Homology, Homotopy and Applications, vol.2, No.1, 2000, pp.1–16 ISSN 1512–0139

STRIPPING AND CONJUGATION IN THE MOD p STEENROD ALGEBRA AND ITS DUAL

DAGMAR M. MEYER

(communicated by Hvedri Inassaridze)

Abstract Let p be an odd prime and A^{∗} the mod p Steenrod algebra. We study the technique known as “stripping” applied to A^{∗} and derive certain conjugation formulas both for A^{∗} and its dual, generalising work of J. H. Silverman for p = 2 (“Conjugation and excess in the Steenrod algebra”, Proc. Am. Math. Soc. 119 (1993), no.2, 657 – 661; “Hit polynomials and conjugation in the dual Steenrod algebra”, Math. Proc. Camb. Philos. Soc. 123 (1998), no.3, 531 – 547) to the case of an odd prime.

1. Introduction and statement of results

In this note we study the technique known as “stripping” applied to the mod p Steenrod algebra A^{∗}, where p is an odd prime, and use the results obtained to prove certain conjugation formulas both in A^{∗} and its dual. This generalises work of Judith Silverman carried out in [S1] and [S3] for p = 2 to the case of an odd prime. More precisely, our results concern Steenrod operations which lie in the sub-Hopf algebra P^{∗} of A^{∗} which is generated by the reduced power operations P(i), i ⩾ 1, in dimensions |P(i)| = 2i(p − 1). We use the convention P(0) := 1.

Of particular interest are the Steenrod operations in P^{∗} which are of the form

P[k; f] := P(p^{k−1}f) · P(p^{k−2}f) · . . . · P(pf) · P(f)

where k ⩾ 1 and f ⩾ 0. Note that P[1; f] is just P(f). Being a sub-Hopf algebra, P^{∗} inherits the canonical anti-automorphism χ of A^{∗}; following notation introduced in [WW], we write θ̂ instead of χ(θ). In particular, P̂(a) = χ(P(a)) and P̂[k; f] = χ(P[k; f]).

For m ⩾ 0 we define

γ(m) := ∑_{i=0}^{m−1} p^{i}.

Our first main result is an explicit conjugation formula for P[k; f] in certain special cases. It generalises Thm. 3.1 in [S1] to odd primes:

Theorem 4.6 For all positive integers s, t and c with 1 ⩽ c ⩽ p the following conjugation formula holds:

P̂[s; cγ(t)] = (−1)^{stc}P[t; cγ(s)]

The main result concerning conjugation in the dual P_{∗} is a conjugation formula for certain elements X_{I}(k), which are defined in Section 5. This formula is the mod p analogue of

During the preparation of this article the author was supported by a “bourse post-doctorale Alfred Kastler” of the French Ministry of Foreign Affairs. Received 11 October 1999; published on 24 January 2000. 1991 Mathematics Subject Classification: 55S10 Key words and phrases: stripping, conjugation, Steenrod algebra, antiautomorphism \begin{matrix}c \\ ◯ \end{matrix} 2000, Dagmar M. Meyer. Permission to copy for private use granted.

Prop. 5.5 in [S3]. A special case states that modulo monomials of length strictly greater than k the operations ξ̂_{i}^{γ(k)} and (−1)^{ik}ξ_{k}^{γ(i)} coincide up to a certain error term; the conjugate of the error term is a sum of monomials of length strictly greater than i:

Theorem 5.6 Let i, k > 0. Modulo monomials of length > k we have

ξ̂_{i}^{γ(k)} ≡ (−1)^{ik}ξ_{k}^{γ(i)} − ∑_{Id_{k}≠τ∈𝔖(k)}sign(τ) ∏_{j=0}^{k−1} ξ̂_{i+τ(j)−j}^{p^{j}} .

Here 𝔖(k) denotes the symmetric group acting on {0, 1, 2, …, k − 1} and ξ̂_{r} := 0 for r < 0. In particular, if f < γ(k + 1) is a non-negative integer then

ξ̂_{k}^{γ(i)} ∩ P[i; f] = (−1)^{ik}ξ_{i}^{γ(k)} ∩ P[i; f] = (−1)^{ik}P[i; f − γ(k)] ,

where we use the notation y ∩ for the stripping operation D(y).

The ideas underlying the proofs of the results in this paper are similar to those of their mod 2 counterparts in [S1] and [S3]. However, getting down to the details we note two major differences that appear in the odd-primary case: first of all, in just about every formula we prove there are some signs involved, and secondly (in Section 4) we have to deal with mod p binomial coefficients which appear as non-trivial coefficients in our formulas. These difficulties cause the generalisation of the mod 2 results to be not quite as straightforward as it may seem at first glance.

Both Thm. 4.6 and Thm. 5.6 are essential ingredients for the work carried out in [M]. There the Steenrod operations P̂[k; f] are studied further; in particular the excess of these operations is determined. In fact, that project was one of the main motivations for the work on the problems discussed in the present paper.

Acknowledgements: I would like to thank Judith Silverman for many helpful comments on the subject of this paper, in particular for supplying the elegant proof of Lemma 4.4 which is reproduced here. I also thank Ken Monks for pointing out to me the work of Ismet Karaca on related problems and for sending me copies of [Ka2], [Ka3].

2. Preliminaries

Let S denote the additive monoid of sequences of non-negative integers almost all of which are 0, with componentwise addition. We write 0_{S} for the trivial element. Throughout we shall use capital letters to denote sequences in S and small letters for their coordinates; e.g. S = (s_{1}, s_{2}, …). If S has s_{i} = 0 for i > L, we write S as (s_{1}, s_{2}, …, s_{L}). The degree of an element S ∈ S is defined to be |S| = ∑_{i⩾1} s_{i}(p^{i} − 1), its length as len(S) = min{i ⩾ 0 | s_{j} = 0 ∀j > i}, and its excess as ex(S) = ∑_{i⩾1} s_{i}. It will be convenient to adjoin an extra element ∗ to S with the property that ∗ + x = x + ∗ = ∗ for all x ∈ S ∪ {∗} =: S^{∗}. We also define sequences B(j) for any j ∈ ℤ: if j ⩾ 0 then B(j) is the sequence with b(j)_{i} := δ_{ij}, if j < 0 we set B(j) := ∗.

