LAMBDA ALGEBRA UNSTABLE COMPOSITION PRODUCTS
AND THE EHP SEQUENCE
WILLIAM RICHTER
Abstract. Simple combinatorial proofs are given of various Lambda
algebra results, mostly due to the MIT school [B-C-K& , Cu1, Pr], but
also the unstable composition formulas of Wang, Mahowald and Singer,
which imply the folklore EHP sequence.
1. Introduction
Mahowald [Ma1 , Ma2 ] initiated a öl w-techä pproach to the unstable
Adams spectral sequence, using a purely algebraic treatment of the Lambda
algebra , and ad-hoc tower constructions. However, full details have not
yet appeared. A few such details, combinatorial proofs, are given here.
The power of Mahowald's approach is shown by his [Ma1 ] EHP se-
quence chain-level map P : (2n + 1) ! (n), defined by composing with
d(~n) 2 2,n+1(n). But the geometric analogue P : 2S2n+1 ! Sn is only
composition with the Whitehead product ['n, 'n] under the double suspen-
sion. His computation [Ma1 , Prop. 3.1] of the Hopf invariant of P is the
the analogue of the author's result [Ri1 ]. Mahowald's P uses Ä dams-
filtration betterü nstable compositions, due to Wang [Wa ], and codified
(without proof) by Singer [Si]. Singer's formulas are proved here, first:
Proposition 1.1 (Singer). Composition in restricts to an unstable com-
position pairing, written as a cup product:
s,t(n) (n + t)! (n),
ff fiØ//_ff ^ fi.
Singer's result follows by easy induction from the special case s = 1
of Mahowald [Ma2 , Lem. 3.5], or Wang's [Wa , Lem. 1.8.1] special case
involving 1,*(n + t). Wang deduced [Wa , Thm. 1.8.4] the MIT school's
result [B-C-K& ], that (n) is a subcomplex, and his proof showed the
folklore (see Remark 3.3) result that H : (n + 1) ! (2n + 1) is a chain
map. Curtis [Cu2 ] first stated without proof the EHP sequence, which
any careful reader could've deduced from these papers [Wa , Ma2 , Si]:
____________
1991 Mathematics Subject Classification. 55T15, 55Q40, 55Q25.
Thanks to Paul Burchard for the diagram package, which uses XY-pic arrows.
1
2 WILLIAM RICHTER
Theorem 1.2. There's an exact sequence of complexes and a chain map P
(n) //___//_E (n + 1) _____////_H (2n + 1), (2n + 1) _____//_P (n),
where H and P are defined by H(~nff) = ff and P (ff) = d(~n) ^ ff, for
ff 2 (2a + 1), and H( (n)) = 0. P induces the cohomology boundary.
Bousfield and Kan [B-K ] construct unstable cohomology compositions1,
which they proved are compatible with the geometric compositions:
(1) Hs,t-s (n) H* (n + t - s) ! H* (n),
ßt-s+nSn ß*Sn+t-s ! ßt-s+nSn.
Since the differential d of preserves the t-degree, Proposition 1.1 im-
mediately implies an Ä dams-filtration better" improvement of (1):
Corollary 1.3 (Singer). Unstable composition induces the pairing
Hs,t (n) H* (n + t) ! H* (n).
An EHPss approach to [Wa ] yields the 3-lines and relations on the 4-
lines for H*( (n)). That's basically how Wang (who never mentions H)
proves (cf. [Ko ]) the Adams differential d(hn) = h0h2n-1for n > 3. This
systematization of Wang's work will appear later, as part of the author's
work with Mahowald on 3-cell Poincare complexes and Unell's theorem.
The admissible monomial basis of MIT school [B-C-K& , Pr] is proved
in section 5, verifying Mahowald's conjecture (cf. [Ma1 , p. 78]) that a com-
binatorial proof exists. Also proved (section 4) is the MIT school's related
result, that d is well-defined (i.e. preserves Adem relations). In section 3,
we prove Theorem 1.2, and in section 6, we prove the Mahowald-Singer
Hopf invariant formula, and explain how [Ma1 , Prop. 3.1] motivates [Ri1 ].
In section 7, we reprove Wang's result on the equivalence of the admissible
and symmetric Adem relations, by Tangora's recursion relation.
This paper is part of an investigation of geometric applications of [Ma3 ]
with Mark Mahowald, who I'd like to thank, especially for his guidance on
the basis. Thanks to Paul Goerss for many helpful tutorials about and
the uAss. Thanks to Pete Bousfield for 2 very interesting and encouraging
discussions. Thanks to Halvard Fausk, who listened to an early version of
the paper and encouraged me to write it up. Thanks to Charles Rezk, for
explaining that [Pr] is a purely algebraic treatment, using nothing of the sim-
plicial Lie algebras of [B-C-K& ]. Thank to Stewart Priddy, who explained
that genealogy [C-M ] strongly indicates that unstable Lambda composition
should be in the same order as composition in unstable homotopy groups.
