
*diately
from Lemma 4.5. Thus
Y Y
(xh0; xh-1 1; : :;:x0h) (bxh0; bxh-1;1: :;:bx0h) (mod 2)
h h
8 KENNETH G. MONKS
Finally let Sq(t1; : :;:tm+n ) be the summand of Sq(R)Sq(S) associated with X a*
*nd let
Sq(Tb) = Sq(bt1; : :;:btm+n) be the summand of (Sq(R))(Sq(S)) associated with b*
*X. Then
by (4.4) for h < m + n
X
bth= bxij
i+j=h0 1
X
= 2 @ xijA+ 1
i+j=h
= 2th + 1
and tm+n = bxmn= 2xmn - 1 = 2tmn - 1. Thus Sq(Tb) = (Sq(T )) which completes t*
*he
proof.
Proof of Theorem 2.7. Let R = (r1; : :;:rm ). It suffices to show that Sq(R)k *
*= 0 if
rm < 2(k-1)m+1 - 1. Let Sq(T ) = Sq (t1; : :;:tp) be any summand of Sq(R)k-1.*
* By
Lemma 4.6 p = (k - 1)m. Let X = (xij) be any Sq(R)Sq(T )-allowable matrix. As s*
*hown
in the proof of Lemma 4.6, xijis odd if i = m and j (k - 1)m. Combining this w*
*ith
(4.2) we have
(k-1)mX (k-1)mX
rm = 2jxmj 2j = 2(k-1)m+1 - 1:
j=0 j=0
Therefore if rm < 2(k-1)m+1 - 1 there are no Sq(R)Sq(T )-allowable matrices, an*
*d hence
Sq(R)k = 0.
Before continuing we would like to outline an alternate proof of Theorem 3 t*
*hat lends
some insight into what is going on at the cost of being much more tedious.
Let Qt-1 = Pt0. It is quite easy to see from the product formula that
QiQj = QjQi for all i; j 2 N (4:*
*8)
Q2i= 0 for all i 2 N (4:*
*9)
and that for any Sq(r1; : :;:rm ) with ri even for all 1 i m
Xm
Sq (s1; : :;:sm )Qi= Qj+iSq(s1; : :;:sj - 2i+1; : :;:sm ) (4:1*
*0)
j=0
where we define Sq(t1; : :;:tm ) to be zero if ti< 0 for any i. Notice that (4.*
*10) gives us a
way to shift Qi's from the right side of a Milnor basis element with even entri*
*es to the left
side. Also notice that the largest Qj obtainable on the left by shifting a Qi v*
*ia (4.10) is
Qm+i and that this can only occur if sm 2i+1. For any Sq(r1; : :;:rm ) 2 O we *
*can write
Sq(r1; : :;:rm ) = Q0Q1 . .Q.m-1Sq(r1 - 1; : :;:rm - 1)
NILPOTENCE IN THE STEENROD ALGEBRA 9
and hence
(Sq(r1; : :;:rmk))= (Q0Q1 . .Q.m-1Sq(r1 - 1; : :;:rm - 1))k (4:1*
*1)
Applying (4.8), (4.9), and (4.10) repeatedly to the right hand side of (4.11) i*
*n order to
collect all of the Qion the left and computing the effect on the mthposition in*
* the resulting
Milnor elements yields the desired result. We leave this verification to the in*
*terested reader.
Proof of Theorem 2.9. Let k be the largest integer such that n -1 (mod 2k). W*
*e can
write n uniquely in the form n = 2ka - 1 for some odd integer a 1. For each 1 *
* h k
define an h-tuple Rn;h= (rn;h;1; rn;h;2; : :;:rn;h;h) by
ae2k-ia + 1 if 1 i < h
rn;h;i=
2k-i+1a - 1 if i=h
For example, for n = 47 we have
R47;1 = (47)
R47;2 = (25; 23)
R47;3 = (25; 13; 11)
R47;4 = (25; 13; 7; 5)
We now wish to show that Sq(Rn;h) is a summand of Sq(n)h for 1 h k , and thus*
* that
Sq(n)k 6= 0.