There are many interesting bases for A^{∗} and hence for P^{∗}; the most important and most commonly used are the basis of admissible monomials (“admissible basis”) and the Milnor basis. Recall that the monomial P(a_{1}) · . . . · P(a_{n}) with a_{n} > 0 is admissible if a_{r} ⩾ pa_{r+1} for all 1 ⩽ r < n; we also define P (0) = 1 to be admissible. The admissible basis of P^{∗} can be parameterised in terms of the numbers s_{i} = a_{i} − pa_{i+1}; that is, given a sequence S ∈ S of length n > 0, we define the admissible element E[S] := P(a_{1}) · . . . · P(a_{n}) by setting a_{n} = s_{n} and a_{i} = pa_{i+1} + s_{i} for 1 ⩽ i ⩽ n − 1. We also set E[0_{S}] := P (0) = 1. For example, if S = (0, …, 0, f) has length k then E[S] = P(p^{k−1}f) · . . . · P(f) = P[k; f].

For the Milnor basis of P^{∗} consider the dual Hopf algebra P_{∗}. This is a polynomial algebra over 𝔽_{p} on generators ξ_{i} (i ⩾ 1) in dimension 2(p^{i} − 1); we use the convention ξ_{0} := 1. For S ∈ S we write ξ[S] for the monomial ∏_{i⩾1} ξ_{i}^{s_{i}}. In particular, ξ[B_{j}] = ξ_{j} for any j ⩾ 0. The Milnor basis of P^{∗} itself is the basis dual to the basis of P_{∗} consisting of all the monomials ξ[S] with S ∈ S; the element dual to ξ[S] will be denoted by M[S].

We further set M[∗] = 0 = E[∗] and ξ[∗] = 0, and we adopt the convention that M[S] = 0 = E[S] and ξ[S] = 0 if S is a finite sequence of integers which does not belong to S, i.e. with at least one negative entry. In particular, ξ_{i} := 0 if i < 0.

For any S ∈ S we define length and excess of the monomial ξ[S] as len(S) and 2ex(S) respectively. Likewise, for the admissible and the Milnor basis we define

len_{E}(E[S]) :=len(S) =: len_{M}(M[S]) ,
ex_{E}(E[S]) :=2ex(S) =: ex_{M}(M[S]) .

More generally, suppose θ is any homogeneous element of P^{∗} with a basis representation given by θ = ∑_{i=1}^{n} α_{i}B[S_{i}], where B stands for either E or M. Then we set

len_{B}(θ) := max_{i}{len_{B}(B[S_{i}])} = max_{i}{len(S_{i})}
ex_{B}(θ) := min_{i}{ex_{B}(B[S_{i}])} = 2 min_{i}{ex(S_{i})} .

The excess of any operation θ in P^{∗} can also be defined as ex(θ) := min {n | θ(ι_{n}) ≠ 0 ∈ H^{∗}(K(ℤ/p, n); 𝔽_{p})}, where ι_{n} ∈ H^{∗}(K(ℤ/p, n); 𝔽_{p}) is the fundamental class. In fact, all the different definitions of excess that we have given coincide (cf. [Kr]); in particular ex_{E}(θ) = ex(θ) = ex_{M}(θ).

By [Mi], the change-of-basis matrix in each dimension between the admissible and the Milnor basis is upper triangular with diagonal entry ±1, if for both bases we use the order induced by the right-lexicographical order on S. From this it follows that for any S ∈ S we have len_{E}(M[S]) = len_{E}(E[S]) = len(S) and len_{M}(E[S]) = len_{M}(M[S]) = len(S), and one easily sees that this implies len_{M}(θ) = len_{E}(θ) for any θ ∈ P^{∗}. Henceforth we denote this common value simply by len(θ).

3. Stripping in P^{∗}

3.1. Recollections about the stripping technique

Much recent progress on problems related to the structure of the Steenrod algebra has been made by applying a tool that has become known as “stripping technique” (for a detailed account see [W]). This technique applies to any Hopf algebra, so in particular to the cocommutative, connected Hopf algebra P^{∗}.

Let Δ^{∗} denote the diagonal map of P^{∗} and ⟨ , ⟩ the inner product. We consider the natural action of the dual Hopf algebra P_{∗} on P^{∗} which is given for each ξ ∈ P_{∗} by

D(ξ) : P^{∗} ⟶^{Δ^{∗}} P^{∗} ⊗ P^{∗} ⟶^{id⊗⟨ξ, ⟩} P^{∗};

this action satisfies

⟨ξ · ψ, θ ⟩ = ⟨ψ, D(ξ)θ ⟩  (1)

for all ψ ∈ P_{∗}, θ ∈ P^{∗}. The operation D(ξ) : P^{∗} ⟶ P^{∗} is called “stripping by ξ” and can be considered as a kind of cap-product. For this reason the notation

D(ξ)θ =: ξ ∩ θ

has become customary.

For the reader’s convenience we now recall some important properties of the stripping operation (cf. [S2]):

Let Δ_{∗} denote the product of P_{∗} and ϕ_{∗} the comultiplication; the canonical anti-automorphism of P_{∗} will again be denoted by χ, with χ(y) =: ŷ. In what follows let ϕ_{∗}(y) =: ∑ y′ ⊗ y′′ and Δ^{∗}(θ) =: ∑ θ′ ⊗ θ′′. We write D for the 𝔽_{p}-vector space with basis {D(ξ[S]) | S ∈ S}.

The maps χ : P_{∗} ⟶ P_{∗}, Δ_{∗} : P_{∗} ⊗ P_{∗} ⟶ P_{∗} and ϕ_{∗} : P_{∗} ⟶ P_{∗} ⊗ P_{∗} induce maps

\begin{matrix}χ : D ⟶ D, & D(y) ↦ D(ŷ) \\ Δ_{∗} : D ⊗ D ⟶ D, & D(y_{1}) ⊗ D(y_{2}) ↦ D(y_{1} · y_{2}) \\ ϕ_{∗} : D ⟶ D ⊗ D, & D(y) ↦ ∑ D(y′) ⊗ D(y′′) . \end{matrix}

Proposition 3.1. The following formulas hold:

1. (y_{1} + y_{2}) ∩ θ = y_{1} ∩ θ + y_{2} ∩ θ

2. (y_{1} · y_{2}) ∩ θ = (y_{2} · y_{1}) ∩ θ = y_{1} ∩ (y_{2} ∩ θ) = y_{2} ∩ (y_{1} ∩ θ)

3. y ∩ (θ_{1} · θ_{2}) = ∑(y′ ∩ θ_{1}) · (y′′ ∩ θ_{2})

4. ŷ ∩ (θ_{1} · θ_{2}) = ∑(ŷ′′ ∩ θ_{1}) · (ŷ′ ∩ θ_{2})

5. ŷ ∩ θ̂ = { \atop y ∩ θ}

□

3.2. Stripping in the Milnor basis and in the admissible basis

The effect of stripping by an element y ∈ P_{∗} on a Milnor basis element can easily be described by writing y as a sum of basis elements ξ[R]. In fact, recall that the comultiplication Δ^{∗} of P^{∗} is determined by the formula

Δ^{∗}(M[S]) = ∑_{S′+S′′=S} M[S] ⊗ M[S′′]

([Mi]). From this and the definition of stripping one easily sees that

ξ[R] ∩ M[S] = M[S − R] .

