CHANGE OF BASIS, MONOMIAL RELATIONS, AND PtsBASES
FOR THE STEENROD ALGEBRA
KENNETH G. MONKS
July 1995
Abstract. The relationship between several common bases for the mod 2 Ste*
*enrod
algebra is explored and a new family of bases consisting of monomials in *
*distinct
Pst's is developed. A recursive change of basis formula is produced to c*
*onvert
between the Milnor basis and each of the bases for which the change of ba*
*sis matrix
in every grading is upper triangular. In particular, it is shown that the*
* basis of
admissible monomials, the new Pstbases, and two bases due to D. Arnon, are
all bases having this property, and the corresponding change of basis for*
*mula is
produced for each of them. Some monomial relations for the mod 2 Steenrod
algebra are then obtained by exploring the change of basis transformation*
*s.
1. Introduction
There are many descriptions of bases for the mod 2 Steenrod algebra, A; in the
literature. In addition to the classical basis of admissible monomials, there a*
*re the
bases developed by Milnor [Mil] and Wall [Wall] as well as the more recent bases
developed by D. Arnon [Arn ] and R. Wood [Wo ]. In this article we investigate
the relationship between these bases, and add a family of new bases consisting *
*of
monomials in distinct Pts's to the existing collection. These bases are all des*
*cribed
in detail in Section 3.
Given so many different bases, a natural question to ask is: how can we conve*
*rt
from one basis to the other? Since almost all of the bases under consideration*
* are
described in terms of unevaluated products of A, such simple linear algebraic i*
*nfor-
mation actually can yield information about the product structure of A as well.
All of the bases we consider can be described in terms of unevaluated monomia*
*ls in
Milnor basis elements. Thus it is a simple matter to convert from one of these *
*bases,
call it B, to the Milnor basis, BMil, by using the product formula (3.1) develo*
*ped by
Milnor [Mil]. A difficulty arises when trying to convert in the other direction*
*: from
___________
1991 Mathematics Subject Classification. Primary 55S10, 55S05; Secondary 57T0*
*5.
Key words and phrases. Steenrod Algebra.
1
2 KENNETH G. MONKS
the Milnor basis to back the basis B. Having such a formula for every basis wou*
*ld
then allow us to convert between any two bases, indirectly, via the Milnor basi*
*s.
A brute force approach might be to compute the change of basis matrix, M, from
B to BMil in a given grading using the Milnor product formula and compute M-1
to obtain the change of basis matrix in the opposite direction. But this appro*
*ach
is extremely inefficient and is unworkable in all but the lowest gradings where*
* the
vector space dimension is quite small.
Suppose, however, that we have the following situation.
Definition 1.1.Suppose there exists orderings, and _; of bases B and BMil re-
spectively, such that the change of basis matrix M (with respect to these order*
*ings) is
upper triangular in every grading. In this situation we say the basis B is tria*
*ngular
with respect to the Milnor basis.
This will be the situation if and only if there is an order preserving biject*
*ion
fl : BMil ! B such that fl () is the -largest summand of Milnor element when
expressed in basis B.
Remark 1.1. If B is triangular with respect to the Milnor basis, then we have *
*a well
defined recursive formula to convert a Milnor basis element to the basis B: Nam*
*ely,
for any 2 BMil, we have
X
(1.1) B = fl () + (i)B
i
P
where xB denotes the representation of x in basis B and fl () = + ii is the
Milnor representation of fl ()Mil obtained via the Milnor product formula.
This is a well defined recursive formula because all of the Milnor basis elem*
*ents i
must be strictly _-less than and so the recursion must eventually end when we *
*reach
elements for which fl () = holds. Since A is finite dimensional in each gradin*
*g, we
must have fl () = for the which is the _ -smallest Milnor basis element in a *
*given
grading.
Thus in order to show that B is triangular with respect to the Milnor basis a*
*nd
determine the change of basis formula (1.1) for converting an element from the *
*Milnor
basis to basis B it suffices to:
1. Define a bijection fl : BMil ! B.
2. Define the ordering _ on BMil. Then let be the unique ordering of B such
that fl is order preserving.
3. Prove that fl () is the -largest summand of the representation of B for a*
*ny
2 BMil.
We will follow this procedure several times in what follows. Note that requir*
*ement
#3 can also be satisfied by showing fl-1 () is the _-largest Milnor basis summ*
*and
of the Milnor basis representation of for any 2 B; since fl is order preservi*
*ng and
CHANGE OF BASIS IN THE STEENROD ALGEBRA 3
the inverse of a triangular matrix is also triangular. Also in place of require*
*ment #2
we can define the ordering on B and then let _ be the unique ordering of BMil
such that fl is order preserving.
In this article we will accomplish three things. First, we will construct a *
*new
family of bases for the Steenrod algebra A consisting of monomials in distinct *
*Pts's
and add these new bases to the list of bases being considered in this article. *
*Second,
we will determine which of the bases being considered are triangular with respe*
*ct
to the Milnor basis, and determine the change of basis formula of the form (1.1*
*) for
each basis that is. Finally, we will show how such information may lead to prod*
*uct
information by determining an infinite family of elements which are both admiss*
*ible
monomial and Milnor basis elements.
2. Summary of Main Results
In this section we give a general overview of the main results which are cont*
*ained
in this paper. The details, notation, background and proofs will be presented l*
*ater
in the paper. Our first result is the construction of an infinite family of new*
* bases
for A. Let R denote right lexicographic order (Definition 3.1).
Theorem 2.1. The set, BPR , of all monomials of the form Pts00Pts11.P.s.ptpsuc*
*h that
(s0; t0)R (s1; t1)R . .R. (sp; tp)is a basis for A. In addition, any set BP
obtained by changing the order of the factors of any of the monomials in BPR is*
* also
a basis for A:
Adding these bases to the list of bases mentioned above (those of Wall, Arnon*
*, and
Wood and the basis of admissible monomials) we can completely determine which of
these bases are triangular with respect to the Milnor basis and determine the c*
*hange
of basis formula of the form (1.1) by specifying the required fl:
Theorem 2.2.
(1) The following bases are triangular with respect to the Milnor basis and h*
*ave
change of basis formula (1.1) for the value of fl shown in the table:
_____________________________________|| |
|| Basis ||fl required for||
|______________________|formula(1.1)__|
| Admissible monomials |Definition 4.1 |
|| s || ||
||Any Pt basis |Definition|5.1 ||
||Arnon C ||Definition 6.1||
|_Arnon_A_____________|Definition__7.1|___
(2) Wall's basis, Wood's Y basis, and Wood's Z basis are not triangular with
respect to the Milnor basis.
4 KENNETH G. MONKS
It should be noted that in each case there is a simple heuristic for computin*
*g fl
which makes these change of basis formulas quite easy to use in practice. We g*
*ive
both these heuristics and sample calculations along with the proofs in later se*
*ctions
of the article.
Of some interest in its own right is the unusual ordering of the Milnor basis*
* elements
used in the proof of Theorem 2.2 for the Arnon A basis. This ordering is given*
* in
Definition 7.3.
One way to improve on the recursive change of basis formulas given in Theorem
2.2, would be to determine explicit non-recursive formulas. As a first step in*
* this
direction one might ask what elements two bases have in common. For example, it
is well known that the Sq (n)= Sqn are common to both the Milnor and admissible
monomial bases. Our final result determines an infinite family of elements whic*
*h are
common to these two bases.
Theorem 2.3. If ri -1 mod 2!(ri+1)for all 1 i < m then Sq(r1; : :;:rmi)s an e*
*l-
ement of both the Milnor and admissible monomial bases. In this case Sq(r1; : :*
*;:rm=)
Sqt1Sqt2. .S.qtmwhere tm = rm and ti= ri+ 2ti+1 for 1 i < m.
We point out that this linear algebra result is actually providing us with in*
*formation
about monomial products in A. Based on computer calculations we conjecture that
these are the only elements common to the Milnor and admissible monomial bases.
It is hoped that results of this sort would provide the first step in determini*
*ng non-
recursive change of basis formulas for these bases.
3. Bases for A: Old and New
We begin by describing the bases to be discussed in this article. Algebraica*
*lly
the Steenrod algebra can be described as the quotient of the free associative g*
*raded
algebra over the field with two elements, F2, on symbols Sqn in grading n; by t*
*he
ideal generated by the Adem relations:
ba_2cX !
b - n - 1 a+b-n n
SqaSq b= Sq Sq (fora < 2b)
n=0 a - 2n
where the binomial coefficients are taken mod 2 and Sq0 = 1, the multiplicati*
*ve
identity.
In order to describe the bases we wish to consider we first define the follow*
*ing.
