DSpace at VNU: On the behavior of the algebraic transfer

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On the Behavior of the Algebraic Transfer Author(s): Robert R Bruner, Lê M Hà and Nguyễn H V Hung Source: Transactions of the American Mathematical Society, Vol 357, No (Feb., 2005), pp 473487 Published by: American Mathematical Society Stable URL: http://www.jstor.org/stable/3845233 Accessed: 30-12-2015 08:43 UTC Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at http://www.jstor.org/page/ info/about/policies/terms.jsp JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range of content in a trusted digital archive We use information technology and tools to increase productivity and facilitate new forms of scholarship For more information about JSTOR, please contact support@jstor.org American Mathematical Society is collaborating with JSTOR to digitize, preserve and extend access to Transactions of the American Mathematical Society http://www.jstor.org This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions TRANSACTIONS OF THE SOCIETY AMERICAN MATHEMATICAL Volume 357, Number 2, Pages 473-487 S 0002-9947(04)03661-X Article electronically published on May 28, 2004 OF THE ALGEBRAIC ON THE BEHAVIOR TRANSFER ROBERT R BRUNER, LE M HA, AND NGUYEN H V HUNG Dedicated to Professor Huynh Mui on the occasion of his sixtiethbirthday ABSTRACT Let Trk : F2 GLk PH (BVk) -> Ext k+ (F2F2)F be the alge- braic transfer,which is defined by W Singer as an algebraic version of the geometrical transfertrk : 7rS((BVk)+) -+ 7rS(S?) It has been shown that the algebraic transferis highly nontrivial and, more precisely,that Trk is an isomorphismfor k = 1,2,3 However, Singer showed that Tr5 is not an epimorphism In this paper, we prove that Tr4 does not detect the nonzero element gs C E xt4122 (F2, F2) for every s > As a consequence, the localized (Sq?)-lTr4 given by invertingthe squaring operation Sq? is not an epimorphism This gives a negative answer to a predictionby Minami INTRODUCTION AND STATEMENT OF RESULTS The subject of the presentpaper is the algebraictransfer Trk : F2 ( GLk PHi(BVk) -3 Extk+i (F2,F2), whichis definedby W Singeras an algebraicversionof the geometricaltransfer trk 7rs(S?) to the stable homotopy groups of spheres Here 7rrs((BVk)+) denotes a k-dimensional F2-vectorspace, and PH*(BVk) is the primitivepart Vk consistingof all elementsin H (BVk) that are annihilatedby everypositive-degree operationin the mod Steenrodalgebra,A Throughoutthe paper, the homology in F2 is takenwithcoefficients It has been proved that Trk is an isomorphismfor k = 1,2 by Singer [14] and fork = by Boardman [1] These data togetherwith the fact that Tr = (see [14]) showthatTrk is highlynontrivial Dk>0Trk is an algebrahomomorphism to be a usefultool forstudyingthe transfer is considered the Therefore, algebraic mysteriouscohomology of the Steenrod algebra, Ext*5 (F2, F2) In [14], Singer also gave computationsto show that Tr4 is an isomorphismup to a range of internal degrees.However,he provedthat Tr5 is not an epimorphism Based on thesedata, we are particularlyinterestedin the behaviorof the fourth theoremis the main resultof this paper algebraictransfer.The following Received by the editors June 18, 2003 2000 Mathematics Subject Classification Primary 55P47, 55Q45, 55S10, 55T15 Key words and phrases Adams spectral sequences, Steenrod algebra, invarianttheory,algebraic transfer The third author was supported in part by the Vietnam National Research Program, Grant N0140801 The computer calculations herein were done on equipment supplied by NSF grant DMS-0079743 02004 American Mathematical 473 This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions Society 474 ROBERT LE M HA, AND NGUYEN R BRUNER, H V HUNG Theorem 1.1 For each s > 1, thenonzeroelementgs C Ext,12 (F2,F2) is not in the image ofTr4 The readeris referred to May [11] forthe generatorgl and to Lin [8] or [9] for the generatorsgs As a consequence,we get a negativeanswerto a predictionby Minami [13] Corollary 1.2 The localizationof thefourthalgebraictransfer (Sq)-Tr4 : (Sq?)-1F2 - PH*(BV4) GL4 (Sq?)