ON CONJUGATION INVARIANTS IN
THE DUAL STEENROD ALGEBRA
M. D. Crossley and Sarah Whitehouse
3 April 1998
Abstract. We investigate the canonical conjugation, O, of the mod 2 dual*
* Steenrod
algebra A*, with a view to determining the subspace, AO*, of elements in*
*variant under
O. We give bounds on the dimension of this subspace for each degree and *
*show that,
after inverting 1, it becomes polynomial on a natural set of generators.
1. Introduction.
The mod 2 dual Steenrod algebra, A*, being a connected commutative Hopf
algebra, has a canonical conjugation or anti-automorphism O. This map was first
studied by Thom [T] but most of what we know today about O is due to Milnor
[M]. Our aim is to study the subspace of A* consisting of elements invariant u*
*nder
this conjugation map. We denote this subspace by AO*. While we are unable to
give a complete description of AO*, we have established bounds on its dimension
in each degree (Theorem 3.1), and we can show that, after inverting the element
1 2 A*, the invariant subspace becomes polynomial (Theorem 4.1). Finally, our
investigations have also lead us to conjecture a set of generators for AO*(Con*
*jecture
5.3).
We begin by recalling the structure of A* and certain well known facts abou*
*t the
map O, while in section 2 we deduce some elementary properties of O. In sectio*
*n 3
we derive our bounds on the dimension of the subspace of invariants and in sec*
*tion
4 we study the result of inverting 1. The final section contains our conjectu*
*res
about generators for the invariants.
The structure of the Hopf algebra A* was determined by Milnor [M]. As an
algebra, A* = F2[1; 2; 3; : :]:, where the degree of i is 2i - 1. We use the
notation (r1; r2; : :;:rk) to denote the monomial r11r22: :r:kk. The coproduct*
* OE is
determined by the formula
Xk i
OE(k) = 2k-i i;
i=0
where 0 is interpreted as 1.
From [MM] we know that any connected Hopf algebra, H, has a unique bijective
linear map, O : H ! H, called the `conjugation', with the following properties:
1) O(1) = 1,
Typeset by AM S-*
*TEX
1
2 M. D. CROSSLEY AND SARAH WHITEHOUSE
2) O(xy) = O(y)O(x)P(i.e. O is an anti-automorphism), P
3) If OE(a) = a0i a00iwhere a 2 H+ (i.e. deg a > 0), then a0iO(a00i) = *
*0.
(Since the coproduct OE always satisfies the identity OE(a) a1+1a mod H+
H+ , the last property determines O inductively.)
Furthermore, if the Hopf algebra is either commutative or cocommutative, th*
*en
O2 is the identity homomorphism.
In the case of the dual Steenrod algebra, property 3 leads inductively to t*
*he
following formula for O (Lemma 10 of [M]).
1.1 Lemma. In the dual Steenrod algebra A*,
X l(ff)Yoe(i)
O(n) = 2ff(i);
ff2Part(n)i=1
where P art(n) denotes the set of all ordered partitions of n; and for a given*
* or-
deredPpartition ff = (ff(1); ff(2); : :;:ff(l)) 2 P art(n), oe(i) denotes the *
*partial sum
i-1
j=1ff(j).
This lemma enables us to determine O on an arbitrary element of A*, by virt*
*ue
of multiplicativity (which follows from property 2 since A* is commutative) and
linearity.
We end this introduction with a few comments on motivation for the problems
discussed in this paper. Expressions like Hm (n; ss*(E^n )) arise in spectral*
* se-
quences for gamma cohomology of an E1 -ring spectrum E [RW]. For E suitably
nice, this is Hm (n; (E*E)(n-1) ), the n action here being described in [W]; *
*for
n = 2, 2 acts by the usual conjugation on E*E. This paper is therefore concern*
*ed
with the very special case of this problem where E = HF2 and n = 2.
