STRIPPING AND CONJUGATION IN THE STEENROD ALGEBRA
JUDITH H. SILVERMAN
January 10, 1996
Abstract.Let S(k; f) = Sq(2k1f) . Sq(2k2f) . : :.:Sq(2f) . Sq(f) in the m*
*od2 Steenrod
algebra A*, and let O denote the canonical antiautomorphism of A*. Given *
*positive integers
k, and j with 1 j , we prove that
OS(k; 2 )j= S(  (j  1); 2j1(2k)1). OS(k; 2j1)j;
generalizing formulae of Davis and the author.
Given a positive integer f, denotePby (f) the minimal number of summand*
*s in any
representation of f in the form (2ik 1). The antiautomorphism formula *
*above implies
that for f = 2 j, 1 j +2, the excess of OS(k; f) satisfies ex(OS(k; f)*
*)= (2k1)(f)
for all k, confirming the conjecture of the author (Proc. Amer. Math. Soc*
*., 119(2):657661,
1993) for such f. We also prove that ex(OS(k; f)) (2k 1)(f) for all f an*
*d k.
1.Introduction
1.1. Informal statement of results. The mod2 Steenrod algebra A* is multiplica*
*tively
generated by the Steenrod squares Sq(i) in dimension Sq(i) = i, 0 i < 1. The*
* product
Sq(a1) . : :.:Sq(an) is admissible if ai 2ai+1for i < n, and an > 0 if n > 1; t*
*he admissible
elements form an additive basisPof A*. The excess of the admissible element Sq(*
*a1).: :.:Sq(an)
of dimension d is given by n1i=1(ai 2ai+1)=+2ana  d [SE62 ]. In general, t*
*he excess of
a sum of admissibles is the minimum of the excesses of the summands.
We write
S(k; f) = Sq(2k1f) . Sq(2k2f) . : :.:Sq(2f) . Sq(f)
and apologize for the change in notation from [Sil93]. The dimension S(k; f) *
*= (2k  1)f
and the excess ex(S(k; f))= f.
The Steenrod algebra is a connected Hopf algebra, and as such has a unique an*
*tiautomor
phism, commonly denoted by O [Mil58]. Following [WW94 ], we write ^for O. In p*
*articular,
S^q(a) = OSq(a) and ^S(k; f) = OS(k; f).
In Section 6 below, we prove the following antiautomorphism formula, generali*
*zing results
of [Dav74 ] and [Sil93]:
Theorem 1.1. Let k and be positive integers, and suppose that 1 j . Then
^S(k; 2  j)= S(  (j  1); 2j1(2k  1)) . ^S(k; 2j1 j):
__________
1991 Mathematics Subject Classification. Primary 55S05,55S10; Secondary 57T05.
Key words and phrases. Steenrod algebra, antiautomorphism, stripping, excess.
1
2 JUDITH H. SILVERMAN
The Steenrod algebra A* acts on F2[x1; : :;:xs] according to wellknownPrules*
*. The image
of this action, i.e. the set of polynomials which can be written F = i>0Sq(i*
*)Fi, is re
lated to various entities of importance in algebraic topology; among these are *
*ExtsA*(F2; F2)
[Sin89] and cobordism classes of closed manifolds [Pet89]. In addition, this im*
*age contains
information about the simple representations of the general linear group GL(s; *
*F2) [Woo89a ].
In [Sil93], we discuss the connection between this image and the excess of the *
*Steenrod op
erations S(k; f), and frame a conjecture which would permit the argument of [Wo*
*o89b ] to
prove the conjecture of [SS95] concerning this image. A stronger version of thi*
*s conjecture
appears in [Sil] and is equivalent to Conjecture 1.2 as stated in Section 1.4. *
* In order to
state our present results without introducing further notation, we recall here *
*the original
statement of the conjecture:
Conjecture 1.2, weak version Let f be a positive integer. Then for all positive*
* integers
k we have
i j
ex(S^(k; f))=(2k  1)ex(S^q(f))= (2k  1)ex(S^(1; f)):
Theorem 1.1 will imply
Theorem 1.3 Conjecture 1.2 is true for f satisfying 2  ( + 2) f 2  1 for *
*some
0.
