CHANGE OF BASIS, MONOMIAL RELATIONS, AND Pt FOR THE STEENROD ALGEBRA. July 1995

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1 CHANGE OF BASIS, MONOMIAL RELATIONS, AND Pt s FOR THE STEENROD ALGEBRA BASES KENNETH G MONKS July 1995 Abstract The relationship between several common bases for the mod 2 Steenrod algebra is explored and a new family of bases consisting of monomials in distinct Pt s s is developed A recursive change of basis formula is produced to convert between the Milnor basis and each of the bases for which the change of basis matrix in every grading is upper triangular In particular, it is shown that the basis of admissible monomials, the new Pt s bases, and two bases due to D Arnon, are all bases having this property, and the corresponding change of basis formula is produced for each of them Some monomial relations for the mod 2 Steenrod algebra are then obtained by exploring the change of basis transformations 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 are 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 Pt s s to the existing collection These bases are all described in detail in Section 3 Given so many different bases, a natural question to ask is: how can we convert 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 information 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 monomials in Milnor basis elements Thus it is a simple matter to convert from one of these bases, call it B, to the Milnor basis, B Mil, by using the product formula (31) developed by Milnor [Mil] A difficulty arises when trying to convert in the other direction: from 1991 Mathematics Subject Classification Primary 55S10, 55S05; Secondary 57T05 Key words and phrases Steenrod Algebra 1

2 2 KENNETH G MONKS the Milnor basis to back the basis B Having such a formula for every basis would then allow us to convert between any two bases, indirectly, via the Milnor basis A brute force approach might be to compute the change of basis matrix, M, from B to B Mil 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 approach 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 11 Suppose there exists orderings, and, of bases B and B Mil respectively, such that the change of basis matrix M (with respect to these orderings) is upper triangular in every grading In this situation we say the basis B is triangular with respect to the Milnor basis This will be the situation if and only if there is an order preserving bijection γ : B Mil B such that γ (θ) isthe -largest summand of Milnor element θ when expressed in basis B Remark 11 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 Namely, for any θ B Mil, we have (11) θ B = γ (θ)+ (θ i ) B i where x B denotes the representation of x in basis B and γ (θ) =θ + i θ i is the Milnor representation of γ (θ) Mil obtained via the Milnor product formula This is a well defined recursive formula because all of the Milnor basis elements θ i must be strictly -less than θ and so the recursion must eventually end when we reach elements for which γ (θ) =θ holds Since A is finite dimensional in each grading, we must have γ (θ) =θ for the θ 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 and determine the change of basis formula (11) for converting an element from the Milnor basis to basis B it suffices to: 1 Define a bijection γ : B Mil B 2 Define the ordering on B Mil Thenlet be the unique ordering of B such that γ is order preserving 3 Prove that γ (θ) isthe -largest summand of the representation of θ B for any θ B Mil We will follow this procedure several times in what follows Note that requirement #3 can also be satisfied by showing γ 1 (θ) isthe -largest Milnor basis summand of the Milnor basis representation of θ for any θ B, since γ is order preserving and

3 CHANGE OF BASIS IN THE STEENROD ALGEBRA 3 the inverse of a triangular matrix is also triangular Also in place of requirement #2 we can define the ordering on B and then let be the unique ordering of B Mil such that γ 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 Pt s 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 respect to the Milnor basis, and determine the change of basis formula of the form (11) for each basis that is Finally, we will show how such information may lead to product information by determining an infinite family of elements which are both admissible monomial and Milnor basis elements 2 Summary of Main Results In this section we give a general overview of the main results which are contained in this paper The details, notation, background and proofs will be presented later 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 31) Theorem 21 The set, B PR, of all monomials of the form P s 0 t 1 P sp such that (s 0, ) R (s 1,t 1 ) R R (s p, ) is a basis for A In addition, any set B P obtained by changing the order of the factors of any of the monomials in B PR 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 change of basis formula of the form (11) by specifying the required γ Theorem 22 (1) The following bases are triangular with respect to the Milnor basis and have change of basis formula (11) for the value of γ shown in the table: Basis γ required for formula(11) Admissible monomials Definition 41 Any Pt s basis Definition 51 Arnon C Definition 61 Arnon A Definition 71 (2) Wall s basis, Wood s Y basis, and Wood s Z basis are not triangular with respect to the Milnor basis

4 4 KENNETH G MONKS It should be noted that in each case there is a simple heuristic for computing γ which makes these change of basis formulas quite easy to use in practice We give both these heuristics and sample calculations along with the proofs in later sections 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 22 for the Arnon A basis This ordering is given in Definition 73 One way to improve on the recursive change of basis formulas given in Theorem 22, 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) =Sq n are common to both the Milnor and admissible monomial bases Our final result determines an infinite family of elements which are common to these two bases Theorem 23 If r i 1mod2 ω(ri+1) for all 1 i<mthen Sq (r 1,,r m ) is an element of both the Milnor and admissible monomial bases In this case Sq (r 1,,r m )= Sq t 1 Sq t2 Sq tm where t m = r m and t i = r i +2t i+1 for 1 i<m We point out that this linear algebra result is actually providing us with information 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 determining nonrecursive 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 Algebraically the Steenrod algebra can be described as the quotient of the free associative graded algebra over the field with two elements, F 2,onsymbolsSq n in grading n, by the ideal generated by the Adem relations: Sq a Sq b = a 2 n=0 ( ) b n 1 Sq a+b n Sq n a 2n (for a<2b) where the binomial coefficients are taken mod 2 and Sq 0 = 1, the multiplicative identity In order to describe the bases we wish to consider we first define the following Definition 31 Let R = r 1,,r m and S = s 1,,s n be finite sequences of integers Write R R S if R is less than S in lexicographic order from the right, ie if either m<nor else m = n and there exists i such that r i <s i and r j = s j 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, ie R L S if there exists i such that r i <s i and r j = s j