Definition 3.1.Let R = r1; : :;:rm and S = s1; : :;:sn be finite sequences of i*
*nte-
gers. Write R R S if R is less than S in lexicographic order from the right, i.*
*e. if
either m < n or else m = n and there exists i such that ri < si and rj = sj for*
* all
j > i . If R R S we will say R is rlex less than S. We make a similar definition
for left lexicographic order, i.e. R L S if there exists i such that ri< si and*
* rj = sj
CHANGE OF BASIS IN THE STEENROD ALGEBRA 5
for all j < i (where we take rk = 0 for k > m and sk = 0 for k > n). If R L S we
will say R is llex less than S.
The bases we consider in this article are:
(1) Admissible monomials: A monomial of the form Sqt1Sqt2. .S.qtmis said to
be admissible if ti 2ti+1for 1 i < m: The set of all admissible monomials
forms a basis for A which we will denote by BAdm : Whether Sqt1Sqt2. .S.q*
*tm
is admissible or not, we will often abbreviate Sqt1Sqt2. .S.qtmby Sqt1;::*
*:;tmand
in addition if T = t1; : :;:tm we will write Sqfor Sqt1;:::;tm.
(2) Milnor [Mil]: Milnor showed that A is also a Hopf algebra whose dual, A*,*
* is
the polynomial algebra F2[1; 2; : :]:on generators n in grading 2n - 1. T*
*he
basis of A which is dual to the basis of monomials in A* is called the Mi*
*lnor
basis and will be denoted BMil. The element dual to r11r22. .r.mmin this *
*basis
is denoted Sq(r1; : :;:rm.)Comparing with the notation given above we have
Sq (n)= Sqn : If R = r1; : :;:rm is a finite sequence of nonnegative inte*
*gers,
we will often use multi-index notation and write Sq for the Milnor ba*
*sis
element Sq (r1; : :;:rm.)
The algebra structure on A in this basis can be described by the product
formula given by Milnor. Namely,
X
(3.1) Sq (r1; r2; :S:):q(s1; s2;=: :):Sq(t1; t2; : :):
X
where the sum is taken over all matrices X = satisfying:
X
(3.2) xij= sj
X i
(3.3) 2jxij= ri
j
Y
(3.4) (xh0; xh-1;1; : :;:x0h) 1 (mod 2)
h
where (n1; : :;:nmi)s the multinomial coefficient (n1+. .+.nm )!= (n1! . *
*.n.m!).
(The value of x00is never used an may be taken to be 0.) Each such allowa*
*ble
matrix produces a summand Sq (t1; t2; :g:):iven by
X
(3.5) th = xij:
i+j=h
In such a situation we say that X is a Sq Sq ~~-allowable matrix which
produces Sq : We will also find it convenient to say that X produces *
*the
sequence T if T satisfies (3.5) regardless of whether or not X is allowab*
*le.
n 2n-1 2k
(3) Arnon A [Arn , Theorem 1A]: Define Xnk= Sq2 Sq . .S.q . Then the set
of all monomials of the form Xn0k0Xn1k1.X.n.pkpsuch that (n0; k0)L (n1; k*
*1)L
. .L.(np; kp)forms a basis for A which we will denoted by BArA:
6 KENNETH G. MONKS
k 2k+1 2n
(4) Wall [Wall, pg. 433]: Define Qnk= Sq2 Sq . .S.q : Then the set of all
monomials of the form Qn0k0Qn1k1.Q.n.pkpsuch that (np; kp)L (np-1; kp-1)L
. .L.(n0; k0)forms a basis for A which we will denoted by BWall: This bas*
*is
was also discussed in [Arn , Theorem 1B].
(5) Arnon C [Arn , Theorem 1C]: A monomial of the form Sqtm Sqtm-1. .S.qt1
is said to be C-admissible if ti+1 2ti for 1 i < m and ti is divisible *
*by
2i-1: The set of all C-admissible monomials forms a basis for A which we *
*will
denote by BArC:
n(2k+1-1)
(6) Wood Y [Wo , Theorem 1]: Define Ykn = Sq2 . Then the set of all
monomials of the form Ykn00Ykn11. .Y.npkpsuch that (np; kp)L (np-1; kp-1)L
. .L.(n0; k0)forms a basis for A which we will denote by BWdY : Wood shows
that this basis has a nice property with respect to the Hopf subalgebras *
*An of
i
A generated by the Sq2 with i n . Namely if any factor of any summand
of WdY is not in An then itself is not in An.
n(2k+1-1)
(7) Wood Z [Wo , Theorem 2]: Let Ykn = Sq 2 as above. Then the
set of all monomials of the form Ykn00Ykn11. .Y.npkpsuch that (np + kp; n*
*p)L
(np-1+ kp-1; np-1)L . .L.(n0 + k0; n0)forms a basis for A which we will
denoted by BWdZ : Wood shows that this basis also has the same nice prope*
*rty
with respect to the Hopf subalgebras An that was mentioned above for the Y
basis.
(8) Pts-bases: In this article we will prove that the following is a basis fo*
*r A: Let
Pts= Sq(r1; : :;:rt)where rt= 2s and ri= 0 for i < t . For each finite se*
*t, S;
of Pts's choose an ordering of the elements of S; and let M(S) be the mon*
*omial
formednby taking the productoof the elements of S in increasing order, i*
*.e. if
S = Pts00; Pts11; : :;:Ptsppand we order the elements of S in the order *
*shown
then M (S)= Pts00Pts11.P.s.ptp: The monomials M (S) form a basis for A: T*
*his
gives us an infinite family of bases, one for each choice of ordering the*
* sets S
(not all of them are distinct, of course).
For example, the set of all monomials of the form Pts00Pts11.P.s.ptpsuch *
*that
(s0; t0)R (s1; t1)R . .R.(sp; tp)is one such basis which we will denote by
BPR :
Before leaving this section we give a few elementary definitions and notation*
* that
will be needed later on.
Definition 3.2.If BName is one of the bases of A described above and 2 A then
Name will denote the representation of in that basis.
i j
For example, Sq2Sq 1 Mil= Sq(3) + Sq(0; 1)while Sq (0; 1)Adm= Sq2Sq 1+ Sq3:
For any Milnor basis consists of element, Sq (r1; : :;:rm;)itPis clear from t*
*he defi-
nition that the grading or degree of Sq (r1; : :;:rmi)s mi=1(2i- 1)ri: For an*
*y of the
CHANGE OF BASIS IN THE STEENROD ALGEBRA 7
other bases, the degree of a monomial is thePsum of the degrees of its Milnor b*
*asis
factors. The excess of Sq (r1; : :;:rmi)s mi=1riPand its length is m. The e*
*xcess
of an admissible monomial Sqt1;:::;tmis tm + m-1i=1(ti- 2ti+1): We will denot*
*e the
excess of 2 BMil by ex() : Note that Sq (r1; : :;:rmi)s not uniquely determined
by its degree, excess, and length as can be seen by the elements Sq (0; 1; 2; 0*
*;a1)nd
Sq(2; 0; 0; 1;:1)
We can extend the definitions of left and right lexicographic order to both M*
*ilnor
basis elements and monomials in Sqn in the obvious manner, i.e. if R R S then
Sq R Sq ~~~~ and SqR R Sq Sand similarly if R L S then Sq L Sq ~~~~
and SqR L SqS .
For any positive integerPn, let ffi(n) be the coefficient of 2i in the binary*
* expan-
sion of n, i.e. n = 1i=0ffi(n)2i and ffi(n) 2 {0; 1}for all i. We say that m *
*and n
are disjoint and write m n if ffi(m) + ffi(n) 1 for all i. It isiwelljknown *
*that
this is equivalent to the condition that the binomial coefficient m+nm is odd.*
* Con-
sequently, the multinomial coefficient (n1; : :;:nmi)s odd if and only if the i*
*ntegers
n1; : :;:nm are pairwise disjoint. This fact is used frequently throughout the *
*article
when evaluating condition (3.4).
We often write 2i 2 n for ffi(n) = 1 since the meaning is clear from the cont*
*ext.
The following fact will be used implicitly several times and is an elementary e*
*xercise
in binary arithmetic. Let 0 b < 2t. Then
(3.6) 2l2 b , 2l2 2ta + b and l < t:
Finally let (n)be the largest integer such that n 0 mod 2(n) (and take (0)=
1 ). Let ! (n)be the smallest integer such that 2!(n)> n: Notice that for n > 0*
* we
always have 2(n) 2 n and also that (n)< ! (n).
4.Milnor vs. Admissible
We begin by focusing on the relationship between BMil and BAdm . The elements
Sq(n) (= Sq n) are common to both the Milnor and admissible monomial bases.
Therefore to express an admissible monomial in the Milnor basis, we only need u*
*se
the product formula (3.1) for multiplying Milnor basis elements.
To convert a element from the Milnor basis to the basis of admissible monomia*
*ls
we now show that the basis of admissible monomials is triangular with respect t*
*o the
Milnor basis and define the fl and ordering needed for the recursive formula (*
*1.1).
To satisfy requirement #1 following (1.1) we make the following definition.