We also prove that
Theorem 1.4 One of the inequalities of Conjecture 1.2 is true for all f;
ex(S^(k; f)) (2k  1)ex(S^q(f))
for all k; f 0.
In the remainder of this section, we introduce enough notation to state Conje*
*cture 1.2 in
full.
1.2. Enotation for admissibles. Denote by S the set of finite sequences of non*
*negative
integers.P We define the dimension of an element S = (s1;Ps2; : :;:sn) 2 S to b*
*e S =
isi(2i 1), its length l(S) = n, and its excess ex(S)= isi.
For our purposes, it will be convenient to parametrize the admissible basis i*
*n terms of the
numbers si= ai 2ai+1, the contributions to excess at each stage. That is, give*
*n a sequence
S = (s1; : :;:sn) 2 S, we define the admissible element E(S) = Sq(a1) . : :.:Sq*
*(an) where
an = sn and ai = 2ai+1+ si for 1 i n  1. For example, if S = (0; : :;:0; f)*
*, then
j
E(S) = S(j; f) as defined in Section 1.1. We have that ex(E(S) )= ex(S)and E(S*
*)  = S.
1.3. Elements of minimal excess. We now single out particular basis elements in*
* each
degree. Given a positive integer f, we denote by (f) the least excess of all se*
*quences in S
of dimension f. Let (f) = max{ : 2  1 f} . In [Sin91], Singer observes that *
*for any
f there exists a unique sequence R1(f) = (r1; : :;:r(f)) 2 S of dimension f suc*
*h that ri 1
for all i, except that the first nontrivial riis 2; this sequence has ex(R1(f*
*))= (f). The
STRIPPING AND CONJUGATION IN THE STEENROD ALGEBRA 3
corresponding admissible element E(R1(f)) is thus of minimal excess among all e*
*lements of
the admissible basis in dimension f.
The sequence R1(f) may be constructed inductively by increasing the (f)th en*
*try of
R1(f  (2(f)  1)) by 1.
Example. We have
R1(2  1)= (0; : :;:0; 1)
8
>>>(0; : :;:0; 2 ; 1; 1; :2:;:1 )j
>>> j+1 1
>><
R1(2  j)= > (1; 1; : :;:1 ) j = + 1
>>> 1
>>>
>:(0; 1; : :;:1 ) j = + 2:
1
Consequently,
8
>  1 j = + 1
:  2 j = + 2:

For k 1, let Rk(f) = ((2k  1)r1; (2k  1)r2; : :;:(2k  1)r(f)). Then Rk(f) *
*and the
corresponding admissible basis element E(Rk(f)) have dimension (2k  1)f and ex*
*cess
(2k  1)(f).
1.4. Results. As we shall see in Section 3.3, the admissible element E(R1(f)) a*
*ppears in
S^q(f) = ^S(1; f), and so ex(S^(1; f))= (f). Conjecture 1.2 below purports to *
*generalize
these phenomena to k > 1.
Conjecture 1.2. [Sil] Let f be a positive integer. Then for all positive integ*
*ers k we have
(i)the element E(Rk(f)) is a (nontrivial) summand in the admissiblebasis r*
*epresen
tation of ^S(k; f), and
(ii)its excess is minimal among all such summands, so that ex(S^(k; f))= ex(*
*E(Rk(f)))=
(2k  1)(f).
In this paper, we prove
Theorem 1.3. Conjecture 1.2 is true for numbers of the form f = 2  j, 1 j *
* + 2.
and
Theorem 1.4. Part (i) of Conjecture 1.2 holds for all pairs (f; k) of positive*
* integers, and
consequently ex(S^(k; f)) (2k  1)(f).
1.5. Acknowledgements. I thank David Carlisle, Grant Walker, and Reg Wood for m*
*any
helpful conversations and manuscripts, and the referee for the careful reading *
*and detailed
comments.
4 JUDITH H. SILVERMAN
2. Hopf algebras
Recent work on nilpotence in A* has exploited the "stripping" action of the d*
*ual Steenrod
algebra A* on A* itself ([WW94 ], [Mon94 ]). In this section, we identify the *
*stripping action
as a general Hopf algebra phenomenon. I thank Bill Schmitt, Grant Walker, and R*
*eg Wood
for elucidating this point of view.