____________
1Actually somewhat less, due to "fringing" problems, which Bousfield says we*
*re later
overcome. Bousfield and Kan work for all spaces, not just spheres, and not actu*
*ally use ,
but a description of H* (n) as an Extgroup in a category of unstable A-modules.
LAMBDA ALGEBRA EHP SEQUENCE 3
2. unstable Lambda algebra composition products
The Lambda algebra is generated by {~i : i 0}, and has relations the
admissible Adem relations
X ` n - k - 1 '
(2) ~p~2p+1+n + ~p+n-k ~2p+1+k, p, n 0.
k 0 k
A monomial ~(a1, . .,.as) is admissible iff ai 2ai-1 for 1 < i s.
Adem relations reduce the right-lexicographical order, while fixing s and
the t-degree a1 + . .+.as + s, so by induction the admissible monomials
span. The MIT school [B-C-K& ] showed the harder fact that has a basis
of admissible monomials, and also that (n) is a subcomplex, where
(n) has the basis of admissible monomials ~I(a1, . .,.as) with a1 < n.
To motivate Proposition 1.1, let's ask how we could construct Bousfield
and Kan's unstable compositions (1)in . Their geometric compatibility
(same order e.g.) shows us we need ~(a1, . .,.as) to belong to (n), for
any sequence (a1, . .,.as) satisfying the inequalities
a1 < n a2 < n + a1 . . .as < n + a1 + . .+.as-1.
And it's not hard to see this is true, because Adem relations preserve these
inequalities, and then we'd have (n) . (n + t - s) (n). But we note:
(1) Left multiplication by ~-1 is more or less d, and this [B-K ] type
unstable composition isn't enough to prove (n) is a subcomplex.
(2) Performing an Adem relation improves the above inequalities.
This leads us to stronger inequalities:
Definition 2.1. APmonomial ~(a1, . .,.as) is called n-pseudo-admissible if
ai < n + i - 1 + j0 k
Then d(~n)J 2 (n) by Proposition 1.1:
2,n+1(n) . (2n + 1) (n)
Now we develop the EHP sequence of the MIT school [Cu1 ]. Note that
this follows from our proof of Corollary 3.1, but not the statement itself.
Clearly the inclusion E : (n) ! (n + 1) is a chain map, so we have
some EHP sequence, but we want to a better grip on the quotient complex
(n + 1)= (n). Recall that the Hopf invariant
H : (n + 1) ! (2n + 1)
is defined to annihilate E, and H(~nJ) = J, for admissible monomials
J 2 (2n + 1). Now we have
Corollary 3.2. The linear map H : (n + 1) ! (2n + 1) is a chain map.
Proof. It suffices to show dH = Hd holds for an element ~nJ, for an ad-
missible monomial J 2 (2n + 1), since d (n) (n), by Corollary 3.1.
But replicating the proof of Corollary 3.1, we have
Hd(~nJ) = H(d(~n)J + ~nd(J)) = 0 + d(J) = dH(~nJ)
since d(~n)J 2 (n), and d(J) 2 (2n + 1).
6 WILLIAM RICHTER
Mahowald's description [Ma2 ] of P is now immediate, and we've proved
Proposition 1.2 of the introduction.
Remark 3.3. Ravenel and Kochman [Ra , Ko ] implausibly assert that Corol-
laries 3.2 and 3.1 follow immediately from Formulas (2)and (3). Curtis and
Mahowald [C-M , p. 128] implausibly offer no proof or citation for these 2
results. Curtis [Cu1 , sec. 11] fails to prove that H is a chain map, first
by merely citing [Cu2 ], and second by an error in his proof that H is in-
duced by the geometric Hopf invariant. Curtis claims that a sum of maps
induce an isomorphism on E1 terms, but clearly each map induces zero, as
they're all Whitehead product, with positive Adams filtration. I give Curtis
credit for his bold attempt, and I think a version of his argument works with
Mahowald's [Ma1 ] äm pping cone" construction for an unstable Adams
resolution over the fiber of a map, in this case E :Sn ! Sn+1 , although I
think we'd have to abandon the Lower Central Series filtration.
Bousfield and Curtis [B-C , Rem. 5.3] construct a long exact cohomology
EHP sequence, using unstable A-modules, but I believe that one cannot
glean a proof of Theorem 1.2 from their argument, but instead, that they
use [B-C , Lem. 3.5] Corollary 3.2. Singer [Si, top p. 380] reconstructs
the long exact cohomology EHP sequence, and it's clear that his proof that
h : Ls(Sn) ! Ls-1(S2n-1) is a chain map uses Corollary 3.2, which of
course he could've proved himself. Wang [Wa , Prop. 1.8.3] "immediately"
deduced that d(~n) . (2n + 1) (n), and therefore Corollary 3.1, from
his special case s,t(n) . ~m (n) for m < n + t of Proposition 1.1. I
contend that Wang's leap shows the importance of stating Singer's result,
from which his result does follows immediately. Wang could easily have
deduced Corollary 3.2 from his Prop. 1.8.3, and he point out its obvious
corollary, that H(x) 2 (2n + 1) is a cycle if x 2 (n + 1) is a cycle.
