THE NILPOTENCE HEIGHT OF Pts
KENNETH G. MONKS
June 1994
Abstract. The method of Walker and Wood is used to completely determine t*
*he
nilpotence height of the elements Pstin the Steenrod algebra at the prime*
* 2. In
particular, it is shown that (Pst)2bs=tc+2= 0 for all s 0, t 1. In addi*
*tion, several
interesting relations in A are developed in order to carry out the proof.
1. Introduction and Main Results
In [WW ] Grant Walker and Reg Wood show that Sq(2s)2s+2= 0 for all s 0 in
the mod 2 Steenrod algebra, A, thereby settling a conjecture which dates back to
1975. Using their method, along with previous results of Andrew Gallant, Judith
Silverman, and some new results, we prove the following theorem. Recall that Pt*
*sis
the Milnor basis element Sq(r1; : :;:rt) with rt= 2s and ri= 0 for i < t.
Theorem 1.1. For all s 0, t 1, (Pts)2bs=tc+2= 0.
This establishes a conjecture of the author [M1 , Conjecture 3.1] and extends*
* the
result discussed above which proves Theorem 1.1 for the case t = 1. It was show*
*n in
[M1 ] that (Pts)2bs=tc+16= 0. Thus Theorem 1.1 completely determines the nilpot*
*ence
height of the elements Pts. The problem of determining the nilpotence of eleme*
*nts
of A has been of much recent interest and we refer the reader to [M1 ] and [WW *
* ] for
more background on the general question.
In order to extend the proof of [WW ] to arbitrary values of t we must deri*
*ve
appropriate generalizations of the propositions and lemmas used in their proof,*
* many
of which are interesting in their own right. These are discussed in detail in *
*the
following sections.
2.Relations in A and the Antiautomorphism
In this section we prove some of the preliminary propositions needed for the *
*proof
of Theorem 1.1.
___________
1991 Mathematics Subject Classification. Primary 55S10, 55S05; Secondary 57T0*
*5.
1
2 KENNETH G. MONKS
Much of the notation we will use follows [WW ]. For 2 A write b for O() wh*
*ere
O denotes the canonical antiautomorphism of A. Following [M2 ], let Sqt(r1; : :*
*;:rm )
be the Milnor basis element Sq(s1; : :;:stm) where sti= ri and sj = 0 if t does*
* not
divide j. In particular, Sqt(2s) = Ptsand Sq1(n) = Sq(n). If R = (r1; r2; : :*
*):is a
sequence of nonnegative integers, we writePSqt for the corresponding Milnor *
*basis
element.P The degree of Sqt is |R|t = (2it- 1)ri, and the excess of Sqt*
* is
ex(R) = ri.
For any positivePinteger n, let ffi(n) be the coefficient of 2iin the binary *
*expansion
of n, i.e. n = 1i=0ffi(n)2i. We say that m and n are disjoint if ffi(m)i+ ffi*
*(n)j 1 for
all i. This is equivalent to the condition that the binomial coefficient m+nm *
*is odd
[L ], which is in turn equivalent to the condition that m + n dominates m (writ*
*ten
m + n< m) in the notation of [S1]. If m and n are disjoint we will find it conv*
*enient
to write m n. If either m or n is negative we will also say that m and n are n*
*ot
disjoint. We will also write 2i2 n for ffi(n) = 1 since the meaning will be cle*
*ar from
the context.
In what follows will make frequent use of the following trivial facts which w*
*e state
without proof. Let 0 b < 2t.
(2.1) 2l2 a () 2l+t2 2ta + b
(2.2) 2l2 b () 2l2 2ta + b and l < t
P n-1 it
As in [M2 ] we let flt(n) = i=02 (take flt(0) = 0) for any integers n 0 a*
*nd
t 1. It follows that
(2.3) flt(m + 1) = 2tflt(m) + 1:
We begin by proving a generalization of the Adem relation Sq(2m+1 -1)Sq(2m ) *
*= 0:
Proposition 2.1. For any t 1, m 0, and i < t
Sqt(2iflt(m + 1))Ptmt+i= 0:
P t
Proof. By the Milnor product formula [Mil] Sqt(a)Sqt(b) = Sqt(a+b-(2 +1)j; *
*j)
where the sum is taken over all j such that a - 2tj b - j. Thus it suffices to*
* show
that 2iflt(m + 1 ) - 2tj is not disjoint from 2mt+i- j for any 0 j 2iflt(m). *
*We will
prove something stronger, namely
Lemma 2.2. For all t 1, m 0, 0 i < t, and 0 j 2iflt(m) there exists
k 0 such that
2kt+i2 2iflt(m + 1 ) - 2tj and 2kt+i2 2mt+i- j
Proof. We will proceed by induction on m.