-lExt4A4+ (F2, F2) givenbyinverting Sq? is not an epimorphism It is well known(see [10]) that thereare squaringoperationsSq' (i > 0) acting on the cohomologyof the Steenrodalgebra, whichshare most of the properties with Sqi on the cohomologyof spaces However,Sq? is not the identity.We refer to Section forthe precise meaningof the operationSq? on the domain of the algebraictransfer We next explainthe idea of the proofof Theorem1.1 Let Pk '= H*(BVk) be the polynomialalgebra of k variables,each of degree1 Then, the domain of Trk, F2 PH,(BVk), is dual to (F2 ( Pk)GLk In orderto A GLk proveTheorem1.1, it sufficesto showthat (F2 P4)12L_4 = 0, foreverys > A Direct calculationof (F2 0P4)12.2s-4 is difficult, as P4 in degree 12 25 - is A a huge F2-vectorspace, e.g its dimensionis 1771 fors = To computeit, we observethat the iterateddual squaringoperation (SqO) : (F2 0P4)l2.2A -4 (F2 A P4)8 is an isomorphism of GL4-modulesforany s > This isomorphism is obtainedby the applyingrepeatedly followingproposition Proposition 1.3 Let k and r be positiveintegers.Suppose thateach monomial Xl * xk of Pk in degree2r + k withat least one exponent it even is hit Then Sq : (F2 Pk)2r+k -t (F2 0Pk)r A A is an isomorphismofGLk-modules Here, as usual, we say that a polynomialQ in Pk is hitifit is A-decomposable Further,we show that (F2 OP4)8 is an F2-vectorspace of dimension55 Then, A a specificbasis of it, we provethat (F2 pP4)GL4 = As a conseby investigating A quence, we get (F2 P4)G2L._4 = foreverys > A The readerwho does not wishto followthe invarianttheorycomputationabove weakertheorem,and thenwouldnot need to read may be satisfiedby the following the paper's last sections Theorem 1.4 Tr4 is not an isomorphism This theoremis provedby observingthat,on the one hand, (IF2 A P4)2L4 (F2 P4)GL A This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC TRANSFER 475 and on the otherhand, 20(F F) = F2 -91 Ext44+8 (F, F The paper is dividedintosix sectionsand organizedas follows.Section2 starts with a recollectionof the squaring operationand ends with a proofof the isoExt4 - (F2 OPk)8 morphism (F2 OP4)12.2s-4 A Theorem 1.4 is proved in Section We A compute(F2 0P4)8 and its GL4-invariantsin Section4 We proveTheorem1.1 in A Section5 Finally,in Section6, we describethe GL4-modulestructureof (F2 (P4)8 A A SUFFICIENT CONDITION FOR THE SQUARING OPERATION TO BE AN ISOMORPHISM This sectionstartswitha recollectionof Kameko's squaringoperation Sq? : F2 ? PH*(BVk) -> F2 GLk GLk PH*(BVk) The mostimportantpropertyofKameko'sSq? is thatit commuteswiththeclassical Sq? on Ext((F2, F2) (definedin [10]) throughthe algebraictransfer(see [1], [13]) This squaringoperationis constructedas follows As is well known,H*(BVk) is the polynomialalgebra,Pk := F2[xl, ,xk], on k generatorsx1, , Xk,each of degree1 By dualizing, H,(BVk) - r(a, ., ak) is the divided power algebra generatedby al, , ak, each of degree 1, whereai is dual to xi C H1(BVk) Here the dualityis taken with respectto the basis of H*(BVk) consisting of all monomials in x1, ,Xk In [6] and [7] Kameko defineda homomorphism Sq: - H, (BVk) (ii al H* (BVk), a (2il+l) ak(2ik+l) ak * * * a a1 ak is dual to x1 xk k The following lemma is well known We to a make self-contained the give proof paper where a(i1 Lemma 2.1 Sq? is a GLk-homomorphism Proof We use the explanation of Sq? by Crabb and Hubbuck [3], which does not depend on the chosen basis of H*(BVk) The element a(Vk) = a ak is nothing but the image of the generatorof Ak(k) underthe (skew) symmetrization map Ak(Vk) Hk(Bk) = rk() = (V k vk)Sk, k times where the symmetricgroup Sk acts on Vk ) ***0Vk by permutations of the factors Let c: H,(BVk) H*(BVk) be the degree-halving whichis dual to epimorphism, the Frobenius monomorphism F : H*(BVk) any x We have Sq?