k. It is shown in [K ] t*
*hat
thePexcess of Milnor basis elements can easily be computed by ex(Sq (r1; : :;:r*
*m))=
m
i=1ri, and that for an arbitrary 2 A, ex() is the minimum of the excesses of*
* the
Milnor basis summands of .
For any finite sequence R = (r1; : :;:rmo)f non-negative integers let Sqt(R) *
*de-
note the Milnor basis element Sqt(r1;P: :;:rm.)The dimension of an arbitrary Mi*
*l-
nor basis element SqP(r1; : :;:rmi)s mi=1(2i- 1)ri. Consequently the dimensio*
*n of
Sqt(r1; : :;:rmi)s mi=1(2ti- 1)ri. As in the introduction, for a fixed t let *
*Bt be the
vector subspace of A with basis the set of all Sqt(R) . One can easily verify *
*[Gal ],
either directly or by using the results in [AD ], that Bt is a Hopf subalgebra *
*of A.
We note that B1 = A and that all of the non-zero elements of Bt are in dimensio*
*ns
divisible by (2t- 1).
Notice that for any t and k
i j i j
(2.2) 2t- 1 flt(k) = 2tk- 1 .
P m t P m i tk j
Consequently, n = i=1flt(ki)if and only if (2 - 1)n = i=1 2 i- 1.
Let Sqt(r1; : :;:rmb)e any Milnor basis element in dimension (2t- 1)n. Then by
the dimension formula
i j mX i j
2t- 1 n = 2ti- 1 ri
i=1
(2.3) i j mX riX
= 2t- 1 flt(i)
i=1j=1
and so
mX riX
(2.4) n = flt(i).
i=1j=1
P m
This expression provides us with a t-decomposition of n of length i=1ri, the *
*excess
of Sqt(r1; : :;:rm.)It is elementary to check that this correspondence between *
*Milnor
basis elements in Bt in dimension (2t- 1)n and t-decompositions of n of length
ex() is one to one (modulo reordering of the t-decompositions and interspersion*
* of
zeros). Thus the value of t(n) is exactly the excess of the element of Btin dim*
*ension
(2t- 1)n having least excess, i.e.
n i j o
(2.5) t(n) = min ex() | 2 Bt; dim() = 2t- 1 n .
4 KENNETH G. MONKS
Accordingly, we will call t the minimum excess function (for the subalgebra Bt).
In particular, all elements of A in dimension n have excess greater than or equ*
*al to
1(n).
We now establish some number theoretic properties of the minimum excess funct*
*ion
that will be needed for our proofs of the main results. We begin by proving the*
* fact
referred to in the introduction.
Lemma 2.1. t(n) (2t- 1)1(n) for all t; n 1.
Proof.By the definition of 1(n) there are positive integers k1; k2; : :;:k1(n) *
*such
that
1(n)X
(2.6) n = fl1(ki).
i=1
For each ki let ki= tqi+ ri where qiand ri are non-negative integers and 0 ri<*
* t.
Then using (2.2) we have
1(n)X
n = fl1(tqi+ ri)
i=1
1(n)Xi j
= 2tqi+ri- 1
i=1 i j
(2.7) 1(n)X2i j(2tqi- 1) 2t(qi+1)- 13
= 4 2t- 2ri _________t+ (2ri- 1)____________t5
i=1 (2 - 1) (2 - 1)
1(n)X22t-2riX 2ri-1X 3
= 4 flt(qi)+ flt(qi+ 1)5
i=1 j=1 j=1
the last expression yielding a t-decomposition of n having at most (2t- 1) 1(n)
terms. Hence, t(n) (2t- 1)1(n). __|_ |
We will require more information about t.
Lemma 2.2. t(n) = n if n 2t.
Proof.Since n 2t we have n 2t < 2t+ 1 = flt(2). Since flt is strictly increas*
*ing,
the only possible t-decomposition of n is a sequence of n ones. __|_ |
We now need some new notation. Let K = (k1; : :;:kmb)e any sequence of non-
negative integers. Define
mX
(2.8) |K|= ki
i=1
POLYNOMIAL MODULES OVER THE STEENROD ALGEBRA 5
(2.9) (K) = max {ki}
i
and
mX
(2.10) Vt(K) = flt(ki).
i=1
Thus every such sequence K is a t-decomposition of Vt(K) :
Now assume that k1 k2 . . .km (so that (K) =k1) and also that |K| 1.
For such a sequence define
(2.11) ffi (K) = (k01; : :;:k0m)
by
8