Let A* be a Hopf algebra over a field K with diagonal_*, multiplication OE*, *
*and conjuga
tion O [MM65 ]. We continue to write O as ^or depending on the typographical *
*complexity
of . Let A* be the dual Hopf algebra andPwrite < ; > :PA* A* ! K for the inner*
* product.
In what follows, yi2 A*, j 2 A*, * = 0 00and OE*y = y0 y00; we write 12 for
OE*(1 2) and y1y2 for *(y1 y2).
There is a natural action of A* on A* in which y 2 A* acts via
* * * 1*
(1) D(y) : A* ! A A  ! A :
Following [CW94 ] and [WW94 ], we refer to D(y) as the operation of stripping*
* by y. Evidently
D(y1 + y2) = D(y1) + D(y2); coassociativity of A* gives D(y1y2) = D(y1) O D(y2)*
*. Let
D = {D(y) : y 2 A*}. Define the maps
O : D ! D
D(y) 7! D(^y)
* : D D ! D
D(y1) D(y2) 7! D(*(y1 y2))
OE* : D ! D D
X
D(y) 7! (D D)(OE*y) = D(y0) D(y00):
Henceforth we write ^D(y) for O(D(y)). If A* is cocommutative, a determined cha*
*se of the
defining diagrams (see for example Sections 4 and 8 of [MM65 ]) reveals the fo*
*llowing three
equations:
X
(2) D(y) O OE* = OE*O (OE*D(y));i.e.D(y)(12) = D(y0)1. D(y00)2:
______
(3) ^D(y) = O O D(y) O O;i.e.D^(y) = D(y)^:
X
(4) ^D(y) O OE* = OE*O ((O O)T OE*D(y));i.e.^D(y)(12) = D^(y00)1. ^D(y0)2:
In Sections 3 and 4 below, we discuss the stripping action in the case where *
*A* = A*, the
mod2 Steenrod algebra.
STRIPPING AND CONJUGATION IN THE STEENROD ALGEBRA 5
3.The Milnor basis
3.1. Notation. The dual Hopf algebra A* of A* is a polynomial algebra over F2 o*
*n gen
erators i in dimension 2i 1, 1 i < 1 [Mil58]. For S = (s1; : :;:sn) 2 S,Pwri*
*te (S)
for the monomial s11. .s.nn; evidently the dimension of this monomial is isi(*
*2i 1) = S.
The Milnor basis of A* itself is the basis dual to the monomials inPthe i; we d*
*enote the
dual element to (S) by M(S). Then M(S)  = E(S)  and ex(M(S) )= si = ex(E(*
*S) )
[Kra71].
In [Mon95 ], Monks shows that for all S 2 S, each of E(S) and M(S) appears in*
* the
representation of the other in the appropriate basis. Moreover, the difference *
*ffi(S) = E(S) 
M(S) satisfies ex(ffi(S)) ex(S)and also ffi(S) 1, we have
D^(k)Sq(f) = D(2k11)Sq(f) = Sq(f  (2k  1))
for all f and k.
In Section 4.3, we generalize Part (ii) of Consequence 4.3 with a formula parti*
*ally describing
the effect of ^D(jk) on S(; f) for ; j 2.
STRIPPING AND CONJUGATION IN THE STEENROD ALGEBRA 7
4.3. Stripping S(; f) by ^jk.
Lemma 4.4. For any 2 A*, we have ^D(k)[S(2; f) . ] = Sq(2f) . [D^(k)Sq(f) . ].
P k 2j
Proof. The comultiplication in A* is given by OE*(k) = j=0kj j [Mil58], and*
* conse
quently (4) implies
Xk
^D(k)[S(2; f) . ]= Sq(2f) . [D^(k)Sq(f) . ] + D^(j)Sq(2f) . [D^(2jkj)Sq(f*
*) . ]:
j=1
Since ^D(j)Sq(2f) = Sq(2f  (2j 1)) = ^D(2j1)Sq(2f  1) by Part (ii) of Conse*
*quence 4.3,
we have
Xk
D^(k)[S(2; f) . ]= Sq(2f) . [D^(k)Sq(f) . ] + D^(2j1)Sq(2f  1) . ^D((2j1k*
*1(j1))2)[Sq(f) . ]
j=1
= Sq(2f) . [D^(k)Sq(f) . ] + ^D(2k1)[Sq(2f  1) . Sq(f) . ]:
But Sq(2f  1) . Sq(f) = 0, proving the lemma. 