Definition 4.1.Let Sq = Sq (r1; : :;:rmb)e a Milnor basis element. Define
fl (Sq(r1; : :;:rm))= Sqt1Sqt2. .S.qtmwhere
mX
(4.1) ti= 2k-irk:
k=i
8 KENNETH G. MONKS
(Abbreviation: we will sometimes write fl for fl () ).
Note that the ti can quickly be computed by starting with tm = rm and then
applying the simple recursion
(4.2) ti= ri+ 2ti+1:
It follows immediately that fl Sq is an admissible monomial for any Sq 2 *
*BMil:
The map fl is clearly a bijection on A in each degree and preserves both excess*
* and
rlex order. So we take both and _ to be R in this case to satisfy requirement *
*#2
following (1.1). So in order to satisfy requirement #3 following (1.1) we show:
Theorem 4.1. fl Sq is the rlex-largest summand of Sq Adm.
Hence BAdm is triangular with respect to BMil: As a result we have a recursive
formula of the form (1.1) for converting an element of A from the Milnor basis *
*to the
basis of admissible monomials.
P
Corollary 4.2. Let Sq 2 BMil and suppose fl (Sq)Mil= Sq + iSq :
Then
X
SqAdm = fl Sq + SqAdm
i
is a well defined recursive formula for computing Sq Adm.
Note that fl (Sq)Mil can easily be obtained from the Milnor product formula
(3.1). All of the elements Sq are strictly rlex-less than Sq which is*
* why
the recursive formula is well defined. This makes the formula quite easy to us*
*e in
practice.
For example, to convert Sq(2; 2)to the basis of admissible monomials using Co*
*rol-
lary 4.2 we first compute fl Sq(2; 2)= Sq 6Sq2. By the Milnor product formula,
Sq(6)Sq (2) = Sq (2; 2)+ Sq (5; 1): The error term, Sq (5; 1)is smaller than the
original term Sq (2; 2)in rlex order and we invoke Corollary 4.2 again. This t*
*ime
fl Sq(5; 1)= Sq7Sq 1; but by the Milnor product formula we find that Sq (7)Sq(1*
*)=
Sq(5; 1): Thus we have shown that
Sq (2; 2)= Sq6Sq 2+ Sq7Sq1
which provides the conversion we desired.
In order to prove these results we begin by proving a useful lemma.
CHANGE OF BASIS IN THE STEENROD ALGEBRA 9
Let Sq (r1; : :;:rm,)Sq (s1; : :;:sn)2 BMil: Let X = be the matrix
______*______||0_||0_||._.|.|0_
______r1______||0_||0_||._.|.|0_
.. . . . .
_______.______||..||..||..||.._
_____rm-1_____||0_||0_||._.|.|0_X
rm - 2ksk ||s1s||2||. .|.|sn
We will call X the rlex champion matrix for Sq (r1; : :;:rmS)q(s1; : :;:sn).
Lemma 4.3. If the rlex champion matrix for Sq Sq~~~~ produces T , then eve*
*ry
other Sq Sq~~~~-allowable matrix produces a sequence which is rlex-less than*
* T:
Proof . By (3.2) and (3.3), any other such matrix must have xij6= 0 for some
0 i < m and 1 j n: Let j be the largest such value. Then by (3.2) we
have xmj < sj: Therefore the element Sq(u1; : :;:um+np)roduced by the new matrix
must have uk = sk = tk for m + j < k m + n and um+j = xmj < sj = tj: Thus
Sq~~__ R Sq :
As an immediate consequence we have
Corollary 4.4. Let T be the sequence produced by the rlex champion matrix for
Sq Sq__~~ . If U R S then for every Milnor summand Sq of the product
Sq Sq~~__ we have V R T:
This follows from Lemma 4.3 and the fact that the sequence T 0produced by the
rlex champion matrix for Sq Sq ____ is easily seen to be rlex less than T:
We are now ready to prove Theorem 4.1.
Proof of Theorem 4.1. We wish to show that fl Sq is the rlex-largest summa*
*nd
of Sq Adm . Since fl is bijective and preserves rlex order, itisufficesjto s*
*how that
for any admissible monomial Sqthe Milnor basis element fl-1 Sq is the rl*
*ex
i j
largest summand of Sq Mil:
i *
* j
Let T = t1; : :;:tm be an admissible sequence. Recall that by (4.2) fl-1 Sq<*
*T> =
Sq(r1; : :;:rmw)here rm = tm and ri= ti- 2ti+1for i < m: We proceed by induction
on m . i j i j
If m = 1 then Sq Mil= Sq (t1)and fl-1 Sq = Sq (t1), so the base case
holds.
Now for the inductive hypothesisiassumejthat for any admissible monomialiSq__~~j
of length less than m , fl-1 Sq~~~~ is the rlex-largest summand of Sq~~~~Mil. T*
*hen in
i j i *
* j
particular, fl-1 Sqt2;:::;tm= Sq(r2; : :;:rmi)s the rlex-largest summand of S*
*qt2;:::;tmMil:
So we can write
i j X
Sqt2;:::;tmMil= Sq(r2; : :;:rm+) Sq
10 KENNETH G. MONKS
where Sq R Sq (r2; : :;:rmf)or all i:
Thus we have
i j i j
Sqt1;:::;tmMil= Sq (t1) Sqt2;:::;tmMil
i X j
= Sq (t1) Sq(r2; : :;:rm+) Sq
X
= Sq (t1)Sq(r2; : :;:rm+) Sq(t1)Sq :
The rlex champion matrix X for Sq (t1)Sq(r2; : :;:rmi)s
________*________||0_||0_||._.|.|0__P
t1 - mk=22k-1rkr||2||r3.||.|.|rm:
Clearly X is admissible. To see that it produces Sq (r1; : :;:rmw)e need only v*
*erify
that
mX mX
t1 - 2k-1rk= t1 - 2 2k-2rk
k=2 k=2
= t1 - 2t2
= r1:
Therefore by Lemma 4.3 every other Sq (t1)Sq(r2; : :;:rm )-allowable matrix pro-
duces Milnor elements which are rlex-less than Sq (r1; : :;:rm.) So Sq (r1; : :*
*;:rm )
is the rlex-largest summand of Sq(t1)Sq(r2; : :;:rm ). In addition, every summa*
*nd of
Sq(t1)Sq is rlex less than Sq (r1; : :;:rmb)y Corollary 4.4.
5. Milnor vs. Pts
We now turn our attention to the relationship between BMil and BPR . All of t*
*he
results and arguments in this section carry over to any Ptsbasis, but we illust*
*rate
them for this particular ordering of the monomial factors. The elements Ptsare
common to both the Milnor and Ptsbases. Therefore to express an element of BPR *
*in
the Milnor basis, we only need use the product formula (3.1) for multiplying Mi*
*lnor
basis elements.
Notice that we have not yet shown that BPR is a basis for A although we have
defined it as a set. To see that BPR is in fact a triangular basis with respect*
* to the
Milnor basis we begin by defining a grading preserving bijection fl : BMil ! BP*
*R .
Definition 5.1.Let Sq (r1; : :;:rm2)BMil. Define
fl (Sq(r1; : :;:rm))= Pts00Pts11.P.s.ptp
where the right hand side is the unique monomial in BPR satisfying
Pjiis a factor ofPts00Pts11.P.s.ptp, ffi(rj)= 1
for all i and j:
CHANGE OF BASIS IN THE STEENROD ALGEBRA 11
The map fl is clearly a bijection on A in each grading.
There is a useful heuristic device for computing the fl Sq(r1; : :;:rm:)We de*
*fine
the binary chart of Sq (r1; : :;:rmt)o be the array:
.. . . .
. || .. .. ..
2 ||ff2(r1)ff2(r2)ff2(r3) . . .
s 1 ||ff1(r1)ff1(r2)ff1(r3) . . .
0 ||ff0(r1)ff0(r2)ff0(r3)_._._.
1 2 3 . . .
t
In other words simply write the binary expansions or the numbers r1; : :;:rm ve*
*rti-
cally next to each other. Then Ptsis a factor of fl Sq(r1; : :;:rmi)f and only *
*if there
is a 1 in location (s; t)in the binary chart. The factors are then multiplied *
*in the
correct order for whichever Ptsbasis we are considering.
For example, to compute fl Sq(2; 5; 1)we make the chart:
1
1 0
0 1 1
and read off the factors P11; P20; P22, and P30: Multiplying them in the correc*
*t order
for BPR we get fl Sq(2; 5; 1)= P11P20P22P30:
Now define an ordering E on BMil as follows.
Definition 5.2.For any Sq ; Sq~~~~ 2 BMil; we say Sq E Sq ~~~~ if
ex(Sq )< ex(Sq ~~~~)
or else
ex (Sq) = ex(Sq ~~~~)and Sq R Sq ~~~~:
The second condition is simply used to make a total ordering out of the parti*
*al
ordering induced by excess and is never used.