4. d preserves the Adem relations
Before proving the admissible monomial basis, we'll prove an easier
result of the MIT school [B-C-K& , Pr]:
Proposition 4.1. The differential d : ! is well-defined.
Proof. is a tensor algebra modulo the 2-sided ideal generated by the
Adem relations. The Leibniz rule defines d on the tensor algebra, but we
must show that d sends Adem relations to the 2-sided ideal. To prove this,
we'll expand the tensor algebra to include ~-1, well-known to be related to
d, and use what Pete Bousfield calls "pension operators", i.e. selfmaps of
tensor powers which preserve Adem relations.
Let V be the Z=2 vector space with basis {~p : p -1}. Let e be
the selfmap of V given by e(~p) = ~p+1. Define the selfmap of V 2 by
LAMBDA ALGEBRA EHP SEQUENCE 7
D = e 1 + 1 e. As Mahowald recommends (cf. [Ma1 , p. 78]), we'll use
the original [B-C-K& ] symmetric Adem relations, for p -1, n 0:
X ` n'
(4) [p, n] := Dn(~p ~2p+1) = ~p+i ~2p+1+j 2 V 2.
i+j=n i
The (original [B-C-K& ] symmetric) differential on comes from p = -1:
(5) d(~n) = [-1, n + 1] + ~-1~n + ~n~-1 2 V 2, n 0.
Now define the selfmap C = e e2 of V 2. Then C preserves Adem
relations as well, and we have
C([p, n]) = [p + 1, n], D([p, n]) = [p, n + 1].
It's well known that all the Adem relations are obtained from [-1, -1] by
applying powers of C and D. Call I2 = 1 1 the identity selfmap of V 2.
Now we'll define selfmaps of V 3, and we'll apply them to
~p ~2p+1 ~4p+3 = [p, 0] ~4p+3 = ~p [2p + 1, 0].
We'll also call D the selfmap D = e I2 + 1 e 1 + I2 e of V 3, so
D = D 1 + I2 e = 1 D + e I2.
We venture into new territory with the selfmap E of V 3 defined by
E = e e2 1 + e 1 e2 + 1 e e2
= C 1 + D e2 = 1 C + e D2.
We've written both E and D as the sum of 2 commuting operators on V 3,
in 2 different ways, so the binomial theorem computes powers of Dm and
En, just as with [p, n] above. Let's define, for p -1, n, m 0, elements
[p, n, m] := Dm En([p, 0] ~4p+3) = Dm En(~p [2p + 1, 0]) 2 V 3.
By the binomial theorem, [p, n, m] has 2 expressions. Equating them gives
X ` n' ` m ' ` [p + i, j + s] ~4p+3+2j+t'
(6) = 0.
i+j=n,s+t=m i s + ~p+j+t [2p + 1 + i, 2j + s]
Now we specialize to p = -1, and assume n > 0, and project this equation
onto the positive part of V 3. I.e. we throw out the terms containing ~-1.
This will be our equation for why d preserves Adem relations.
The terms in Equation (6)containing ~-1 come from either j = t = 0 or
i = 0, and add up to
X ` m ' ` [-1, n + s] ~2n+t-1'
~-1 [n-1, m]+[n-1, m] ~-1+ .
s+t=m s + ~n+t-1 [-1, 2n + s]
8 WILLIAM RICHTER
By formula (5), the Leibniz rule, and switching s and t in the second part,
the positive projection of this expression is
X ` m ' ` d(~n+s-1) ~2n+t-1'
= d([n - 1, m]),
s+t=m s + ~n+t-1 d(~2n+s-1)
So the positive projection of Equation (6)shows, for n > 0, m 0, that
d([n - 1, m]) is the sum of the positive Adem relations
X ` n' `m ' ` [i - 1, j + s] ~2j+t-1'
(7)
i+j=n, s+t=m i s +~j+t-1 [i - 1, 2j + s]
i>0, (j,t)6=(0,0)
5.the admissible monomial basis
Let W V be the subvectorspace with basis {~p : p 0}, and let
R W 2be the subvectorspace Z=2{[p, n] : p, n 0}. Then with I the
2-sided ideal generated by R, we have
= T (W )=I, I = T (W ) . R . T (W )
We now prove the MIT school's result [B-C-K& , Pr]
Proposition 5.1. has a basis of the admissible monomials.
