Turk J Math
24 (2000) , 359 – 365.
c TÜBİTAK
On Conjugation in the Mod-p Steenrod algebra
Ismet Karaca and Ilkay Yaslan Karaca
Abstract
In this paper we prove a formula involving the canonical anti-automorphism χ
of the mod-p Steenrod algebra.
Key Words: Steenrod algebra, anti-automorphism, Milnor basis
1. Introduction and Main Result
Let A be a mod-p Steenrod algebra. Let R = (r1 , r2 , . . . ) be a sequence of nonnegative
integers with finitely many nonzero terms. Let P(R) denote the corresponding Milnor
basis element in A so that the elements P(R) form an additive basis for the subalgebra
∞
P
pi − 1 r i
Ap of A generated by the Steenrod powers P i , i ≥ 0. We define |R| =
i=1
and e(R) =
∞
P
ri . Thus, considered as a mod-p cohomology operation, P(R) raises the
i=1
dimension of a cohomology class by 2 |R| and has excess 2e(R). The anti-automorphism
χ of Ap plays a fundamental part in our argument, and we find it convenient to write
θb = (−1)dimθ χ(θ)
for every element θ ∈ Ap .
We are interested in an explicit conjugation formula for the Steenrod operations of
Ap in the form
X(k, n) = P(pk n)P(pk−1 n) · · · P(pn)P(n),
AMS Mathematics Subject Classification: Primary 55S10, 55S05
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KARACA, KARACA
where k and n are nonnegative integers. So the following formula is the mod-p analogue
of Theorem 3.1 in [6].
Theorem 1.1 For all positive integers j and i, we have
b pi+1 − 1) = X(i, pj+1 − 1).
X(j,
We will introduce the following useful notation: each natural number a has a unique
p-adic expansion
a=
∞
X
αi (a)pi
i=0
with 0 ≤ αi (a) < p. It is a fact that
Y
∞ αi (a)
a
.
≡
αi (b)
b
i=0
(1)
Using Davis’ method [1] we can derive the following formulae.
Proposition 1.2
b
P(u) · P(v)
=
X |R| + e(R)
pu
R
b
P(u)
· P(v) =
P(R)
X e(R)
P(R),
v
(2)
(3)
R
where the sum is taken over all R for which |R| = (p − 1)(u + v).
Proof.
See [2] for the proof of (2) and look at [4] for the proof of (3).
Using these formulae, we can prove the following proposition.
Proposition 1.3 For nonnegative integers k, l, m, n and k > l, suppose that
(1) m + n = pk − pl
(2) m < pk−1
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(3) m ≡ 0 mod pl .
Then
(i) When l = 0, we have
b
b − (p − 1)m − 1)P(pm + 1)
P(m) · P(n)
= P(n
(ii) When l > 0, we have
p−1 X p−1 l
l−1
b
b
b
))P(pm+(p−1)pl )−
).
P(m+wpl−1 )·P(n−wp
P(m)·P(n)
= P(n−(p−1)(m+p
w
w=1
Proof.
(i) Let l = 0. Using Proposition 1.3, we have
X |R| + e(R)
P(R),
pm
b
P(m) · P(n)
=
R
and
b − (p − 1)m − 1) · P(pm + 1) =
P(n
X e(R) P(R),
pm + 1
R
where |R| = (p − 1)(pk − 1) and 1 ≤ e(R) ≤ pk − 1. In order to prove these sums are
equivalent in mod-p, we need to show that their binomial coefficients are equivalent in
mod-p, i.e.
We know that |R| =
∞
P
|R| + e(R)
pm
≡
e(R)
pm + 1
ri (pi − 1) and e(R) =
i=1
∞
P
mod p.
ri . Using these facts, we have
i=1
∞
pk − 1 =
X
|R|
= r1 +
ri (pi−1 + pi−2 + · · · + p + 1)
p−1
i=2
= e(R) −
∞
X
i=2
= e(R) +
∞
X
ri +
∞
X
ri (pi−1 + pi−2 + · · · + p + 1)
i=2
ri p(pi−2 + pi−3 + · · · + p + 1).
i=2
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Since
∞
P
ri p(pi−2 + pi−3 + · · · + p + 1) ≡ 0 mod p, e(R) ≡ p − 1 mod p. Using Equation
i=2
(1), and the upper bounds of e(R) and m, we have
|R| + e(R)
pm
=
e(R)
(p − 1)(pk − 1) + e(R)
≡
mod p.
pm
pm + 1
This completes the proof of part (i).
(ii) Let l > 0. Again using Proposition 1.3, we have
b
P(m) · P(n)
=
X |R| + e(R)
P(R),
pm
R
P(m + wp
l−1
b − wpl−1 ) =
) · P(n
X |R| + e(R)
pm + wpl
R
P(R),
and
b − (p − 1)(m + pl )) · P(pm + (p − 1)pl ) =
P(n
X
R
e(R)
P(R)
pm + (p − 1)pl
where |R| = (p − 1)(pk − pl ) and 1 ≤ e(R) ≤ pk − pl . In order to prove the sums in part
(ii) are equivalent, we need to show that
p−1
X p − 1|R| + e(R) e(R)
|R| + e(R)
≡
mod p.
+
w
pm + wpl
pm + (p − 1)pl
pm
w=1
Case 1: 0 ≤ αl (e(R)) < p − 1. Then
e(R)
pm+(p−1)pl
are equivalent to zero in mod-p.