Finally let be the ordering induced on BPR induced by the bijection fl and E*
* :
Then we have:
Theorem 5.1. fl Sq is the -largest summand of Sq PR.
It follows immediately that the elements of BPR are linearly independent in e*
*ach
grading and since fl is a bijection, BPR is, indeed, a basis as claimed. Furthe*
*r, with
this definition of fl and E we have satisfied requirements #1-3 in Section 1 an*
*d so
BPR is triangular with respect to BMil: As a result we have a recursive formula*
* of
the form (1.1) for converting an element of A from the Milnor basis to the basi*
*s of
admissible monomials.
12 KENNETH G. MONKS
P
Corollary 5.2. Let Sq 2 BMil and suppose fl (Sq)Mil= Sq + iSq :
Then
X
SqPR = fl Sq + SqPR
i
is a well defined recursive formula for computing Sq PR.
Note that fl (Sq)Mil can easily be obtained from the Milnor product formula
(3.1). All of the elements Sq are strictly E -less than Sq which is why*
* the
recursive formula is well defined.
For example, to convert Sq(4; 2)to the basis BPR using Corollary 5.2 we first*
* com-
pute fl Sq(4; 2)= P12P21. By the Milnor product formula, P12P21= Sq (4)Sq(0; 2)=
Sq(4; 2)+ Sq(0; 1; 1): The error term, Sq (0; 1; 1)is smaller than the original*
* term
Sq(4; 2)in E order and so we invoke Corollary 5.2 again. This time fl Sq(0; 1; *
*1)=
P20P30; but by the Milnor product formula we find that P20P30= Sq(0; 1)Sq(0; 0;*
* 1)=
Sq(0; 1; 1): Thus we have shown that
Sq(4; 2)= P12P21+ P20P30
which provides the conversion we desired.
In order to prove these results we begin by proving a few useful lemmas.
Let Sq (r1; : :;:rm,)Sq (s1; : :;:sn)2 BMil: Let X = be the matrix
_*__||s1s||2||._.|.|sn_
_r1_||0_||0_||._.|.|0_
.. . . . .
__._||..||..||..||.._
rm ||0 ||0.||.|.|0
We will call X the excess champion matrix for Sq (r1; : :;:rmS)q(s1; : :;:sn).
Let R = r1; r2; : :a:nd S = s1; s2; : :.:Then we define the obvious sum
R + S = r1 + s1; r2 + s2; : :;:ri+ si; : :::
In this notation we see that the excess champion matrix for Sq Sq ~~~~ produc*
*es
Sq : Notice that ex(Sq )= ex(Sq )+ ex(Sq ):
Lemma 5.3. If X is an allowable Sq Sq~~~~ matrix which produces Sq then
ex(Sq)< ex(Sq ):
P
Proof . Since the excess of Sq = Sq (t1; t2; :i:):s ti and by (3.5)Peac*
*h ti is
the sum of the ithdiagonal of X = , it follows that ex(Sq )= xij; i.*
*e. it is
i;j
the sum of all of the entriesPof the matrix. By (3.2) the sum of the entries in*
* columns
to the right of column 0; j>0xijmust equal ex (Sq~~~~): By (3.3) xi0 ri for ea*
*ch
i so that thePentries in column 0 must have a sum less than or equal to the exc*
*ess of
Sq, i.e. j=0xij ex(Sq ): But since X is not the excess champion matrix,
CHANGE OF BASIS IN THE STEENROD ALGEBRA 13
we must have xuv 6= 0 for some u > 0 and v > 0: But by (3.3) it follows that xu*
*0 < ru
and so the sum of column 0 is strictly less than ex(Sq ): Hence
X
ex (Sq) = xij
i;j
X X
= xi;j+ xi;j
j=0 j>0
< ex (Sq) + ex(Sq ~~~~)
= ex (Sq)
as claimed.
As an immediate consequence we have
Corollary 5.4. If ex(Sq ~~__) < ex(Sq ) then every Milnor summand Sq of
the product Sq ____Sq __~~ (or Sq ~~~~Sq ~~__) has excess less than ex(Sq ):
We are now ready to prove Theorem 5.1.
Proof of Theorem 5.1. We wish to show that fl Sq is the -largest sum-
mand ofiSq PR. Itjsuffices to show that for any elementiPts00Pts11.P.s.ptpi*
*njBPR ,
fl-1 Pts00Pts11.P.s.ptpis the E -largest summand of Pts00Pts11.P.s.ptpMil: We*
* will show
i sj *
* i sj
something slightly stronger, namely that fl-1 Pts00Pts11.P.t.ppis a summand of*
* Pts00Pts11.P.t.ppMil
i sj
and every other Milnor summand of Pts00Pts11.P.t.ppMilwill have excess strictl*
*y less
i i sjj
than ex fl-1 Pts00Pts11.P.t.pp:
i sj
Let = Pts00Pts11.P.s.ptp2 BPR and let Sq(r1; : :;:rm=)fl-1 Pts00Pts11.P.t.p*
*pwhere
P s
ri= tj=i2 j. We proceed by induction on p .
If p = 0 then (Pts00)Mil= Pts00and fl-1 (Pts00)= Pts00, so the base case hold*
*s.
Now for the inductive hypothesis assume that for any element 2 BPR having fe*
*wer
than p + 1 factors, fl-1 () is the rlex-largest summand of Mil. Then in particu*
*lar,
i s j
fl-1 Pts00Pts11.P.t.p-1p-1= Sq(r1; : :;:rm-1 ; rm - 2sp)
i sj
is a summand of Pts00Pts11.P.t.ppMiland every other summand has excess less th*
*an
ex(Sq(r1; : :;:rm-1 ; rm - 2sp)): So we can write
i sj X
Pts00Pts11.P.t.ppMil= Sq(r1; : :;:rm-1 ; rm - 2sp)+ Sq
where ex(Sq )< ex(Sq (r1; : :;:rm-1 ; rm -)2sp)for all i:
14 KENNETH G. MONKS
Thus we have
i sj i s j s
Pts00Pts11.P.t.ppMil= Pts00Pts11.P.t.p-1p-1MilPtpp
i X j s
= Sq(r1; : :;:rm-1 ; rm - 2sp)+ SqPtpp
X sp
= Sq(r1; : :;:rm-1 ; rm - 2sp)Ptspp+ SqPtp:
P sp
Now each of the Milnor summands of Sq Ptp must have excess strictly less *
*than
ex(Sq(r1; : :;:rm))by Corollary 5.4. Every summand of Sq(r1; : :;:rm-1 ; rm - 2*
*sp)Ptspp
other than Sq (r1; : :;:rmm)ust have excess strictly less than ex(Sq (r1; : :;:*
*rm))by
Lemma 5.3. Finally, it is easy to see that the excess champion matrix associate*
*d with
Sq(r1;i: :;:rm-1j; rm - 2sp)Ptsppis allowable and thus Sq (r1; : :;:rmi)s a sum*
*mand of
Pts00Pts11.P.s.ptpMil.
6. Milnor vs. Arnon C
We now turn to the relationship between BMil and BArC. In many ways this
relationship is similar to the situation we find for BAdm : The elements Sq(n) *
*are
again common to both bases, so to express 2 BArC in the Milnor basis, we only
need use the product formula (3.1).
To convert a element from the Milnor basis to the Arnon C basis we follow the*
* now
familiar path of showing that the basis of C-admissible monomials is triangular*
* with
respect to the Milnor basis by defining the appropriate fl and ordering needed*
* for
the recursive formula of the form (1.1).
Definition 6.1.Let Sq = Sq (r1; : :;:rmb)e a Milnor basis element. Define
fl (Sq(r1; : :;:rm))= SqtmSq tm-1. .S.qt1where
Xm
(6.1) ti= 2i-1 rk:
k=i
Note that fl (Sq(r1; : :;:rm))can easily be computed by the following heurist*
*ic.
First, write the sequence r1; : :;:rm in a vertical column with r1 on top. Then*
* working
to the left, construct the following triangular shaped diagram in which each co*
*lumn
contains entries which are twice the entry to its right:
r1
2r2 r2
.. .
. ..
2m-2 rm-1 . . .2rm-1 rm-1
2m-1_rm___2m-2_rm__._._._2rm____rm___
tm tm-1 . . . t2 t1
the value of ti is then simply the sum of the ithcolumn from the right as indic*
*ated.
CHANGE OF BASIS IN THE STEENROD ALGEBRA 15
It is clear from the definition that ti is divisible by 2i-1and also that
(6.2) ti+1= 2ti- 2iri
holds for 1 i < m. Hence ti+1 2ti so that fl Sq is indeed in BArC.
The map fl is a bijection on A in each grading and by (6.2)
i j
fl-1 Sqtm ;:::;t1= Sq(r1; : :;:rm )
where ri= 2ti-ti+1_2ifor 1 i < m and rm = _tm_2m-1:
For this basis we choose R for the ordering of BMil and let be the ordering
induced by fl on BArC: Then we have
Theorem 6.1. fl Sq is the -largest summand of Sq ArC.