We proceed by finite induction on h. If h = 1 then Sq(Rn;1) = Sq(n), which i*
*s clearly a
summand of Sq(n)1. Assume as the induction hypothesis that Sq(Rn;h) is a summan*
*d of
Sq(n)h where h < k. Suppose Sq(Rn;h+1) is a summand of Sq(n)Sq(T ) for some sum*
*mand
Sq(T ) = Sq(t1; : :;:th) of Sq(n)h. Let X = (xij) be the associated matrix. The*
*n by (4.4)
x1h = rn;h+1;h+1= 2k-ha - 1
and from (4.2)
Xh
n= 2jx1j
j=00 1
h-1X
= @ 2jx1jA + 2hx1h
j=0
0 1
h-1X
= @ 2jx1jA + 2h 2k-ha - 1
j=0
0 1
h-1X
= @ 2jx1jA + n - 2h + 1:
j=0
10 KENNETH G. MONKS
From which we obtain
h-1X
2jx1j= 2h - 1
j=0
But once again using the fact from the proof of Lemma 4.6 that x1j is odd for 1*
* j h
we conclude that (4.2) is satisfied if and only if x1j = 1 for 1 j < h (assumi*
*ng x1h =
2k-ha - 1). But from (4.1) and (4.4) with 1 j < h we have
tj = x0j+ x1j
= (rn;h+1;j- 1) + 1
= 2k-ja + 1
= rn;h;j
and
th= x0h + x1h
= (rn;h+1;h- 1) + rn;h+1;h+1
= ((2k-ha + 1) - 1) + (2k-(h+1)+1a - 1)
= 2k-h+1 a - 1
= rn;h;h
Thus we have shown that if Sq(Rn;h+1) is a summand of Sq(n)Sq(T ) for some summ*
*and
Sq(T ) = Sq(t1; : :;:th) of Sq(n)h then Sq(T ) = Sq(Rn;h). But by our very cons*
*truction
the matrix X satisfies (4.1) and (4.2) for the product Sq(n)Sq(Rn;h). It also s*
*atisfies (4.3)
as rn;h;jis always odd and therefore the multinomial coefficient (1; rn;h;j- 1)*
* is odd also.
Thus X is Sq(n)Sq(Rn;h)-allowable and Sq(Rn;h+1) is a summand of Sq(n)h+1, comp*
*leting
the induction and the proof.
Proof of Theorem 2.10. By Theorem 2.7 we have Nil(Sq(2k - 1)) k + 1 and by The*
*orem
2.9 Nil(Sq(2k - 1)) > k. Therefore Nil(Sq(2k - 1)) = k + 1. By (2.5)
k+1 m-1 k+1 m-1
(m-1) Sq (2 - 1) = Sq 2 (2 - 1) - (2 - 1)
m k
= Sq 2 (2 - 1) + 1)
for every m 1; k 0. Thus
m k i(m-1) k+1 j
Nil Sq 2 (2 - 1) + 1 = Nil Sq 2 - 1
k+1
= Nil Sq 2 - 1
= k + 2
NILPOTENCE IN THE STEENROD ALGEBRA 11
In order to prove Theorem 3.2 we must first recall the following information*
* from [Mi].
Let A* be the Hopf dual of A. A * is isomorphic to the polynomial algebra Z2[1;*
* 2; : :]:
on generators i in dimension 2i- 1. If R = (r1; : :;:rm ) we will write R to m*
*ean the
monomial r11. .r.mm. The basis of monomials R in A* is dual to the Milnor basis*
* for A .
As is common we will write for the evaluation of y 2 A* on x 2 A. Thus
ae1 if R = S
=
0 otherwise
The algebra homomorphism OE : A* ! A* A* by
X j
OE(k) = 2i j
i+j=k
is the dual of the product map in A.
Let E be the exterior subalgebra of A generated by {Qi| i 2 N }. There is a *
*doubling
isomorphism D : A ! A ==E given by
D (Sq(s1; s2; : :):)= [Sq(2s1; 2s2; : :):]
where [x] denotes the equivalence class in A ==E of x 2 A.