5 CHANGE OF BASIS IN THE STEENROD ALGEBRA 5 for all j<i(where we take r k =0for k>mand s k =0for 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 Sq t 1 Sq t2 Sq tm is said to be admissible if t i 2t i+1 for 1 i<mthe set of all admissible monomials forms a basis for A which we will denote by B Adm Whether Sq t 1 Sq t2 Sq tm is admissible or not, we will often abbreviate Sq t 1 Sq t2 Sq tm by Sq t 1,,t m and in addition if T = t 1,,t m we will write Sq T for Sq t 1,,t m (2) Milnor [Mil]: Milnor showed that A is also a Hopf algebra whose dual, A,is the polynomial algebra F 2 [ξ 1,ξ 2,] on generators ξ n in grading 2 n 1 The basis of A which is dual to the basis of monomials in A is called the Milnor basis and will be denoted B Mil The element dual to ξ r 1 1 ξ r 2 2 ξm rm in this basis is denoted Sq (r 1,,r m ) Comparing with the notation given above we have Sq (n) =Sq n If R = r 1,,r m is a finite sequence of nonnegative integers, we will often use multi-index notation and write Sq R for the Milnor basis element Sq (r 1,,r m ) The algebra structure on A in this basis can be described by the product formula given by Milnor Namely, (31) Sq (r 1,r 2,)Sq(s 1,s 2,)= Sq (t 1,t 2,) X where the sum is taken over all matrices X = x ij satisfying: (32) x ij = s j i (33) 2 j x ij = r i j (34) (x h0,x h 1,1,,x 0h ) 1(mod2) h where (n 1,,n m ) is the multinomial coefficient (n 1 + +n m )!/ (n 1! n m!) (The value of x 00 is never used an may be taken to be 0) Each such allowable matrix produces a summand Sq (t 1,t 2,)givenby (35) t h = x ij i+j=h In such a situation we say that X is a Sq R Sq S -allowable matrix which produces Sq T We will also find it convenient to say that X produces the sequence T if T satisfies (35) regardless of whether or not X is allowable (3) Arnon A [Arn, Theorem 1A]: Define Xk n =Sq 2n Sq 2n 1 Sq 2k Then the set of all monomials of the form X n 0 such that (n 0, ) L (n 1,k 1 ) L L (n p, ) forms a basis for A which we will denoted by B ArA

6 6 KENNETH G MONKS (4) Wall [Wall, pg 433]: Define Q n k =Sq2k Sq 2k+1 Sq 2n Then the set of all monomials of the form Q n 0 Q n 1 k 1 Q np such that (n p, ) L (n p 1, 1 ) L L (n 0, ) forms a basis for A which we will denoted by B Wall This basis was also discussed in [Arn, Theorem 1B] (5) Arnon C [Arn, Theorem 1C]: A monomial of the form Sq tm Sq t m 1 Sq t 1 is said to be C-admissible if t i+1 2t i for 1 i<mand t i is divisible by 2 i 1 The set of all C-admissible monomials forms a basis for A which we will denote by B ArC (6) Wood Y [Wo, Theorem 1]: Define Yk n =Sq 2n (2 k+1 1) Then the set of all monomials of the form Y n 0 Y n 1 k 1 Y np such that (n p, ) L (n p 1, 1 ) L L (n 0, ) forms a basis for A which we will denote by B WdY Wood shows that this basis has a nice property with respect to the Hopf subalgebras A n of A generated by the Sq 2i with i n Namely if any factor of any summand of θ WdY is not in A n then θ itself is not in A n (7) Wood Z [Wo, Theorem 2]: Let Yk n = Sq 2n (2 k+1 1) as above Then the set of all monomials of the form Y n 0 Y n 1 k 1 Y np such that (n p +,n p ) L (n p 1 + 1,n p 1 ) L L (n 0 +,n 0 ) forms a basis for A which we will denoted by B WdZ Wood shows that this basis also has the same nice property with respect to the Hopf subalgebras A n that was mentioned above for the Y basis (8) Pt s -bases: In this article we will prove that the following is a basis for A Let Pt s =Sq(r 1,,r t )wherer t =2 s and r i =0fori<t For each finite set, S, of Pt s s choose an ordering of the elements of S, and let M(S) be the monomial formed by taking the product of the elements of S in increasing order, ie if S = { } P s 0, t 1,,P sp and we order the elements of S in the order shown then M (S) =P s 0 The monomials M (S) form a basis for A This 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 P s 0 such that (s 0, ) R (s 1,t 1 ) R R (s p, ) is one such basis which we will denote by B PR Before leaving this section we give a few elementary definitions and notation that will be needed later on Definition 32 If B Name is one of the bases of A described above and θ A then θ Name will denote the representation of θ in that basis For example, ( Sq 2 Sq 1) Mil =Sq(3)+Sq(0, 1) while Sq (0, 1) Adm =Sq2 Sq 1 +Sq 3 For any Milnor basis consists of element, Sq (r 1,,r m ), it is clear from the definition that the grading or degree of Sq (r 1,,r m )is m i=1 (2 i 1)r i For any of the

7 CHANGE OF BASIS IN THE STEENROD ALGEBRA 7 other bases, the degree of a monomial is the sum of the degrees of its Milnor basis factors The excess of Sq (r 1,,r m )is m i=1 r i and its length is m The excess of an admissible monomial Sq t 1,,t m is t m + m 1 i=1 (t i 2t i+1 ) We will denote the excess of θ B Mil by ex (θ) Note that Sq (r 1,,r m ) is not uniquely determined by its degree, excess, and length as can be seen by the elements Sq (0, 1, 2, 0, 1) and Sq (2, 0, 0, 1, 1) We can extend the definitions of left and right lexicographic order to both Milnor basis elements and monomials in Sq n in the obvious manner, ie if R R S then Sq R R Sq S and Sq R R Sq S and similarly if R L S then Sq R L Sq S and Sq R L Sq S For any positive integer n, letα i (n) be the coefficient of 2 i in the binary expansion of n, ie n = i=0 α i (n)2 i and α i (n) {0, 1} for all i We say that m and n are disjoint and write m n if α i (m) +α i (n) 1 for all i It is well known that this is equivalent to the condition that the binomial coefficient ( ) m+n m is odd Consequently, the multinomial coefficient (n 1,,n m ) is odd if and only if the integers n 1,,n m are pairwise disjoint This fact is used frequently throughout the article when evaluating condition (34) We often write 2 i n for α i (n) = 1 since the meaning is clear from the context The following fact will be used implicitly several times and is an elementary exercise in binary arithmetic Le b<2 t Then (36) 2 l b 2 l 2 t a + b and l<t Finally let ν (n) be the largest integer such that n 0mod2 ν(n) (and take ν (0) = ) Let ω (n) be the smallest integer such that 2 ω(n) >nnotice that for n>0we always have 2 ν(n) n and also that ν (n) <ω(n) 4 Milnor vs Admissible We begin by focusing on the relationship between B Mil and B Adm 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 use the product formula (31) for multiplying Milnor basis elements To convert a element from the Milnor basis to the basis of admissible monomials we now show that the basis of admissible monomials is triangular with respect to the Milnor basis and define the γ and ordering needed for the recursive formula (11) To satisfy requirement #1 following (11) we make the following definition Definition 41 Let Sq R = Sq(r 1,,r m ) be a Milnor basis element γ (Sq (r 1,,r m )) = Sq t 1 Sq t2 Sq tm where m (41) t i = 2 k i r k k=i Define