First we prove an analogue of Proposition 4.1:
Lemma 5.2. For p, n, m 0, we can rewrite ~p [2p + 1 + n, m] as a sum
X X
~p [2p + 1 + n, m] = ~xi [pi, ni] + [qj, mj] ~yj 2 W 3
i j
where for each i, the triple (xi, pi, 2pi+ 1 + ni) has lower right-lex order
than (p, 2p + 1 + n, 4p + 3 + 2n + m).
Proof. Equation (6)simplifies to
X `n '` m '
(8) ~p+j+t [2p + 1 + i, 2j + s] 2 R W W 3.
i+j=n,s+t=m i s
The (i, s)-term produces the triple (p+j +t, 2p+1+i, 4p+3+2n+s), and
the maximum right-lex order occurs uniquely at s = m and i = n, which
corresponds to the term ~p [2p + 1 + n, m].
Remark 5.3. We proved what we will use below, but here's a more straight-
forward analogue of Proposition 4.1. Define the excess of (a, b) to be
b - 2a - 1. Then the excess of (p + j + t, 2p + 1 + i) is i - 2j - 2t n, and
the maximum n is achieved only for j = t = 0. So Formula (8)rewrites
~p [2p+1+n, m] as an element of R W plus a sum of elements ~a [b, c]
LAMBDA ALGEBRA EHP SEQUENCE 9
with b - 2a - 1 < n. By induction ~p [2p + 1 + n, m] is an element of
R W plus a sum of elements ~a [b, c] with each (a, b) is admissible.
Proof of Proposition 5.1.The problem is that 2-sided ideal I is öt o big".
We first define a sub-vectorspace J of I so that T (W )=J has a basis of the
admissible monomials. J will be the sub-vectorspace I that's defined by
the algorithm of performing an Adem relation on the left-most inadmissible
pair of a monomial. Formally, let K I be the subvectorspace with basis
{~(a1, ..., as)[p, n] : p, n, s 0, ~(a1, ..., as) admissible, p 2as if}s,> 0
and define J = K . T (W ). It's obvious that T (W )=J has a basis of the
admissible monomials. We'll use to Lemma 5.2 to show I = J.
I is spanned by spanning elements
ff = ~(a1, ..., as)[p, n]OE, ai, p, n, s 0, OE 2 T (W ).
By abuse of notation, let's call s the Adams filtration of ff. We'll say that f*
*f is
an admissible spanning element if (a1, . .,.as, p) is admissible. Of course,
ff 2 J if ff is admissible. If ff is inadmissible, we'll perform reductions
until ff is a sum of admissible spanning elements, and then ff 2 J.
We need an ordering on the spanning element, derived from the orderings
Priddy [Pr] and Mahowald [Pr, Ma1 , Prop. 5.5] used in their cohomological
proofs of this basis result. We order the spanning elements ff of a given
word-length N = s + 2 + r and a given stem degree
a1 + . .+.as + p + (2p + 1 + n) + b1 + . .+.br
first by the Adams filtration s and then by right lexicographical order on the
N-tuple (a1, . .,.as, p, 2p + 1 + n, b1, . .,.br). We can now induct because
there are only a finite number of elements with lower filtration than ff. We're
going to perform a sequence of reductions until ff is a sum of admissible
spanning elements, and then ff 2 J. Our two reduction moves are:
(1) Apply a symmetric Adem relations [q, n] to any inadmissible pair in
~(a1, . .,.as)
(2) Apply a higher Adem relations Dm En(q 2q + 1 4q + 3) to
as[p, n], if (as, p) is inadmissible.
We'll see that both moves strictly lower the filtration order. It will be ob-
vious that both moves preserve the word-length and the stem degree.Then
I = J by the same inductive argument that proves why admissibles span
: keep applying moves in any order until (by finiteness) we have a sum of
admissible spanning elements.
10 WILLIAM RICHTER
Let's illustrate the type (1) move for s = 2. If (a1, a2) = (q, 2q + 1 + m)
is inadmissible, then
X `m '
ff = [q, m]([p, n]OE) + ~(q + i, 2q + 1 + j)[p, n]OE,
i+j=m, j 1 follows by easy induction by the strictly associativity of the formula.
First some obvious properties of unstable composition, involving asso-
ciativity, suspension naturality, Sq0 and admissible concatenation:
0,t0 0
Lemma 6.2. If ff 2 s,t(n), fi 2 s (n + t), and fl 2 (n + t + t ), then
ff ^ (fi ^ fl)= (ff ^ fi) ^ fl 2 (n),
E(ff ^ fi)= E(ff) ^ E(fi) 2 (n + 1),
0,2(t+t0)
Sq0(ff ^ fi)= Sq0(ff) ^ Sq0(fi) 2 s+s (2n).