2. Elementary Properties of conjugation.
Now we make some elementary observations on the properties of the conjugati*
*on
O. For a connected Hopf algebra H, we denote by HO the invariant elements of
H under the conjugation map O : H ! H. The identity homomorphism will be
denoted by 1, so that HO = Ker(O - 1).
2.1 Lemma. If H is commutative, HO is a subring of H.
Proof. The conjugation O is always an anti-automorphism. So, when H is commu-
tative, it is a ring homomorphism.
We denote by Hd the degree d part of a graded object (e.g. Hopf algebra) H.
2.2 Lemma. If H is a commutative or cocommutative Hopf algebra over F2, then
dim HOd dim Hd=2.
Proof. We have O2 = 1 by the (co)commutativity hypothesis. Since we are working
over F2, this gives (O - 1)2 = 0, and so Im (O - 1) Ker (O - 1). Since dim Hd*
* =
dim (Im (O - 1)d) + dim(Ker (O - 1)d), we have the result.
CONJUGATION INVARIANTS 3
2.3 Remark. In the case of the mod 2 (dual) Steenrod algebra, the above dimen-
sion constraint is sharp in low degrees, although not in general. It first fai*
*ls in
degree 42. More details are in section 3, particularly Example 3.4.
2.4 Lemma. In a commutative Hopf algebra H over F2, Im (O - 1) is an ideal in
Ker (O - 1). In particular, Im (O - 1) is a subring of H.
Proof. Let x 2 Ker (O - 1). Then x(O(y) - y) = xO(y) - xy = O(x)O(y) - xy =
(O-1)(xy). The second part can be proved directly: (O(x)-x)(O(y)-y) = O(z)-z
mod 2, where z = xy + xO(y).
In this situation, the higher cohomology Hn (2; H) = Ker(O - 1)= Im(O - 1) f*
*or
n > 0 (where 2 acts on H by conjugation) has an algebra structure.
We now fix our attention on the dual Steenrod algebra, A* = F2[1; 2; : :]:. *
*We
shall frequently need to order the monomials in A* and the right lexicographic
ordering turns out to be the most useful.
2.5 Proposition. With respect to right lexicographic ordering of the monomial
basis, the matrix of the conjugation map in each degree, O : (A*)d ! (A*)d, is
unitriangular.
Proof. It follows from Lemma 1.1 that O(k) = k + Pk where Pk is a polynomial
in 1; : :;:k-1 and hence strictly lower than k. Consequently for any monomial
M in 1; : :;:k, we have that O(M) = M + Q where Q is strictly lower than M.
This is because the right lexicographic ordering has the property that if x < x*
*0 and
y y0 then xy < x0y0. It then follows that the matrix is unitriangular.
2.6 Remark. This unitriangularity property also holds for certain other orderi*
*ngs.
A simple example is left lexicographic ordering. More interestingly, if we defi*
*ne the
weight w of the monomial r11: :r:kkto be r1 + r2 + . .+.rk then we may obtain
further orderings by combining weight and lexicographic orderings. For example,
the `weight/reverse-left lex' order is defined by a b if either w(a) < w(b) or
w(a) = w(b) and a follows b in left lexicographic ordering. For all such order*
*ings
the above proof can be easily modified. In fact, all the proof needs is that fo*
*r all k,
(O-1)(k) is strictly lower than k (or that (O-1)(k) is always strictly higher) *
*and
that the ordering is `multiplicative', i.e. x < x0 and y y0 implies that xy < *
*x0y0.
3. Bounds on dimension.
In this section we state and prove the following theorem which gives bounds *
*on
the dimension of the invariant subspace AO*in a given degree.
3.1 Theorem. Let the dimension of the (dual) Steenrod algebra in degree d be
denoted Dd. Then Dd-1 =2 dim (O - 1)(A*)d Dd=2 and hence
Dd=2 dim (AO*)d Dd - (Dd-1 =2) :
In fact the upper bound on dim (AO*)d can be improved upon - see Lemma 3.3
and Example 3.4.
The lower bound on dim (AO*)d was given by Lemma 2.2; the rest of the theorem
is a combinatorial corollary of the following result.