Hence BArC is triangular with respect to BMil and we have a recursive formula*
* of
the form (1.1) for converting an element of A from the Milnor basis to the basi*
*s of
C-admissible monomials.
P
Corollary 6.2. Let Sq 2 BMil and suppose fl (Sq)Mil= Sq + iSq :
Then
X
Sq ArC= fl Sq + SqArC
i
is a well defined recursive formula for computing Sq ArC.
Note that once again fl (Sq)Milcan easily be obtained from the Milnor prod*
*uct
formula (3.1) and all of the elements Sq are strictly rlex-less than Sq *
* which
is why the recursive formula is well defined.
For example, to convert Sq (3; 2)to the basis of C-admissible monomials using
Corollary 6.2 we first compute fl Sq(3; 2)= Sq4Sq 5. By the Milnor product form*
*ula,
Sq(4)Sq (5) = Sq (3; 2)+ Sq (6; 1): The error term, Sq (6; 1)is smaller than the
original term Sq (3; 2)in rlex order and so we invoke Corollary 6.2 again. This*
* time
fl Sq(6; 1)= Sq2Sq 7; but by the Milnor product formula we find that Sq (2)Sq(7*
*)=
Sq(6; 1): Thus we have shown that
Sq (3; 2)= Sq4Sq 5+ Sq2Sq7
which provides the conversion we desired.
We are now ready to prove Theorem 6.1.
Proof of Theorem 6.1. Once again it sufficesitojshow that for any C-admissib*
*le
monomial Sqthe Milnor basis element fl-1 Sq is the rlex largest summand *
*of
i j
SqMil:
Let Sq= Sqtm ;:::;t1be a C-admissible monomial. Then
i j
fl-1 Sqtm ;:::;t1= Sq(r1; : :;:rm )
16 KENNETH G. MONKS
where ri= 2ti-ti+1_2ifori1j i < m and rm = _tm_2m-1:iWejproceed by induction on*
* m .
If m = 1 then Sq Mil= Sq (t1)and fl-1 Sq = Sq (t1), so the base case
holds.
Now for the inductive hypothesisiassumejthat for any admissible monomialiSq__~~j
of length less than m , fl-1 Sq~~~~ is the rlex-largest summand of Sq~~~~ Mil. *
*Then
in particular, we can compute
i j
fl-1 Sqtm-1;:::;t1= Sq(r1; : :;:rm-2 ; rm-1 + rm )
i j
which must be the rlex-largest summand of Sqtm-1;:::;t1Mil: So we can write
i j X
Sqtm ;:::;t1Mil= Sq(r1; : :;:rm-2 ; rm-1 + rm+) Sq
where Sq R Sq (r1; : :;:rm-2 ; rm-1 +frmo)r all i:
Thus we have
i j i j
Sqtm ;:::;t1Mil= Sq(tm ) Sqtm-1;:::;t1Mil
i X j
= Sq(tm ) Sq(r1; : :;:rm-2 ; rm-1 + rm+) Sq
X
= Sq(tm )Sq(r1; : :;:rm-2 ; rm-1 + rm+) Sq(tm )Sq:
The rlex champion matrix for Sq (tm )Sq(r1; : :;:rm-2 ; rm-1 +irms)not allowa*
*ble
in this case so instead we let X be the matrix
_*_||r1.||.|.|rm-2r||m-1___
0 ||0 ||. .|.|0 ||rm:
Clearly X is allowable and produces Sq (r1; : :;:rm:)To see that X is indeed a
Sq(tm )Sq(r1; : :;:rm-2 ; rm-1 + rmm)atrix we need only note that rm = _tm_2m-1:
Now every other Sq (tm )Sq(r1; : :;:rm-2 ; rm-1 +-rma)llowable matrix produces
Milnor elements which are rlex-less than Sq (r1; : :;:rms)ince by (3.3) any oth*
*er
such matrix must produce Sq (t1; : :;:tmw)ith tmP< rm :
So it remains to show that every summand of Sq (tm )Sqis rlex less than
Sq(r1; : :;:rm.) Let Sq = Sq (u1; : :;:un)be one of the summands. We know
Sq is rlex less than Sq (r1; : :;:rm-2 ; rm-1 +.rmI)f n < m - 1 then every
summand of Sq (tm )Sqhas length less than m and is therefore rlex less than
Sq(r1; : :;:rm.) On the other hand, if n = m - 1 then there exists j such that
uj < rj and
Sq (u1; : :;:um-1=)Sq(u1; : :;:uj; rj+1; : :;:rm-2 ; rm-1:+ rm )
In this case the matrix
_*_||u1_||._.|.|ujrj+1._._.rm-2_||rm-1__
0 ||0 ||. .|.|0 0 . . . 0 ||rm
CHANGE OF BASIS IN THE STEENROD ALGEBRA 17
produces the sequence (u1; : :;:uj; rj+1; : :;:rm-1(;wrmh)ether or not it is al*
*lowable)
which is clearly rlex less than (r1; : :;:rm:)By the same argument as above, an*
*y other
Sq(tm )Sq-allowable matrix must produce Sq for which V R U which is in
turn rlex less than (r1; : :;:rm.)
7. Milnor vs. Arnon A
The strangest of the bases discussed here which are triangular with respect t*
*o the
Milnor bases has to be BArA due to the unusual ordering _ on BMil that is used
for the proof. Since elements of BArA are monomials in the elements Sq (2n);we *
*can
express an admissible monomial in the Milnor basis by using the product formula
(3.1) for multiplying Milnor basis elements.
To convert a element from the Milnor basis to the basis of admissible monomia*
*ls
we show that the Arnon A basis is triangular with respect to the Milnor basis a*
*nd
define the fl and ordering _needed for the recursive formula of the form (1.1)*
*. For
fl we make the following:
Definition 7.1.Let Sq = Sq (r1; : :;:rmb)e a Milnor basis element. Define
fl (Sq(r1; : :;:rm))= Xn0k0Xn1k1.X.n.pkpwhere
(1) (n0; k0)L (n1; k1)L . .L.(np; kp)and
(2) Xnkis a factor of Xn0k0Xn1k1.X.n.pkpif and only if ffk (rn-k+1)= 1.
A heuristic for easily computing this gamma is very similar to that used for *
*BPR :
First, write down the binary chart for Sq (r1; : :;:rm.)Then for each chart loc*
*ation
(i; j) where there is a 1; we have an associated factor Xi+j-1jof fl Sq(r1; : :*
*;:rm.)
These factors are then multiplied in the correct order.
For example, to compute fl Sq(2; 5; 6)we make the chart:
1
1 0
0 1 1
and read off the factors X11; X10; X32, and X20: Multiplying them in the correc*
*t order
we get fl Sq(2; 5; 6)= X10X11X20X32:
The order _ on BMilwhich we require is quite unusual. We begin with an orderi*
*ng
on pairs of integers.
Definition 7.2.Define an ordering on N x N by (a; b) (c; d)if
(1) a + b < c + d or
(2) a + b = c + d and b < d:
For example, (0; 0)is the smallest element in this ordering and the ordering *
*begins
with
(0; 0) (1; 0) (0; 1) (2; 0) (1; 1) (0; 2) (3; 0) . .:.
18 KENNETH G. MONKS
The purpose of this ordering is to order the entries on our binary charts which*
* will
then provide an ordering on BMil:
Definition 7.3.Let Sq (r1; : :;:rma)nd Sq (s1; : :;:sn)be elements of BMil: We *
*say
Sq(s1; : :;:sn)A Sq (r1; : :;:rmi)f there exists (h; k)such that
(1) ffi(rj)= ffi(sj)for all (i; j) (h; k)and
(2) ffk (rh)< ffk (sh):
In other words, we compare the entries of the binary charts of Sq (r1; : :;:r*
*ma)nd
Sq(s1; : :;:sn)in increasing order until we find the first location (h; k)wher*
*e they
differ. Whichever element has the 0 at (h; k)is the larger element. (Note that *
*the
second condition is equivalent to the condition ffk (rh)= 0 and ffk (sh)= 1:)
Armed with this fl and ordering A on BMil we can now prove:
Theorem 7.1. Sq (r1; : :;:rmi)s the A-largest summand of fl Sq(r1; : :;:rmM)il:
Thus the Arnon A basis is triangular with respect to the Milnor basis and we *
*have
the recursive formula (1.1) for converting an element of A from the Milnor basi*
*s to
the basis BArA.
P
Corollary 7.2. Let Sq 2 BMil and suppose fl (Sq)Mil= Sq + iSq :
Then
X
Sq ArA= fl Sq + SqArA
i
is a well defined recursive formula for computing Sq ArA.
