The Cohomology of the Steenrod Algebra and Representations of the General Linear Groups Author(s): Nguyễn H V Hung Source: Transactions of the American Mathematical Society, Vol 357, No 10 (Oct., 2005), pp 4065-4089 Published by: American Mathematical Society Stable URL: http://www.jstor.org/stable/3845119 Accessed: 07-03-2015 06:03 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 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions OF THE TRANSACTIONS AMERICAN MATHEMATICAL SOCIETY Volume 357, Number 10, Pages 4065-4089 S 0002-9947(05)03889-4 Article electronically published on May 20, 2005 THE COHOMOLOGY REPRESENTATIONS ALGEBRA AND OF THE STEENROD OF THE GENERAL LINEAR GROUPS NGUYEN H V HUNG Dedicated to Professor Nguye^nHtruAnh on the occasion of his sixtiethbirthday ABSTRACT Let Trk be the algebraic transferthat maps fromthe coinvariants of certain GLk-representationsto the cohomology of the Steenrod algebra This transferwas defined by W Singer as an algebraic version of the geometrical transfertrk?: 7rS((BVk)+) - irS(S?) It has been shown that the algebraic transferis highlynontrivial,more precisely,that Trk is an isomorphism fork = 1, 2, and that Tr = Qk Trk is a homomorphismof algebras In this paper, we firstrecognize the phenomenonthat if we start fromany degree d and apply Sq? repeatedly at most (k- 2) times, then we get into the region in which all the iterated squaring operations are isomorphismson the coinvariantsof the GLk-representations As a consequence, every finite Sq?-family in the coinvariants has at most (k- 2) nonzero elements Two applications are exploited The firstmain theorem is that Trk is not an isomorphism for k > Furthermore,forevery k > 5, there are infinitelymany degrees in which Trk is not an isomorphism.We also show that if Tre detects a nonzero element in certain degrees of Ker(Sq?), then it is not a monomorphismand further,for each k > ?, Trk is not a monomorphismin infinitelymany degrees The second main theorem is that the elements of any Sq?-family in the cohomologyof the Steenrod algebra, except at most its first(k - 2) elements, are eitherall detected or all not detected by Trk, forevery k Applications of this study to the cases k = and show that Tr4 does not detect the three familiesg, D3 and p', and that Tr5 does not detect the family{hn+lgnl n > 1} INTRODUCTION AND STATEMENT OF RESULTS Therehave been severalefforts, implicitor explicit,to analyzethe Steenrodalgeofthe generallineargroups (See Muii[22,23, bra byusingmodularrepresentations Adams-Gunawardena-Miller [27], [3],Priddy-Wilkerson 24], Madsen-Milgram[19], Peterson[25],Wood [32],Singer[28],Priddy[26],Kuhn [15]and others.) In particular, one of the most directattemptsin studyingthe cohomologyof the Steenrod of the generallineargroupswas the algebra by means of modularrepresentations the so-called a homomorphism, work W which introduced surprising Singer, [28]by Received by the editors November 13, 2003 2000 Mathematics Subject Classification Primary 55P47, 55Q45, 55S10, 55T15 Key words and phrases Adams spectral sequences, Steenrod algebra, modular representations, invarianttheory This work was supported in part by the National Research Program, Grant No 140 804 )2005 by Nguyen H V Hrung, Nguyen H V Khue and Nguyen My Trang 4065 This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions NGUYEN 4066 H V HUNG of the algebraictransfer, mappingfromthe coinvariantsof certainrepresentations linear to the of the Steenrod algebra general group cohomology Let Vk denote a k-dimensionalF2-vectorspace, and let PH,(BVk) denotethe primitivesubspace consistingof all elementsin H,(BVk) that are annihilatedby everypositive-degreeoperationin the mod Steenrodalgebra, A Throughout in F2 The generallineargroup the paper, the homologyis taken withcoefficients on on the homologyand cohomolacts and therefore := GLk regularly Vk GL(Vk) of the two actions of and Since