4 M. D. CROSSLEY AND SARAH WHITEHOUSE
3.2 Proposition. Let a monomial (r1; r2; : :;:rk) (i.e. r11r22: :r:kk) be c*
*alled
`uniterminal' if rk = 1. In (A*)d, the uniterminal monomials have linearly in*
*de-
pendent images under O - 1.
Proof. We use right lexicographic ordering and claim that the lowest unitermin*
*al
monomial which has (r1+1; r2; : :;:rk-2 ; rk-1 +2) as a summand in its image u*
*nder
O-1 is (r1; : :;:rk-1 ; 1). Assuming this claim we argue as follows. Let Q be *
*a linear
combination of images under O-1 of uniterminal monomials. So we may write Q as
Q = (O-1)(P ) where P is a linear combination of uniterminal monomials. Suppose
that (r1; : :;:rk-1 ; 1) is the highest monomial which appears in P . Then our*
* claim
shows that (O - 1)(P ) will have (r1 + 1; r2; : :;:rk-2 ; rk-1 + 2) as a summa*
*nd, and
so cannot be zero. The proposition then follows.
Now to prove the claim. From Lemma 1.1,
k-1
O(k) = k + 2k-11 + k-1 21 + Pk-2 ;
where Pk-2 is some polynomial in 1; : :;:k-2 . Looking at the second term on
the right hand side of the above expression, we see that the largest monomial *
*in
(O-1)(r1; : :;:rk-1 ; 1) which does not contain a k is (r1+1; r2; : :;:rk-2 ; *
*rk-1 +2).
Now we need to see that no earlier uniterminal monomial has this term in its
image under O - 1. Suppose it does appear in the image of (j1; : :;:jk0-1; 1),*
* with
(j1; : :;:jk0-1; 1) < (r1; : :;:rk-1 ; 1) so, in particular, k0 k. If k0 k *
*- 1 then
(O - 1)(j1; : :;:jk0-1; 1) will have no summands with k0-exponent greater than*
* 1,
so cannot have (r1 + 1; r2; : :;:rk-1 + 2) as a summand. If k0 = k then the f*
*act
that (j1; : :;:jk0-1; 1) < (r1; : :;:rk-1 ; 1) implies that (j1 + 1; j2; : :;:*
*jk-1 + 2) <
(r1; r2; : :;:rk-1 + 2). In this case, the image of (j1; : :;:jk0-1; 1) canno*
*t contain
(r1; r2; : :;:rk-1 + 2) as a summand.
So dim (O - 1)(A*)d Rd where Rd is the number of uniterminal monomials
in degree d. In order to complete the proof of Theorem 3.1 we now obtain some
information about Rd.
3.3 Lemma.
1) Rd Dd-1 =2 ,
2) Rd = Dd-1 - Rd-1 .
Proof. We pair up each uniterminal monomial in degree d with another degree d
monomial that is not uniterminal by the pairing
(r1; r2 : :;:rk-2 ; rk-1 ; 1) ! (r1 + 1; r2; : :;:rk-2 ; rk-1 + 2) :
The monomials left unpaired are characterised by the fact that they begin with*
* zero
and are not uniterminal. The number of these is clearly less than or equal to *
*the
total number beginning zero, which is Dd - Dd-1 . This gives the first claim. *
*The
actual number unpaired is given by (Dd - Dd-1 ) - (Rd - Rd-1 ), since the numb*
*er
of uniterminal monomials starting with zero in degree d is Rd - Rd-1 . This gi*
*ves
the second part.
Note that this gives a recursive formula for the Rd's in terms of the Dd's.*
* Alter-
natively, lettingPP ibe the polynomial algebra F2[1; : :;:i] (P 0= F2), it is *
*clear
that Rd = i1 dim(P i-1)d-(2i-1).
CONJUGATION INVARIANTS 5
3.4 Example. In degree 42, A* has dimension 92 while the dimension of the O-
invariants is 47. The number of uniterminal monomials, R42, is 44 while D41 is *
*86.