Finally, let An be the subalgebra of A generated by Sq 2i | i n .
Proof of Theorem 3.2. Let n; t 2 N, t 6= 0. For each i 2 N let ji and ffli be *
*the unique
integers satisfying i = 2ji+ ffli where ffli2 {0; 1}. Define an integer sequence
Rn;t(i) = (ri;1; ri;2; ri;3; : :):
recursively on i so that it satisfies the three conditions
Rn;t(1) = (2nt; 0; 0; : :): (4:1*
*2)
ae2-tr if i is even
ri;k= i-1;k-1 for k > 1 and i > 1 (4:1*
*3)
ri-1;k if i is odd
ji+1X
ri;k= 2nt+ffli: (4:1*
*4)
k=1
Notice that (4.14) is used to compute ri;1after obtaining ri;kfor k > 1 from (4*
*.13). For
example, for n = 3 and t = 2 (dropping trailing zeros)
R3;2(1)= (64)
R3;2(2)= (48; 16)
R3;2(3)= (112; 16)
R3;2(4)= (32; 28; 4)
R3;2(5)= (96; 28; 4)
R3;2(6)= (32; 24; 7; 1)
R3;2(7)= (96; 24; 7; 1)
12 KENNETH G. MONKS
We will require the following implication of (4.14) for odd i.
ji+1X
ri;1= 2nt+1 - ri;k
k=2
ji+1X
= 2nt+ 2nt- ri-1;k
k=2
ji+1X ji+1X
= 2nt+ ri-1;k- ri-1;k
k=1 k=2
= 2nt+ ri-1;1
Q ji+1ri;k
Define the monomial Rn;t(i)2 A* by Rn;t(i)= k=1 kt (notice this is not the s*
*ame as
the definition of R given before because of the kt subscript). Then for i > 1
i j ji+1Y !
OE Rn;t(i)= OE ri;kkt
k=1
ji+1Y
= OE (kt)ri;k
k=1
ji+1Yi t j ri;k
= kt 1 + 2(k-1)t t+ S1
k=1
ji+1Yi t j ri;k
= kt 1 + 2(k-1)t t + S2
k=1
where S1 is a sum of termsnoftthe form a b with b =2{1; t}and S2 is a sum of t*
*erms of
the form a b with b 6= 2t .
Continuing this derivation with i even yields
i j ji+1Yt ! nt
OE Rn;t(i)= 2(ri;kk-1)t 2t + S3
k=1
ji+1Y ! nt
= ri-1;k-1(k-1)t 2t + S3
k=1
jiY ! nt
= ri-1;kkt 2t + S3
k=1
nt
= Rn;t(i-1) 2t + S3
P ji+1
where S3 is a sum of terms of the form a b with b 6= 2nttbecause k=1 ri;k= 2*
*nt.
NILPOTENCE IN THE STEENROD ALGEBRA 13
On the other hand, continuing the derivation with i odd yields
i j ji+1Yi t j ri;k
OE Rn;t(i) = (t 1 + 1 t) ri;1 kt 1 + 2(k-1)t t + S2
k=2
nt ri-1;1ji+1Yi 2t j ri;k
= (t 1 + 1 t) 2 (t 1 + 1 t) kt 1 + (k-1)t t + S2
k=2
i nt ntj ji+1Yi t j ri-1;k
= 2t 1 + 1 2t (t 1 + 1 t) ri-1;1 kt 1 + 2(k-1)t t + S2
k=2
ji+1Y ! nt ntji+1Yt ! nt
= ri-1;1t ri-1;kkt 2t + 2t 2(ri-1;kk-1)t 2t + S4
k=2 k=2
ji+1Y ! nt nt t ji+1Yt ! nt
= ri-1;kkt 2t + 2t +2 ri-1;2 2(ri-1;kk-1)t 2t + S4
k=1 k=3 !
nt 2nt+1 ri-3;1ji+1Y2tri-1;k2nt
= Rn;t(i-1) 2t + t t (k-1)t t + S4
k=3
P ji+1
where S4 is a sum of terms of the form a b with b 6= 2nttbecause k=1 ri-1;k=*
* 2nt
and in the last equality we have used the fact that 2tri-1;2= ri-2;1= 2nt+ ri-3*
*;1(taking
r0;1= 0).