Base Case: If m = 0 then 2iflt(m + 1 ) - 2tj = 2mt+i- j = 2i so we can take k*
* to
be 0.
THE NILPOTENCE HEIGHT OF Pst 3
Inductive Step: Assume that for all t 1, 0 i < t, and 0 j 2iflt(m - 1)
there exists k 0 such that 2kt+i2 2iflt(m) - 2tj and 2kt+i2 2(m-1)t+i- j.
Choose t 1, i < t, and 0 j 2iflt(m) . To simplify the notation we let X =
2iflt(m + 1 ) - 2tj and Y = 2mt+i- j.
If j = 0 then X = 2mt+i+ 2(m-1)t+i+ . .+.2i and Y = 2mt+i so that 2mt+i 2 X
and 2mt+i2 Y . Thus in this case we can take k to be m.
Now suppose j 6= 0. We consider two cases.
Case 1: 2i2=j - 1
Using (2.3) to replace flt(m + 1) in X yields
X = 2i(2tflt(m) + 1) - 2tj
= 2t(2iflt(m) - j) + 2i
Since 2i< 2t it follows from (2.2) that 2i2 X
Since 2i =2j - 1 we can express j - 1 = 2i+1q + r where 0 r < 2i. Substituti*
*ng
for j in Y yields
Y = 2mt+i- (2i+1q + r + 1)
= 2i+1[2mt-1 - (q + 1)] + 2i+ [2i- (r + 1)]
Since r < 2i it follows that the term [2i- (r + 1)] is nonnegative and less tha*
*n 2i.
Therefore 2i2 Y by (2.1) and (2.2).
So 2i2 X and 2i2 Y . Thus in this case we can take k to be 0.
Case 2: 2i2 j - 1
In this case we let
(2.4) j - 1 = 2tq + r where 0 r < 2t:
Indeed, since 2i2 j - 1 we must have 2i r < 2t by (2.2). Replacing j via (2.4) *
*and
flt(m) via (2.3) in the assumption that j 2iflt(m) and solving for q + 1 shows
2t+ 2i- (r + 1)
q + 1 2iflt(m - 1) + _______________:
2t
t+2i-(r+1) i i
But 2________2t< 1 since r 2 . Thus q + 1 2 flt(m - 1) and so by the inductive
hypothesis there exists k 0 such that 2kt+i2 2iflt(m) - 2t(q + 1)and 2kt+i2
2(m-1)t+i- (q + 1). Let k be any such value. We will now show that 2(k+1)t+i2 X
and 2(k+1)t+i2 Y which will complete the induction and hence the proof.
To see 2(k+1)t+i2 X, we use (2.3) and (2.4) to obtain
X = 2i(2tflt(m) + 1) - 2t(2tq + r + 1)
i j
(2.5) = 2t [2iflt(m) - 2t(q + 1)] + [(2t- 1) -+r]2i
4 KENNETH G. MONKS
Now 2i 2 j - 1 =) 2i 2 r =) 2i =2[(2t- 1) - r]. Thus since (2t- 1) - r < 2t
it follows that [(2t - 1) - r] + 2i < 2t also. Writing 2iflt(m) - 2t(q + 1)in *
*the
form 2t[2iflt(m - 1) - (q + 1)] + 2i, we see that 2kt+i2 2iflt(m) - 2t(q + 1)im*
*plies
2kt+i2 2iflt(m) - 2t(q + 1)+ [(2t- 1) - r] which in turn implies 2(k+1)t+i2 X by
(2.1), (2.2), and (2.5).
Finally, to see 2(k+1)t+i2 Y , we use (2.4) to obtain
Y = 2mt+i- (2tq + r + 1)
= 2t[2(m-1)t+i- (q + 1)] + [2t- (r + 1)]
But [2t- (r + 1)] < 2t. Thus by (2.1), 2kt+i2 2(m-1)t+i- (q + 1) implies 2(k+1)*
*t+i2
Y . __|__|_|_ |
The following result of Andrew Gallant [G , Corollary 1a] describes the produ*
*ct of
an arbitrary element Sqt(u) with OSqt(v).
X
(2.6) Sqt(u)S"qt(v)= Sqt : |R|t= (2t- 1)(u + v); |R|t+ ex(R)< 2tu
This generalizes a formula of Don Davis [D ]. Using Davis' method one can der*
*ive
the analogous formula for "Sqt(u)Sqt(v),
(2.7) S"qt(u)Sqt(v) = X Sqt : |R|t= (2t- 1)(u + v); ex(R)< v:
Using these formulae one can prove the following.