So it is enough to show that
p−1
X p − 1|R| + e(R)
|R| + e(R)
≡0
+
w
pm + wpl
pm
w=1
i.e.
1+
p−1 X
p − 1 1 + αl (e(R))
w=1
362
w
w
mod p
≡ 0 mod p
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for 0 ≤ αl (e(R)) < p − 1. The following equivalent holds
X p − 11 + αl (e(R))
p−1
w=0
w
w
≡
p + αl (e(R))
p−1
mod p
by considering coefficient of xp−1 in the binomial expansion of
(x + 1)p+αl (e(R)) = (x + 1)p−1 (x + 1)1+αl (e(R)).
Since 0 ≤ αl (e(R)) < p − 1,
p−1
X
p+αl (e(R))
p−1
w=0
i.e.
1+
p−1
w
is equivalent to zero in mod-p. Hence
1 + αl (e(R))
w
≡0
p−1 X
p − 1 1 + αl (e(R))
w=1
w
w
mod p
≡ 0 mod p
So the result holds.
Case 2: αl (e(R)) = p − 1. Then
p−1 X
p − 1 |R| + e(R)
w=1
and
w
pm + wpl
≡ 0 mod p
|R| + e(R)
e(R)
mod p.
≡
pm
pm + (p − 1)pl
2
Therefore the result holds.
2. Proof of Main Result
Proof of Theorem 1.1 We are going to prove the theorem by induction on i under the
b p − 1) = P(pj+1 − 1) = X(0, pj+1 − 1) by
assumption that i ≤ j. For i = 0 and all j , X(j,
Davis’ formula in [1]. Assume that for all î ≤ i − 1 and all j, and for î = i and ĵ ≤ j − 1,
b î, pĵ+1 − 1) = X(ĵ, pî+1 − 1).
X(
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The inductive proof will draw on the following remark: under the above assumptions,
b l−1 ) · X(l − 1, pi+1 − 1) = 0
P(cp
(4)
where c is a unit in mod-p and 1 ≤ l ≤ i. Indeed, by induction on l, we have
z
}|
{
\
b − 1, pi+1 − 1) · P(cpl−1 )]
b l−1 ) · X(l − 1, pi+1 − 1) = [X(l
P(cp
}|
{
z
\
= [X(i, pl − 1) · P(cpl−1 )]
}|
{
z
\
= [X(i − 1, p(pl − 1))P(pl − 1) · P(cpl−1 )].
By Adem relations, P(pl − 1) · P(cpl−1 ) = 0. Therefore this verifies Equation (4).
We claim that for 0 ≤ l ≤ j
b i (pj+1 − 1)) = P(p
b i (pj+1 − pl+1 )) · X(l, pi+1 − 1).
X(l, pi − 1) · P(p
(5)
The case l = 0 follows from Proposition 1.4 (i). Suppose that the statement is true
for l − 1. Then by induction on l and Proposition 1.4 (ii), we have
b i (pj+1 − 1)) = P(pl (pi − 1)) · P(p
b i (pj+1 − pl )) · X(l − 1, pi+1 − 1)
X(l,pi − 1) · P(p
b i (pj+1 − pl+1 )) · P(pl (pi+1 − 1))] · X(l − 1, pi+1 − 1)
= [P(p
p−1 X
p−1
b i+j+1 − pl+i − wpl−1 )] · X(l − 1, pi+1 − 1).
P(pl (pi − 1) + wpl−1 ) · P(p
−[
w
w=1
From Equation (4),
b i+j+1 − pl+i − wpl−1 ) · X(l − 1, pi+1 − 1) = 0
P(p
for every w = 1, 2, 3, . . . , p − 1. Hence we have
b i (pj+1 − 1)) = P(p
b i (pj+1 − pl+1 )) · X(l, pi+1 − 1).
X(l, pi − 1) · P(p
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This proves our claim. Finally, taking l = j, we find that
b − 1, pj+1 − 1)) · P(p
b i (pj+1 − 1))
b pj+1 − 1) = X(i
X(i,
b i (pj+1 − 1))
= X(j, pi − 1) · P(p
b
= P(0)
· X(j, pi+1 − 1).
2
This completes the proof.
Acknowledgement
We would like to sincerely thank Referee for many helpful comments on this paper.
References
[1] D.M. Davis The Anti-automorphism of the Steenrod Algebra Proc. Amer. Math. Soc. 44,
(1974), 235–236.
[2] A.M. Gallant Excess and Conjugation in the Steenrod Algebra Proc. Amer. Math. Soc. 76,
(1979) 161–166.
[3] I. Karaca On the action of Steenrod algebra on polynomial algebra Turkish Journal of
Mathematics 22, (1998), 163–170.
[4] I. Karaca The nilpotence height of Pts for odd primes Trans.Amer. Math. Soc. 351, (1999),
547–558.
[5] J. Milnor The Steenrod Algebra and Its Dual Ann. of Math. 67, (1958), 150–171.
[6] J.H. Silverman Conjugation and Excess in the Steenrod Algebra Proc. Amer. Math. Soc.
119, (1993), 657–661.
Ismet KARACA and Ilkay Yaslan KARACA
Received 18.03.1999
Department of Mathematics Ege University,
35100 İzmir-TURKEY
e-mail: [email protected] and
e-mail: [email protected]
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