So the bounds in Theorem 3.1 tell us that 46 dim (AO*)42 49, whereas if we
are prepared to put in more effort (calculating Rd precisely) then we shall obt*
*ain
a sharper upper bound of 48. Note also that the lower bound here is not sharp; *
*in
fact 42 is the first degree in which this happens.
4. Conjugation invariants with 1 inverted.
In this section we adjoin a formal inverse to 1, denoting the resulting obje*
*ct by
A*[-11]. This is regarded as containing A* in the usual way. Since 1 is invaria*
*nt,
A*[-11] inherits an action of O and the subspace of invariants turns out to be *
*much
more manageable. In fact we show (Theorem 4.1) that it is a polynomial ring on
certain natural invariant elements, ffl2; ffl3; . ...Clearly AO*is the intersec*
*tion of A*
and A*[-11]O. So one might conclude that, since we have simple descriptions of *
*A*
and A*[-11]O, we can easily obtain a description of AO*. However, this turns o*
*ut
to be far from the case - the problem of finding the highest power of 1 dividin*
*g a
given polynomial in 1; ffl2; ffl3; . .s.eems to be difficult in general. In fac*
*t, low degree
calculations quickly reveal that the algebra AO*is complicated; in particular i*
*t is
far from being polynomial.
At the end of this section we show how Theorem 4.1 generalises nicely to giv*
*e a
description of the invariants of (A*=<1; : :;:n-1 >)[-1n].
Note that A*[-11]O = AO*[-11]. That is to say, if we adjoin -11 to the ring *
*AO*,
we obtain the same object as if we take the O-invariants of the ring A*[-11].
4.1 Theorem. Let k = F2[1; -11]. Then
AO*[-11] = k[ffl2; ffl3; : :]:;
where ffl2 = 2O(2) and, for n 3, ffln = 2n + O(2n). Furthermore
Hi(2; A*[-11]) = 0 for i > 0:
The proof of the second statement is straightforward : in A* we have (O-1)2 =
31. It follows that in A*[-11], 1 is the image under O - 1 of -312. But Im (O -*
* 1)
is an ideal in Ker (O - 1) and so, since 1 2 Im (O - 1), these two objects must*
* be
equal.
Now we consider the first statement. It is evident that the elements ffln *
*are
invariant and we claim that they are also algebraically independent. This fol-
lows from the fact that, for each n 2, ffln has a summand involving n, whereas
ffl2; : :;:ffln-1 have no such summands. (The highest right lexicographic monom*
*ial of
ffln is 31n if n 3 and 22if n = 2.) Thus, since the invariants form a subalgeb*
*ra,
k[ffl2; ffl3; : :]: AO*[-11]. The opposite inclusion follows directly from Pro*
*position
4.3.
We use the right lexicographic ordering on monomials in the 's and polynomia*
*ls
in the 's inherit a total order. This is by comparing first their highest (with
respect to right lexicographic ordering) monomials; if these are equal their ne*
*xt
highest monomials are compared and so on. For example, (1; 2) < (2; 0; 1) and
(1; 2) + (1; 1) < (2; 0; 1) + (1).
6 M. D. CROSSLEY AND SARAH WHITEHOUSE
4.2 Proposition.
1) Consider a monomial (r1; r2; : :;:rk). If r2 is odd then the image of this *
*mono-
mial cannot be expressed as the image of a linear combination of lower mono*
*mi-
als.
2) Let x 2 A* be invariant and let (r1; r2; : :;:rk) be the highest monomial a*
*ppearing
in x. Then r2 is even.