In particular, stripping does not increase length.

Determining the effect of D(ξ[R]) on a given admissible monomial is more involved. More generally, let P(a_{1}) ⋯ P(a_{n}) be any (not necessarily admissible) monomial in P^{∗}. For n ⩾ k, we define V_{n,k} to be the set of all sequences (v_{1}, …, v_{n}) in which the non-zero elements form exactly the subsequence (p^{k−1}, …, p, 1). For example, V_{3,2} consists of (0, p, 1), (p, 0, 1), and (p, 1, 0). For n < k, we define V_{n,k} := ∅.

Proposition 3.2. With this notation

ξ_{k} ∩ (P(a_{1}) · . . . · P(a_{n})) = ∑_{V ∈V_{n,k}} P(a_{1} − v_{1}) · . . . · P(a_{n} − v_{n}) .

Proof. The proof is analogous to that of Prop. 3.1 in [S3]. Alternatively, see [CWW, Section 2]. □

We note the following consequences:

Corollary 3.3. 1. If θ ∈ P^{∗} has length n, then ξ[S]∩θ = 0 for any S ∈ S of length greater

than n; in particular ξ_{k} ∩ θ = 0 for any k > n .

2. If P(a_{1}) · . . . · P(a_{k}) is admissible of excess 2e, then

ξ_{k} ∩ (P(a_{1}) · . . . · P(a_{k})) = P(a_{1} − p^{k−1}) · P(a_{2} − p^{k−2}) · . . . · P(a_{k} − 1) ,

which is again admissible and has excess 2e − 2. Consequently, if R = (r_{1}, …, r_{k}) ∈ S, then ξ_{k} ∩ E[R] = E[(r_{1}, …, r_{k−1}, r_{k} − 1)].

3. In particular,

ξ_{k} ∩ P[k; f] = P[k; f − 1] and ξ̂_{k} ∩ P̂[k; f] = P̂[k; f − 1] ,

where the second equation follows from Prop. 3.1(5).

□

The next thing we determine is the action of D(ξ̂[R]) on a given element θ ∈ P^{∗}. By [Mi], conjugation in P_{∗} is determined by

ξ̂_{k} = ∑_{α∈Part(k)}(−1)^{l(α)} ∏_{i=1}^{l(α)} ξ_{α_{i}}^{p^{σ_{i}(α)}}  (2)

where α runs through all ordered partitions α = (α_{1}, α_{2}, …, α_{l(α)}) of k, l(α) is the length of the partition α, and σ_{i}(α) is the partial sum ∑_{j=1}^{i−1} α_{j}.

Consequences 3.4. 1. The excess of ξ_{k} = ξ[B_{k}] is 2 for any k, so the summand with the largest excess in formula (2) is the monomial corresponding to the partition α of length l(α) = k with α_{i} = 1 for 1 ⩽ i ⩽ k, i.e. the summand

(−1)^{k} ∏_{i=1}^{k} ξ_{1}^{p^{i−1}} = (−1)^{k}ξ_{1}^{γ(k)}

which has excess 2γ(k). Hence stripping by ξ̂_{k} reduces excess by no more than 2γ(k).

2. Since ξ_{l} ∩ P(f) = 0 for all l > 1, we have

ξ̂_{k} ∩ P(f) = (−1)^{k}ξ_{1}^{γ(k)} ∩ P(f)
= \{ \begin{matrix}(−1)^{k}P(f − γ(k)) & if f ⩾ γ(k) \\ 0 & otherwise. \end{matrix}

3.3. Stripping P[Λ; f] by ξ̂_{k}^{j}

We will be mostly concerned with the special Steenrod operations P[Λ; f]. Therefore we take a closer look at the action of the stripping operations D(ξ̂_{k}^{j}) on these elements.

Lemma 3.5. For any θ in P^{∗} we have

ξ̂_{k} ∩ (P[2; f] · θ) = P(pf) · (ξ̂_{k} ∩ (P(f) · θ)) .

Proof. The proof is analogous to the proof of Lemma 4.4 in [S2]: recall that the comultiplication in P_{∗} is given by

ϕ_{∗}(ξ_{k}) = ∑_{j=0}^{k} ξ_{k−j}^{p^{j}} ⊗ ξ_{j}  (3)

([Mi]). Hence by Prop. 3.1(3) we obtain

ξ̂_{k} ∩ (P[2; f] · θ) = P(pf) · (ξ̂_{k} ∩ (P(f) · θ)) + ∑_{j=1}^{k} (ξ̂_{j} ∩ P(pf)) · (ξ̂_{k−j}^{p^{j}} ∩ (P(f) · θ)) .

Cons. 3.4(2) implies that ξ̂_{j} ∩ P(pf) = −ξ̂_{j−1}^{p} ∩ P(pf − 1), thus

∑_{j=1}^{k} (ξ̂_{j} ∩ P(pf)) · (ξ̂_{k−j}^{p^{j}} ∩ (P(f) · θ))

= − ∑_{j=1}^{k} (ξ̂_{j−1}^{p} ∩ P(pf − 1)) · ((ξ̂_{(k−1)−(j−1)}^{p})^{p^{j−1}} ∩ (P(f) · θ))
= −ξ̂_{k−1}^{p} ∩ (P(pf − 1) · P(f) · θ) .

But by the Adem relations P(pf − 1) · P(f) = 0, which proves the claim. □

The following more general result is now easily proved by induction on Λ, the case Λ = 2 being given by Lemma 3.5:

Proposition 3.6. For Λ ⩾ 2 and any θ in P^{∗} we have

ξ̂_{k} ∩ (P[Λ; f] · θ) = P[Λ − 1; pf] · (ξ̂_{k} ∩ (P(f) · θ)) .
□

Finally, we investigate what happens if we strip P[Λ; f] by ξ̂_{k} a total of j times. We note that only the right-most j places are affected:

Proposition 3.7. Suppose Λ > j ⩾ 1. Then

ξ̂_{k}^{j} ∩ P[Λ; f] = P[Λ − j; p^{j}f] · (ξ̂_{k}^{j} ∩ P[j; f]) .