For example, to compute Sq (2; 2)ArAwe first compute fl Sq(2; 2)= X11X21: By *
*the
Milnor product formula
X11X21 = Sq (2)Sq(4) Sq(2)
= Sq (2; 2)+ Sq(5; 1):
Applying Corollary 7.2 to the error term we find fl Sq(5; 1)= X00X10X22: So by *
*the
product formula
X00X10X22 = Sq (1)Sq (2)Sq(1)Sq (4)
= Sq (5; 1):
Thus we have the desired answer
Sq (2; 2)= X11X21+ X00X10X22:
In order to prove these results we begin by defining some notation that will *
*be
convenient.
Definition 7.4.Let Sq(r1; : :;:rmb)e any Milnor basis element. We say Sq(r1; : *
*:;:rm )
is zero up to (h; k)if ffi(rj)= 0 for all (i; j) (h; k):
CHANGE OF BASIS IN THE STEENROD ALGEBRA 19
Definition 7.5.Let Sq(r1; : :;:rmb)e any Milnor basis element. We say Sq(r1; : *
*:;:rm )
has a 1 at (h; k)if ffk (rh)= 1:
Clearly if Sq is zero through (h; k)and Sq ~~~~is not, then Sq ~~~~A Sq :
This notation is very intuitive when considering the ordering A and the binary
charts of Milnor basis elements, as in the following technical lemma.
Lemma 7.3. Let = Xn0k0Xn1k1.X.n.pkp2 BArA and let Sq (r1; : :;:rm=)fl-1 () :
Then
(1) Sq (r1; : :;:rmh)as a 1 at (n0 - k0 + 1; k0):
(2) Sq (r1; : :;:rmi)s zero up to (n0 - k0 + 1; k0)
(3) Xn0k0Xn1k1.X.n.pkp= Sq (2n0)Xn0-1k0Xn1k1.X.n.pkpand Xn0-1k0Xn1k1.X.n.pkp2*
* BArA
(takeiXn0-1k0= 1 if n0j= k0). i *
* j
(4) fl-1 Xn0-1k0Xn1k1.X.n.pkp= Sq r1; : :;:rh-1; rh + 2k0; rh+1 - 2k0; rh+2*
*; : :;:rm
whereih = n0 - k0 ( if h =j0 we interpreting the right hand expression as
Sq r1 - 2k0; r2; : :;:rm).
Proof: (1) By definition of fl; Xn0k0being a factor of implies that ffk0(rn0*
*-k0+1)= 1
which in turn implies that Sq (r1; : :;:rmh)as a 1 at (n0 - k0 + 1; k0):
(2) Assume the contrary. Then there must be (i; j) (n0 - k0 + 1; k0)such that
Sq(r1; : :;:rmh)as a 1 at (i; j); i.e. such that ffj(ri)= 1: Thus by definition*
* of fl;
Xi+j-1jmust be a factor of : Now (i; j) (n0 - k0 + 1; k0)implies that either i+*
*j <
n0+1 or else i+j = n0+1 and j < k0 so that in either case (i + j - 1; j)L (n0; *
*k0):
But this contradicts the factor Xn0k0must be the smallest factor in left lexico*
*graphic
order of its indices by definition of BArA:
(3) This follows trivially from
(n0 - 1; k0)L (n0; k0)L . .L.(nm ; km )
i j
and Xn0k0= Sq(2n0)Sq (2n0-1). .S.q2k0 = Sq(2n0)Xn0-1k0:
(4) Since Xn0k0Xn1k1.X.n.pkpand Xn0-1k0Xn1k1.X.n.pkponlyidifferjin the first *
*factor, then
by definition of fl, Sq (r1; : :;:rma)nd fl-1 Xn0-1k0Xn1k1.X.n.pkpmust have id*
*entical
binary charts with the exception of the 1's corresponding to theileading factor*
*s.j
By (1), Sq (r1; : :;:rmh)as a 1 at (n0 - k0 + 1; k0)and fl-1 Xn0-1k0Xn1k1.X.n.*
*pkphas
a 1 at (n0 - k0; k0): But byi(2) Sq (r1; : :;:rmd)oesjnot have a 1 at (n0 - k0;*
* k0)
and it is clear that fl-1 Xn0-1k0Xn1k1.X.n.pkpdoes not have a 1 at (n0 - k0 + *
*1; k0)
sinceiXn0k0is not a factorj(remembering that (n0; k0)L (n1; k1)). Thus to obta*
*in
fl-1 Xn0-1k0Xn1k1.X.n.pkpfrom Sq(r1; : :;:rmw)e simply remove the 1 at (n0 - k*
*0 + 1; k0)
by subtracting 2k0from rn0-k0+1and create a 1 at (n0 - k0; k0)by adding 2k0to r*
*n0-k0:
20 KENNETH G. MONKS
Thus
i n j i *
* j
fl-1 Xn0-1k0Xn1k1.X.p.kp= Sq r1; : :;:rn0-k0+ 2k0; rn0-k0+1- 2k0; : :;:rm
as required.
We are now ready to prove Theorem 7.1. i j
Proof of Theorem 7.1. It suffices to show that fl-1 Xn0k0Xn1k1.X.n.pkpis th*
*e A-
i n j n
largest summand of Xn0k0Xn1k1.X.p.kpMil: So let Xn0k0Xn1k1.X.p.kp2 BArA: Then
i n j
fl-1 Xn0k0Xn1k1.X.p.kp= Sq (r1; : :;:rmw)here ffk (rn-k+1)= 1 if and only if X*
*nk
P p
is a factor of Xn0k0Xn1k1.X.n.pkp. Let q = i=0(ni- ki+ 1) which is the total *
*number
of factors of the form Sq (2i)in the product Xn0k0Xn1k1.X.n.pkpwhen expanded us*
*ing
the definition of Xnk. We proceed by inductionion q. j
If q = 1 then p = 0 and n0 = k0 so that Xn0k0Xn1k1.X.n.pkpMil= Sq (2n0)Mil=
i j
Sq(2n0)and fl-1 Xn0n0= Sq(2n0) and hence the theorem holds.
Assume that the theorem is true for all in BArA having less than q factors o*
*f the
form Sq (2i). Then by Lemma 7.3
i n j i j
fl Xn0-1k0Xn1k1.X.p.kp= Sq r1; : :;:rh + 2k0; rh+1 - 2k0; : :;:rm
where h = n0 - k0. For brevity let
"R= r1; : :;:rh + 2k0; rh+1 - 2k0; : :;:rm :
So by our inductive hypothesis we have
i n j i n j
Xn0k0Xn1k1.X.p.kpMil= Sq(2n0)Xn0-1k0Xn1k1.X.p.kpMil
i D E X j
= Sq(2n0) Sq "R + Sq
D E X
= Sq(2n0)Sq "R + Sq(2n0)Sq
D E
where Sq A Sq "R for all i.
Now the matrix X
_*_||r1.||.|.|rhr||h+1-_2k0.||.|.|rm_
(7.1)
0 ||0 ||. .|.|2k0 ||0 ||. .|.|0
D E
is Sq (2n0)Sq R" -admissible since it clearly satisfies (3.2), (3.3) (because 2*
*h2k0 =
2n0-k02k0 = 2n0), and (3.4) (since 2k0 2 rh+1 by Lemma 7.3 (1) it follows that *
*2k0 =2
rh+1 - 2k0 ). The matrix X produces Sq (r1; : :;:rma)s desired.
CHANGE OF BASIS IN THE STEENROD ALGEBRA 21
D E
Let X be any other Sq(2n0)Sq "R -allowable matrix which produces Sq(t1; : *
*:;:tn).
Then X has the form
_*__||x1x||2||.|.|.xw_
(7.2) :
y0 ||y1 ||y2.||.|.|yw
We consider two cases: h 6= 0 and h = 0.
Case 1: h 6= 0: P
Since X is admissible, 2iyi= 2n0 and hence yh 2k0: But yh 6= 2k0 sin*
*ce we
are assuming this matrix is not the same as (7.1). So yh < 2k0: Thus by*
* (3.2)
xh = rh+2k0-yh: But since yh < 2k0there exists u k0 such that 2u 2 xh *
*(this
follows from the fact that 2i2=rh for i k0 since Sq (r1; : :;:rmi)s ze*
*ro up to
(h + 1; k0)by Lemma 7.3). But also th = xh + yh-1 and so by (3.4) 2u 2 *
*th
also. Thus Sq(t1; : :;:tn)has a 1 at (h; u): But h + u h + k0 < h + k0*
*+ 1 so
that (h; u) (h + 1; k0): But Sq (r1; : :;:rmi)s zero up to (h + 1; k0)s*
*o that
Sq (t1; : :;:tn)A Sq (r1; : :;:rm:)
Case 2: h = 0: P
In this case n0 = k0. Since X is admissible, 2iyi = 2n0 and thus the*
*re
must be some v such that yv 6= 0 and consequently some u such that 2u 2*
* yv
with u + v n0: Notice also that we have u < k0 since we are assuming t*
*his
matrix is not the same as (7.1). By (3.4) 2u 2 yv implies 2u 2 tv+1: T*
*hus
Sq (t1; : :;:tn)has a 1 at (v + 1; u): But u + v + 1 n0 + 1 and u < k0*
* so
that (v + 1; u) (1; k0): By Lemma 7.3 Sq (r1; : :;:rmi)s zero up to (1;*
* k0)
so that Sq (t1; : :;:tn)A Sq (r1; : :;:rm:)
D E
So in both cases we have shown that any other Sq (2n0)Sq "R-allowable matr*
*ix
other than (7.1) produces Sq(t1; : :;:tn)which is strictly ADlessEthan Sq(r1;*
* : :;:rm:)
Thus Sq (r1; : :;:rmi)s a summand of the product Sq (2n0)Sq "R :
So all that remains to be demonstrated is that Sq (r1;D:E:;:rmi)s not a sum-
mand of Sq (2n0)Sqfor any of the terms Sq A Sq "R . So let Sq =
Sq(s1; : :;:sn)be any such term. We again consider two cases.