(c(y)) for y E H*(BVk) -* H*(BVk) defined by F(x) = x2 for = a(k)y, To prove that this is well definedwe need to show that if c(y) = 0, then a(Vk)y = Indeed, c(y) = implies (c(y),x) = (y,x2) = for every x e H*(BVk) Here (., ) denotes the dual pairing between H*(BVk) and This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions ROBERT R BRUNER, LE M HA, AND NGUYEN H V HUNG 476 a(ik, thenthereis at least one it whichis H*(BVk) So, ifwe writey = Eai) odd in each termof the sum Therefore, a(i) a(ik)) = 0, a(Vk)y = a ak( because atait) = forany odd it So, Sq? is well defined As c is a GLk-epimorphism, the map SqO is a GLk-homomorphism The lemmais proved O Further,it is easy to see that cSq2t+l = O, cSq2t = Sqtc So we have Sq,2t+lSqo = 0, Sq2tSq0 Sq?S0qt (See [4] foran explicitproof.) Therefore,Sq? maps PH,(BVk) to itself Kameko's Sq? is definedby Sq? = Sq?: F2 PH,(BVk) - F2 ( PH*(BVk) GLk GLk GLk The dual homomorphism Sq : Pk -+ Pk of Sq? is obviouslygivenby Sq(Igl *tXlfk ) ? 31-1 3k-1 ,Jk odd, otherwise jl, X xi'1 0, Hence Ker(Sq? : Pk -* Pk) = Even, whereEven denotesthe vectorsubspace of Pk spannedby all monomialsx1l withat least one exponentit even Let s: Pk -i Pk be a rightinverseof Sq? definedas follows: s(ixl "V l ik Xk; xk 2ik+l =271+1 Xk XI It should be notedthat s does not commutewiththe doublingmap on A, that is, in general Sq2ts sSqt However,in one particularcircumstancewe have the following Lemma 2.2 Underthe hypothesis of Proposition1.3, the map s : (F2 OPk)r A s[X] -* (F2 0Pk)2r+k, A = [sX] is a well-defined linearmap Proof We startwithan observationthat Im(Sq2ts - sSqt) C Even We provethis by showingequivalentlythat SqO(Sq2ts - sSqt) = O Indeed, SqO(Sq2ts - sSqt) = SqSq2ts = Sqtqs = Sqt id - id Sqt = - SqsSqt - Sq?sSqt This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC TRANSFER As a consequence, s maps (A+Pk)r to (A+Pk + Even)2r+k 477 Here and in what follows,A+ denotesthe submoduleofA consistingofall positivedegreeoperations Further,by the hypothesisof Proposition2.3, we have (A+Pk Hence, s maps (A+Pk)r to linearmap, as s is The lemma is proved C (A+Pk)2r+k + Even)2r+k So the map s is well defined Then it is a (A+Pk)2r+k O The followingpropositionis also numberedas Proposition1.3 Proposition 2.3 Let k and r be positiveintegers.Suppose that each monomial Xi xk of Pk in degree 2r + k with at least one exponent it even is hit Then - Sq* : (F2?Pk)2r+k A (F2 0Pk)r A is an isomorphismofGLk-modules Proof On the one hand, we have Sqs =- id(2 followsthat Sq s[X] = Sq[sX] gk) A [Ss Indeed,fromSq?s = idpk, it = [X], X] forany X in degreer of Pk On the other hand, we have sSq? = id(lF2?k)2r+k Indeed, by the hypoth- A esis, any monomial with at least one even exponent representsthe class in (F2 0Pk)2r+k, so we need onlyto checkon the classes of monomialswithall expoA nentsodd We have - [s (xi [x 2i+l xik)] .2ik+l] ~2ik+1r?kfk] ~~2i1+1 for any x21 x2ik+1 in degree 2r + k of Pk Combining the two equalities, SqOs = id(F2 Pk)r and sSq we see that SqO : (F2 (Pk)2r+k (F2 OPk)r A A (IF20Pk) 2r+k A -* A = id(l (Pk)2r+ A (F2 OPk)r is an isomorphismwith inverse~s A The propositionis proved D The targetof this sectionis the following Lemma 2.4 For everypositiveintegers, (Sqo)s : (F2 P4)12.2 (]F2 A is an isomorphismofGL4-modules A P4)8 Proof By usingProposition2.3 repeatedly,it sufficesto show that any monomial of P4 in degreem = 12 *2s - with at least one even exponentis hit Since m is even,the numberof even exponentsin such a monomialmust be either2 or If all exponentsof the monomialare even, then it is hit by Sq1 Hence we need only to considerthe case of a monomialR with exactlytwo even exponents(and so exactlytwo odd exponents).Wood proves([15]) that if a(m + oo(R)) > ao(R) This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions ROBERT 478 R BRUNER, LE M HA, AND NGUYEN H V HUNG thenR is hit,whereao(R) is the numberofodd exponentsin the monomialR, and a(n) is the numberof ones in the binaryexpansionof n We have ao(R) = and a(m + aoQ(R))= a(12 25 - 2) = s + 2, so Wood's criterionis met,and R is hit D The lemmais proved THE FOURTH ALGEBRAIC TRANSFER IS NOT AN ISOMORPHISM The targetof this sectionis to provethe followingtheorem,whichis also numbered as Theorem1.4 Theorem 3.1 Tr4 : F2 ( GL4 is not an isomorphism PHi(BV4) Ext 4+i (F2, F2) Proof For any r, we have a commutativediagram (IF2X PHi(BV4))r Tr4 Ext 44+r( GL422 SqO (F2 F) SqO ( PHi(BV4))2r+4 Tr4 Ext4,8+2r(F2, (F F2) GL4 wherethe firstverticalarrowis the Kameko Sq? and the secondverticalone is the classical