8 8 KENNETH G MONKS (Abbreviation: we will sometimes write γθ for γ (θ) ) Note that the t i can quickly be computed by starting with t m applying the simple recursion = r m and then (42) t i = r i +2t i+1 It follows immediately that γ Sq R is an admissible monomial for any Sq R B Mil The map γ 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 (11) So in order to satisfy requirement #3 following (11) we show: Theorem 41 γ Sq R is the rlex-largest summand of Sq R Adm Hence B Adm is triangular with respect to B Mil As a result we have a recursive formula of the form (11) for converting an element of A from the Milnor basis to the basis of admissible monomials Corollary 42 Let Sq R B Mil and suppose γ (Sq R ) Mil =Sq R + i Sq R i Then Sq R Adm = γ Sq R + i Sq R i Adm is a well defined recursive formula for computing Sq R Adm Note that γ (Sq R ) Mil can easily be obtained from the Milnor product formula (31) All of the elements Sq R i are strictly rlex-less than Sq R which is why the recursive formula is well defined This makes the formula quite easy to use in practice For example, to convert Sq (2, 2) to the basis of admissible monomials using Corollary 42 we first compute γ Sq (2, 2) = Sq 6 Sq 2 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 42 again This time γ Sq (5, 1) = Sq 7 Sq 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) = Sq 6 Sq 2 +Sq 7 Sq 1 which provides the conversion we desired In order to prove these results we begin by proving a useful lemma

9 CHANGE OF BASIS IN THE STEENROD ALGEBRA 9 Let Sq (r 1,,r m ), Sq (s 1,,s n ) B Mil Let X = x ij be the matrix * r r m r m 2 k s k s 1 s 2 s n We will call X the rlex champion matrix for Sq (r 1,,r m )Sq(s 1,,s n ) Lemma 43 If the rlex champion matrix for Sq R Sq S produces T, then every other Sq R Sq S -allowable matrix produces a sequence which is rlex-less than T Proof By (32) and (33), any other such matrix must have x ij 0 for some 0 i<mand 1 j n Let j be the largest such value Then by (32) we have x mj <s j Therefore the element Sq (u 1,,u m+n ) produced by the new matrix must have u k = s k = t k for m + j<k m + n and u m+j = x mj <s j = t j Thus Sq U R Sq T As an immediate consequence we have Corollary 44 Let T be the sequence produced by the rlex champion matrix for Sq R Sq S If U R S then for every Milnor summand Sq V of the product Sq R Sq U we have V R T This follows from Lemma 43 and the fact that the sequence T produced by the rlex champion matrix for Sq R Sq U is easily seen to be rlex less than T We are now ready to prove Theorem 41 Proof of Theorem 41 We wish to show that γ Sq R is the rlex-largest summand of Sq R Adm Since γ is bijective and preserves rlex order, it suffices to show that for any admissible monomial Sq T the Milnor basis element γ ( 1 Sq T ) is the rlex largest summand of ( Sq T ) Mil Let T = t 1,,t m be an admissible sequence Recall that by (42) γ ( 1 Sq T ) = Sq (r 1,,r m )wherer m = t m and r i = t i 2t i+1 for i<mwe proceed by induction on m If m =1then ( Sq T ) =Sq(t 1 )andγ ( 1 Sq T ) =Sq(t 1 ), so the base case Mil holds Now for the inductive hypothesis assume that for any admissible monomial Sq S of length less than m, γ ( 1 Sq S ) is the rlex-largest summand of ( Sq S ) Thenin Mil particular, γ ( ) 1 Sq t 2,,t m =Sq(r2,,r m ) is the rlex-largest summand of ( Sq t 2,,t m )Mil So we can write ( ) Sq t 2,,t m =Sq(r 2,,r m )+ Sq R i Mil

10 10 KENNETH G MONKS where Sq R i R Sq (r 2,,r m ) for all i Thus we have ( ) Sq t 1,,t m = Sq(t 1 ) ( ) Sq t 2,,t m Mil Mil = Sq(t 1 ) ( Sq (r 2,,r m )+ Sq R i ) = Sq(t 1 )Sq(r 2,,r m )+ Sq (t 1 )Sq R i The rlex champion matrix X for Sq (t 1 )Sq(r 2,,r m )is * t 1 m k=2 2 k 1 r k r 2 r 3 r m Clearly X is admissible To see that iroduces Sq (r 1,,r m ) we need only verify that m m t 1 2 k 1 r k = t k 2 r k k=2 = t 1 2t 2 = r 1 Therefore by Lemma 43 every other Sq (t 1 )Sq(r 2,,r m )-allowable matrix produces Milnor elements which are rlex-less than Sq (r 1,,r m ) So Sq (r 1,,r m ) is the rlex-largest summand of Sq (t 1 )Sq(r 2,,r m ) In addition, every summand of Sq (t 1 )Sq R i is rlex less than Sq (r 1,,r m ) by Corollary 44 5 Milnor vs P s t We now turn our attention to the relationship between B Mil and B PR All of the results and arguments in this section carry over to any Pt s basis, but we illustrate them for this particular ordering of the monomial factors The elements Pt s are common to both the Milnor and Pt s bases Therefore to express an element of B PR in the Milnor basis, we only need use the product formula (31) for multiplying Milnor basis elements Notice that we have not yet shown that B PR is a basis for A although we have defined it as a set To see that B PR is in fact a triangular basis with respect to the Milnor basis we begin by defining a grading preserving bijection γ : B Mil B PR Definition 51 Let Sq (r 1,,r m ) B Mil Define k=2 γ (Sq (r 1,,r m )) = P s 0 where the right hand side is the unique monomial in B PR satisfying for all i and j Pj i is a factor of P s 0 α i (r j )=1