Proof. The proof is by induction on j, starting with j = 1 where the result is provided by Prop. 3.6. □

4. Conjugation formulas for P^{∗}

In this section we establish some useful formulas involving conjugation of elements in P^{∗}. In particular, we determine the simple formula for P̂[s; cγ(t)] with 1 ⩽ c ⩽ p that was announced in the introduction.

Suppose that y is a non-negative integer. We use the notation α_{i}(y) for the coefficient of p^{i} in the p-adic expansion of y, i.e. y =: ∑_{i⩾0} α_{i}(y)p^{i}.

The following lemma will be needed for the proof of Prop. 4.3.

Lemma 4.1. Suppose that k, l, c, m and e are non-negative integers with

1. k > l,

2. 1 ⩽ c ⩽ p − 1,

3. m < p^{k−1},

4. m ≡ 0 mod p^{l}.

Then the following relation mod p holds:

\binom{c(p^{k} − p^{l}) + e}{pm} ≡ − ∑_{i=1}^{c} \binom{c}{i}\binom{c(p^{k} − p^{l}) + e}{pm + ip^{l}} + \binom{e}{pm + cp^{l}}  (4)

Proof. The proof relies on the fact that mod p we have the relation \binom{x}{y} ≡ ∏_{i⩾0} \binom{α_{i}(x)}{α_{i}(y)}. There

are three cases: (I) α_{l}(e) = c, (II) 0 ⩽ α_{l}(e) ⩽ c − 1, and (III) c + 1 ⩽ α_{l}(e) ⩽ p − 1. If we are in case (I) then the first term on the right of (4) is 0 and

\binom{c(p^{k} − p^{l}) + e}{pm} ≡ \binom{e}{pm + cp^{l}}

as required.

If we are in case (II) then the second term on the right of (4) is zero and so we have to show that

\binom{c(p^{k} − p^{l}) + e}{pm} ≡ − ∑_{i=1}^{c} \binom{c}{i}\binom{c(p^{k} − p^{l}) + e}{pm + ip^{l}} ,

i.e. that

1 ≡ − ∑_{i=1}^{c} \binom{c}{i}\binom{p − c + α_{l}(e)}{i}

for 0 ⩽ α_{l}(e) ⩽ c − 1. Setting a := p − c + α_{l}(e) this amounts to showing that

∑_{i=0}^{c} \binom{c}{i}\binom{a}{i} ≡ 0

for all p − c ⩽ a ⩽ p − 1. In order to show this equivalence, note that

∑_{i=0}^{c} \binom{c}{i}\binom{a}{i} ≡ ∑_{i=0}^{c} \binom{c}{i}\binom{a}{a − i} ≡ \binom{c + a}{c}  (5)

as one sees by considering the coefficient of x^{c} in the binomial expansion of (x + 1)^{c+a} = (x + 1)^{c}(x + 1)^{a}. Now the claim follows since \binom{c+a}{c} ≡ 0 for p − c ⩽ a ⩽ p − 1.

Case (III) is similar. □

We will need the following multiplication formulas:

Lemma 4.2. Let u and v be non-negative integers. Then

P(u) · P̂(v) = (−1)^{v} ∑_{R} \binom{|R| + ex(R)}{pu}_{p} M[R]  (6)

and

P̂(u) · P(v) = (−1)^{u} ∑_{R} \binom{ex(R)}{v}_{p} M[R]  (7)

where the sum ranges over all sequences R in S with |R| = (p − 1)(u + v) and ( )_{p} denotes mod p binomial coefficients.

Proof. The proof of (6) can be found in [G]. The other equality, (7), can be extracted from [Ka1]. □

Remark. In [Ka1], our Lemma 4.2 is stated (wrongly) without any minus signs. Unfortunately, Karaca does not explicitly say what his definition of P̂(u) is. Instead, for the special Milnor basis elements M[(0, …, 0, r_{t} = p^{s})] =: P_{t}^{s} he defines P̂_{t}^{s} as (−1)^{s}χ(P_{t}^{s}). Since there exists a basis of P^{∗} which consists of certain monomials in elements of the form P_{t}^{s}, it is possible to figure out what the expression P̂(u) should mean according to Karaca’s definition, assuming that { \atop P_{t}^{s} · P_{u}^{v}} := P̂_{u}^{v} · P̂_{t}^{s}. However, doing this translation one easily sees that there should be some non-trivial coefficients in his formula. The correct result can nevertheless easily be deduced from the argument given in [Ka1].

After these preparations we are in a position to prove the following “hat-passing formula”, which is a slightly generalised odd prime version of the formula given in [S1, Lemma 2.3]:

Proposition 4.3. Suppose that k, l, c, m and n are non-negative integers with 1. k > l,

2. 1 ⩽ c ⩽ p − 1,

3. m + n = cp^{l}γ(k − l),

4. m < p^{k−1},

5. m ≡ 0 mod p^{l}.

We use the convention P̂(s) := 0 if s < 0. Then for l = 0 we have

P(m) · P̂(n) = (−1)^{c}P̂(m + n − pm − c) · P(pm + c)

and for l > 0 we have

P(m) · P̂(n) = ∑_{i=1}^{c}(−1)^{i+1}\binom{c}{i}_{p}P(m + ip^{l−1}) · P̂(n − ip^{l−1})
+ (−1)^{c}P̂(m + n − pm − cp^{l}) · P(pm + cp^{l}) .

Proof. In order to see that for l = 0 only one term in the expression for P(m) · P̂(n) appears, note that |R| = (p − 1)cγ(k) = c(p^{k} − 1), so that by applying Equation (6) in Lemma 4.2 we obtain

P(m) · P̂(n) = (−1)^{n} ∑_{|R|=c(p^{k}−1)} \binom{c(p^{k} − 1) + ex(R)}{pm}_{p}M[R] .

Now recall that ex(R) = ∑_{i⩾1} r_{i}. Dividing |R| by (p − 1) and substituting ex(R) − ∑_{i⩾2} r_{i} for r_{1}

we have

cγ(k) = \frac{|R|}{p − 1} = ∑_{i⩾1} r_{i}γ(i) = ex(R) + ∑_{i⩾2} r_{i}pγ(i − 1) .

Thus we see that ex(R) ≡ c mod p. Now we apply Lemma 4.1 with e = ex(R); we have just seen that we are always in case (I) so that

\binom{c(p^{k} − 1) + ex(R)}{pm} ≡ \binom{ex(R)}{pm + c} .