Sq? The dual statementof Lemma 2.4 fors = claims that Sq?: (F2 PHi(BV4))8 GL4 - (F2 ( GL4 PHi(BV4))20 is an isomorphism.On the otherhand, it is known(May [11]) that = Ext4,4+20(2, IF2)=F2 Ext44+8 ~A (F2 IF) ~A 9gi This impliesthat Tr4 is not an isomorphism.The theoremis proved ] or Remark3.2 This proofdoes not showwhetherTr4 failsto be a monomorphism failsto be an epimorphism.We will see that actuallyTr4 is not an epimorphism in Section below GL4-INVARIANTS OF THE INDECOMPOSABLES OF P4 IN DEGREE From now on, let us writex = x1, y -= 2, z = x3 and t = x4 and denotethe monomialxaybzctdby (a, b,c, d) forabbreviation Proposition 4.1 (F2?P4)8 is an F2-vectorspace of dimension55 witha basis A consistingof the classes represented bythefollowingmonomials: (A) (7,1,0,0), (7,0,1,0), (7,0,0,1), (1,7,0,0), (1,0,7,0), (1,0,0,7), (0, 7,1,0), (0,7, 0,1), (0, 1,7,0), (0,1, 0,7), (0, 0,7,1), (0,0, 1,7), (B) (3, 3,1, 1), (3, 1,3, 1), (3, 1,1, 3), (1, 3, 3, 1), (1, 3,1, 3), (1, 1,3, 3), This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC TRANSFER47 479 (C) (6,1,1,0),(6,1,0,1),(6,0,11,1)1(1,6,1,0),(1,6,0,1)1(1,11,60),) (1,1,0,6), (1,0,6,1), (1,0,1,6), (0,6,1,~1), (0, 1,6,1), (0,1,1,6), (D) (5, 3,0,0),1(5, 0,3,0), (5, 0,0,3), (0,15,3,0), (0, 5,~0,3), (0, 0,5,3), (E) (5, 2,1,10),(5, 2,0,l1),(5, 0,2,1),~(2, 5,1,10),(2, 5,0,1), (2,1,15,0), (2,1,~0,~5),(2, 0,5,1), (2, 0,1,15),(0, 5,2, 1), (0, 2,5,1), (0, 2,1,15), (F) (5, 1,1,1), (1, 5,1,1),(1,1,5,1), (1,1,1, 5), The propositionis provedby combinling a couple of lemmas Lemma 4.2 (F2 OP4)8 A is generatedbythe55 elements listed in Proposition Proof It is easy to see that everymonomial(a, b,c, d) with a, b,c, d all even is hit (morepreciselyby Sq1) The only monomials(a, b,c, d) in degree8 with at least one of a, b,c, d odd are the followingup to permutationsof the variables: (7, 1,0,~0),1 (3,3,1,1), (6, 1,1,10),1 (5,3,0,0), (5,2,1,0), (5,1,1, I~1), (4, 2,1,11), (4, 3,1,~0),1 (3, 3,2,0), (3, 2,2,1) The last monomialsand theirpermutationsare expressedin termsof the first monomialsand theirpermutationsas follows: (4,3,1,0) = (3, 3,2,0) = (2,5,1,0)?Sq 4(1, 2,110)?+Sq2(2,3,1,0), (5,2,1,0)?+(2,5, 1,0) +Sq 4(2,1,1, 10)?+Sq4(1,2,1,0) ?Sq (3,2,1,~0) + Sq2(2, 3,1,10)+ Sq1(3, 3,1,0), (3,2,2, 1) = (5,1,1,1) + (4,2,1,1) + (4,11,2,1) +Sq (3,1,1,11) ? Sq'(4, 1,1,11)+ Sql(3,221, 1) ? Sq1(3,1,12,1) Hence, (F2 0&P4)8 is generatedby the following7 monomialsand theirpermutaA tions: By the familyof a monomial(a, b,c, d) we mean the set of all monomialswhich are obtainedfrom(a, b,c, d) by permutationsof the variables The monomialsin the familiesabove whichare not in Proposition4.1 can be expressedin termsof the 55 elementslisted thereas follows (We give only one expressionfromeach symmetry class.) (3, 5,0,~0) - (5,1,2,0) (4,1,1,2) (2,4, 1,1) (2,1,1,4) - (5,3, 0,0) +Sq 4(2,2, 0,0)?+Sq 2(31310,0), (6,1 1,10)?+(5,2, 1,0)?+Sq1(5, 1,110), (4,2, 1,1)-i- (4,1,2, 1)?+Sq1 (4,1,1, 1), (4,2,1 1)?+Sq4(1,11,11)-i-Sq 2(2 2,1,11), (4,2,1,1) +(4,1,2,1) +Sq4(1, 1,1,~1) + Sq2(2, 1,1,2) + Sql(4, 1,1,1), This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions 480 ROBERT (1,4,1,2) = (1,2,1,4) = (1,1,4,2) = (1,1,2,4) = R BRUNER, LE M HA, AND NGUYEN H V HUNG (4,2,1,1)+(1,4,2,1) +Sq4(, 1,1, 1) +Sq2(,2,2,1,1)+ Sq1(1,4,1,1), (4,2,1,1)+ (1,4,2,1) +Sq2(2,2, 1, 1) + Sq2(1, 2, 1,2) + Sq(1,4, 1, 1), (4,1,2,1)+(1,4,2,1) +Sq2(2,1,2,1) + Sq2(1, 2,2,1) + Sq(1, 1,4,1), (4,1,2,1) + (1,42,,1) +Sq4(1,1,1,1) +Sq2(2,1,2,1) + Sq2(1, 2,2,1) + Sq2(1,1,2,2) + Sql(1, 1,4,1) D The lemma is proved in Lemma 4.3 The 55 elementslistedin Proposition4.1 are linearlyindependent (F2 ?P4)8 A Proof We will use an equivalencerelationdefinedby sayingthat, fortwo polynomials P and Q, P is equivalentto Q, denotedby P ~ Q, ifP - Q is hit If X is one of the lettersfromA to G, let Xi be the i-th elementin familyX accordingto the orderlistedin Proposition4.1 (This is the lexicographicalorder in each family.) Suppose thereis a linearrelationbetweenthe 55 elementslistedthere, 12 12 Za,A,A + ybiBz i=l + i= i= ciCi + i= 12 diDi + E i= eiEi i= +fiFi + i=l giGi = 0, are where a,, b,,ci, di,e%,f, gi E F2 We need to show that