11 CHANGE OF BASIS IN THE STEENROD ALGEBRA 11 The map γ is clearly a bijection on A in each grading There is a useful heuristic device for computing the γ Sq (r 1,,r m ) We define the binary chart of Sq (r 1,,r m ) to be the array: 2 α 2 (r 1 ) α 2 (r 2 ) α 2 (r 3 ) s 1 α 1 (r 1 ) α 1 (r 2 ) α 1 (r 3 ) 0 α 0 (r 1 ) α 0 (r 2 ) α 0 (r 3 ) t In other words simply write the binary expansions or the numbers r 1,,r m vertically next to each other Then Pt s is a factor of γ Sq (r 1,,r m ) if 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 Pt s basis we are considering For example, to compute γ Sq (2, 5, 1) we make the chart: and read off the factors P 1 1,P 0 2,P 2 2,andP 0 3 Multiplying them in the correct order for B PR we get γ Sq (2, 5, 1) = P 1 1 P 0 2 P 2 2 P 0 3 Now define an ordering E on B Mil as follows Definition 52 For any Sq R, Sq S B Mil, we say Sq R E Sq S if or else ex (Sq R ) < ex (Sq S ) ex (Sq R ) =ex(sq S ) and Sq R R Sq S The second condition is simply used to make a total ordering out of the partial ordering induced by excess and is never used Finally let be the ordering induced on B PR induced by the bijection γ and E Then we have: Theorem 51 γ Sq R is the -largest summand of Sq R PR It follows immediately that the elements of B PR are linearly independent in each grading and since γ is a bijection, B PR is, indeed, a basis as claimed Further, with this definition of γ and E we have satisfied requirements #1-3 in Section 1 and so B PR is triangular with respect to B Mil As a result we have a recursive formula of the form (11) for converting an element of A from the Milnor basis to the basis of admissible monomials

12 12 KENNETH G MONKS Corollary 52 Let Sq R B Mil and suppose γ (Sq R ) Mil =Sq R + i Sq R i Then Sq R PR = γ Sq R + i Sq R i PR is a well defined recursive formula for computing Sq R PR Note that γ (Sq R ) Mil can easily be obtained from the Milnor product formula (31) All of the elements Sq R i are strictly E -less than Sq R which is why the recursive formula is well defined For example, to convert Sq (4, 2) to the basis B PR using Corollary 52 we first compute γ Sq (4, 2) = P1 2P 2 1 By the Milnor product formula, P 1 2P 2 1 =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 52 again This time γ Sq (0, 1, 1) = P2 0P 3 0, but by the Milnor product formula we find that P 2 0P 3 0 =Sq(0, 1) Sq (0, 0, 1) = Sq (0, 1, 1) Thus we have shown that Sq (4, 2) = P 2 1 P P 0 2 P 0 3 which provides the conversion we desired In order to prove these results we begin by proving a few useful lemmas Let Sq (r 1,,r m ), Sq (s 1,,s n ) B Mil Let X = x ij be the matrix * s 1 s 2 s n r r m We will call X the excess champion matrix for Sq (r 1,,r m )Sq(s 1,,s n ) Let R = r 1,r 2, and S = s 1,s 2, Then we define the obvious sum R + S = r 1 + s 1,r 2 + s 2,,r i + s i, In this notation we see that the excess champion matrix for Sq R Sq S produces Sq R + S Notice that ex (Sq R + T ) =ex(sq R )+ex(sq T ) Lemma 53 If X is an allowable Sq R Sq S matrix which produces Sq T then ex (Sq T ) < ex (Sq R + S ) Proof Since the excess of Sq T =Sq(t 1,t 2,)is t i and by (35) each t i is the sum of the i th diagonal of X = x ij, it follows that ex (Sq T ) = x ij, ie it is i,j the sum of all of the entries of the matrix By (32) the sum of the entries in columns to the right of column 0, j>0 x ij must equal ex (Sq S ) By (33) x i0 r i for each i so that the entries in column 0 must have a sum less than or equal to the excess of Sq R, ie j=0 x ij ex (Sq R ) But since X is not the excess champion matrix,

13 CHANGE OF BASIS IN THE STEENROD ALGEBRA 13 we must have x uv 0 for some u>0andv>0 But by (33) it follows that x u0 <r u and so the sum of column 0 is strictly less than ex (Sq R ) Hence ex (Sq T ) = i,j x ij = x i,j + x i,j j=0 j>0 < ex (Sq R )+ex(sq S ) = ex(sq R + S ) as claimed As an immediate consequence we have Corollary 54 If ex (Sq U ) < ex (Sq R ) then every Milnor summand Sq T of the product Sq U Sq S (or Sq S Sq U ) has excess less than ex (Sq R + S ) We are now ready to prove Theorem 51 Proof of Theorem 51 We wish to show that γ Sq R is the -largest summand of Sq R PR It suffices to show that for any element P s 0 in B PR, γ ( ) 1 P s 0 is the E -largest summand of ( P s 0 We will show )Mil something slightly stronger, namely that γ ( ) ( ) 1 P s 0 is a summand of P s 0 Mil and every other Milnor summand of ( P s 0 will have excess strictly less )Mil than ex ( γ ( )) 1 P s 0 Let θ = P s 0 t 1 r i = t j =i 2 s j We proceed by induction on p P sp B PR and let Sq (r 1,,r m )=γ 1 ( P s 0 t 1 P sp ) where If p =0then(P s 0 ) Mil = P s 0 and γ 1 (P s 0 )=P s 0, so the base case holds Now for the inductive hypothesis assume that for any element θ B PR having fewer than p +1factors,γ 1 (θ) is the rlex-largest summand of θ Mil Then in particular, γ ( 1 P s 0 t 1 P s ) p 1 1 =Sq(r1,,r m 1,r m 2 sp ) P sp )Mil is a summand of ( P s 0 t 1 ex (Sq (r 1,,r m 1,r m 2 sp )) So we can write and every other summand has excess less than ( P s 0 )Mil =Sq(r 1,,r m 1,r m 2 sp )+ Sq R i where ex (Sq R i ) < ex (Sq (r 1,,r m 1,r m 2 sp )) for all i