A GLk upon H*(BVk) commutewith ogy BVk each other,thereare inheritedactionsof GLk on F2 0H*(BVk) and PH,(BVk) A In [28],W Singerdefinedthe algebraictransfer Trk : F2 PHd(BVk) GLk - Extk k+d( as an algebraic version of the geometrical transfertrk F2) 7rs((BVk)+) -* 7rS(S) to the stable homotopygroupsof spheres It has been provedthat Trk is an isomorphismfork = 1,2 by Singer[28] and fork = by Boardman [4] Amongotherthings,thesedata togetherwiththe fact that Tr = (k Trk is an algebra honmomorphism (see [28]) showthat Trk is highly nontrivial.Therefore,the algebraictransferis expectedto be a usefultool in the studyof the mysteriouscohomologyof the Steenrodalgebra,ExtI* (F2, F2) Directlycalculatingthe value ofTrk on any nonzeroelementis difficult (see [28], [4], [11]) In this paper, our main idea is to exploitthe relationshipbetweenthe algebraictransferand the squaringoperationSq? It is well known(see [18]) that thereare squaringoperationsSqi (i > 0) actingon the cohomologyofthe Steenrod algebra that share most of the propertieswith Sqi on the cohomologyof spaces However,Sq? is not the identity.On the otherhand,thereis an analogoussquaring and one, actingon the domainofthe algebraictransfer operationSq?, the Kamneko transfer commutingwiththe classical Sq? on Extk(F2, F2) throughthe algebraic We referto Section2 forits precisemeaning The keypointis that the behaviorsof the twosquaringoperationsdo not agree in infinitely manycertaindegrees,called k-spikes A k-spikedegreeis a number that can be written as (2n1 - 1) + + (2nk - 1), but cannot be written as a sum of less than k termsof the form(2n - 1) (See a discussionof this notionafter Definition3.1.) The followingresultis originallydue to Kameko [13]: If m is a k-spike,then Sq : PH,(BVk) r,,2 -> PH, (BVk)m such is an isomorphism of GLk-modules,whereSq is certainGLk-hornomorphism that Sq? = GLk Sq (See Section for an explanation of Sq ) We recognizetwo phenomenaon thleuniversalityand the stabilityof k-spikes: First, if we start fromany degree d that can be writtenas (2n - 1) + + (2nk- 1), and apply the function6k with 6k(d) = 2d + k repeatedlyat most (k - 1) times, then we get a k-spike;second, k-spikesare mapped by 6k to k-spikes.Therefore, we have Theorem 1.1 Let d be an arbitrarynonnegative integer Then (Sq )i-k+2 : PH*(BVk)2k-2d+(2^-2-1)k - PH*(BVk)2id+(21-)k is an isomorphism of GLk-modules for every i > k- This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4067 Fromthe resultofCarlisleand WBood[8]on the boundednessconjecture,one can see that,forany degreed, thereexistst such that -_ (Sq )i : PH,(BVk)2td+(2,-l)k - PH*(BVk)2ld+(2-1)k is an isomorphismof GLk-modulesforeveryi > t However,this resultdoes not confirmhow large t should be Theorem 1.1 shows that a rathersmall number t k- commonlyservesforeverydegreed It will be pointedout in Remark6.5 that t = k- is the minimumnumberforthis purpose An inductivepropertyof k-spikes,whichwill also play a key role in the paper, is that ifm is a k-spike,then (2n - + m) is a (k + 1)-spikeforn big enough Two applicationsof the studywill be exploitedin this paper The firstapplication is the followingtheorem,whichis one of the paper's main results Theorem 1.2 Trk is not an isomorphism for k > Furthermore, for every k > 5, there are infinitelymany degrees in which Trk is not an isomorphism That Tr5 is not an isomorphismin degree9 is due to Singer[28] In orderto prove this theorem,using the notion of k-spike,we introducethe conceptof a criticalelementin Extk (F2, F2) in such a way that ifd is the stem of a criticalelement,then Trk is not an isomorphismeitherin degreed or in degree 2d + k Further,we showthat ifx is critical,thenso is hnXforn big enough Our inductiveprocedurestartswiththe initialcriticalelementPh2 fork = CombiningTheorem1.2 and theresultsby Singer[28],Boardman[4]and BrunerHa-Htrng[7],we get Corollary 1.3 (i) Trk is an isomorphism for k = 1, and