all these coefficients zero The proofis dividedinto steps Step We call a monomiala spikeif each of its exponentsis of the form2n - forsome n It is well knownthat spikesdo not appear in the expressionof SqiY forany i positiveand any monomialY, since the powersx2 -1 are not hit in the of any spikeis zero in everylinearrelation one variablecase Hence,the coefficient in F2O Pk A Amongthe 55 elementsof Proposition4.1, the classes of familiesA and B are spikes So = bj = 0, foreveryi and j Then, we get 12 i=l 12 ciCi +, diD + i=l eiEi + 2fiFi i=l giGi = + i=l i=l F2 OP4 -* F2 ?P2 inducedby the projection Step Considerthe homomorphism A A the image of the above linear P4 - P4/(z, t) - P2 Under this homomorphism, relationis d1(5, 3) = In orderto showthat d1 = 0, we need to provethat (5,3) is nonzeroin F2 ?P2 A x F- x, y - x + y sends (5, 3) to (8, 0) + (7, 1) + (6, 2) + The lineartransformation (5, 3) ~ (7, 1) + (5, 3) As the action of the Steenrodalgebracommuteswithlinear maps, if (5, 3) is hit thenso is (7, 1) + (5, 3) But it is impossible,because (7, 1) is a spike Hence, (5,3) in F2 ?P2 and dl = A This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC 481 TRANSFER Similarly,usingall the projectionsof P4 to its quotientsby the ideals generated by each pair of the fourvariables,we get dai -0 foreveryi So we get 12 12 i c +Z~~~~~~~~~~eiEi ? ZfiFi ZeiCi Step Considerthe homomorphism IF2OP4 P4 ~~j=0 =i1 i1 ii F2 OP3 inducedby the projection A A P4/(t) N P3 Underthis homomorphism, the linearrelationabove is sent to ci(6, 1,1) + c4(1,6, 1) + C6(1,1,6) + ei(5,2, 1) + e4(2,5, 1) + e6(2,1, 5)z Applying the linear map x H-? x, y t-* x, z H-+ y to this relation, we obtain (ci + C4+ e1 + e4)(7,1)+ c6(2,6) + e6(3,5) = (cl C4+ el + e4)(7, 1) + e6(3,5) = Since (7, 1) is a spike,(cl + c4 ? e1 + e4) = 0, hence e6(3, 5) = As for(5, 3), we can showthat (3,5) $y0 E F2 OP2 and get e6 = A = By similar arguments,we have eI = e4 = e6 = The equality (cl+c4+eI+e4) showsthat cl + C4 = or c1 = C4 By similararguments,cl = c4 = c6 We denote this commoncoefficient by c and get c{(6, 1, 1) + (1.6,1) ? (1 11,6)}= 0 Suppose the We provethat c - by showingthat (6,1, 1) +-(1,6, 1)?(1, 1, 6) is the unstable that hit Then, by propertyof contrary, (6,1,1) + (1, 6,1) + (1, 1,6) the action of A on the polynomialalgebra,we have (6,1,1)?(1,6,1)+ (1,1,6) = Sql(P)+ Sq2(Q)+ Sq4(R), f and this forsome polynomialsP, Q, R By the degreeinformation, R2 Sq4(R) to assume (6, 1, 1) + (1,6,1) + (1, 1,6) elementis hit by Sq1 Therefore, it suffices ? Sq1(P) Sq2(Q) Let Sq2Sq2Sq2 act on the both sides ofthisequality.The righthand side is sent to zero,as Sq2 Sq2 Sq2 annihilatesSq1 and Sq2 On the otherhand, Sq2Sq2Sq2 {(6, 1,1) + (1,6, 1) + (1,1, 6)} This is a contradiction.So, it implies(6, 1, 1) We get fjFj + i l Step Apply the linear map x fi(5,3) + (f2 ? (8,4,2) ? symmetries Z = and c = (1, 6,1) + (1, 1,6) and we have = H-4 x, y gjG i Gl ~-+ y,z H y,t f3 + fh + 93)(1, F-* 7) + (g, + y to the above equality, 92)(4, 4) = fi(5,3) + (f2 + f3+ f4 + 93)(1, 7) = As (7, 1) is a spike,we obtain (f2+ f3+ f4+ 93) - Oand fi(5, 3) = As (5, 3)#f0, it yieldsfi = Next, apply the linear map x F-* x, y H-+ y, z x, t H-4 x to the equality Z,f1f7F,+ E3 I1iGi = 0, and we have f2(3,5) + (fh + f4 + 92)(7, 1) + gi(6, 2) + 93(4,4) - f2(3,5) + (f3 + f4 + 92)(7, 1) = This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions ROBERT 482 LE M HA, AND NGUYEN R BRUNER, As (7,1) is a spike, we get (f3 + f4 + implies f2 = 92) H V HUNG = and f2(3, 5) = Since (3, 5) : 0, it Similarly,apply the linear map x -* x, y H- x, z 1+ f4F4 + f3F3 f gGi = 0, and we have - y,t -* x to the equality f3(3,5) + (f4 + gi)(7, 1) + (92 + 93)(6, 2) = f3(3,5) + (f4+ gi)(7, 1) =0 As (7,1) is a spike,we get f4 + 91 = and thenf3 = Finally, apply the linear map x i- x,y i- x,z i- x,t f4F4 + E3i= gGi - 0, and we have ) y to the equality f4(3, 5) + (g9 + 92 + 93)(7, 1) = f4(3,5) + (9 + 92 + 93)(7,1) = As (7,1) is a spike, we get 91 + 92 + 93 = and then f4 = Substituting fi = f2 = f3 = f4 = into the equations (f2 + f3 + f4 + 93) = 0, (f3 + f4 + 92) = 0, f4 + 91 = 0, we get 91 = 92 = g3 = of an arbitrarylinearrelationbetweenthe We have shownthat all coefficients D 55 elementslistedin Proposition4.1 are zero The lemmafollows CombiningLemmas 4.2-4.3, we get Proposition4.1 Proposition 4.4 (F2 ?P4)GL4 A = Proof If X is one ofthe lettersA, B, C, D, E, F, G, let L(X) be thevectorsubspace of (IF2 P4)8 spannedbythe elementsoffamilyX in Proposition4.1 Let Sk denote A the symmetricsubgroupof GLk Accordingto the relationslisted in the proof of Lemma 4.2, L(A), ?