Proposition 4.5.For 2 and for any 2 A*,
D^(k)[S(; f) . ]= S(  1 ; 2f) . ^D(k)[Sq(f) . ]:
Proof. The proof is by induction on . The case = 2 follows from the proof of *
*Lemma 4.4.
Suppose that the result is known for  1. Then
Xk
D^(k)[S(; f) . ] = ^D(j)S(2; 22 f) . ^D(2jkj)[S(  2 ; f) . ] (4)
j=0
kX j
= Sq(21 f) . [D^(j)Sq(22 f)] . ^D(2kj)[S(  2 ; f) . ] (Lemma 4*
*.4)
j=0
= Sq(21 f) . ^D(k)[S(  1 ; f) . ]
ind=Sq(21 f) . S(  2 ; 2f) . ^D(
k)[Sq(f) . ]
= S(  1 ; 2f) . ^D(k)[Sq(f) . ]:

Next we generalize Proposition 4.5 to show that applying ^D(k) to S(; f) a to*
*tal of j
times affects only the rightmost j places:
Proposition 4.6.D^(jk)S(; f) = S(  j ; 2jf) . ^D(jk)S(j; f).
8 JUDITH H. SILVERMAN
Proof. Proposition 4.5 gives the case j = 1. Suppose that the result is known *
*for j0< j
and all 0, and for j and 0< . We have
D^(jk) S(; f) = D^(k) [D^(j1k)S(; f)]
ind=^D( j1 ^ j1
k) [S(  (j  1); 2 f) . D(k )S(j  1; f)]
= S(  j ; 2jf) . ^D(k)[Sq(2j1f) . ^D(j1k) S(j  1; f)] (Pro*
*position 4.5)
ind=S(  j ; 2jf) . ^D(j
k)S(j; f):

4.4. Application. In [Sil93], we prove the following conjugation formula, which*
* grounds
the inductive proof of Theorem 1.1:
Theorem 4.7. [Sil93] For all positive integers k and , we have S^(k; 2 *
* 1) =
S(; 2k  1).
Stripping both sides of this equation by ^k, using (7) on the left and Propos*
*ition 4.5 on
the right, we find that
^S(k; 2  2)=S(  1 ; 2(2k  1)) . ^D(k)Sq(2k  1):
Since ^D(k)Sq(2k  1) = Sq(0) = 1 by (6), we find that
(8) ^S(k; 2(21  1))=S(  1 ; 2(2k  1));
a formula conjectured in [Sil93] which is a special case of Theorem 1.1 below. *
* As a con
sequence, we see that ^S(k; 2(2  1)) and ^S(k; 2  1) are both of length exa*
*ctly , where
length is as defined in Section 3.2 above.
Recall now from Section 3.4 that stripping operations do not increase length.*
* The result
below, conjectured in [Sil], follows by a sandwich argument from (7) and the co*
*nclusion of
the previous paragraph.
Theorem 4.8. If 2  1 f < 2+1  1, then the elements ^S(k; f) are of length*
* exactly
independently of k.
5. Proof of Theorem 1.4
We are now ready to prove
Theorem 1.4 Part (i) of Conjecture 1.2 is always true, and consequently for all*
* f 1 we
have ex(S^(k; f)) (2k  1)(f) for all k 1.