is not an isomorphism for k > (ii) Trk k For and = for each k > 5, there are infinitelymany degrees in which (iii) is not an Trk isomorphism Remarkably,we not knowwhetherthe algebraictransferfailsto be a monomorphismor failsto be an epimorphismfork > Therefore,Singer'sconjecture is stillopen foreveryk Conjecture 1.4 ([28]) Trk is a monomorphism The followingtheoremis relatedto this conjecture Theorem 1.5 If Tre detects a critical element, then it is not a monomorphism and further,for each k > ?, there are infinitelymany degrees in which Trk is not a monomorphism A family{a| i > 0} of elementsin Extk (F2,IF2) is called a Sq?-family if = (Sq?)'(ao) (or in IF2 for every i > GLk PH,(BVk)) Recall that, if a C Extt (F2, F2), thent- k is called the stemof a, and denotedby Stem(a) The root degreeofa is the maximumnonnegativeintegerr such that Stem(a) can be written in the formStem(a) - 2rd+ (2' - 1)k, forsome nonnegativeintegerd The second applicationof our studyis the followingtheorem,whichis also one of the paper's main results Theorem 1.6 Let {aiI i > 0} be an Sq?-family in Extk(F2,F2) and let r be the root degree of ao If Trk detects an for some n > max{k-r-2, then it detects 0}, > n and r i detects modulo 2, 0}< j < n for every Ker(Sq?)n-j for max{k aj This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions NGUYEN 4068 H V HUNG An Sq0-familyis called finiteifit has onlyfinitely manynonzeroelements.The existenceof finiteSq?-familiesin Ext (IF2,IF2)is well known,and that of finite Sq?-familiesin F2 PH (BVk) will be shownin Section9 GLk The followingis a consequenceof Theorem1.1 and Theorem1.6 Corollary 1.7 (i) EveryfiniteSq?-familyin F2 PH*(BVk) has at most GLk (k - 2) nonzeroelements thenit does not detectany elementof a finite (ii) If Trk is a monomorphism, Sq?-familyin Extk(F2,F2) withat least (k - 1) nonzeroelements The followingis an applicationof Theorem1.6 intothe investigation of Tr4 Proposition 1.8 Let {bil i > 0} and {bi i > 0} be theSq?-familiesin ExtA(F2,IF2) withbo one of the usual five elementsdo,eo,Po,D3(O),pO, and bo one of the usual two elementsfo,g (i) If Tr4 detectsbn for some n > 1, thenit detectsbi for everyi > (ii) If Tr4 detectsbnfor some n > 0, thenit detectsbi for everyi > Based on this event,we provethe followingtheoremby showingthat Tr4 does not detect gl, D3(1), p' Theorem 1.9 Tr4 does not detectany elementin the threeSq?-families{gil i > 1}, {D3(i)l i > 0} and {p' i > 0} This theoremgives furthernegativeinformation on Minami's ([21]) conjecture that the localization of the algebraic transfergiven by invertingSq? is an isomorphism The firstnegative answer to this conjecturewas given in BrunerHa-Hirng [7] by showingthat the elementin (Sq?)-1Ext (F2,F2) representedby the family{gil i > 1} is not detected by (Sqo)-lTr4 From Theorem 1.9, the two elementsin (Sq)-lExt (IF2, F2) representedrespectivelyby the two families {D3(i)l i > 0} and {pi[ i > 0} are also not detectedby (Sq?)-1Tr4 Recently,T N Nam informedthe author about his claim that Tr4 does not detectD3(0) Conjecture 1.10 Tr4 is a monomorphismthat detects all elements in Ext4(F2,F2) except the ones in the threeSq?-families{gil i > 1}, {D3(i)l i > 0} and {p'i i > 0} The following theoremwouldcompleteourknowledgein Corollary1.3 on whether Tr5 is not an isomorphismin infinitely manydegrees Theorem 1.11 If h,n+lg is nonzero,thenit is not detectedbyTr5 It has been claimedby Lin [16] that hn+lgn is nonzeroforeveryn > The paper is dividedinto nine sectionsand organizedas follows.Section is a recollectionof the Kameko squaringoperation.In Section3, we explainthe notion of k-spikeand then study the Kameko squaringand its iteratedoperationsin kspike degrees Section4 deals withan inductiveway of producingk-spikes,which plays a keyrolein the proofsof Theorems1.1, 1.2, 1.5 and 1.6 In Section5, based on the concept of criticalelement,we prove Theorems1.2 and 1.5 Section is devotedto