(B), L(C), L(D), L(F), L(G) are S4-submodules The subspace L(E) is not an S4-submodule.However,the sum C(C, E) = L(C) ( (E) is We have a decompositionof S4-modules (F2 ?P4)8 - L(A) LC(B)f ?(C, E) A 12(D) ?C(F) ?2(G) Let a be an arbitraryGL4-invariantin (F2 OP4)8 It can uniquelybe writtenin the form A a = aA + COB+ aC,E + aD + aOF + acG, whereax E ?(X) forX E {A, B, D, F, G}, and aC,E E )(C, E) Each termofthis sum is S4-invariant thenall Note that ifa linearcombinationofelementsin a familyis S4-invariant, are equal, because each elementin thefamilycan be obtainedfrom ofits coefficients any otherby a suitable permutation.Let sx denote the sum of all the elements in the family X listed in Proposition 4.1 Then, we have aA = asA, aB = bsB, where a, b,c, d, e, f, g E F2 aD = dsD, aF = fSF, aG = 9SG, and ac,E = csc+esE, Let p be the transpositiongivenby p(x) = y, p(y) = x, p(z) = z, p(t) = t It is easy to see that p(2,1,0,5) = (1,2,0,5) (2,1,0,5)+(1,1,0,6), = p(2,1,5,0) (1,2,5,0)=(2,1,5,0)+(1,1,6,0) This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC TRANSFER 483 from(2, 1,0, 5) and (2, 1,5,0) in familyE are Further,the 10 elementsdifferent divided into pairs with p acting on each pair by twisting So, p(sE) = SE + (1,1,0, 6) ? (1,1,6, 0) On the other hand, as the familyC is full,in the sense that it containsall the variablepermutationsof a monomial,we have p(sc) = sc Hence,we get - p(cSC p(ac,E) + eSE) - cSC + eSE + e(1, 1,O, 6) + e(1, 1, 6, 0) As aC,E is S4-invariant, e(l, 1, 0, 6) ? e(1, 1, 6, 0) = So e = 0, because the two elementsare linearlyindependentby Lemma 4.3 We obtain a = aA + aB + aC + aD + aF + aG, whereac = ac,E - csc Let us now consider the transvection co given by yv(x) - x, yO(y) = y, 0p(z) = x, cp(t)= x + t A routinecomputationshows = A ? (7,1,0,0) + (7,0,10) (7,0,0,1) + (1,7,0,0) + (1,0,7,0) (P(SA) ?(6, 1,0,1) ? (6, 0,1,1) + (1,1,0, 6) ? (1,0,1,6), SB + (6,1,1,0) + (1,6,1,0) ? (1,1,6,60)- (2, 5,1,0) + (2,1, 5, 0) = cp(SB) ?(5, 1,1,1) + (1,5,1,11)+ (1, 1,5,1) ? (4,2,1,) 1? (4, 1,2,1) ?(3,3,1,1) + (3,1,3, 1), co(sc) = sc + (6, 1,1,0) + (1,6,1,0) + (1, 1,6,0), = SD + (7, 0, 0,1) ? (1,6, 0,1) ? (1,0, 6,1) + (5,3, 0,0) + (5,0,3, 0), (O(SD) ? (2, 5,1,0) ? (2,1,5,0) ? (5,,1,11) + (4, 2,1,1) ? (4,1,2,1), SO(SF) = SF cp(SG) = Sc + (6, 1, 1,0) Let rx = yp(sx)- sx whereX is one ofthe lettersA, B, C, D, F, G The equality a is rewrittenas co(a)= cp(aSA + bSB + cSc + dSD? or equivalently f SF + gSG) = aSA + bSB + cSC + dSD + fSF + gSG, arA + brB + crc + drD + frF + grG = In this linearcombination,rB and rD are the only termscontaining(3,3, 1, 1) in familyB and (5,3,0,0) in familyD respectively.From Lemma 4.3, we get b - d - 0, and therefore arA ? crC + frF + grG = In the new linearcombination,as rA, rc and rF are the only termscontaining (7, 1,0,0) in familyA, (1, 6, 1,0) in familyC and (4, 2, 1, 1) in familyF respectively, we have a = c = f = As a consequence,grG - 0, so we finallyget g - In summary,we have shown that everyGL4-invarianta in (F2 0P4)8 equals A D zero The propositionis proved THE FOURTH ALGEBRAIC TRANSFER IS NUT AN EPIMORPHISM The goal ofthispaper is to provethe following theorem,whichis also numbered as Therein1.1 Theorem 5.1 For each s > 1, Tr4 :F2 PH (BV4) GL4 ExtA4 (F2,F2) does not detectthe nonzeroelementg, G Ext 412-2' (F2, F2) This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions ROBERT 484 R BRUNER, LE M Hk, AND NGUYEN H V HUNG Proof CombiningLemma 2.4 and Proposition4.4 we get (F2 A = 0, )P4)G24 foreverynonnegativeintegers On the otherhand, it is wellknownthat Ext424(IF2,F2) is spannedby the generator gl (see May [11]) Further,gs = (Sq0)S-l(gl) is nonzeroin Ext4l12'2 (F2,F2) (see Lin [8] and also [9]) As F2 PH12.2 4(BV4) GL4 Tr4: JF2( GL4 is dual to (F2 ?P4)2GL A PH12.25 4(BV4) - Ext 4i ,122 (IF2, 2) does not detectthe generatorgs, foreverynonnegativeintegers The theoremis proved D As a consequence,we get a negativeanswerto a predictionby Minami [13] (This corollaryis also numberedas Corollary1.2.) Corollary 5.2 The localizationof thefourthalgebraictransfer (Sq?)