For fi 2 (2n + 1), we have ~nfi = ~n ^ E(fi) 2 (n + 1).
Proof. We must only check that all of the unstable compositions are defined,
since unstable composition is just the multiplication desuspended to the
appropriate sub-vectorspace (i) .
Proof of Proposition 6.1.We'll prove Equation (9) by induction on s, the
Adams filtration of the first argument ff.
First we'll do s = 1, and be very pedantic about unstable products. So
ff = ~a, with 0 a n. For fi 2 (n + a + 2), we need
(10) EH(~a ^ fi) = ffia,nfi + ~2a+1 ^ EH(fi) 2 (2n + 2)
Let's write m = n + a + 1, so fi 2 (m + 1).
12 WILLIAM RICHTER
Assume a < n. Write ~a 2 1,a+1(n), and ~2a+1 2 1,2a+2(2n + 1).
Then ~2a+1 ^ H(fi) 2 (2n + 1), since (2n + 1) + (2a + 2) = 2m + 1,
and H(fi) 2 (2m + 1). So Equation (10)desuspends to
H(E(~a) ^ fi) = ~2a+1 ^ H(fi) 2 (2n + 1).
Let's write fi = ~m ^ E(x) + E(y) in admissible form, for x 2 (2m + 1),
and y 2 (m). Now let's write the Adem relation for ~a~m as
E(~a) ^ ~m = ~n~2a+1+E(Ra,m) 2 (n+1), for Ra,m 2 2,m+a+2(n).
Then Ra,m ^ x 2 (n), since n+m+a+2 = 2m+1, and ~a ^ y 2 (n),
since n + a + 1 = m. Then we have
E(~a) ^ fi= ~n ^ E(~2a+1 ^ x) + E(Ra,m ^ x + ~a ^ y), so
H(E(~a) ^ fi) = ~2a+1 ^ x = ~2a+1 ^ H(fi) 2 (2m + 1).
This finishes the case a < n.
Now assume a = n. Then write fi 2 (2n + 2) in admissible form as
fi = ~2n+1 ^ EH(fi) + E(y), for y 2 (2n + 1). Since ~n~2n+1 = 0, we
have ~n ^ fi = ~ny, and the case s = 1 is concluded by
EH(~n ^ fi) = E(y) = fi + ~2n+1 ^ EH(fi) 2 (2n + 2).
The induction step follows from the strict associativity of the RHS. Take
0,t0 0
ff fi fl 2 s,t(n + 1) s (n + t + 1) (n + t + t + 1).
Assuming the result for s and s0, the Adams filtrations of ff and fi, we'll
show it's true for ff ^ fi in the first argument. Using Lemma 6.2, we have
EH((ff ^ fi) ^ fl) = EH(ff ^ (fi ^ fl))
= EH(ff) ^ fi ^ fl + Sq0(ff) ^ EH(fi ^ fl)
0
= EH(ff) ^ fi ^ fl + Sq0(ff) ^ EH(fi) ^ fl + Sq (fi) ^ EH(fl)
0 0 0
= EH(ff) ^ fi + Sq (ff) ^ EH(fi) ^ fl + Sq (ff) ^ Sq (fi) ^ EH(fl)
= EH(ff ^ fi) ^ fl + Sq0(ff ^ fi) ^ EH(fl).
So Equation (9)is true with ff ^ fi in the first argument. This completes
our induction, since every ffP2 s,t(n + 1) is a sum of such products. Just
write ff admissibly as ff = ni=0~i^ E(xi), for xi 2 s-1,t-i-1(2i + 1),
and we've proved the result for Adams filtration 1 and s - 1.
There are two important special cases when Proposition 6.1 desuspends.
First, when the second argument fi desuspends, we have [Si, Prop. 5.2]
Corollary 6.3 (Singer). For ff 2 s,t(n + 1) and fi 2 (n + t), we have
H(ff ^ E(fi)) = H(ff) ^ fi 2 (2n + 1).
LAMBDA ALGEBRA EHP SEQUENCE 13
That is, letting m = n + t, the diagram commutes:
H id
s,t(n + 1) (m) ___________________//s-1,t-n-1(2n_+ 1) (m)
| |
id E | |^
fflffl| fflffl|
^
s,t(n + 1) (m + 1) ____//(n_+ 1) __________//_(2nH+ 1)
We only need observe that both sides actually desuspend. Proposition 6.1
also implies the desuspension when the first argument ff desuspends:
Corollary 6.4. For ff 2 s,t(n) and fi 2 (n + t + 1), we have
H(E(ff) ^ fi) = E(Sq0(ff)) ^ H(fi) 2 (2n + 1).
As Singer explains [Si], we can now perform analogues of various
geometric EHP construction that Toda, Barratt and others used.
Consider Toda's calculation [To ] of ßs7= Z=16, generated by oe 2 ß15S8.