14 14 KENNETH G MONKS Thus we have ( ) P s 0 Mil = ( P s 0 t 1 P s ) p 1 t P sp p 1 t Mil p = ( Sq (r 1,,r m 1,r m 2 sp )+ Sq R i ) P sp = Sq(r 1,,r m 1,r m 2 sp ) P sp + Sq R i P sp Now each of the Milnor summands of Sq R i P sp must have excess strictly less than ex (Sq (r 1,,r m )) by Corollary 54 Every summand of Sq (r 1,,r m 1,r m 2 sp ) P sp other than Sq (r 1,,r m ) must have excess strictly less than ex (Sq (r 1,,r m )) by Lemma 53 Finally, it is easy to see that the excess champion matrix associated with Sq (r 1,,r m 1,r m 2 sp ) P sp is allowable and thus Sq (r 1,,r m ) is a summand of ( P s 0 )Mil 6 Milnor vs Arnon C We now turn to the relationship between B Mil and B ArC In many ways this relationship is similar to the situation we find for B Adm The elements Sq(n) are again common to both bases, so to express θ B ArC in the Milnor basis, we only need use the product formula (31) 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 γ and ordering needed for the recursive formula of the form (11) Definition 61 Let Sq R = Sq(r 1,,r m ) be a Milnor basis element γ (Sq (r 1,,r m )) = Sq tm Sq t m 1 Sq t 1 where m t i =2 i 1 (61) r k k=i Define Note that γ (Sq (r 1,,r m )) can easily be computed by the following heuristic First, write the sequence r 1,,r m in a vertical column with r 1 on top Then working to the left, construct the following triangular shaped diagram in which each column contains entries which are twice the entry to its right: r 1 2r 2 r 2 2 m 2 r m 1 2r m 1 r m 1 2 m 1 r m 2 m 2 r m 2r m r m t m t m 1 t 2 t 1 the value of t i is then simply the sum of the i th column from the right as indicated

15 CHANGE OF BASIS IN THE STEENROD ALGEBRA 15 It is clear from the definition that t i is divisible by 2 i 1 and also that (62) t i+1 =2t i 2 i r i holds for 1 i<m Hence t i+1 2t i so that γ Sq R is indeed in B ArC The map γ is a bijection on A in each grading and by (62) γ 1 ( Sq tm,,t 1) =Sq(r1,,r m ) where r i = 2t i t i+1 for 1 i<mand r 2 i m = tm For this basis we choose R for the ordering of B Mil and let be the ordering induced by γ on B ArC Then we have Theorem 61 γ Sq R is the -largest summand of Sq R ArC Hence B ArC is triangular with respect to B Mil and we have a recursive formula of the form (11) for converting an element of A from the Milnor basis to the basis of C-admissible monomials Corollary 62 Let Sq R B Mil and suppose γ (Sq R ) Mil =Sq R + i Sq R i Then 2 m 1 Sq R ArC = γ Sq R + i Sq R i ArC is a well defined recursive formula for computing Sq R ArC Note that once again γ (Sq R ) Mil can easily be obtained from the Milnor product formula (31) and all of the elements Sq R i are strictly rlex-less than Sq R 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 62 we first compute γ Sq (3, 2) = Sq 4 Sq 5 By the Milnor product formula, 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 62 again This time γ Sq (6, 1) = Sq 2 Sq 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) = Sq 4 Sq 5 +Sq 2 Sq 7 which provides the conversion we desired We are now ready to prove Theorem 61 Proof of Theorem 61 Once again it suffices to show that for any C-admissible monomial Sq T the Milnor basis element γ 1 ( Sq T ) is the rlex largest summand of ( Sq T ) Mil Let Sq T =Sq tm,,t 1 be a C-admissible monomial Then γ 1 ( Sq tm,,t 1) =Sq(r1,,r m )

16 16 KENNETH G MONKS where r i = 2t i t i+1 for 1 i<mand r 2 i m = tm We proceed by induction on m 2 m 1 If m =1then ( Sq T ) =Sq(t 1 )andγ ( 1 Sq T ) =Sq(t 1 ), so the base case Mil holds Now for the inductive hypothesis assume that for any admissible monomial Sq S of length less than m, γ ( 1 Sq S ) is the rlex-largest summand of ( Sq S ) Then Mil in particular, we can compute γ 1 ( Sq t m 1,,t 1 ) =Sq(r1,,r m 2,r m 1 + r m ) which must be the rlex-largest summand of ( Sq t m 1,,t 1 So we can write )Mil ( ) Sq t m,,t 1 =Sq(r 1,,r m 2,r m 1 + r m )+ Sq R i Mil where Sq R i R Sq (r 1,,r m 2,r m 1 + r m ) for all i Thus we have ( ) Sq t m,,t 1 = Sq(t m ) ( ) Sq t m 1,,t 1 Mil Mil = Sq(t m ) ( Sq (r 1,,r m 2,r m 1 + r m )+ Sq R i ) = Sq(t m )Sq(r 1,,r m 2,r m 1 + r m )+ Sq (t m )Sq R i The rlex champion matrix for Sq (t m )Sq(r 1,,r m 2,r m 1 + r m ) is not allowable in this case so instead we let X be the matrix * r 1 r m 2 r m r m Clearly X is allowable and produces Sq (r 1,,r m ) To see that X is indeed a Sq (t m )Sq(r 1,,r m 2,r m 1 + r m ) matrix we need only note that r m = tm 2 m 1 Now every other Sq (t m )Sq(r 1,,r m 2,r m 1 + r m )-allowable matrix produces Milnor elements which are rlex-less than Sq (r 1,,r m ) since by (33) any other such matrix musroduce Sq (t 1,,t m )witht m <r m So it remains to show that every summand of Sq (t m )Sq R i is rlex less than Sq (r 1,,r m ) Let Sq R i =Sq(u 1,,u n ) be one of the summands We know Sq R i is rlex less than Sq (r 1,,r m 2,r m 1 + r m ) If n < m 1thenevery summand of Sq (t m )Sq R i has length less than m and is therefore rlex less than Sq (r 1,,r m ) On the other hand, if n = m 1 then there exists j such that u j <r j and Sq (u 1,,u m 1 )=Sq(u 1,,u j,r j+1,,r m 2,r m 1 + r m ) In this case the matrix * u 1 u j r j+1 r m 2 r m r m