Thus in both cases weihave shownjthati j
nt
OE Rn;t(i) = Rn;t(i-1)+ 0 2t + S5
where S5 is a sum of terms of the form ab with b 6= 2nttand 0is divisible by 2n*
*t+1tso that
its evaluation on all elements of At(n+1)-1is zero. This shows that for any 1 *
*i 2n + 1
D i E D i-1 E D ntE D i-1 E
Ptnt ; Rn;t(i)= Ptnt ; Rn;t(i-1). Ptnt; 2t = Ptnt ; Rn;t(i-1):
(4:1*
*5)
ff D ntE
Noting that Ptnt; Rn;t(1)= Ptnt; 2t = 1 we can use (4.15) and finite inducti*
*on on i
to see that D E
Ptnti ; Rn;t(i) for all 1 i 2n + 1:
Thus (Ptnt)2n+16= 0 for all n; t 2 N, t 6= 0.
Invoking the doubling isomorphismiwejnoticehthat D (Pts)= Pts+1. Since D i*
*s an
s+1ii i i ij
algebra isomorphism we have D (Pts)i= Pt . Thus (Pts)6= 0 =) D (Pts) 6=
h ii i
0 =) Pts+1 6= 0 =) Pts+1 6= 0. So by induction on w, (Pts)i6= 0 =)
s+w i
Pt 6= 0 for all w 2 N. Since any s can be written uniquely as s = nt + w*
* with
nt+w 2n+1 2bs=tc+1
n = bs=tc we see that (Ptnt)2n+16= 0 =) Pt 6= 0 =) (Pts) 6= 0 f*
*or all
s; t 2 N, t 6= 0.
Finally, we can prove Theorem 3.3 by the following Lemma.
14 KENNETH G. MONKS
Lemma 3. Let s; t 2 N with bs=tc = 1 and let w = s - t. Then
(1) (Pts)2= Sq(t1; t2; : :):where
8
><2w (2t- 1) if i = t
ti= > 2w if i = 2t
: 0 otherwise
(2) (Pts)3= Sq(t1; t2; : :):where
8
><2w (2t+1- 1) if i = t
ti= > 2w if i = 2t
: 0 otherwise
(3) (Pts)4= 0
The proof of this Lemma is an elementary, though tedious, exercise in using *
*the product
formula and we shall not present it here. Computer calculations indicate that a*
*n analogous
method should work for the case bs=tc = 2 but that this method will not work fo*
*r the case
bs=tc = 3.
Acknowledgement
I would like to express my sincere gratitude to D. Davis. Since introducing*
* me to
this problem he has been a tireless source of encouragement and support. Our nu*
*merous
mathematical discussions were quite beneficial. In addition, his proof of the c*
*ase t = 1 of
Theorem 3.2 provided the model from which I was able to construct the proof of *
*Theorem
3.2. This paper surely would not exist without his help.
References
[Conf]N. Ray and Y. Rudyak, List of Problems, J. F. Adams Symposium Proceedings*
* (to appear).
[Da] D. M. Davis, On the Height of Sq(2i), unpublished (1985).
[Ma] H. Margolis, Spectra and the Steenrod Algebra, vol. 29, North-Holland Math*
* Library, 1983.
[Mi] J. Milnor, The Steenrod Algebra and Its Dual, Ann. of Math. 67 (1958), 150*
*-171.
[Mo] K. G. Monks, Nilpotence and Torsion in the Steenrod Algebra and Its Cohomo*
*logy, Ph.D. thesis,
Lehigh University (1989).
Kenneth G. Monks
Department of Mathematics
University of Scranton
Scranton, PA 18510
E-mail: kgm303@jaguar.uofs.edu