Proof. We claim that, if r2 is odd, then the monomial (r1; r2; r3; : :;:rk) is*
* the first
to have (r1 + 3; r2 - 1; r3; : :;:rk) as a summand of its image under O - 1. F*
*irstly,
these two elements are adjacent in the right lexicographic ordering, so by 2.5*
* this
is the first image in which the latter element can occur. Secondly,
O(r1; r2; r3; : :;:rk)= O(r1; 0; r3; : :;:rk)O(2)r2
= O(r1; 0; r3; : :;:rk)(2 + 31)r2 :
r
From this, and the fact that 12 1 mod 2, it is clear that (O-1)(r1; r2; r3;*
* : :;:rk)
does contain a summand (r1 + 3; r2 - 1; r3; : :;:rk). (Comparison of exponent*
*s of
k; k-1 ; : :;:3 shows that this term cannot arise in any other way.)
The second part follows from the first, for suppose x is invariant, that is*
* (O -
1)(x) = 0, with highest monomial (r1; r2; : :;:rk). Then (O - 1)(r1; r2; : :;:*
*rk) can
be expressed as the image of a linear combination of lower monomials, namely
x - (r1; r2; : :;:rk). If r2 is odd then this contradicts the first part.
4.3 Proposition. If x 2 AO* then there exists some integer s such that s1x 2
F2[1; ffl2; ffl3; : :]:.
Proof. We assume that x is homogeneous since the general result clearly follows
from this case. We proceed by a recursion on a well-founded relation. It is ev*
*ident
that < is well-founded (by which we mean that any non-empty set of polynomials
has a minimal element) and we claim that whenever x is such that the propositi*
*on
is true for all x0 < x then it is true for x. Now consider the set of polynomi*
*als for
which the proposition is false. If this set is not empty then it must have a m*
*inimal
element, whose existence contradicts our claim. Hence the proposition is true*
* for
all x 2 AO*.
We now prove the claim. Let l = (r1; r2; : :;:rk) be the leading monomial o*
*f x. By
4.2, r2 is even. Let z = r11fflr2=22fflr33: :f:flrkk. Recalling the definition*
* of the ffln's, we see
that the leading monomial in z is r1+t1r22r33: :r:kk, where t = 3(r3 + r4 + . *
*.+.rk).
This is the leading term of t1x and we set x0= t1x+z, noting that this is inva*
*riant,
as it is the sum of two invariants. Since the leading monomial of z is equal t*
*o that
of t1x, the leading term of x0 will be strictly lower. Denote the leading term*
* of x0
by l0. Homogeneity of x implies that x0 will have the same degree as t1x and *
*it
is straightforward to see that since l0 < t1l and deg l0 = deg(t1l), we have l*
*0 < l.
Thus x0< x and, by the hypothesis of the claim, the proposition is true for x0*
*, say
s1x02 F2[1; ffl2; ffl3; : :]:. Then s+t1x = s1x0+ s1z is also in F2[1; ffl2; f*
*fl3; : :]:and the
claim is proved.
CONJUGATION INVARIANTS 7
4.4 Remark. The only properties of ffln used in the proof are that ffln is inv*
*ariant
and has a certain highest term. In particular we could replace ffl2 by "ffl2= *
*(O -
1)(-113). This gives an alternative proof that Im (O - 1) = Ker (O - 1) once 1 *
*is
inverted, since "ffl22 Im (O - 1) and ffln 2 Im (O - 1) for all n 3 and we can*
* re-run
the whole programme of the proof of 4.3 with `Im (O - 1)' in place of `Ker (O -*
* 1)'.
Now we mention a generalisation of Theorem 4.1. We consider the situation
where we kill off the first n - 1 generators, 1; : :;:n-1 of A* and invert the*
* n-th
generator : let A* denote (A*=<1; : :;:n-1 >)[-1n], where <1; : :;:n-1 > is *
*the
ideal generated by 1; : :;:n-1 . Note that n is invariant in A*=<1; : :;:n-1 >,*
* so
A* inherits a well defined map O. (In fact, n; n+1 : :2:n-1are all invariant*
*.)