STRIPPING AND CONJUGATION IN THE STEENROD ALGEBRA 9
Proof. The cases f = 2  1 and f = 2(2  1) follow from Theorem 4.7 and (*
*8)
respectively. We assume inductively that the result is known for f 2  1 and *
*prove it for
2 f < 2(2  1). Observe that
D^(2k1)S^(k; 2(2  1))=S^(k; 2  1) (7)
= S(; 2k  1) Theorem 4.7
k1 k
= D(2 )S(; 2(2  1)) Corollary 4.2 (ii)
k1
= D(2 )S^(k; 2(2  1))Theorem 4.7,
1 2k1
i.e. D^(2k ) and D( ) agree on ^S(k; 2(2  1)). Moreover, since stripping *
*operations
commute with each other, these two operations also agree on all elements of the*
* form
D(y)S^(k; 2(2  1)). In particular, they agree on
S^(k; f)= ^D(2(2k1)f) ^S(k; 2(2  1))
for all f 2(2  1). Thus by (7) we have
k1 2 1
(9) D(2 )S^(k; f) =^D(k )S^(k; f)= ^S(k; f  (2  1)):
By Theorem 4.8, all admissible elements appearing in ^S(k; f) have length , as*
* those in
S^(k; f  (2  1)) have length  1. By Proposition 4.1, those in ^S(k; f) of*
* length <
k1
vanish when stripped by 2 . By Part (i) of Corollary 4.2, those of the form E*
*(r1; : :;:r )
either vanish (if r < 2k  1) or map to the single admissible element E(r1; : *
*:;:r1 ; r 
(2k  1)). Therefore (9) implies that the map
E(s1; : :;:s1 ) 7! E(s1; : :;:s1 ; 2k  1)
assigns to each admissible summand of ^S(k; f  (2  1)) an admissible summand*
* of ^S(k; f)
of length with last entry 2k  1. Since by assumption E(Rk(f  (2  1))) app*
*ears in
S^(k; f  (2  1)), we find from the inductive construction of R1(f) and the d*
*efinition of
Rk(f) in Section 1.3 that indeed E(Rk(f)) appears in ^S(k; f). As the excess of*
* E(Rk(f)) is
(2k  1)(f), this completes the inductive step in our proof. 
6.Proof of Theorems 1.1 and 1.3
We now apply results of Section 4.3 to prove Theorems 1.1 and 1.3, restated b*
*elow for
convenience. We will make use of the following observation, immediate from the *
*definition
of excess and the Adem relations governing multiplication in A*:
Observation 6.1. If a 2 N and Sq(a) . has degree d, then ex(Sq(a) .) 2a  d.
Theorem 1.3 for f = 2  1 is an immediate consequence of Theorem 4.7 above, *
*which
states that ^S(k; 2  1) = S(; 2k  1) for all k and , generalizing a result o*
*f [Dav74 ] to
k > 1. Theorem 1.1, from which the rest of Theorem 1.3 will follow, likewise g*
*eneralizes
another result of [Dav74 ]:
Theorem 1.1 ^S(k; 2  j) = S(  (j  1); 2j1(2k  1)) . ^S(k; 2j1 j) for 1*
* j .
10 JUDITH H. SILVERMAN
Proof. For such j, we have
__________________
^S(k; 2  j)= Dj1(k)S(k; 2  1) Corollary 4.2 (i*
*i)
= ^D(j1k)S^(k; 2  1) (3)
= ^D(j1k)S(; 2k  1) Theorem 4.7
= S(  (j  1); 2j1(2k  1)) . ^D(j1k) S(j  1;P2kro1)position*
* 4.6
= S(  (j  1); 2j1(2k  1)) . ^D(j1k) ^S(k; 2j1T1)heorem 4.7
= S(  (j  1); 2j1(2k  1)) . ^S(k; 2j1 j):

Before recalling the statement of Theorem 1.3, we recall from Section 1.3 that
8
>  1 j = + 1
:  2 j = + 2:
We are now ready to prove
Theorem 1.3 Conjecture 1.2 is true for all f satisfying 2  ( + 2) f 2  1 *
*for some
0.
Proof. By Theorem 1.4, we need only show that for 1 j + 2, we have
ex(S^(k; 2 )j) (2k  1)(2  j)
for all k 0. If 1 j , then by Observation 6.1 and Theorem 1.1 we have
ex(S^(k; 2 )j) 2 . 2j . 2j1(2k  1)  (2k  1)(2  j)
= (2k  1)[2  (2  j)]
= (2k  1) . j
= (2k  1)(2  j) (10):
The result for j = + 1 and j = + 2 follows from the case j = and Part (i) of
Consequence 4.3. 
STRIPPING AND CONJUGATION IN THE STEENROD ALGEBRA 11
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Indiana UniversityPurdue University at Columbus
Email address: jsilver@indyvax.iupui.edu