Equation (7) in Lemma 4.2 now implies that

P(m) · P̂(n) = (−1)^{n} ∑_{|R|=c(p^{k}−1)} \binom{c(p^{k} − 1) + ex(R)}{pm}_{p} M[R]

= (−1)^{c}(−1)^{m+n−pm−c} ∑_{|R|=c(p^{k}−1)} \binom{ex(R)}{pm + c}_{p} M[R]
= (−1)^{c}P̂(m + n − pm − c) · P(pm + c) .

The formula for l > 0 easily follows from Lemma 4.1, carefully keeping track of any minus signs that enter into the formula. □

In order to arrive at the simple description of P̂[s; cγ(t)] that will be obtained in Theorem 4.6 we need yet another lemma. The elegant proof given here, due to Judith Silverman, is a nice application of the “stripping technique” discussed in Section 3 and replaces the original, more complicated proof which didn’t use stripping at all.

Lemma 4.4. Let c and l be positive integers with 1 ⩽ c ⩽ p − 1. Then P(cγ(l)) · P(ap^{l−1}) = 0 for any a which satisfies p − c ⩽ α_{0}(a) ⩽ p − 1.

Proof. The lemma is proved by downward induction on c. We start with the case c = p − 1 so that 1 ⩽ α_{0}(a) ⩽ p − 1. Then by the Adem relations we have

P(p^{l} − 1) · P(ap^{l−1})

= ∑_{z=0}^{p^{l−1}}(−1)^{p^{l}−1+z}\binom{(p − 1)(ap^{l−1} − z) − 1}{p^{l} − 1 − pz}_{p}P(p^{l} − 1 + ap^{l−1} − z) · P(z) .

We show that the mod p binomial coefficients appearing in this formula are all 0. First consider the case z = 0: since 1 ⩽ α_{0}(a) ⩽ p − 1 we have 0 ⩽ α_{l−1}((p − 1)ap^{l−1} − 1) ⩽ p − 2, but α_{l−1}(p^{l} − 1) = p − 1 and so \binom{(p−1)ap^{l−1}−1}{p^{l}−1} ≡ 0. On the other hand, if z ≠ 0 then there exists some index j_{0} with 0 ⩽ j_{0} ⩽ l − 2 such that 1 ⩽ z_{j_{0}} ⩽ p − 1 but z_{j} = 0 for all 0 ⩽ j < j_{0}. Hence 1 ⩽ α_{j_{0}}((p − 1)z) = p − z_{j_{0}} ⩽ p − 1 and so 0 ⩽ α_{j_{0}}((p − 1)(ap^{l−1} − z) − 1) ⩽ p − 2. But α_{j_{0}}(p^{l} − 1 − pz) = p − 1 and so again \binom{(p−1)(ap^{l−1}−z)−1}{p^{l}−1−pz} ≡ 0.

Now let 1 ⩽ c < p − 1 and suppose that the lemma has been shown to be true for all ĉ with c < ĉ ⩽ p−1. Choose a with p−c ⩽ α_{0}(a) ⩽ p−1 (which implies p−(c+1) ⩽ α_{0}(a−1) ⩽ p−1 and p − (c + 1) ⩽ α_{0}(a) ⩽ p − 1). The lemma for c + 1 guarantees that

P((c + 1)γ(l)) · P(ap^{l−1}) = 0  (8)

and

P((c + 1)γ(l)) · P((a − 1)p^{l−1}) = 0 .  (9)

Using Equation (3), Prop. 3.1(4) and Cons. 3.4(2) we strip Equation (8) by ξ̂_{l} to obtain

0 = ξ̂_{l} ∩ [P((c + 1)γ(l)) · P(ap^{l−1})]
= [ξ̂_{l} ∩ P((c + 1)γ(l))] · P(ap^{l−1})
+ ∑_{i=0}^{l−1} [ξ̂_{i} ∩ P((c + 1)γ(l))] · [ξ̂_{l−i}^{p^{i}} ∩ P(ap^{l−1})]  (10)
= (−1)^{l}P(cγ(l)) · P(ap^{l−1}) + E,

where E is defined to be the big sum in (10). It remains to show that E = 0. We fix i with 1 ⩽ i ⩽ l − 1 and observe that for any b ⩾ 0 we have

ξ̂_{l−i}^{p^{i}} ∩ P(b) = (−1)^{l−i}P(b − p^{i}γ(l − i)) = −ξ̂_{l−i−1}^{p^{i}} ∩ P(b − p^{l−1}) .

Setting b = ap^{l−1}, we find that E can be rewritten as

E = − ∑_{i=0}^{l−1} [ξ̂_{i} ∩ P((c + 1)γ(l))] · [ξ̂_{l−i−1}^{p^{i}} ∩ P((a − 1)p^{l−1})]
= −ξ̂_{l−1} ∩ [P((c + 1)γ(l)) · P((a − 1)p^{l−1})] .  (11)

But by (9), the product in (11) is 0. Consequently E = 0 as desired. □

The next lemma establishes the basis of induction for Theorem 4.6.

Lemma 4.5. Let c be an integer with 1 ⩽ c ⩽ p − 1. Then

P̂(cγ(s)) = (−1)^{sc}P[s; c] .

Proof. The case s = 1 is clear: by [Mi] we have

P̂(c) = (−1)^{c} ∑_{|Q|=c(p−1)} M[Q] = (−1)^{c}P(c) ,

and in general

P̂(cγ(s)) = (−1)^{cγ(s)} ∑_{|Q|=c(p^{s}−1)} M[Q] .  (12)

By induction and Equation (6) we obtain

(−1)^{sc}P[s; c] = (−1)^{sc}P(p^{s−1}c) · P[s − 1; c]
= (−1)^{c}P(p^{s−1}c) · P̂(cγ(s − 1))
= (−1)^{cγ(s)} ∑_{|R|=c(p^{s}−1)} \binom{|R| + ex(R)}{cp^{s}}_{p} M[R] ,

so that by (12) it only remains to show that \binom{|R|+ex(R)}{cp^{s}} ≡ 1 for all R with |R| = c(p^{s} − 1).