Case 1: h = 0.
In this case n0 = k0. Let X be a SqP(2n0)Sq-allowable matrix (which
must be of the form (7.2)). Since 2iyi= 2n0 there must be some v such*
* that
yv 6= 0 and consequently some u such that 2u 2 yv with u + v n0: By (3*
*.4)
2u 2 yv implies 2u 2 tv+1: Thus Sq (t1; : :;:tn)has a 1 at (v + 1; u): *
*Since
u + v n0 it follows that u + v + 1 n0 + 1:
Case 1.1:u + v + 1 < n0 + 1 or u + v + 1 = n0 + 1 and u < n0:
In this case (v + 1; u) (1; n0): But Sq(r1; : :;:rmi)s zero up to *
*(1; n0)
so that Sq (t1; : :;:tn)A Sq (r1; : :;:rm:)
Case 1.2:u + v + 1 = n0 + 1 and u = n0.
22 KENNETH G. MONKS
In this case v = 0: Since X is allowable, 2n0 =2s1 (by (3.4)). Th*
*us X
produces Sq (t1; : :;:tn)where t1 = s1 + 2n0; and ti= si for i > 1*
*: Thus
it is easy to see that the binary chart of Sq (t1; : :;:tn)is iden*
*tical to
that of Sq (s1; : :;:sn)with the exception of the 1 at location (1*
*; n0)of
Sq(t1; : :;:tn): Similarly,DtheEbinary chart of Sq (r1; : :;:rmi)s*
* identi-
cal to that of Sq "R with the exception of the 1 at location (1; *
*n0)of
D E
Sq(r1; : :;:rm()since Sq "R is zero up to (n1 - k1 + 1; k1) (1; k*
*0)).
D E
Then the fact that Sq (s1; : :;:sn)A Sq "R implies that there ex*
*ists
D E
(a; b)such that the binary charts of Sq(s1; : :;:sn)and Sq "R mat*
*ch at
all locationsD(i;Ej) (a; b)and that Sq (s1; : :;:sn)has a 1 at (a;*
* b)
while Sq "R has a 0 at (a; b): Simply changing the 0 at (1; n0)on
both charts to a 1 does not affect this situation so that once aga*
*in
Sq(t1; : :;:tn)A Sq (r1; : :;:rm:)
Case 2: h > 0:
Let X be a Sq (2n0)Sq-allowable matrix which producesPSq (t1; : :;:*
*tn)
(and must be of the form (7.2)). Once again since 2iyi = 2n0 there m*
*ust
be some v such that yv 6= 0 and consequently some u such that 2u 2 yv w*
*ith
u + v n0: By (3.4) 2u 2 yv implies 2u 2 tv+1: Thus Sq (t1; : :;:tn)has*
* a 1 at
(v + 1; u):
Case 2.1:u + v < n0 or (u + v = n0 and u < k0).
In this case (v + 1; u) (n0 - k0 + 1; k0): But Sq(r1; : :;:rmi)s z*
*ero up
to (n0 - k0 + 1; k0)so that Sq (t1; : :;:tn)A Sq (r1; : :;:rm:)
Case 2.2 u + v = n0 and u k0: Then X has the form
_*_||s1.||.|.|sv-1s||v-_2us||v+1||.|.|.sn_
0 ||0 ||. .|.|0 ||2u ||0 ||. .|.|0
Case 2.2.1:2u =2sv.
In this case 2u 2 sv-2u which implies that 2u 2 tv: Thus Sq(t1*
*; : :;:tn)
has a 1 at (v; u): But u + v = n0 < n0 + 1 so that (v; u)
(n0 - k0 + 1; k0): But Sq (r1; : :;:rmi)s zero up to (n0 - k0 *
*+ 1; k0)
so that Sq (t1; : :;:tn)A Sq (r1; : :;:rm:)
Case 2.2.2:2u 2 sv:
In this case we have
Sq (t1; : :;:tn)= Sq(s1; : :;:sv - 2u; sv+1 + 2u; sv+2; :::*
*;:sn)
Notice that 2u =2sv+1 since X is admissible, so that the only *
*differ-
ence between the binary charts of Sq(t1; : :;:tn)and Sq(s1; : *
*:;:sn)
is that the 1 at (v; u)in Sq(s1; : :;:sn)is moved to location *
*(v + 1; u)
in Sq (t1; : :;:tn): Also the difference between the binary ch*
*arts of
CHANGE OF BASIS IN THE STEENROD ALGEBRA 23
D E *
*D E
Sq "R and Sq(r1; : :;:rmi)s that the 1 at location (h; k0)in Sq*
* R"
is moved to location (hD+E1; k0)in Sq (r1; : :;:rm:)By definition
Sq (s1; : :;:sn)A Sq "R implies that there exists (a; b)such th*
*at
D E
the binary charts of Sq (s1; : :;:sn)and Sq "R match at all loc*
*a-
tionsD(i;Ej) (a; b)and that Sq (s1; : :;:sn)has a 1 at (a; b)whi*
*le
Sq "R has a 0 at (a; b):
Case 2.2.2.1:u = k0.
In this case (v; u)= (h; k0). It is clear that simply moving*
* the
1 at (h; k0)to location (h + 1; k0)on both binary charts to a
1 does not affect the fact that (a; b)is the first location *
*where
the charts differ and does not change the values of the char*
*ts
at (a; b), so that once again Sq (t1; : :;:tn)A Sq (r1; : :;*
*:rm:)
Case 2.2.2.2:u > k0:D E
Since Sq "R has a 1 at (h; k0)and is zero up to (h; k0)and *
*also
D E
Sq A Sq "R then either Sq (s1; : :;:sn)is not zero up to
(h; k0)or else it is and it has a 1 at (h; k0)also.
Case 2.2.2.2.1:Sq(s1; : :;:sn)is not zero up to (h; k0).
In this case there is a 1 at (i; j)for some (i; j) (h; k0)
(v; u): As the binary charts of Sq(t1; : :;:tn)and Sq(s1; : *
*:;:sn)
only differ at locations (v; u)and (v + 1; u), Sq(t1; : :;:t*
*n)must
also have a 1 at (i; j) (h; k0) (h + 1; k0): Thus since
Sq (r1; : :;:rmi)s zero up to (h + 1; k0)we have Sq(t1; : :;*
*:tn)A
Sq (r1; : :;:rm:)
Case 2.2.2.2.2:Sq(s1; : :;:sn)is zero up to (h; k0)and has a 1 at (h; k0).
Since the binary charts of Sq (t1; : :;:tn)and Sq (s1; : :;:*
*sn)
only differ at locations (v; u)and (v + 1; u)and (h; k0) (v;*
* u),
Sq (t1; : :;:tn)must also have a 1 at (h; k0) (h + 1; k0):
Thus since Sq (r1; : :;:rmi)s zero up to (h + 1; k0)we have
Sq (t1; : :;:tn)A Sq (r1; : :;:rm:)
8. Non-triangular Bases
The remaining bases, BWall, BWdY , and BWdZ are not triangular with respect *
*to
the Milnor basis. There is an interesting relationship between the BWalland BWdZ
bases however, which we note in this section.