the proofsof Theorems1.1 and 1.6 Sections7 and are applications to the study of the fourthand the fifthalgebraic transfers.Final remarksand conjecturesare givenin Section This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions AND GENERAL ALGEBRA STEENROD ON THE PRELIMINARIES LINEAR GROUPS 4069 OPERATION SQUARING this sectionis a recollectionof the Kameko To make the paper self-contained, squaringoperationSq? on F2 PH,(BVk) The most importantpropertyof the GLk Kameko Sq? is that it commuteswiththe classical Sq? on Ext, (F2, F2) (definedin [18]) throughthe algebraictransfer(see [4], [21]) This squaringoperationis constructedas follows As is well known,H*(BVk) is the polynomialalgebra,Pk := F2[x, ,xk], on k generatorsx, xk, each of degree1 By dualizing, H*(BVk) = (a, .,ak) is the divided power algebra generatedby a1, , ak, each of degree 1, where is dual to xi E H1(BVk) Here the dualityis taken with respectto the basis of H*(BVk) consistingof all monomialsin xl, ,Xk In [13] Kameko defineda homomorphism Sq: a where a(il ik) * - H*(BVk) (i) H*(BVk), ak(2ik+l) a (2i1+1) (ik) ak k is dual to x1 x i The followinglemma is well known -0 Lemma 2.1 Sq is a homomorphism ofGLk-modules See e.g [7] fora proof.Further,thereare two well-known relations, See [10] for an explicit proof Therefore, Sq maps PH,(BVk) The Kameko Sq? is definedby 0 Sq? = Sq GLk : F2 ( GLk -0 The dual homomorphism Sq Sq(x'2 * )= PH*(BVk) : Pk = Sq Sqt q2tq =0, 2t+1Sq -? ~0 Pk of Sq Jl X1 O 0, - k-1 "Xk F2 -* GLk to itself PH,(BVk) is obviously given by , li,.-.,Jk odd, otherwise Hence Ker(Sq.) = Even, whereEven denotesthe vectorsubspace ofPk spannedby all monomialsx~1.x.k withat least one exponentit even The followinglemmais moreor less obvious Lemma 2.2 ([7]) Let k and d be positiveintegers.Suppose that each monomial x xi * of Pk in degree 2d + k with at least one exponent it even is hit Then Sq : (F2 0Pk)2d+k A (F2 0Pk)d A is an isomorphism of GLk-modules This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions 4070 NGUYEN H V HUNG Here, as usual, a polynomialis called hitifit is A-decomposablein Pk A proofof this lemmais sketchedas follows Let s: Pk -> Pk -0 be a right inverse of Sq, defined by 5(2ii S(X * 2i XI ^ik Xk )) = 1+1 2ik+l Xk It should be noted that s does not commute with the doubling map on A, that is, in general, However, Im(Sq2ts- Sq2s sSqt sSqt) C Even Let A+ denotethe ideal of A consistingof all positivedegreeoperations.Under the hypothesisof the lemma,we have + Even)2d+k (A+Pk C (A+Pk)2d+k Therefore,the map S:(F2 A Pk)d - (F2 ?Pk)2d+k s[X] = [sX] A is a well-defined linearmap Further,it is the inverseof -0 Sq : (F2 ?Pk)2d+k A (F2 ?Pk)d A Therefore,Sq, is an isomorphismin degree2d + k THE ITERATED SQUARING OPERATIONS IN k-SPIKE DEGREES The followingnotion,whichis originallydue to Kraines [14], formulatessome special degreesthat we will mainlybe interestedin Definition 3.1 A naturalnumberm is called a k-spikeif (a) n = (2n - 1) + + (2nk 1) with n, , nk > 0, and m cannot be written as a sum of less than k termsof the form(2' - 1) (b) Note that k-spikeis our terminology Otherauthorswrite/,(m)= k to say in is a k-spike.(See e.g Wood [33, Definition4.4].) One easily checkse.g that 20 is a 4-spike,27 is a 5-spikeand 58 is a 6-spike Let a(m) denotethe numberofones in the dyadicexpansionofm The following two lemmasare moreor less obvious,but usefullater Lemma 3.2 Condition (a) in Definition 3.1 is equivalent to a(rn + k) < k < rn, In _ k (mod 2) Proof Suppose In = (2'n - 1) + + (2nk 1) with nl, , nk > Then m > k = (21 - 1)+ + (2 - 1) (k terms) In addition, from7n+ k = 2T' + + 2n with 71, , nk > 0, it implies a(n + k) < k and n- k (mod 2) The equality a(mr + k) = k occurs if and only if n., , nk are differentfrom each other Conversely,suppose that a(m + k) < k < Im and m = k (mod 2) Let i a(m + k) Then we have m + k = 2"1 + + 2"2, This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4071 