4.5 Theorem. Let k = F2[n; -1n]. Then
A*O = k[ffln+1 ; ffln+2 ; : :]:;
where fflm = 2nm + O(2nm ) if m 6= 2n and ffl2n = 2nO(2n). Furthermore
Hi(2; A*) = 0 for i > 0:
The proof of this result is entirely analogous to that of Theorem 4.1. Forn*
*the
second part, it is sufficient to note that the image under O - 1 of 2n is 2n+1
(modulo 1; : :;:n-1 ). For the first part, one observes that the fflm 's ar*
*e all
invariant and algebraically independent and so k[ffln+1 ; ffln+2 ; : :]: *
*A*O.
Then one shows that if = en+1n+1en+2n+2:e:k:kwhere e2n is even, then there exi*
*sts an
r such that rn can be expressed as the leading term of a monomial in the ffl*
*m 's
(modulo 1; : :;:n-1 ). Using this, one runs through the recursion argument of
Proposition 4.3 with the following lemma in place of Proposition 4.2.
4.6 Lemma. If is the leading term of a polynomial in A*=<1; : :;:n-1 > which
is invariant then the exponent of 2n in is even.
The argument is broadly the same as that of Propositionn4.2 : if the exponent
of 2n in is odd then (O - 1) contains the summand 2n+1 -12n and one checks
that there cannot exist any monomial 0 < such that (O - 1)0 also contains this
summand.
5. Some conjectures.
A natural question to ask is : what is the first (i.e. lowest degree) inva*
*riant
polynomial which involves n ? If n = 1 it is clearly 1 and if n = 2 one can see
that it is ffl2 = 2O(2). For n 3, ffln = (O - 1)(2n) is an obvious candidate,
being in degree 2n + 2. However, we shall see that, at least for n 7, we can f*
*ind
an invariant in degree 2n + 1 involving n. The following lemma shows that there
are no such invariants in degrees 2n - 1 or 2n.
5.1 Lemma. The monomials n and 1n are not summands of any invariant
elements.
Proof. The image of n contains the monomial 12n-1, which immediately precedes
it in the right lexicographic ordering. Since n is the highest monomial in its *
*degree,
8 M. D. CROSSLEY AND SARAH WHITEHOUSE
nothing else can have this monomial 12n-1 in its image. Hence n cannot be a
summand of an invariant polynomial (c.f. the proof of Proposition 4.2). Simila*
*rly,
1n has 212n-1in its image and is the only monomial which does so.
However, the above argument cannot be applied to 21n. The highest term in
its image is 312n-1, but this does not immediately precede 21n in the ordering*
* -
22n-1 occurs between them and is seen to also have 312n-1 in its image. In fact
21n is a summand (and hence the leading term) of an invariant for small n and
we make the following
5.2 Conjecture. For each n 3, there exists a polynomial dn in degree 2n + 1,
invariant under O and with leading term 21n.
We have been able to construct such elements for n = 3; 4; 5; 6; 7. We desc*
*ribe
below one particular choice of such elements but first we need some more notat*
*ion.
Notation. We define certain elements of A*, which are evidently invariant.
an = nO(n ) ;
bm1;m2;:::;mn= (O - 1)(m1 m2 : :m:n) :
The following elements are examples of the dn's of Conjecture 5.2.
d3= b2;3=1 ;
d4= (b2;4+ a32)=1 ;
d5= (b2;5+ a2a23+ a22b3;4)=1 ;
d6= (b2;6+ a2a24+ a23b3;5+ a62a4 + a42a33+ a112)=1 :
(We have not yet found an expression for d7 that is small enough to write on o*
*ne
line.) Note that the above expressions are all quotients by 1 of some sum star*
*ting
with b2;n. In fact a conjecture equivalent to 5.2 is
5.2' Conjecture. For each n 3, there exists an invariant polynomial xn, not
equal to b2;n, but such that
xn b2;nmod <1> ;
where <1> denotes the ideal generated by 1.
We would then obtain dn by setting dn = (b2;n+ xn)=1. (The invariance of xn
implies that of dn since a quotient of one invariant by another is invariant.)
Supposing Conjecture 5.2 to be true, we would hope to describe the generato*
*rs
of the algebra AO*as follows.