17 CHANGE OF BASIS IN THE STEENROD ALGEBRA 17 produces the sequence (u 1,,u j,r j+1,,r m 1,r m ) (whether or not it is allowable) which is clearly rlex less than (r 1,,r m ) By the same argument as above, any other Sq (t m )Sq R i -allowable matrix musroduce Sq V for which V R U which is in turn rlex less than (r 1,,r m ) 7 Milnor vs Arnon A The strangest of the bases discussed here which are triangular with respect to the Milnor bases has to be B ArA due to the unusual ordering on B Mil that is used for the proof Since elements of B ArA are monomials in the elements Sq (2 n ),we can express an admissible monomial in the Milnor basis by using the product formula (31) for multiplying Milnor basis elements To convert a element from the Milnor basis to the basis of admissible monomials we show that the Arnon A basis is triangular with respect to the Milnor basis and define the γ and ordering needed for the recursive formula of the form (11) For γ we make the following: Definition 71 Let Sq R = Sq(r 1,,r m ) be a Milnor basis element Define γ (Sq (r 1,,r m )) = X n 0 where (1) (n 0, ) L (n 1,k 1 ) L L (n p, ) and (2) Xk n is a factor of X n 0 if and only if α k (r n k+1 )=1 A heuristic for easily computing this gamma is very similar to that used for B PR First, write down the binary chart for Sq (r 1,,r m ) Then for each chart location (i, j) where there is a 1, we have an associated factor X i+j 1 j of γ Sq (r 1,,r m ) These factors are then multiplied in the correct order For example, to compute γ Sq (2, 5, 6) we make the chart: and read off the factors X1 1,X1 0,X3 2,andX2 0 Multiplying them in the correct order we get γ Sq (2, 5, 6) = X0X 1 1X 1 0X The order on B Mil which we require is quite unusual We begin with an ordering on pairs of integers Definition 72 Define an ordering on N 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 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 B Mil Definition 73 Let Sq (r 1,,r m ) and Sq (s 1,,s n ) be elements of B Mil We say Sq (s 1,,s n ) A Sq (r 1,,r m ) if there exists (h, k) such that (1) α i (r j )=α i (s j ) for all (i, j) (h, k) and (2) α k (r h ) <α k (s h ) In other words, we compare the entries of the binary charts of Sq (r 1,,r m )and Sq (s 1,,s n ) in increasing order until we find the first location (h, k) where they differ Whichever element has the 0 at (h, k) is the larger element (Note that the second condition is equivalent to the condition α k (r h )=0andα k (s h )=1) Armed with this γ and ordering A on B Mil we can now prove: Theorem 71 Sq (r 1,,r m ) is the A -largest summand of γ Sq (r 1,,r m ) Mil Thus the Arnon A basis is triangular with respect to the Milnor basis and we have the recursive formula (11) for converting an element of A from the Milnor basis to the basis B ArA Corollary 72 Let Sq R B Mil and suppose γ (Sq R ) Mil =Sq R + i Sq R i Then Sq R ArA = γ Sq R + i Sq R i ArA is a well defined recursive formula for computing Sq R ArA For example, to compute Sq (2, 2) ArA we first compute γ Sq (2, 2) = X1 1X2 1 By the Milnor product formula X1 1 X1 2 = Sq(2)Sq(4)Sq(2) = Sq(2, 2) + Sq (5, 1) Applying Corollary 72 to the error term we find γ Sq (5, 1) = X0 0X1 0 X2 2 So by the product formula X0X 0 0X = Sq (1) Sq (2) Sq (1) Sq (4) = Sq(5, 1) Thus we have the desired answer Sq (2, 2) = X1 1 X2 1 + X0 0 X1 0 X2 2 In order to prove these results we begin by defining some notation that will be convenient Definition 74 Let Sq (r 1,,r m ) be any Milnor basis element We say Sq (r 1,,r m ) is zeroupto(h, k) if α i (r j )=0for all (i, j) (h, k)

19 CHANGE OF BASIS IN THE STEENROD ALGEBRA 19 Definition 75 Let Sq (r 1,,r m ) be any Milnor basis element We say Sq (r 1,,r m ) has a 1 at (h, k) if α k (r h )=1 Clearly if Sq R is zero through (h, k) and Sq S is not, then Sq S A Sq R 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 73 Let θ = X n 0 B ArA and let Sq (r 1,,r m )=γ 1 (θ) Then (1) Sq (r 1,,r m ) has a 1 at (n 0 +1, ) (2) Sq (r 1,,r m ) is zero up to (n 0 +1, ) (3) X n 0 =Sq(2 n 0 ) X n 0 1 (take X n 0 1 =1if n 0 = ) (4) γ ( 1 X n 0 1 k 1 and X n 0 1 B ArA X np ) =Sq ( r1,,r h 1,r h +2,r h+1 2,r h+2,,r m ) where h = n 0 ( if h =0we interpreting the right hand expression as Sq ( r 1 2,r 2,,r m ) ) Proof: (1) By definition of γ, X n 0 being a factor of θ implies that α k0 (r n0 +1) =1 which in turn implies that Sq (r 1,,r m )hasa1at(n 0 +1, ) (2) Assume the contrary Then there must be (i, j) (n 0 +1, ) such that Sq (r 1,,r m ) has a 1 at (i, j), ie such that α j (r i )=1 Thus by definition of γ, X i+j 1 j must be a factor of θ Now (i, j) (n 0 +1, ) implies that either i+j < n 0 +1 or else i+j = n 0 +1 and j< so that in either case (i + j 1,j) L (n 0, ) But this contradicts the factor X n 0 must be the smallest factor in left lexicographic order of its indices by definition of B ArA (3) This follows trivially from (n 0 1, ) L (n 0, ) L L (n m,k m ) and X n 0 =Sq(2 n 0 )Sq(2 n0 1 ) Sq ( ) 2 =Sq(2 n 0 ) X n 0 1 (4) Since X n 0 and X n 0 1 only differ in the first factor, then by definition of γ, Sq(r 1,,r m )andγ ( ) 1 X n 0 1 must have identical binary charts with the exception of the 1 s corresponding to the leading factors By (1), Sq (r 1,,r m )hasa1at(n 0 +1, )andγ ( ) 1 X n 0 1 has a 1 at (n 0, ) But by (2) Sq (r 1,,r m ) does not have a 1 at (n 0, ) and it is clear that γ 1 ( X n 0 1 k 1 since X n 0 ) X np does not have a 1 at (n0 +1, ) is not a factor (remembering that (n 0, ) L (n 1,k 1 )) Thus to obtain γ ( ) 1 X n 0 1 from Sq (r1,,r m ) we simply remove the 1 at (n 0 +1, ) by subtracting 2 from r n0 +1 and create a 1 at (n 0, ) by adding 2 to r n0