-1Tr4: (Sq?)-1F2 PH*(BV4) GL4 (Sq?)-1Ext4j4+*(F2,IF2) givenbyinvertingSq? is not an epimorphism Proof Indeed, it does not detect the nonzeroelementg, whichis representedby D the family(gs)s>o withgs = (SqO)s-1(gi) The corollaryfollows Remark5.3 Our resultdoes not affectSinger'sconjecturethat the k-thalgebraic foreveryk (See [14].) transferis a monomorphism FINAL REMARK: GL4-MODULE STRUCTURE Boardman's studyof the variableproblemshowsthat the GLk modulestructureof F2 Pk may be a usefultool In thisveinwe close witha descriptionofthe A module (F2?P4)8 A as a GL4-module From the "Modular Atlas" [5] we findthat thereare irreduciblemodulesforGL4 in characteristic2, of dimensions1, 4, 4, 6, 14, 20, 20, and 64 With a littlecalculationwe findthe following descriptionof them: 1: the trivialmoduleF2, N: the naturalmoduleF4, N*: the dual of the naturalmodule, A: the alternatingsquare of N or N*, S: the nontrivialconstituentof N ( N*, whichhas compositionfactors1,S, 1, T: a constituentof N A, whichhas compositionfactorsN* and T, T*: a constituentof N* A, whichhas compositionfactorsN and T*, St: the Steinbergmodule Using a "meataxe" programwrittenin MAGMA, togetherwitha MAGMA programto computeBrauer characters,we have foundthat (F2 P4)8 is an extension A A This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC TRANSFER 485 A 49 N M S 45 35 31 31 / 30' 25 30 ~1 \ A 24 20 T FIGURE Some GL4-submodules of (F2 A P4)8s wherethe 25-dimensionalmodule M is an extension O - >iA - -M N S The correspondinglattice of submodulesof (F2 OP4)8 is shownin Figure We A name the submodulesby theirdimension,using a prime to distinguishthe two submodulesof dimension30 We label the edges by the correspondingquotient module In it, intersections are shown,but sums are omittedforclarity.That is, the intersectionof the submodules30' and 35 is the submodule24, but the sum of 30' and 35 (a submoduleof dimension41) is not shown The two extensions above can be seen in the lattice,in the sense that, forexample,the submoduleof dimension24 is the directsum of the submodulesof dimensions4 and 20, since theirintersectionis trivial Further,the quotientof 55 by 24 is the directsum of the quotientsof 30' by 24 and of 49 by 24 The generatorsforthesesubmodulesare providedbythesame computerprogram used to findthisdecompositionand are listedbelow When all themonomialsin one of the seven familieslistedin Proposition4.1 appear, we simplywritethe name of thefamily, so that,forexample,all the monomialsin familyA are in the submodule This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions 486 ROBERT R BRUNER, LE M HA, AND NGUYEN H V HUNG of dimension20 Also, recallthe element SG = (4, 2, 1, 1) + (4, 1,2, 1)+ (1, 4, 2, 1) used in the proofof Proposition4.4 Finally note that elementswhichformbases forthe subquotientscan be read offby comparingthese lists of generators.For example,the quotientofthe module30 by the submodule24 is A, and the elements of familyD generateit 4: (6, 1,1,0)+ (1,6,1,0) + (1,1,6,0), (6,1,0,1) + (1,6,0,1) + (1,1,0,6), (6,0,1,1) + (1,0,6,1)+ (1,0,1,6), (0,6,1,1)+ (0,1,6,1)+ (0,1,1,6) 20: (A), (6,11,1,0)+ (1,1,6,0), (6,1,0,1)+ (1, 1,0,6), (6,0,1,1) + (1,0,1,6), (1,6,1,0) + (1,1,6,0), (1,6,0,1)+ (1,1,0,6), (1,0,6,1)+ (1,0,1,6), (0,6,1,1) + (0,1,1,6), (0,1,6,1)+ (0,11,1,6) 24: (A) and (C) 25: (A), (C), and sG 30: (A), (C), and (D) 30': (A), (C) and (5,1,1,1) + (1,5,1,1) + SG+(3,3,1,1), 1, 3, 1), (5, 1,1, 1)+(1, 111, 5)+G+(3, (5,1,1, 1)+(1, 1,5, 1)+sG+(3, 1,1,3), 3, 3, 1), (1, 5, 1, 1)+(1, 1, 1, 5)+SG+(1, 3,1,3), (1, 5, 1, 1)+(1, 1, 5, 1)+SG+(1, (1,1, 5,1) + (1,1,1,5)+ sG + (1,1,3,3) 31: (A),(C), (D) and sG 35: (A), (C), (D), SG and (5,2,1,0) + (5,2,0,1)+ (5,0,2,1)+ (5,1,1,1), (2,5,1,0)+(2,5,0,1)+ (0,5,2,1)+(1,5,1,1), 