The problem is to construct his elements oe0, oe00, oe000which are born on S7,
S6, and S5 respectively, and are stably 2oe, 4oe, and 8oe respectively, with
Hopf invariants j, j2 & j3 respectively. oe000is a Toda bracket < , 8', >,
used in the constructing the Adams selfmap [Ad ]. But oe0 and oe00are more
mysterious, not expressed as Toda brackets.
In , the oe, oe0, oe00, oe000story is easy. Starting with the cycle ~7 2 (8*
*),
with H(~7) = * 2 (15), Proposition 1.1 and Corollary 6.4 imply
~0~7 2 (7), H(~0~7) = ~1,
~20~72 (6), H(~20~7) = ~21,
~30~72 (5), H(~30~7) = ~31.
Of course, these equations are trivial to verify by hand. Note that ~30~7
therefore is a cycle with leading term 4111. Compare [Ra , Ex. 3.3.11],
where 4111 is completed to a cycle by the Curtis algorithm.
~30~7 brings up an obvious corollary of Proposition 1.1 and Corollary 3.1:
Corollary 6.5. For ff 2 s,t(n) and fi 2 (n + t + 1), we have
d(ff ^ fi) = d(ff) ^ fi + ff ^ d(fi) 2 (n).
Now take ff = d~n 2 2,n+1(n) and note H(ff) = (n - 1)~0. Propo-
sition 6.1 and Theorem 1.2 immediately imply Mahowald's result [Ma1 ,
Prop. 3.1]: The composition
(2n + 1) -P! (n) -H! (2n - 1) -E! (2n)
sends fi to (n-1)~0 ^ fi +Sq0(d~n) ^ H(fi). Then d~2n+1 = ESq0(d~n),
and specializing to n even, Mahowald observed that the composition
2
(4n + 1) -P! (2n) -H! (4n - 1) -E! (4n + 1)
14 WILLIAM RICHTER
sends fi to ~0 ^ fi + d(~4n+1) ^ H(fi). Recall the Hilton-Hopf expan-
sion [B-S , Wh ] James used for his 2-primary exponent [Ja, Co , B-C-G& ]:
(11) 2' . ff = ff . 2' + ['n, 'n] . H(ff), for ff 2 ß*(Sn).
It's well-known that d(~n) corresponds to ['n, 'n], and ~0 corresponds to 2'.
Assuming this, Bousfield and Kan's (1), leads us to expect that left/right
composition with ~0 corresponds to left/right geometric composition by 2'.
Mahowald then observed the following result:
Proposition 6.6 (Mahowald). The composition
2
(4n + 1) -P! (2n) -H! (4n - 1) -E! (4n + 1)
induces a selfmap of H* (4n + 1), which is E2 . H . P (fi) = fi~0.
Proof. We only need to prove the analogue of Equation (11). Singer [Si,
Thm. 4.1] proves the full analogue of the Barratt-Toda commutation for-
mula [To ]: for f 2 ßm+a Sa and g 2 ßn+aSb we have
f . g - (-1)nm g . f = [1a+b-1, 1a+b-1] . H(f) ^ H(g) 2 ßm+n+a+b Sa+b-1.
We'll only prove a special case. For a cycle f 2 (p + 1), we'll show
~0 ^ f + f ^ ~0 = d(~p+1) ^ H(f) 2 H* (p + 1)
To prove this, write f admissibly as f = ~pA + B, for B 2 (p) and
A 2 (2p + 1). Since f is a cycle, A must be a cycle, as Wang (who didn't
use H) observed [Wa , Thm. 1.8.4]: dA = dH(f) = Hd(f) = H(0) = 0.
By Equation (5), commutation with ~-1 is the boundary map d:
df = [f, ~-1] 2 T (V ).
Now we'll extend our operator D to T (V ), so D satisfies the Leibniz rule,
and D(~p) = ~p+1. Writing D(ff) = ff0, we have
(df)0= [f, ~-1]0= [f0, ~-1] + [f, ~0] = d(f0) + [f, ~0] 2 T (W ).
We now pass to , since D, as an operator on T (W ), preserves Adem rela-
tions, i.e. D([a, m]) = [a, m + 1]. Since f is a cycle, i.e. df = 0, that's
d(f0) = [f, ~0] = ~0f + f~0 2 .
We need to show d(f0) is cohomologous to d(~p+1) ^ H(f) 2 (p + 1).
Differentiate the defining equation for f and apply the boundary map d:
f = ~pA + B
f0 = ~p+1A + ~pA0+ B0
(12) ~0f + f~0 = d(f0) = d(~p+1)A + d(~pA0+ B0) 2 ,
since d(A) = 0. We'll show that (~pA0+ B0) 2 (p + 1), because
(13) C 2 (k) =) C0 2 (k + 1)
LAMBDA ALGEBRA EHP SEQUENCE 15
To see this, take an admissible monomial C = ~(a1, . .,.as) 2 (k). Then
C0 is a sum of s terms, each of which is either admissible or zero, since
~a~2a+1 = 0. So the first term ~(a1 + 1, a2, . .,.as) 2 (k + 1), and the re-
maining terms of the sum C0 belong to (k). This proves implication (13).