5.3 Conjecture. The following elements generate the conjugation invariants, A*
*O*:
1) 1,
2) an, for n 2,
3) bm1;m2;:::;mnwhere 2 m1 < m2 < . .<.mn, n 2 and either n > 2 or m1 > 2,
4) dn, for n 3.
In support of this Conjecture, we have the following result showing that th*
*ese
elements are all necessary generators, i.e. none of them are decomposable.
CONJUGATION INVARIANTS 9
5.4 Proposition. The families 1-3 of Conjecture 5.3 are necessary generators.
For each n 3, if dn exists then it is necessary and if it doesn't exist then b*
*2;n is
necessary.
Proof. Evidently 1 isPnecessary, so we turn to family 2. Suppose that an were
decomposable, an = i xiyi. where the xi's and yi's are homogeneous invariant
polynomials of positive degree. Note that an has a summand 2n. For this to occur
in one of the summands xiyi we must have that xi and yi each have a summand
n. However, by Lemma 5.1, there is no invariant polynomial which has n as a
summand. Hence an is not decomposable.
Similarly, looking at the summand 21n of dn we see that if dn were decomposa*
*ble
then that would imply the existence of an invariant polynomial with either n or
1n as a summand, both of which are precluded by Lemma 5.1. Now suppose n
is such that dn does not exist, i.e. there is no invariant polynomial with 21n *
*as a
summand. Then we can apply the same argument to the summand 31n of b2;n to
conclude that b2;n is not decomposable.
Finally consider bm1;m2;:::;mnwhere m1; : :;:mn satisfy the conditions given*
* in
5.3. If m1 = 2 then this has a summand 31m2 : :m:n. The decomposability
of bm1;m2;:::;mnwould then imply the existence of an invariant with a summand
n1 : :n:kwhere k 2 and 1 n1 < . .<.nk. We claim that no such invariant
exists and hence that bm1;m2;:::;mncannot be decomposable if m1 = 2. Before
proving the claim we will show that the indecomposability of bm1;m2;:::;mnin the
case where m1 > 2 also follows from it. For if m1 > 2 then bm1;m2;:::;mnhas a
summand 12m1-1m2 : :m:n. Any attempt to write this summand as a product
of two monomials will involve at least one factor which, by the claim, cannot o*
*ccur
as a summand in an invariant polynomial.
The proof of the claim is just a generalisation of that of Lemma 5.1. One se*
*es eas-
ily that any monomial n1 : :n:k, where k 2 and 1 n1 < . .<.nk, is the highest
monomial in its degree with respect to right lexicographic ordering. The image *
*un-
der O - 1 of this monomial has leading term 12n1-1n2 : :n:k, which immediately
precedes it and hence cannot occur in the image of any other monomial.
We have also verified that the generators of Conjecture 5.3 are sufficient in
degrees up to 40.
Conjecture 5.3 would imply that for k > 0, Hk (2; A*) = AO*=(O - 1)A* is
generated by the following elements : 1, an (for n 2) and dn (for n 3). These
elements determine non-zero classes in AO*=(O-1)A* since they contain the highe*
*st
terms (w.r.t. right lexicographic ordering) in their degrees and so they cannot*
* lie in
the image (O - 1)A*. Conversely, if a product of the elements listed in Conject*
*ure
5.3 has a factor bm1;m2;...;mnthen it will lie in (O - 1)A* as this is an ideal*
* in AO*.
Acknowledgements
We would like to thank Lionel Schwartz and Neil Strickland for helpful con-
versations and we acknowledge the support of TMR grants from the European
Union, held at the Laboratoire d'Analyse, Geometrie et Applications (UMR 7539
au CNRS), Universite Paris-Nord.
10 M. D. CROSSLEY AND SARAH WHITEHOUSE
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Departement de Mathematiques, Institut Galilee, Universite Paris 13, 93430 V*
*il-
letaneuse, France.
E-mail address: crossley@math.univ-paris13.fr, sarah@math.univ-paris13.fr