It follows directly from the definitions that 0 ⩽ ex(R) ⩽ \frac{|R|}{p−1} = cγ(s). On the other hand it is easy to see that the sequence (0, ⋯, 0, r_{s} = c) is of excess c and that this is the minimal excess of any sequence in S of degree c(p^{s} − 1). The inequality c ⩽ ex(R) ⩽ cγ(s) now implies that

cp^{s} ⩽ |R| + ex(R) ⩽ cpγ(s) = cp^{s} + cp^{s−1} + … + cp

so that indeed \binom{|R|+ex(R)}{cp^{s}} ≡ 1 for all R with |R| = c(p^{s} − 1). □

Finally we can prove the conjugation formula announced earlier on, which is a slightly generalised mod p version of [S1, Theorem 3.1]. The proof is similar to the one in the mod 2 case.

Theorem 4.6. For all positive integers s, t and c with 1 ⩽ c ⩽ p the following conjugation formula holds:

P̂[s; cγ(t)] = (−1)^{stc}P[t; cγ(s)]

Proof. We first prove the theorem for 1 ⩽ c ⩽ p − 1. The case c = p will follow from the case c = 1 by a stripping argument.

The proof for 1 ⩽ c ⩽ p − 1 is by induction on t. The basis of induction (i.e. the case t = 1 or equivalently s = 1) has been established in Lemma 4.5. So let us assume that t > 1, s > 1 and that the theorem has been shown to be true for all 1 ⩽ t̂⩽ t − 1, all s and also for t̂= t, all 1 ⩽ ŝ ⩽ s − 1. We begin with the following remark: Remark. Under the above assumptions the following is true: For all non-negative integers a with p − c ⩽ α_{0}(a) ⩽ p − 1 and for all 1 ⩽ l < s we have

P̂(ap^{l−1}) · P[l; cγ(t)] = 0 .

We prove this result as follows: we have

P̂(ap^{l−1}) · P[l; cγ(t)] = χ[P̂[l; cγ(t)] · P(ap^{l−1})] ,

which by induction equals

(−1)^{tlc}χ[P[t; cγ(l)] · P(ap^{l−1})] = (−1)^{tlc}χ[P[t − 1; pcγ(l))] · P(cγ(l)) · P(ap^{l−1})] .

But by Lemma 4.4 the expression P(cγ(l)) · P(ap^{l−1}) vanishes. This proves the remark. Now we get back to the proof of the theorem: by induction we obtain

P̂[t; cγ(s)] = χ(P[t − 1; cγ(s)]) · χ(P(p^{t−1}cγ(s)))
= (−1)^{(t−1)sc}P[s; cγ(t − 1)] · P̂(p^{t−1}cγ(s)) .  (13)

We claim that for 1 ⩽ d ⩽ s the following formula holds:

P[d; cγ(t − 1)] · P̂(p^{t−1}cγ(s)) = (−1)^{dc}P̂(p^{t+d−1}cγ(s − d)) · P[d; cγ(t)]

Proof of the claim: for d = 1 we have to show that

P(cγ(t − 1)) · P̂(p^{t−1}cγ(s)) = (−1)^{c}P̂(p^{t}cγ(s − 1)) · P(cγ(t)) .

This follows immediately from Prop. 4.3 with m = cγ(t − 1), n = p^{t−1}cγ(s), k = t + s − 1 and l = 0. So suppose that 2 ⩽ d ⩽ s, assuming that the claim has been proved for all 1 ⩽ d̂ < d. Then using induction we obtain

P[d; cγ(t − 1)] · P̂(p^{t−1}cγ(s))
= P(p^{d−1}cγ(t − 1)) · P[d − 1; cγ(t − 1)] · P̂(p^{t−1}cγ(s))
= (−1)^{(d−1)c}P(p^{d−1}cγ(t − 1)) · P̂(p^{t+d−2}cγ(s − d + 1)) · P [d − 1; cγ(t)] .

Again, we apply Prop. 4.3, this time to the first two terms, with the parameters m = p^{d−1}cγ(t− 1), n = p^{t+d−2}cγ(s − d + 1), k = t + s − 1 and l = d − 1. We deduce that

P(p^{d−1}cγ(t − 1)) · P̂(p^{t+d−2}cγ(s − d + 1))
= ∑_{i=1}^{c}(−1)^{i+1}\binom{c}{i}_{p}P(p^{d−1}cγ(t − 1) + ip^{d−2}) · P̂((p^{t+d−2}cγ(s − d + 1) − ip^{d−2})
+(−1)^{c}P̂(p^{t+d−1}cγ(s − d)) · P(p^{d−1}cγ(t)) .

By the remark, the terms in the big sum vanish upon multiplication with P[d − 1; cγ(t)] from the right, and so we arrive at

P[d; cγ(t − 1)] · P̂(p^{t−1}cγ(s))
= (−1)^{dc}P̂(p^{t+d−1}cγ(s − d)) · P(p^{d−1}cγ(t)) · P[d − 1; cγ(t)]
= (−1)^{dc}P̂(p^{t+d−1}cγ(s − d)) · P[d; cγ(t)]

which proves the claim.

Setting d = s and substituting back into expression (13) yields

P̂[t; cγ(s)] = (−1)^{(t−1)sc}P[s; cγ(t − 1)] · P̂(p^{t−1}cγ(s))
= (−1)^{tsc}P[s; cγ(t)]

which finishes the proof of the theorem for 1 ⩽ c ⩽ p − 1.

There remains the case c = p. We strip the formula

P̂[s; γ(t + 1)] = (−1)^{s(t+1)}P[t + 1; γ(s)]

(this is the case c = 1 with t + 1 instead of t) by ξ̂_{s}, and by Cor. 3.3(3) we obtain

P̂[s; pγ(t)] = ξ̂_{s} ∩ P̂[s; γ(t + 1)]
= (−1)^{s(t+1)}ξ̂_{s} ∩ P[t + 1; γ(s)]

which by Cons. 3.4(2) and Prop. 3.6 equals

(−1)^{s(t+1)}P[t; pγ(s)] · (ξ̂_{s} ∩ P(γ(s))) = (−1)^{s(t+1)}P[t; pγ(s)] · (−1)^{s}P(0)
= (−1)^{st}P[t; pγ(s)] .

This completes the proof of the theorem. □

We observe the following:

Corollary 4.7. Let s, t and c be non-negative integers with s ⩾ 1 and 1 ⩽ c ⩽ p. Then the operations P̂[s; cγ(t)] have length exactly t independently of s and c. More generally, if γ(t) ⩽ f < γ(t + 1) then the operations P̂[s; f] are all of length exactly t, independently of s.