20 20 KENNETH G MONKS Thus γ ( ) ( ) 1 X n 0 1 =Sq r1,,r n0 +2,r n0 +1 2,,r m as required We are now ready to prove Theorem 71 Proof of Theorem 71 It suffices to show that γ ( 1 X n 0 largest summand of ( ) X n 0 k So le Mil Xn 0 k 1 γ 1 ( X n 0 k 1 is a factor of X n 0 k 1 X np ) is the A - B ArA Then ) X np =Sq(r1,,r m )whereα k (r n k+1 ) = 1 if and only if Xk n i=0 (n i k i + 1) which is the total number X np Let q = p of factors of the form Sq (2 i ) in the product X n 0 k 1 X np the definition of Xk n We proceed by induction on q If q =1thenp = 0 and n 0 = so that ( X n 0 )Mil when expanded using =Sq(2 n 0 ) Mil = Sq (2 n 0 )andγ 1 ( X n 0 n 0 ) =Sq(2 n 0 ) and hence the theorem holds Assume that the theorem is true for all θ in B ArA having less than q factors of the form Sq (2 i ) Then by Lemma 73 γ ( ) ( ) X n 0 1 =Sq r1,,r h +2,r h+1 2,,r m where h = n 0 For brevity let So by our inductive hypothesis we have R = r 1,,r h +2,r h+1 2,,r m ( X n 0 )Mil = ( Sq (2 n 0 ) X n 0 1 )Mil = Sq(2 n 0 ) ( Sq R + Sq Si ) = Sq(2 n 0 )Sq R + Sq (2 n 0 )Sq S i where Sq S i A Sq R for all i Now the matrix X (71) * r 1 r h r h+1 2 r m is Sq (2 n 0 )Sq R -admissible since it clearly satisfies (32), (33) (because 2 h 2 = 2 n 0 2 =2 n 0 ), and (34) (since 2 r h+1 by Lemma 73 (1) it follows that 2 / r h+1 2 ) The matrix X produces Sq (r 1,,r m ) as desired

21 CHANGE OF BASIS IN THE STEENROD ALGEBRA 21 Let X be any other Sq (2 n 0 )Sq R -allowable matrix which produces Sq (t1,,t n ) Then X has the form (72) * x 1 x 2 x w y 0 y 1 y 2 y w We consider two cases: h 0 and h =0 Case 1: h 0 Since X is admissible, 2 i y i =2 n 0 and hence y h 2 But y h 2 since we are assuming this matrix is not the same as (71) So y h < 2 Thus by (32) x h = r h +2 y h But since y h < 2 there exists u such that 2 u x h (this follows from the fact that 2 i / r h for i since Sq (r 1,,r m ) is zero up to (h +1, ) by Lemma 73) But also t h = x h + y h 1 and so by (34) 2 u t h also Thus Sq (t 1,,t n ) has a 1 at (h, u) But h + u h + <h+ +1 so that (h, u) (h +1, ) But Sq (r 1,,r m )iszeroupto(h +1, )sothat Sq (t 1,,t n ) A Sq (r 1,,r m ) Case 2: h =0 In this case n 0 = Since X is admissible, 2 i y i =2 n 0 and thus there must be some v such that y v 0 and consequently some u such that 2 u y v with u + v n 0 Notice also that we have u< since we are assuming this matrix is not the same as (71) By (34) 2 u y v implies 2 u t v+1 Thus Sq (t 1,,t n ) has a 1 at (v +1,u) But u + v +1 n and u< so that (v +1,u) (1, ) By Lemma 73 Sq (r 1,,r m ) is zero up to (1, ) so that Sq (t 1,,t n ) A Sq (r 1,,r m ) So in both cases we have shown that any other Sq (2 n 0 )Sq R -allowable matrix other than (71) produces Sq (t 1,,t n ) which is strictly A less than Sq (r 1,,r m ) Thus Sq (r 1,,r m ) is a summand of the product Sq (2 n 0 )Sq R So all that remains to be demonstrated is that Sq (r 1,,r m ) is not a summand of Sq (2 n 0 )Sq S i for any of the terms Sq S i A Sq R So let Sq Si = Sq (s 1,,s n ) be any such term We again consider two cases Case 1: h =0 In this case n 0 = Let X be a Sq (2 n 0 )Sq S i -allowable matrix (which must be of the form (72)) Since 2 i y i =2 n 0 there must be some v such that y v 0 and consequently some u such that 2 u y v with u + v n 0 By (34) 2 u y v implies 2 u t v+1 Thus Sq (t 1,,t n )hasa1at(v +1,u) Since u + v n 0 it follows that u + v +1 n 0 +1 Case 11: u + v +1<n 0 +1oru + v +1=n and u<n 0 In this case (v +1,u) (1,n 0 ) But Sq (r 1,,r m ) is zero up to (1,n 0 ) so that Sq (t 1,,t n ) A Sq (r 1,,r m ) Case 12: u + v +1=n and u = n 0

22 22 KENNETH G MONKS In this case v =0 Since X is allowable, 2 n 0 / s 1 (by (34)) Thus X produces Sq (t 1,,t n )wheret 1 = s 1 +2 n 0, and t i = s i for i>1 Thus it is easy to see that the binary chart of Sq (t 1,,t n )isidenticalto that of Sq (s 1,,s n ) with the exception of the 1 at location (1,n 0 )of Sq (t 1,,t n ) Similarly, the binary chart of Sq (r 1,,r m )isidentical to that of Sq R with the exception of the 1 at location (1,n0 )of Sq (r 1,,r m )(sincesq R iszeroupto(n1 k 1 +1,k 1 ) (1, )) Then the fact that Sq (s 1,,s n ) A Sq R implies that there exists (a, b) such that the binary charts of Sq (s 1,,s n ) and Sq R match at all locations (i, j) (a, b) and that Sq (s 1,,s n ) has a 1 at (a, b) while Sq R hasa0at(a, b) Simply changing the 0 at (1,n 0 )on both charts to a 1 does not affect this situation so that once again Sq (t 1,,t n ) A Sq (r 1,,r m ) Case 2: h>0 Let X be a Sq (2 n 0 )Sq S i -allowable matrix which produces Sq (t 1,,t n ) (and must be of the form (72)) Once again since 2 i y i =2 n 0 there must be some v such that y v 0 and consequently some u such that 2 u y v with u + v n 0 By (34) 2 u y v implies 2 u t v+1 Thus Sq (t 1,,t n ) has a 1 at (v +1,u) Case 21: u + v<n 0 or (u + v = n 0 and u< ) In this case (v +1,u) (n 0 +1, ) But Sq (r 1,,r m ) is zero up to (n 0 +1, )sothatsq(t 1,,t n ) A Sq (r 1,,r m ) Case 22 u + v = n 0 and u Then X has the form * s 1 s v 1 s v 2 u s v+1 s n u 0 0 Case 221: 2 u / s v In this case 2 u s v 2 u which implies that 2 u t v Thus Sq (t 1,,t n ) hasa1at(v, u) But u + v = n 0 < n so that (v, u) (n 0 +1, ) But Sq (r 1,,r m ) is zero up to (n 0 +1, ) so that Sq (t 1,,t n ) A Sq (r 1,,r m ) Case 222: 2 u s v In this case we have Sq (t 1,,t n )=Sq(s 1,,s v 2 u,s v+1 +2 u,s v+2,,s n ) Notice that 2 u / s v+1 since X is admissible, so that the only difference between the binary charts of Sq (t 1,,t n ) and Sq (s 1,,s n ) is that the 1 at (v, u)insq(s 1,,s n ) is moved to location (v +1,u) in Sq (t 1,,t n ) Also the difference between the binary charts of