Thus B0 2 (p + 1) and A02 (2p + 2), so ~pA0= ~p ^ A02 (p + 1),
by Proposition 1.1. So Equation (12)now reads
~0f + f~0 d(~p+1)A 2 (p + 1).
Since H(f) = A, we've proved our formula: For any cycle f 2 (p + 1),
~0 ^ f + f ^ ~0 = d(~p+1) ^ H(f) 2 H* (p + 1).
Now recall E2 . H . P (fi) = ~0 ^ fi + d(~4n+1) ^ H(fi) 2 (4n + 1).
Mahowald then conjectured the geometric analogue of Proposition 6.6:
(P) 2n H 4n-1 E2 3 4n+1
the composite 3S4n+1 --- ! S -! S -! S is homotopic
to the H-space squaring map on 3. The author [Ri1 ] proved this conjec-
ture, which implies the following infinite statement in homotopy groups:
4n-1
(14) 2ßkS4n+1 E2 ßk-2S , for k 3.
[B-C-G& ] shows that (14) is not due to James or Selick [Ja, Se ], even
though (14)does not improve on the James-Selick 2-primary exponent.
7. symmetric and admissible Adem relations
We'll prove Wang's result [Wa , Thm. 1.6.1] that the admissible Adem
relations (2)are equivalent to the original [B-C-K& ] symmetric Adem rela-
tions (4). First We'll prove the MIT school's result [B-C-K& ] that d2 = 0.
Lemma 7.1. d2(~n) = 0 2 , for n 0.
Proof. We'll use the symmetric Adem relations, and show even more, that
d2(~a) vanishes in the tensor algebra. By formula (5), for n 0, we have
X ` n + 1'
d(~n) = ~i-1~j-1.
i+j=n+1, ij>0 i
Then we instantly derive d2(~n) = 0. Using the Leibniz rule
d(~i-1~j-1) = d(~i-1)~j-1 + ~i-1d(~j-1),
d2(~n) is the sum of two terms, the first of which is
X `n + 1 '
~s-1~t-1~j-1,
s+t+j=n+1, stj>0 s, t, j
16 WILLIAM RICHTER
n+1 i n+1
as we see by using the binomial identity i s = s,t,j, where as usual,
n+1
s,t,j= (n + 1)!=(s! t! j!). But the other term, arising from ~i-1d(~j-1), is
equal, so the sum d2(~n) is zero.
Remark 7.2. En arose in a way showing the power of the symmetric formu-
las: Suppose (p, b) is inadmissible. ThenP~p~b~2b+1 = 0 since ~b~2b+1 = 0,
but perform the Adem relation ~p~b = ~x~y first. Each pair (y, 2b + 1)
is inadmissible, so perform an Adem relation on each one. The basis
requires this sum to vanish in , but why? Using the admissible formulas,
this isn't at all clear.PBut using the symmetric formulas, it's easy to rewrite
this sum as a sum [q, r]s, using identities like ni 2i2a= in,a,b. So we
avoided a relation, and the calculation basically hands us the operator En.
Wang [Wa , Thm. 1.6.1] used formal power series to ä dmissify" the
symmetric formulas. We'll use a simple recursion formula due to Tan-
gora [Ta2 , Ta1]. Define Ca,k2 Z=2, for a 0, k 2 Z, recursively by
(15) C0,k= 0, C1,k= ffik,0, and for a 2, Ca,k= Ca-1,k+ Ca-2,k-1.
Then for p -1, and a 0, let's define
X
(16) (p, a) := ~p ~2p+1+a + Ca,k~p+a-k ~2p+1+k 2 V 2.
k
P
By easy induction on a, we see that k Ca,k~p+a-k ~2p+1+k is a finite
sum of admissibles: Ca,k= 0 for either k < 0 or 2k + 1 > a. This justifies
calling Formula (16)the admissible Adem formulas.
Now we obtain relations between the symmetric and admissible Adem
relations, by the usual procedure of applying D to formula (16):
Lemma 7.3. Assume p -1. Then (p, a) = [p, a] for a = 0, 1, 2, and
(17) (p, a + 1) = D(p, a) + (p + 1, a - 2) 2 V 2, for a 2.