Proposition 2.3. Let a, b, and t be positive integers with a t. Then
cPtaSqt(2a(2b- 1)) = Sqt(2a-t(2b+t- 1))Pda-tt:
Proof. By (2.6) and (2.7) it suffices to show that for any sequence R in deg*
*ree
|R|t= 2a+b(2t- 1), ex(R)< 2a(2b- 1) () 2a+b(2t- 1) + ex(R)< 2a(2b+t- 1). Notice
that ex(R) 2a+b and that equality holds only when R is the sequence (2a+b). But
in this case it is easy to check that neither ex(R)< 2a(2b - 1) nor 2a+b(2t - 1*
*) +
ex(R)< 2a(2b+t- 1). Thus assume ex(R) < 2a+b. Then
ex(R)< 2a(2b- 1) () ex(R) - 2a(2b- 1) 2a(2b- 1)
by (2.2) a b a+b t a b
() ex(R) - 2 (2 - 1) 2 (2 - 1) + 2 (2 - 1)
() 2a+b(2t- 1) + ex(R) - 2a(2b+t- 1) 2a(2b+t- 1)
() 2a+b(2t- 1) + ex(R)< 2a(2b+t- 1)
which completes the proof. __|_ |
Let n, k, and t be positive integers with n = k + mt for some m 0. Extending
the notation of Arnon [Ar ] we define
(Xt)nk= Sqt(2n)Sqt(2n-t) . .S.qt(2k) = PtnPtn-tPtn-2t. .P.kt:
THE NILPOTENCE HEIGHT OF Pst 5
Applying Proposition 2.3 repeatedly yields the following corollary.
Corollary 2.4. Let m 1, c 0, and a = c + mt. Then
(Xct)ac+tSqt(2a(2b- 1)) = Sqt(2c(2b+mt - 1))(Xct)a-tc:
Finally, using (2.6) we can easily prove a generalization of Straffin's formu*
*la [St].
Proposition 2.5. For any s 0, t 1,
Pcst= Pts+ Pts-1[Pts-1
P s t s t t s*
*-1
Proof. By (2.6), Pts-1[Pts-1= Sqt : |R|t= 2 (2 -1); 2 (2 -1)+ex(R)< 2 2 *
* .
SincePS"qt(n)is the sum of all Sqt in degree n(2t- 1) (cf. [M2 ] or [G ]), *
*Pcst=
Sqt : |R|t= 2s(2t- 1). Thus it suffices to show that 2s(2t- 1) + ex(R)< 2s*
*+t-1
if and only if R 6= (2s).
If R = (2s) then 2s(2t- 1) + ex(R) = 2s+t which does not dominate 2s+t-1. If
R 6= (2s) then ex(R) < 2s so that 2s(2t-1)+ex(R) = 2s+t-1+2s+t-2+. .+.2s+ex(R)
clearly dominates 2s+t-1. __|_ |
3.The Kristensen Operations
Kristensen [K ] developed a family of linear operations indexed by the admiss*
*ible
monomials that can be used to derive relations in A from other relations. Follo*
*wing
Kristensen, we define the family of operators : A ! A indexed by the Milnor b*
*asis
elements by
X
(3.1) OE(x) = (x)
where OE : A ! A A is the diagonal map and OE(x) is expressed in the Milnor ba*
*sis
(i.e. the basis for AA is taken to be tensor products of Milnor basis elements)*
*. The
operations have many nice properties and we refer the reader to [Gra ] for de*
*tails.
We will refer to the action of on A as `stripping by ' (cf. [WW ]).
It follows immediately from (3.1) that stripping a Milnor basis element by =
Sq(t1; t2; : :):is given by
(3.2) (Sq(r1; r2; : :):) = Sq(r1 - t1; r2 - t2; : :;:ri- ti; : :):
where the right hand side is taken to be 0 if ri < ti for any i. In particular *
*we see
that stripping Ptsby Ptsyields Sq(0) = 1.
Recall that An is the subalgebra of A generated by Sq(2k) with k n. An has
a graded vector space basis consisting of the Milnor basis elements Sq(r1; : :;*
*:rn+1)
having ri< 2n+2-ifor all i. Thus Pts2 As+t-1\As+t-2and the result of stripping *
*any
element of As+t-2by Ptsis zero by (3.2) and the fact that the Kristensen operat*
*ions are
linear. In particular, since Pts-12 As+t-2, stripping cPtsby Ptsyields 1 by Pro*
*position
2.5.
6 KENNETH G. MONKS
The Steenrod algebra is a Hopf algebra with dual A*. A* is isomorphic to the
graded polynomial algebra Z2[1; 2; : :]:on generators i in dimension 2i - 1. If
R = (r1; : :;:rm ) we will write R to mean 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. The algebra homomorphism
* : A* ! A* A* defined by
X j
(3.3) *(k) = 2i j
i+j=k
is the dual of the product map in A. All of this is described in [Mil].