23 CHANGE OF BASIS IN THE STEENROD ALGEBRA 23 Sq R and Sq (r1,,r m ) is that the 1 at location (h, )insq R is moved to location (h +1, )insq(r 1,,r m ) By definition Sq (s 1,,s n ) A Sq R implies that there exists (a, b) such that the binary charts of Sq (s 1,,s n ) and Sq R match at all locations (i, j) (a, b) and that Sq (s 1,,s n )hasa1at(a, b) while Sq R has a 0 at (a, b) Case 2221: u = In this case (v, u) =(h, ) It is clear that simply moving the 1at(h, )tolocation(h +1, ) 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 charts at (a, b), so that once again Sq (t 1,,t n ) A Sq (r 1,,r m ) Case 2222: u> Since Sq R has a 1 at (h, k0 )andiszeroupto(h, )andalso Sq S i A Sq R then either Sq (s1,,s n ) is not zero up to (h, )orelseitisandithasa1at(h, )also Case 22221: Sq (s 1,,s n ) is not zero up to (h, ) In this case there is a 1 at (i, j) for some (i, j) (h, ) (v, u) As the binary charts of Sq (t 1,,t n ) and Sq (s 1,,s n ) only differ at locations (v, u)and(v +1,u), Sq (t 1,,t n )must also have a 1 at (i, j) (h, ) (h +1, ) Thus since Sq (r 1,,r m )iszeroupto(h +1, )wehavesq(t 1,,t n ) A Sq (r 1,,r m ) Case 22222: Sq (s 1,,s n ) is zero up to (h, ) and has a 1 at (h, ) Since the binary charts of Sq (t 1,,t n ) and Sq (s 1,,s n ) only differ at locations (v, u)and(v +1,u)and(h, ) (v, u), Sq (t 1,,t n ) must also have a 1 at (h, ) (h +1, ) Thus since Sq (r 1,,r m ) is zero up to (h +1, )wehave Sq (t 1,,t n ) A Sq (r 1,,r m ) 8 Non-triangular Bases The remaining bases, B Wall, B WdY,andB WdZ are not triangular with respect to the Milnor basis There is an interesting relationship between the B Wall and B WdZ bases however, which we note in this section To see that B Wall is not triangular with respect to the Milnor basis we consider grading 9 In this grading the elements of B Wall are Sq 8,1,Sq 1,2,4,2,Sq 2,4,1,2,Sq 2,4,2,1,

24 24 KENNETH G MONKS and Sq 4,2,1,2 By the Milnor product formula these equal: Sq 8,1 = Sq (9) + Sq (6, 1) Sq 1,2,4,2 = Sq(3, 2) Sq 2,4,1,2 = Sq(6, 1) + Sq (0, 3) + Sq (3, 2) Sq 2,4,2,1 = Sq(3, 2) + Sq (0, 3) + Sq (2, 0, 1) Sq 4,2,1,2 = Sq(6, 1) + Sq (0, 3) + Sq (2, 0, 1) Clearly, any bijection γ mapping B Wall to B Mil must have γ Sq 1,2,4,2 =Sq(3, 2) Now suppose we want to find an ordering of B Mil in grading 9 and extend γ so that θ Mil = γ (θ)+ Sq R i where each Sq R i γ(θ) 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 γ map θ to the largest summand forces γ Sq 2,4,1,2 = γ Sq 4,2,1,2 =Sq(6, 1) which contradicts the injectivity of γ 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 B Wall is not triangular with respect tot the Milnor basis An exactly analogous argument in grading 9 proves that both B WdY and B WdZ are not triangular with respect to the Milnor basis either The Wood bases are related to each other in the same sense that the Pt s bases 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 relationship of sorts between the Wall basis and the Wood Z basis We have the following: Theorem 81 Y k n k is the E largest summand of (Q n k) Mil The proof of this theorem is very similar to the proof of Theorem 51 and will not be presented here Thus we are naturally led to consider the bijection γ : B Wall B WdZ by γ ( Q n 0 Q n 1 k 1 Q np ) = Y n 0 Y k 1 n 1 k 1 Y kp n p 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 B WdZ as claimed We close this section by commenting that it is conceivable that these three bases are triangular with respect to one another, but knowing this would norovide us with a recursive change of basis formula of the form (11) 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

25 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 like 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 further investigation yields an infinite subset of B Mil B Adm By Theorem 41 any element θ B Mil B Adm must satisfy γ (θ) =θ, ie it must be an eigenvector of γ (extended to a linear transformation of A) Theorem 91 If r i 1mod2 ω(r i+1) for all 1 i<mthen Sq (r 1,,r m ) B Mil B Adm (and in this case Sq (r 1,,r m )=γ Sq (r 1,,r m )) We point out that this linear algebra result is also providing us with information abouroducts, ie Sq t 1 Sq t2 Sq tm =Sq(r 1,,r m )wherer i = t i 2t i+1 (take t m+1 =0)ifr i 1mod2 ω(ri+1) for all 1 i<m We also note out that the condition r i 1mod2 ω(ri+1) can easily be checked by writing the ordinary binary representations of numbers r 1,r 2,,r m in horizontally above one another (with r 1 on top) and checking that no digit ever appears below a 0 This is because of the following trivial fact which we state withouroof: (91) r 1mod2 w 2 k r for all k<w For example, Sq (13, 5, 1) is not equal to an admissible monomial because writing the indices in base 2 yields: 13 = = = 1 2 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 Theorem 91 we deduce that Sq (7, 5, 1) = Sq 21 Sq 7 Sq 1 We will also need to make use of the following fact whose verification is an elementary exercise in binary arithmetic Lemma 92 Let x, y, r, w be nonnegative integers If r 1mod2 w and x + y = r then for any k<weither 2 k x or 2 k y but not both, ie α k (x)+α k (y) =1 We now turn our attention to proving Theorem 91 Proof of Theorem 91 Let R = r 1,,r m be any sequence satisfying r i 1mod2 ω(r i+1) for all i We would like to show that γ Sq R =Sq R We proceed by induction on m If m =1thenγ Sq (r 1 )=Sq(r 1 ) by definition of γ