Proof. The case a = 0, 1, 2 is obvious, and we'll deduce formula (17) by
induction on a 2. First, by replacing k by k - 2, we have
X
(p + 1, a - 2) = ~p+1~2p+1+a + Ca-2,k-2~p+1+a-k ~2p+1+k.
k
Then D(p, a) + (p + 1, a - 2) equals ~p~2p+1+a+1 plus the sum
X
(Ca,k+ Ca,k-1+ Ca-2,k-2) ~p+1+a-k ~2p+1+k
k
X
= (Ca,k+ Ca-1,k-1) ~p+1+a-k ~2p+1+k, by (15)for k - 1
k
X
= Ca+1,k~p+(a+1)-k~2p+1+k, by (15)for k
k
LAMBDA ALGEBRA EHP SEQUENCE 17
So by this double application of the Tangora recursion formula (15), we
have D(p, a) + (p + 1, a - 2) = (p, a + 1).
Now we have
Lemma 7.4. For all p, a 0, (p, a) is the admissible Adem relation (2).
a-k-1
Proof. We must only show that Ca,k= k , for k 0 and 2k + 1 a.
Again this follows from induction:
` ' ` ' ` '
a - k - 1 a - k - 1 a - k
Ca+1,k= Ca,k+ Ca-1,k-1= + =
k k - 1 k
by Pascal's triangle.
Now we'll show that the admissible and symmetric Adem relations imply
each other. Let A W 2be the admissible analogue of R, so A has basis
{(p, a) : p, a 0}. Then we have an immediate corollary of Lemma 7.3:
Lemma 7.5. A = R, so can be defined either by the admissible Adem
relations (2)or the symmetric Adem relations (4).
Proof. Since D(R) R, Lemma 7.3 implies that (p, a) 2 R by induction
on a. So A R. But Lemma 7.3 also implies that D(p, a) 2 A. Thus
D(A) A. Since [p, 0] = (p, 0), and D[p, n] = [p, n + 1], we have R A.
Hence A = R.
Lemma 7.6. The differential d : ! can be defined either by the admis-
sible Adem relations (3)or the symmetric Adem relations (5).
Proof. For p -1 and a 0, we can measure the difference between
(p, a) and [p, a] as follows. Let Xp,a= [p, a] + (p, a) 2 V 2. We can restate
Lemma 7.3 as Xp,a= 0 for a = 0, 1, 2, and
(18) Xp,a+1= DXp,a+ (p + 1, a - 2) 2 V 2, for a 2.
Let [p^, a]and (p^, a)be [p, a] and (p, a) plus ~p ~2p+1+a + ~p+a ~2p+1.
Then clearly Xp,a= [p^,+a]^(p,.a)Now [-^1, a]and (-1^, a)are the formulas
in W 2for the symmetric and admissible Adem relations. So specializing
Equation (18)to p = -1 shows by induction on a that
[-^1, a]+ (-1^, a)= X-1,a2 R.
The Tangora recursion relations (15)are a theoretical improvement over
the usual recursion formula (which arose in the proof of Lemma 7.3)
(19) Ca+1,k= Ca,k+ Ca,k-1+ Ca-2,k-2,
because it was clear that we had a sum of admissibles, and it was easy to
see that Ca,k= a-k-1k.
18 WILLIAM RICHTER
In doing hand calculations, the Tangora recursion relations also give a big
improvement over the usual recursion scheme, because Tangora's involves
2 terms instead of 3, and the calculation stays on the same äp ge." For
instance, we quickly and independently obtain on the ~0 and ~1 pages:
~0~1 = 0 ~1~3 = 0
~0~2 = ~1~1 ~1~4 = ~2~3
~0~3 = ~2~1 ~1~5 = ~3~3
~0~4 = ~3~1 + ~2~2 ~1~6 = ~4~3 + ~3~4
~0~5 = ~4~1 ~1~7 = ~5~3
~0~6 = ~5~1 + ~4~2 + ~3~3 ~1~8 = ~6~3 + ~5~4 + ~4~5
~0~7 = ~6~1 + ~4~3 ~1~9 = ~7~3 + ~5~5
~0~8 = ~7~1 + ~6~2 + ~4~4 ~1~10 = ~8~3 + ~7~4 + ~5~6
~0~9 = ~8~1 ~1~11 = ~9~3
In usual recursive scheme, based on (19), one applies D to each equation to
get the next one. In the Steenrod algebra [M-T ], this works OK. To compute
Sq3Sq4 = Sq7 on the Sq4 page, we need the Sqi page for i = 1, 2, 3, and
this presents no hardship. But in , the ~0 page requires part of the ~1 page,
which requires part of the ~2 page, etc. For instance, to compute ~0~9, we
apply D to the equations for ~0~8, ~1~7 and ~2~6 to obtain
~0~9 = (~8~1 + ~6~3 + ~5~4 + ~4~5) + (~6~3 + ~5~4) + ~4~5 = ~8~1.
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William Richter, Mathematics Department, Northwestern University,
Evanston IL 60208
E-mail address: richter@math.nwu.edu