It can be shown (cf. [Gra , Proposition 28.13]) that for any x; y 2 A and any*
* Milnor
basis element Sq
X I;J
(3.4) Sq(xy) = R Sq*(x)Sq(y)
I;J
where I;JR= Sq; R >. In particular, it follows from (3.3) that
Xt
(3.5) st(xy) = j+st-j(x)sj(y)
j=0
where we abbreviate Pvuas vuand take u0to be the identity map.
For a fixed t let Bt be the vector subspace of A with basis the set of all Sq*
*t(R).
One can easily verify [G ], either directly or by using the results in [AD ], t*
*hat Bt is a
Hopf subalgebra of A. Note that B1 = A and that stripping an element of Bt by a
Milnor basis element not in Bt must yield zero by (3.2).
Therefore if x and y are elements of Bt then stripping by Ptsis a differentia*
*l, i.e.
(3.6) st(xy) = st(x)y + xst(y)
This follows from (3.5) since all of the other terms in the sum correspond to s*
*tripping
by Milnor basis elements which are not in Bt and therefore give zero.
Iterating (3.6) we see that for any x1; : :;:xn 2 Bt
Xn
(3.7) st(x1x2. .x.n) = x1. .x.i-1st(xi)xi+1. .x.n
i=1
Using these facts we can now obtain the following technical lemma which is an*
*al-
ogous to [WW , Lemma 1.4] and whose proof is entirely analogous to the proof *
*of
[WW , Lemma 1.4]. Namely, the second relation follows from the first by stri*
*pping
the first by Ptsusing (3.7), and then multiplying the result by Pts.
Lemma 3.1. Let 2 Bt be an element which gives zero when stripped by Pts. Then
(Pcst)(Pts)2k+1= 0 =) (Pts)2k+2= 0
THE NILPOTENCE HEIGHT OF Pst 7
4. The Doubling Isomorphism
Let E be the exterior subalgebra of A generated by { Pt0| t 1}. Recall that
there is an algebra isomorphism D : A ! A ==E given by D(Sq(s1; s2; : :):) =
[Sq(2s1; 2s2; : :):] where [x] denotes the equivalence class in A ==Eiof x 2 A.*
*jIn partic-
ular, since D(Pts-1) = [Pts] we have (Pts)n = 0 =) [Pts]n = [0] =) D (Pts-1)n=
[0] =) (Pts-1)n = 0 (since D is an isomorphism). Thus, by iterating we see th*
*at
(Pts)n = 0 implies (Pti)n = 0 for all i s. This proves the following.
Lemma 4.1. For t > 1, if Theorem 1.1 holds for all s -1 mod t then it holds*
* for
all s.
5. Proof of the Main Result
Having developed the necessary tools, the proof of Theorem 1.1 now follows by
directly imitating the proof of [WW ] for the case t = 1. Accordingly we proc*
*eed by
proving the following two equations by induction on k for 1 k m .
(5.1) (Xct)mt-1kt-1(Ptmt-1)2k-1=0
(5.2) (Xct)(m-1)t-1kt-1(Ptmt-1)2k=0
To begin the induction we note that (Xct)mt-1t-1= Sqt(2t-1flt(m)) by [S2, The*
*orem
1.1]. Thus equation 5.1 for k = 1 follows from Proposition 2.1.
Equation 5.1 for k is equivalent to the relation
(Xct)(m-1)t-1kt-1"Ptmt-1(Ptmt-1)2k-1= 0
which implies equation 5.2 for k by Lemma 3.1 (interpret (Xct)(m-1)t-1mt-1as 1).
Finally, by Corollary 2.4 with a = mt - 1, b = 1, and c = kt - 1 we have
(Xct)mt-1(k+1)t-1(Ptmt-1)2k+1=(Xct)mt-1(k+1)t-1Ptmt-1(Ptmt-1)2k
= Sqt(z)(Xct)(m-1)t-1kt-1(Ptmt-1)2k
for some integer z so that equation 5.2 for k implies equation 5.1 for k + 1. T*
*hus by
induction on k we see that equations 5.1 and 5.2 both hold for 1 k m . Equa*
*tion
5.2 for k = m is (Ptmt-1)2m = 0 which proves the theorem for all s -1 mod t. *
*So
by Lemma 4.1 this proves Theorem 1.1. __|_ |
6. Acknowledgment
I would like to thank Grant Walker and Reg Wood for sending me a preprint of
[WW ] by which this paper was inspired.
8 KENNETH G. MONKS
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Kenneth G. Monks, Department of Mathematics, University of Scranton, Scran-
ton, PA 18510
E-mail address: monks@uofs.edu
*