where m, , mi > 0, as m + k is even If at least one exponentmj > 1, thenwe write(m + k) as a sum of (i + 1) terms of 2-powersas follows: m+k 2m + -2mJ-1 +2m7-l + + 2m This procedure can be continued if at least one of the exponents mi, , mj - 1, mj- 1, , mi is biggerthan Aftereach step,the numberoftermsin the sum increases by The procedurestopsonlyin the case whenthesum becomesm+k = 2+ +2 withthe numberofterms(m + k)/2 > 2k/2= k In particular,we reachedat some step a sum of exactlyk terms n + k = 2n1 + - + 2n with n1, , nk > 0, or equivalently m =(2 -1) + + (2nk -1) D The lemmais proved The followinglemmahelps to recognizek-spikes Lemma 3.3 A naturalnumberm is a k-spikeif and onlyif (i) a(m + k) < k < m, m - k (mod 2), and (ii) a(m + i) > i for < i < k Proof From Lemma 3.2, if m satisfies (i), then m = (2 - 1) +.* + (2k - 1) with nl, ,nk > Also by Lemma 3.2, ifm satisfies(ii), then it cannot be writtenas a sum of less than k termsof the form(2n - 1) So, ifm satisfies(i) and (ii), then it is a k-spike suppose m is a k-spike.Then (i) holds by Lemma 3.2 Conversely, It sufficesto show (ii) Suppose to the contrarythat ca(m+ i) < i forsome i with < i < k We then have cases (m + i) < i < k < m Let us consider the two Case 1: m - i (mod 2) Then, by Lemma 3.2, we get m = (2n - 1) + (2n - 1) with nl, , ni > This contradicts to the definitionof a k-spike + Case 2: m- i - (mod 2) It impliesi > Indeed, if i = 1, combiningthe hypothesisa(m + 1) < with the fact m + is odd, we get m + = This contradictsthe hypothesisthat m is a naturalnumber By Lemma 4.3 below,we have a(m + (i- 1)) = a(n + i) - < i- As m i- (mod 2), we apply Lemma 3.2 again to see that m can be writtenas a sum of (i- 1) termsof the form(2n - 1) This is also a contradiction Combiningthe two cases, we see that if m is a k-spike,then (i) and (ii) hold D The lemmafollows The followingpropositionis originallydue to Kameko [13] We give a proofof it to make the paper self-contained Proposition 3.4 If m is a k-spike,then Sq, : (F2 ?Pk)m A (F2 A Pk) m-2 is an isomorphismofGLk-modules This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions NGUYEN H V HUNG 4072 to showthatanymonomialR ofPk in degree Proof By usingLemma 2.2, it suffices m with at least one even exponentis hit Such a monomialR can be written,up to a permutationof variables,in the form R =x 1.iQ2, with < i < k, where Q is a monomial in degree (m - i)/2 If i = 0, then R = Q2 is simplyin the image of Sql (It is also in the image of Sq , as R = Q2 = Sqn Q.) So, it sufficesto considerthe case < i < k Let X be the anti-homomorphism in the Steenrodalgebra The so-calledx-trick, whichwas knownto Brownand Petersonin the mid-sixties,states that uSqn(v) - X(Sqn)(u)v mod A+M, foru, v in any A-algebraM (See also Wood [32].) In our case, it claims that R = xl' is hit if and only if x(Sq XiQ2 = xl xiSq Q )(xl ** xi)Q is We will show x(Sq )(xz * **xi) = As A is a commutativecoalgebra,X is a homomorphism of coalgebras(see [20, Then we have the Cartan formula Proposition8.6]) E = x(Sqn)(uv) i+j =n x(Sqi)(u)x(Sqj)(v) it is shownby Brownand Petersonin [5] that Furthermore, X(Sqn')(x) )( X(q - 2, \i 0, forxj in degree1 So, in order to prove x(Sq~2 ~)(x cannot be writtenin the form m = (2e if n =2q - l for some q, otherwise, * xi) we need only to show that m2 1) + +(2 - 1) with ?1, ,/i > This equation is equivalent to m - (21+1- 1) + (2eg+1- 1) Since < i < k, this equalitycontradictsthe hypothesisthat m is a k-spike.The D propositionis completelyproved The followinglemmais the base foran iteratedapplicationof Proposition3.4 Lemma 3.5 If m is a k-spike,thenso is (2m + k) Proof (a) Fromthe definitionof k-spike, m= (2l - 1) + + (2nk-1), for nl, , nk > It implies that 2m + k = (2nl+l - 1) + + (2nk+1- 1) of k-spike So, 2m + k satisfiesthe firstconditionin the definition Also we have this definition, by (b) a(m+ k - j) > k- j, This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4073 for1 < j < k Hence a(2m + k + (k - 2j)) (2m + + (k-2j + 1)) = a(2(m + k-j)) = a(m + k - j) > k-j > k - 2j, = a(2(m + k-j) + 1) = c(2(m + k - j)) + (by Lemma 4.3) = a(m k- j)+ > + (k-j) > -2j+1 Note that each i satisfying < i < k can be written either in the form i = k- 2j (for1 < j < k21) or in the formi = k- 2j + (for1 < < k) So, the above two inequalitiesshowthat a(2 + k+ i) > i, for1 < i < k Thus, 2m + k satisfiesthe second conditionin Definition3.1 Combiningparts (a) and (b), we see that 2m + k is a k-spike D Remark 3.6 The converseof Lemma 3.5 is false For instance,27 is a 5-spike, whereas11 = (27 - 5)/2 is not Proposition 3.7 If m is a k-spike,then : PH*(BVk)m-k (Sq)i+1 " PH*(BVk)2m+(2i- 1)k is an isomorphismofGLk-modulesfor everyi > Proof Ifm is a k-spike,thenbythedual ofProposition3.4, we have an isomorphism of GLk-modules Sq -+ PH*(BVk)m : PH*(BVk)m- On the otherhand, fromLemma 3.5, ifm is a k-spike,thenso is 2im + (2i - l)k foreveryi > Hence, repeatedlyapplyingthe dual of Proposition3.4, we get an isomorphismof GLk-modules (Sq )i+ : PH*(BVk)m-k PH*(BVk)2,m+(2i-1)k D The propositionis proved Corollary 3.8 If m is a k-spike,then (Sq0)i+ : (F2 PH*(BVk))m GLk (IF2 ( GLk PH*(BVk))2Tm+(2i-1)k is an isomorphism for everyi > RECOGNITION OF k-SPIKES In this section,we introducean inductiveway of producingk-spikes,whichwill play a key role in the proofsof Theorems 1.1, 1.2, 1.5 and 1.6 in the next two sections Lemma 4.1 If m is a k-spike,then(2' -1 + m) is a (k + 1)-spikefor everyn with2 > m + k- This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4075 Remark 4.4 Lemma 4.1 cannot be improved in the meaning that the hypothesis 2n+1 > m + k- does not imply (2' - + m) to be a (k + 1)-spike This is the case of k = 5, m = 27 and 2n = 16, because 15 + 27 = 42 is not a 6-spike The followingcorollaryis a keypoint in the proofof Lemma 6.3 and therefore in the proofsof Theorems1.1 and 1.6 Corollary 4.5 2k - k is a k-spikefor everyk > Proof.We provethis by inductionon k The corollaryholds triviallyfork = Suppose indlctivelythat 2k - k is a k-spike Then, as 2k > (2k - k) + k- 1, applyingLemma 4.2 to the case n = k and m = 2k - k, we have 2k1 _ (k + 1) (2k ) + (2k k) D to be a (k + 1)-spike The corollary follows THE ALGEBRAIC TRANSFER IS NOT AN ISOMORPHISM FOR k > We firstbrieflyrecall the definitionof the algebraic transfer Let P1 be the submoduleof F2[xl,x 1] spanned by all powersx1 with i > -1 The usual Aaction on P1 = F2[xl] is canonicallyextendedto an A-action on F2[xI, x1 1] (see Adams [2], Wilkerson[31]) P1 is an A-submoduleof F2[xl,xl1] The inclusion P1 C P1 givesrise to a shortexact sequence of A-modules: O - P1 - P1 E- F2 -> Let e1 be the corresponding element in Ext(E-1F2, Singer set ek P) el * e1 E Ext(E-kF2, Tr : he defined Tor A(F2, -kF2) Pk) (k times) Then, TorA (F2, Pk) F2 ?Pk by Tr(z) =eknz Its image is a submodule of (F2 ?Pk)GLk A The k-th algebraic transferis defined to be the dual of Trk A We will need to apply the followingtheoremby D Davis [9] Let hnbe the nonzeroelementin Ext (F2F2F) Theorem 5.1 ([9]) If x is a nonzeroelementin Ext+k d(F2, 2) with4 < d < 23, thenhnx : for everyn > 2j + The followingconceptplays a keyrole in this section Definition 5.2 A nonzero element x C Ext,(F2, F2) is called critical if (a) Sq?(x) = 0, and (b) 2Stem(x) + k is a k-spike Note that, by Lemma 3.5, if Stem(x) is a k-spike,thenso is 2Stem(x) + k Lemma 5.3 If x C Ext(F2,F2F) is critical,then so is hnx for everyn with 2n > max{4d2,d + k}, whered - Stem(x) Proof First,we show that if x is critical,then Stem(x) > Indeed, suppose to the contrary that Stem(x) - 0; then x = hk As x is critical, Sq?(x) = Sq?