To see that BWall is not triangular with respect to the Milnor basis we consi*
*der
grading 9: In this grading the elements of BWallare Sq8;1, Sq1;2;4;2, Sq2;4;1;2*
*, Sq2;4;2;1,
24 KENNETH G. MONKS
and Sq4;2;1;2. By the Milnor product formula these equal:
Sq 8;1 = Sq (9)+ Sq(6; 1)
Sq1;2;4;2= Sq (3; 2)
Sq2;4;1;2= Sq (6; 1)+ Sq(0; 3)+ Sq(3; 2)
Sq2;4;2;1= Sq (3; 2)+ Sq(0; 3)+ Sq(2; 0; 1)
Sq4;2;1;2= Sq (6; 1)+ Sq(0; 3)+ Sq(2; 0; 1):
Clearly, any bijection fl mapping BWall to BMil must have fl Sq1;2;4;2= Sq (3; *
*2):
Now suppose we wantPto find an ordering of BMil in grading 9 and extend fl so
that Mil = fl () + Sq where each Sq fl () : Then among the elements
Sq(6; 1), Sq(0; 3), and Sq(2; 0; 1)we must decide which element is greatest in *
*terms
of : Suppose we choose Sq (6; 1)to be the largest. Then the condition that fl *
*map
to the largest summand forces fl Sq2;4;1;2= fl Sq4;2;1;2= Sq (6; 1)which contr*
*adicts
the injectivity of fl: A similar argument shows that we cannot choose either Sq*
* (0; 3)
or Sq(2; 0; 1)for the largest element. Thus no such ordering and gamma exist, *
*and
we conclude BWall is not triangular with respect tot the Milnor basis. An exac*
*tly
analogous argument in grading 9 proves that both BWdY and BWdZ are not triangu*
*lar
with respect to the Milnor basis either.
The Wood bases are related to each other in the same sense that the Ptsbases
described above are: one basis can be obtained from the other by simply changing
the order of the factors in the monomials. There also is an interesting relatio*
*nship of
sorts between the Wall basis and the Wood Z basis. We have the following:
Theorem 8.1. Ynk-kis the E largest summand of (Qnk)Mil.
The proof of this theorem is very similar to the proof of Theorem 5.1 and wil*
*l not be
presented here. Thus we are naturally led to consider the bijection fl : BWall!*
* BWdZ
by
i n j k
fl Qn0k0Qn1k1.Q.p.kp= Ynk00-k0Ynk11-k1.Y.n.pp-kp:
It is a simple matter to verify that the order of the factors is such that the *
*right hand
side is indeed an element of BWdZ as claimed.
We close this section by commenting that it is conceivable that these three b*
*ases
are triangular with respect to one another, but knowing this would not provide *
*us
with a recursive change of basis formula of the form (1.1) since this relies on*
* the
Milnor product formula to convert from the given basis to the Milnor basis, and*
* we
have no analogous product formula for these bases.
CHANGE OF BASIS IN THE STEENROD ALGEBRA 25
9. Product Relations
In order to improve on the change of basis formulas derived above, we would l*
*ike
to obtain explicit non-recursive formulas. As a first step in this direction it*
* would be
desirable to know which elements are common to two given bases. For example, it*
* is
well known that the elements Sq (n)are common to both the Milnor and admissible
monomial bases. But are these the only such elements? The answer is no, and fur*
*ther
investigation yields an infinite subset of BMil\ BAdm . By Theorem 4.1 any elem*
*ent
2 BMil\ BAdm must satisfy fl () = , i.e. it must be an eigenvector of fl (ext*
*ended
to a linear transformation of A).
Theorem 9.1. If ri -1 mod 2!(ri+1)for all 1 i < m then Sq (r1; : :;:rm2)
BMil\ BAdm (and in this case Sq (r1; : :;:rm=)fl Sq(r1; : :;:rm)).
We point out that this linear algebra result is also providing us with inform*
*ation
about products, i.e. Sqt1Sqt2. .S.qtm= Sq (r1; : :;:rmw)here ri = ti- 2ti+1 (t*
*ake
tm+1 = 0 ) if ri -1 mod 2!(ri+1)for all 1 i < m:
We also note out that the condition ri -1 mod 2!(ri+1)can easily be checked by
writing the ordinary binary representations of numbers r1; r2; : :;:rm in horiz*
*ontally
above one another (with r1 on top) and checking that no digit ever appears belo*
*w a
0: This is because of the following trivial fact which we state without proof:
(9.1) r -1 mod 2w , 2k 2 r for allk < w:
For example, Sq (13; 5; 1)is not equal to an admissible monomial because writ*
*ing
the indices in base 2 yields:
13 = 1 1 0 12
5 = 1 0 12
1 = 12
and the 0 in the two's column of the 5 is beneath the 0 in the same column for *
*13 .
On the other hand, Sq(7; 5; 1)does satisfy the required condition and so by The*
*orem
9.1 we deduce that Sq (7; 5; 1)= Sq21Sq7 Sq1:
We will also need to make use of the following fact whose verification is an *
*elemen-
tary exercise in binary arithmetic.
Lemma 9.2. Let x; y; r; w be nonnegative integers. If r -1 mod 2w and x + y *
*= r
then for any k < w either 2k 2 x or 2k 2 y but not both, i.e. ffk (x)+ ffk (y)=*
* 1.
We now turn our attention to proving Theorem 9.1.
Proof of Theorem 9.1. Let R = r1; : :;:rm be any sequence satisfying ri
-1 mod 2!(ri+1)for all i: We would like to show that fl Sq = Sq : We pro-
ceed by induction on m.
If m = 1 then fl Sq(r1)= Sq(r1) by definition of fl:
26 KENNETH G. MONKS
Now for the inductive hypothesis assume that for any Sq (s1; : :;:sk)2 BMil w*
*ith
k < m; if si -1 mod 2!(si+1)for all i then fl Sq~~~~ = Sq~~~~ : In particular, we*
* have
fl Sq(r2; : :;:rm=)Sq(r2; : :;:rm:)
Let Sq= fl (R)where T = t1; : :;:tm . Then clearly fl Sq(r2; : :;:rm=)Sqt2*
*. .S.qtm
so that
fl Sq = Sq (t1)Sq(t2). .S.q(tm )
= Sq (t1)fl Sq(r2; : :;:rm )
= Sq (t1)Sq(r2; : :;:rm:)
So it suffices to show that Sq (t1)Sq(r2; : :;:rm=)Sq (r1; : :;:rmb)y the Milnor
product formula.
Let X be a Sq (t1)Sq(r2; : :;:rm-)allowable matrix. Then X is of the form
_*__||x2x||3||||.x.m.||_
y1 ||y2 ||y3||||.y.|.|m
such that for 2 i m
(9.2) xi+ yi= ri
(9.3) xi yi-1
and also satisfying
mX
(9.4) t1 = 2i-1yi:
i=1
Let j > 2 and suppose 2i 2 xj: We would like to show that 2i 2 xj-1 also. Now
2i 2 xj implies that 2i xj rj < 2!(rj)by (9.2) and rj-1 -1 mod 2!(rj)implies
that 2k 2 rj-1 for all k < ! (rj)by (9.1). Combining these facts shows 2i2 rj-1*
*. So
by (9.2) and Lemma 9.2 either 2i 2 xj-1 or 2i 2 yj-1: But 2i =2yj-1 by condition
(9.3), so 2i 2 xj-1 . Thus we have shown that 2i 2 xj implies 2i 2 xj-1 . So *
*by
induction we have 2i2 xj implies 2i2 x2. In particular, (xj) (x2)for all j:
Now suppose x2 6= 0: Then 2(x2) 2 x2: By Lemma 9.2 this implies that 2(x2) 2 *
*r2
which in turn implies that 2(x2) 2 r1 by (9.1). Now since (xj) (x2)if follows
that 2(x2) divides xj for all j; i.e. that
(9.5) xj = 2(x2)hj
CHANGE OF BASIS IN THE STEENROD ALGEBRA 27
for some nonnegative integer hj. Solving (9.4) for y1; substituting for t1 usin*
*g (4.1)
and applying (9.5) gives us
mX
y1 = t1 - 2k-1yk
k=2
Xm mX
= 2k-1rk - 2k-1yk
k=1 k=2
mX
= r1 + 2k-1(rk - yk)
k=2
mX
= r1 + 2k-1xk
k=2
mX
= r1 + 2k-12(x2)hk
k=2
mX !
= r1 + 2(x2)+1 2k-2hk
k=2
Combining this with the fact that 2(x2) 2 r1, it follows by (3.6) that 2(x2) 2 *
*y1:
Thus 2(x2) 2 y1 and 2(x2) 2 x2 which contradicts (9.3). Therefore our assumption
that x2 6= 0 must be false.
So x2 = 0: But (xj) (x2)for all j , so it follows that xj = 0 for all j . H*
*ence
X must be the matrix
_*_||0_||0_||._.|.|0_
r1 ||r2r||3||.|.|.rm
which is clearly admissible and produces Sq (r1; : :;:rm.)
References
[Arn]Arnon, D.; Monomial Bases in the Steenrod Algebra, Journal of Pure and App*
*lied Algebra
96 (1994), 215-223
[Wall]Wall, C. T. C.; Generators and Relations for the Steenrod Algebra, Ann. o*
*f Math. (3) 72
(1960), 429-444
[Wo] Wood, R. M. W.; A note on Bases and Relations in the Steenrod Algebra, to *
*appear in Bull.
London Math. Soc.
[Mil]Milnor, J.; The Steenrod algebra and its dual, Ann. of Math. (2) 67 (1958)*
*, 150-171
Department of Mathematics, University of Scranton, Scranton, PA 18510
E-mail address: monks@uofs.edu
~~