Proof. For t ⩾ 1 the first statement is an immediate consequence of Theorem 4.6; for t = 0 the statement is trivial. The second statement follows since stripping operations cannot increase length (cf. Section 3.2). □

5. Conjugation formulas for P_{∗}

We now turn to conjugation in the dual Steenrod algebra. Let 𝔖(k) be the symmetric group with identity Id_{k} acting on {0, 1, 2, …, k − 1}. For τ ∈ 𝔖(k) and i ⩾ 0 we define

Z_{i}(k; τ) := ∑_{j=0}^{k−1} p^{j}B(i + τ(j) − j) ,

X_{i}(k; τ) := ξ[Z_{i}(k; τ)] = ∏_{j=0}^{k−1} ξ_{i+τ(j)−j}^{p^{j}} ,

and

X_{i}(k) := ∑_{τ∈𝔖(k)}sign(τ) X_{i}(k; τ) .

Observation 5.1. Z_{i}(k; Id_{k}) = γ(k)B(i) and X_{i}(k; Id_{k}) = ξ_{i}^{γ(k)}.

We will need the following lemma:

Lemma 5.2. For k ⩾ 1 we have X_{1}(k) = (−1)^{k}ξ̂_{k}.

Proof. The proof is by induction on k. Let k = 1, then X_{1}(1) = ξ_{1} = −ξ̂_{1}, so the assertion is true in this case. Now suppose the statement has been shown to be true for all 1 ⩽ k̂ < k. Note that if X_{1}(k; τ) ≠ 0 then necessarily τ(j) ⩾ j − 1 for all j. So if X_{1}(k; τ) ≠ 0 then define l by l = τ^{−1}k − 1. If l = k − 1 then we obtain a cycle decomposition of τ as (k − 1)σ for some σ ∈ 𝔖(k − 1). If l ≠ k − 1 then we obtain τ(k − 1) = k − 2, τ(k − 2) = k − 3, …, τ(l + 1) = l, so that τ has a cycle decomposition as (k − 1, k − 2, …, l)σ for some σ ∈ 𝔖(l). In any case we have

X_{1}(k; τ) = X_{1}(l; σ) · ξ_{k−l}^{p^{l}} .

So for 0 ⩽ l ⩽ k − 1 let 𝔖_{l}(k) = {τ ∈ 𝔖(k) | τ(l) = k − 1}; obviously 𝔖(k) = ⋃ 𝔖_{l}(k). Then by induction

X_{1}(k) = ∑_{l=0}^{k−1} ∑_{τ∈𝔖_{l}(k)}sign(τ) X_{1}(k; τ)

= ∑_{l=0}^{k−1} ξ_{k−l}^{p^{l}} · ∑_{σ∈𝔖(l)}(−1)^{k−1−l}sign(σ) X_{1}(l; σ)

= (−1)^{k−1} ∑_{l=0}^{k−1} ξ_{k−l}^{p^{l}} · ξ̂_{l} = (−1)^{k}ξ̂_{k} ,

where in the last line we used Milnor’s recursive formula for the anti-automorphism. □

In analogy to [S3] we make the following more general definitions:

Definition 5.3. For k ⩾ 1, let I(k) be the set of non-decreasing sequences (i_{0}, i_{1}, …, i_{k−1}) of positive integers. For τ ∈ 𝔖(k) and I ∈ I(k) we define

Z_{I}(k; τ) := ∑_{j=0}^{k−1} p^{j}B(i_{τ(j)} + τ(j) − j) ,

X_{I}(k; τ) := ξ[Z_{I}(k; τ)] = ∏_{j=0}^{k−1} ξ_{i_{τ(j)}+τ(j)−j}^{p^{j}} ,

Homology, Homotopy and Applications, vol. 2, No. 1, 2000 13 and

X_{I}(k) := ∑_{τ∈𝔖(k)}sign(τ) X_{I}(k; τ) .

We further define

P_{I}(k; τ) := ∑_{j=0}^{k−1} p^{j+i_{0}}B(i_{τ(j)} + τ(j) − (j + i_{0})) ,
R_{I}(k; τ) := ξ[P_{I}(k; τ)] = ∏_{j=0}^{k−1} ξ_{i_{τ(j)}+τ(j)−(j+i_{0})}^{p^{j+i_{0}}} ,

and

R_{I}(k) := ∑_{τ∈𝔖(k)}sign(τ) R_{I}(k; τ) .

Observations 5.4. 1. If I = (i, i, …, i) ∈ I(k) is a constant sequence then we obtain Z_{I}(k; τ) = Z_{i}(k; τ) and consequently X_{I}(k; τ) = X_{i}(k; τ). Moreover, for such a sequence I and τ ≠ Id_{k} we have P_{I}(k; τ) = ∗ and consequently R_{I}(k) = R_{I}(k; Id_{k}) = 1.

2. If I = (i_{0}, i_{1}, …, i_{k−1}) ∈ I(k) and i_{0} > 1 let I[−1] denote the sequence (i_{0} − 1, i_{1} −

1, …, i_{k−1} − 1) ∈ I(k). Then R_{I}(k) = (R_{I[−1]}(k))^{p}. Theorem 5.5. Let k ⩾ 1. Then X̂_{I}(k) ≡ (−1)^{i_{0}k}ξ_{k}^{γ(i_{0})} · R̂_{I}(k) modulo monomials of length > k. Proof. First recall that we have the following expression for X̂_{I}(k):

X̂_{I}(k) = ∑_{ρ∈𝔖(k)}sign(ρ) ∏_{j=0}^{k−1} ξ̂_{i_{ρ(j)}+ρ(j)−j}^{p^{j}}
= ∑_{ρ∈𝔖(k)}sign(ρ) ξ̂_{i_{ρ(0)}+ρ(0)} · ∏_{j=1}^{k−1} ξ̂_{i_{ρ(j)}+ρ(j)−j}^{p^{j}} .

Applying Milnor’s recursive formula for the anti-automorphism we obtain

−ξ̂_{i_{ρ(0)}+ρ(0)} ≡ ∑_{n=1}^{k} ξ_{n} · ξ̂_{i_{ρ(0)}+ρ(0)−n}^{p^{n}}

modulo monomials of length > k. So we have

X̂_{I}(k) ≡ − ∑_{n=1}^{k} ∑_{ρ∈𝔖(k)}sign(ρ) ξ_{n} · ξ̂_{i_{ρ(0)}+ρ(0)−n}^{p^{n}} · ∏_{j=1}^{k−1} ξ̂_{i_{ρ(j)}+ρ(j)−j}^{p^{j}} .