(ho) hk = This impliesthat k > 4, as h1,h ,h all are nonzero,whereas h4 = However,2Stem(x) + k k is not a k-spikefork > 4, because it can be writtenas a sum k = + + + of (k- 2) termsof the form(2' - 1) This contradicts of a criticalelement the definition This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions NGUYEN 4076 H V HUNG Now we have Stem(x) > Combiningthe factthat Sq? is a monomorphism in < positivestemsof ExtA(F2, F2) fork 4, and that x is critical,we get k > As x is a nonzeroelementof positivestemin ExtA (F2, F2) withk > 4, by the vanishing line theorem(see [1]) we have Stem(x) > Therefore, x satisfiesthe hypothesisof Theorem5.1 that d= Stem(x) > Let j be the smallest positive integersuch that 2J > d Then, the smallest positiveintegeri with2i > d2 shouldbe either2j or 2j- Fromthe hypothesis 2n > 4d2, it impliesthat 2n-2 > d2 Hence, we get n - > i > 2j - 1, or n > 2j + equivalently, Therefore,by Theorem5.1, hnx i if27 > 4d2 As SqO is a homomorphism of algebras,we have Sq?(hnx) = Sq?(hn)Sq?(x) = Sq?(hn)? = Since x is critical,m := 2d + k is a k-spike.We need to showthat 2Stem(hnx)+ (k + 1) is a (k + 1)-spike.We have Stem(hnx) = 2n- + Stem(x) = 2n - + d A routinecalculationshows 2Stem(hnx)+ (k + 1) = 2(2 -1 + d) + (k + 1) 2n+ - + (2d + k) + = 2+ - + m By Lemma 4.1, thisnumberis a (k + 1)-spikeforeveryn with2n+1 > m + k - = 2(d + k) - 1, or equivalently2n > d + k In summary,hnx is criticalforeveryn with 2n > max{4d2,d + k} The lemmais proved D Remark5.4 (a) Suppose hnx ? although2n < 4(Stem(x))2 If x is critical and 2n > Stem(x) + k, thenhnxis also critical in positive (b) Thereis no criticalelementfork < 4, as Sq? is a monomorphism stemsof ExtA(F2, F2) fork < Proposition 5.5 (i) For k = 5, thereis at least one number,whichis the stem of a criticalelement many numbers,whichare stems of (ii) For each k > 5, thereare infinitely criticalelements Proof For k = 5, Ph2 E Ext5l16(F2,IF2) e.g Tangora [30]) that Ext532 (F2, F2) is critical Indeed, it is well known (see , so we get Sq?(Ph2) = Further,by Lemma 3.3, 2Stem(Ph2) + = 27 is a 5-spike We can startthe inductiveargumentof Lemma 5.3 withthe initialcriticaleleD mentPh2 The propositionfollows The followingtheoremis also numberedas Theorem1.2 in the Introduction Theorem 5.6 Trk is not an isomorphismfor k > Furthermore, for every k > 5, thereare infinitely manydegreesin whichTrk is not an isomorphism This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4077 Proof In orderto prove the theorem,by means of Proposition5.5 it sufficesto show that Trk is not an isomorphismeitherin degreed or in degree2d + k, where d denotesthe stemof a criticalelementx E ExtA(F2,F2) We considerthe followingtwo cases: Case 1: x is not in the image of Trk Then, Trk is not an epimorphismin degreed Case 2: x = Trk(y) for some y E F2 PH,(BVk) GLk Fromx 0, it impliesthat y 570 Accordingto Boardman [4, Thm 6.9 and Cor 6.12] and Minami [21, Thm 4.4], we have a commutativediagram (F2 GLk Trk PH(BVk))d Extk,k+d(2 Sqo 2) Sq0 v ? Trk (F2 ? PH*(BVk))2d+k GLk Extk2(kd) ) wherethe leftverticalarrowis the Kameko Sq? and the rightverticalone is the classical squaringoperation As m = 2d+ k is a k-spike,by Corollary3.8 the Kameko Sq? is an isomorphism So, fromy :: 0, we have z = Sq(y) $ of the diagram,we get Now, by the commutativity Trk(z) = Trk(Sq?(y)) = Sq?(Trk(y)) == in degree 2d + k The theoremis This means that Trk is not a monomorphism D completelyproved Remark5.7 (a) We can showthat F2 PH*(BV5)11 = It impliesthat Ph2 is GL5 not detectedby Tr5 (b) By Lemma 5.3, hnPh2 is criticalforeveryn > 9, as Stem(Ph2) + < 4(Stem(Ph2))2 = 112 = 484 < 29 = 512 Also, by Remark 5.4, hnPh2 is critical forn = 4, 5,6, as it is nonzero(see [6]) and 24 > Stem(Ph2) + = 16 R Bruner privatelyclaimed h7Ph2 ~ It seems likelythat h8Ph2 If so, by the same argument,thesetwo elementsare also critical The following corollaryis also numberedas Corollary1.3 in the Introduction Corollary 5.8 (i) Trk is an isomorphism for k = 1, and (ii) Trk is not an isomorphism for k > (iii) For k = and for each k > 5, there are infinitelymany degrees in which Trk is not an