1), (2,1,5,0) + (2,0,5,1)+(0,2,5,1)+(1,1,5, (2,1,0,5) + (2,0,1,5)+ (0,2,1,5)+ (1,1,1,5) 45: (A), (C), (D), (E) and (G) 49: (A), (C), (D), (E), (F) and (G) ACKNOWLEDGMENT The researchwas in progressduringthe secondnamed author'svisitto theIHES (France) and the thirdnamed author's visit to Wayne State University,Detroit (Michigan)in the academic year2001-2002 The second named authorthanksLionel Schwartzforhis supportand encouragement The thirdnamed author is gratefulto Daniel Frohardt,David Handel, Lowell Hansen,JohnKlein, Charles McGibbon,Claude Schochetand all colleaguesat the fortheirhospitalityand for Departmentof Mathematics,WayneState University, the warmworkingatmosphere The authorsexpresstheirheartythanksto Tran N Nam forhelpfuldiscussions REFERENCES [1] J M Boardman, Modular representationson the homologyof powers of real projectivespace, Algebraic Topology: Oaxtepec 1991, M C Tangora (ed.), Contemp Math 146 (1993), 4970 MR 95a:55041 [2] E Brown and F P Peterson, H* (MO) as an algebra over the Steenrodalgebra, Notas Mat Simpos (1975), 11-21 MR 86b:55001 [3] M C Crabb and J R Hubbuck, Representations of the homologyof BV and the Steenrod algebra II, Algebraic Topology: New Trends in Localization and Periodicity (Sant Feliu de Guixols, 1994; C Broto et al., eds.), Progr Math 136, Birkhiiuser,1996, 143-154 MR 97h:55018 This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions THE BEHAVIOR OF THE ALGEBRAIC TRANSFER 487 [4] Nguyen H V Hirng, Spherical classes and the algebraic transfer,Trans Amer Math Soc 349 (1997), 3893-3910 MR 98e:55020 [5] C Jansen, K Lux, R Parker, R Wilson, An atlas of Brauer characters (Appendix by T Breuer and S Norton) London Mathematical Society Monographs New Series, 11 Oxford Science Publications The Clarendon Press, Oxford UniversityPress, New York, 1995 MR 96k:20016 [6] M Kameko, Products of projectzvespaces as Steenrod modules, Thesis, Johns Hopkins University1990 [7] M Kameko, Generators of the cohomologyof BV3, J Math Kyoto Univ 38 (1998), 587-593 MR 2000b:55015 [8] W H Lin, Some differentialsin Adams spectral sequence for spheres, Trans Amer Math Soc., to appear [9] W H Lin and M Mahowald, The Adams spectralsequence for Minami's theorem,Contemp Math 220 (1998), 143-177 MR 99f:55023 [10] A Liulevicius, The factorization of cyclic reducedpowers by secondary cohomologyoperations, Mem Amer Math Soc 42 (1962) MR 31:6226 [11] J Peter May, The cohomologyof restrictedLie algebras and of Hopf algebras; applications to the Steenrod algebra,Ph D thesis, Princeton University,1964 [12] J Milnor and J Moore, On the structureof Hopf algebras,Ann of Math 81 (1965), 211-264 MR 30:4259 [13] N Minami, The iterated transferanalogue of the new doomsday conjecture,Trans Amer Math Soc 351 (1999), 2325-2351 MR 99i:55023 [14] W M Singer, The transferin homological algebra, Math Zeit 202 (1989), 493-523 MR 90i:55035 [15] R M W Wood, Steenrod squares of polynomials and the Peterson conjecture,Math Proc Cambridge Phil Soc 105 (1989), 307-309 MR 90a:55030 DEPARTMENTOF MATHEMATICS, WAYNESTATEUNIVERSITY,656 W KIRBY STREET, DETROIT, MICHIGAN48202 E-mail address: rrbOmath.wayne.edu UNIVERSITEDE LILLE I, UFR DE MATHEMATIQUES, UMR 8524, 59655 VILLENEUVED'ASCQ CEDEX, FRANCE E-mail address: Minh-Ha.Le@math.univ-lillel.fr DEPARTMENTOF MATHEMATICS, VIETNAMNATIONALUNIVERSITY,334 NGUYENTRAI STREET, HANOI,VIETNAM E-mail address: nhvhung@vnu edu.vn This content downloaded from 129.16.69.49 on Wed, 30 Dec 2015 08:43:18 UTC All use subject to JSTOR Terms and Conditions ... to Section forthe precise meaningof the operationSq? on the domain of the algebraictransfer We next explainthe idea of the proofof Theorem1.1 Let Pk '= H*(BVk) be the polynomialalgebra of k variables,each... forexample ,the submoduleof dimension24 is the directsum of the submodulesof dimensions4 and 20, since theirintersectionis trivial Further ,the quotientof 55 by 24 is the directsum of the quotientsof 30'... trivialmoduleF2, N: the naturalmoduleF4, N*: the dual of the naturalmodule, A: the alternatingsquare of N or N*, S: the nontrivialconstituentof N ( N*, whichhas compositionfactors1,S, 1, T: a constituentof N