For each ρ ∈ 𝔖(k) we define ρ′ by

ρ′(l) = \{ \begin{matrix}ρ(0) & if l = k − 1 \\ ρ(l + 1) & if 0 ⩽ l ⩽ k − 2. \end{matrix}

Note that sign(ρ) = (−1)^{k−1}sign(ρ′). So

X̂_{I}(k) ≡ (−1)^{k} ∑_{n=1}^{k} ∑_{ρ′∈𝔖(k)}sign(ρ′) ξ_{n} · ξ̂_{i_{ρ′(k−1)}+ρ′(k−1)−n}^{p^{n}} · ∏_{l=0}^{k−2} ξ̂_{i_{ρ′(l)}+ρ′(l)−(l+1)}^{p^{l+1}}

modulo monomials of length > k.

For the proof of the theorem, we fix k and use induction on i_{0}. First suppose that i_{0} = 1. Then

ξ_{k} · R̂_{I}(k) = ∑_{τ∈𝔖(k)}sign(τ) ξ_{k} · ξ̂_{i_{τ(k−1)}+τ(k−1)−k}^{p^{k}} · ∏_{j=0}^{k−2} ξ̂_{i_{τ(j)}+τ(j)−(j+1)}^{p^{j+1}}

so that

X̂_{I}(k) − (−1)^{k}ξ_{k} · R̂_{I}(k)

≡ (−1)^{k} ∑_{n=1}^{k−1} ∑_{ρ′∈𝔖(k)}sign(ρ′) ξ_{n} · ξ̂_{i_{ρ′(k−1)}+ρ′(k−1)−n}^{p^{n}} · ∏_{l=0}^{k−2} ξ̂_{i_{ρ′(l)}+ρ′(l)−(l+1)}^{p^{l+1}} .  (14)

It can easily be verified that the summand in (14) associated to n and ρ′ is the negative of the term associated to n and ρ′′ where

ρ′′(l) = \{ \begin{matrix}ρ′(l) & if l ≠ n − 1 and l ≠ k − 1 \\ ρ′(n − 1) & if l = k − 1 \\ ρ′(k − 1) & if l = n − 1 \end{matrix}

(note that sign(ρ′) = −sign(ρ′′)). So the difference X̂_{I}(k) − (−1)^{k}ξ_{k} · R̂_{I}(k) vanishes modulo monomials of length > k and the theorem holds for i_{0} = 1.

The proof for general I is similar. By induction we can assume that the statement is true for (i_{0} − 1, i_{1} − 1, …, i_{k} − 1) = I[−1]. By Observation 5.4(2)

ξ_{k}^{γ(i_{0})} · R̂_{I}(k) = (ξ_{k}^{γ(i_{0}−1)} · R̂_{I[−1]}(k))^{p} · ξ_{k}

which modulo terms of length > k is

≡ ((−1)^{k(i_{0}−1)}X̂_{I[−1]}(k))^{p} · ξ_{k}
= (−1)^{k(i_{0}−1)}ξ_{k} · ∑_{τ∈𝔖(k)}sign(τ) ∏_{j=0}^{k−1} ξ̂_{i_{τ(j)}−1+τ(j)−j}^{p^{j+1}}

= (−1)^{k(i_{0}−1)} ∑_{τ∈𝔖(k)}sign(τ) ξ_{k} · ξ̂_{i_{τ(k−1)}+τ(k−1)−k}^{p^{k}} · ∏_{j=0}^{k−2} ξ̂_{i_{τ(j)}+τ(j)−(j+1)}^{p^{j+1}} .

Now one can define ρ′′ as before and proceed as in the case i_{0} = 1 in order to establish the inductive step. □

An especially interesting formula arises from Theorem 5.5 if we set I = (i, i, …, i), a constant sequence:

Theorem 5.6. Let i, k > 0. Modulo monomials of length > k we have

ξ̂_{i}^{γ(k)} ≡ (−1)^{ik}ξ_{k}^{γ(i)} − ∑_{Id_{k}≠τ∈𝔖(k)}sign(τ) ∏_{j=0}^{k−1} ξ̂_{i+τ(j)−j}^{p^{j}} .

In particular, if 0 ⩽ f < γ(k + 1) then

ξ̂_{k}^{γ(i)} ∩ P[i; f] = (−1)^{ik}ξ_{i}^{γ(k)} ∩ P[i; f] = (−1)^{ik}P[i; f − γ(k)] .

Proof. The first part follows immediately from Theorem 5.5 and Observation 5.4(1), so it only remains to prove the second statement. By the part already proved we have the following equality:

ξ̂_{i}^{γ(k)} ∩ P̂[i; f] = (−1)^{ik}ξ_{k}^{γ(i)} ∩ P̂[i; f] − ( ∑_{Id_{k}≠τ∈𝔖(k)}sign(τ) ∏_{j=0}^{k−1} ξ̂_{i+τ(j)−j}^{p^{j}}) ∩ P̂[i; f]

Now observe that for any Id_{k} ≠ τ ∈ 𝔖(k) the product ∏_{j=0}^{k−1} ξ_{i+τ(j)−j}^{p^{j}} is of length strictly greater than i, so for any such τ we get

( ∏_{j=0}^{k−1} ξ̂_{i+τ(j)−j}^{p^{j}}) ∩ P̂[i; f] = χ[( ∏_{j=0}^{k−1} ξ_{s+τ(j)−j}^{p^{j}}) ∩ P[i; f]] = 0 .

Using Cor. 3.3(3) we thus obtain ξ̂_{i}^{γ(k)} ∩ P̂[i; f] = P̂[i; f − γ(k)] = (−1)^{ik}ξ_{k}^{γ(i)} ∩ P̂[i; f]. The claim now follows by application of (−1)^{ik}χ to this formula. □

Finally, we note that Theorem 5.5 provides us with useful information regarding the behaviour of the stripping operations D(X̂_{I}(k)):

Corollary 5.7. 1. If len(θ) < k, then X̂_{I}(k) ∩ θ = 0 for all I ∈ I(k).

2. If len(θ) = k, then X̂_{I}(k) ∩ θ = (−1)^{i_{0}k}R̂_{I}(k) ∩ (ξ_{k}^{γ(i_{0})} ∩ θ).

3. In particular, X̂_{I}(k) ∩ P[k; f] = (−1)^{i_{0}k}R̂_{I}(k) ∩ P[k; f − γ(i_{0})].

Proof. This follows immediately from the theorem by invoking Prop. 3.1 and Cor. 3.3. □

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Dagmar M. Meyer dagmar@math.univ-paris13.fr  meyerd@member.ams.org

LAGA, Institut Galilée, Univ. Paris-Nord