isomorphism This resultis due to Singer[28] fork = 1,2, to Boardman [4] fork = 3, and to Bruner-Ha-Hirng[7] fork = That Tr5 is not an isomorphismin degree9 is also due to Singer[28] The remainingpart is shownby Theorem5.6 Our knowledge'sgap on whetherTr5 is not an isomorphismin infinitely many degreeswill be studiedin Section8 This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions H V HUTNG NGUYEN 4078 The followingtheoremis also numberedas Theorem1.5 in the Introduction Theorem 5.9 If Trt detectsa criticalelement,thenit is not a monomorphism, and further, manydegreesin whichTrk is not a for each k > e, thereare infinitely monomorphism Proof The proofproceedsby inductionon k > ? For k = ?, suppose Tre detects a criticalelementxe E ExtA(F2,F2) Then, in degree by Case in the proofof Theorem 5.6, Tre is not a monomorphism 2Stem(xe) + f By means of this argument,it sufficesto show that if Trk detects a critical elementXk,then Trk+l detectsinfinitely manycriticalelements,whose stemsare different fromeach other Fromthe hypothesis,xk = Trk(yk)forsome YkE F2 PH*(BVk) WithambiGLk guityof notation,let hn also denotethe elementin F2? under Tr1 is the usual hn E Extl(IF2, F2) As Tr - algebras (see [28]), we have GL1 PH,(BV1), whoseimage Trk is a homomorphism of (k Trk+l(hnyk) = Trl(hn)Trk(yk) = hnxk By Lemma 5.3, the element hnXk is critical forevery n with 2n > max{4d2, d + k} By the firstpart of the theorem,since Trk+l detectsthe criticalelementhnXk, in degree2Stem(hnxk)+ (k + 1) foreveryn with2n > it is not a monomorphism in infinitely max{4d2,d + k} Thus, Trk+l is not a monomorphism manydegrees D The theoremfollows THE STABILITY OF THE ITERATED SQUARING OPERATIONS The followingtheorem,whichis also numberedas Theorem1.1 in the Introduction,showsthat SqO is eventuallyisomorphicon F2 PH (BVk) More precisely, GLk it claims that ifwe startfromany degreed of this module,and apply SqO repeatedly at most (k- 2) times,then we get into the region,in whichall the iterated squaringoperationsare isomorphisms 6.1 Let d be an arbitrarynonnegative integer Then Theorem )i (Sq +2 - PH*(BVk)2k-2d+(2k-2-_)k PH*(BVk)2,d+(22-1)k is an isomorphism of GLk-modules for every i > k- In the theorem, for k = we take the convention that 21-2d + (21-2 Let us denote (Sq?) 1(F2 GLk PH*(BVk))d -lim Sq( i (F2 GLk PH*(BVk))2d+(2,-1)k 1)k = d Sq( The followingcorollaryis an immediateconsequenceof Theorem6.1 Corollary 6.2 Let d be an arbitrarynonnegative integer Then, (i) The following iterated operation is an isomorphism for every i > k - 2: (SqO)i-k+2 F2 ( GLk PH,(BVk)2k-2d+(2k-2-)k - F2 GLk PH*(BVk)21d+(2-l)k This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions ) STEENROD AND GENERAL ALGEBRA LINEAR GROUPS 4079 (ii) (Sq?)- (F2 ( GLk (F2 PH,(BVk))2k-2d+(2k-2_1)k PH,(BVk))d GLk (iii) If d = 2k-2d' + (2k-2 - 1)k for some nonnegative integer d', then (Sq0)- (F2 PH*(BVk))d - (F2 PH*(BVk))d GLk GLk In orderto proveTheorem6.1, we need the followinglemma Let Sk denotethe functiongivenby 6k(d) = 2d + k Lemma 6.3 If d is a nonnegativeintegerwitha(d + k) < k, then 5k-l(d) 2k-ld + (2k-l - l)k is a k-spike Proof The lemmaholdstriviallyfork = Indeed,fromthe hypothesisa(d + 1) < it implies that d = - for some n Then 0?(d) - d- 2n - is an 1-spike We now consider the case of k > First, we observe that k < 2k- d + k (mod 2) and (2k- - 1)k a(2k- d + (2k-1 - l)k + k) = a(2k-l(d + k)) = a(d + k) < k By Lemma3.2, k-l(d) = 2k-ld+(2k-~-ll)k satisfiescondition(a) ofDefinition3.1 So, in orderto provethe lemma,it sufficesto showthat -l)k+i) a(2k-ld+(2k-1 >i for1 < i < k We now work modulo 2k-1 First, we have 2kld + (2k-1 - 1)k _ (2k-1 - l)k (mod 2k-1) Let k = 2n + + + 2n be the dyadic expansion of k with nt > > nl We get (2k-1 - 1)k = 2k-1(2nt + + 2n2) + (2k-l+nl - * (2nt + + 2n1)) Thus (2k-1 - l)k - 2k-l+nl - - + 2T1) (mod 2k-l) (2nt + 2k - (2nt + + 2n1) (mod 2k-1) - 2k-1- k(mod2k-l), where2k-1 - k > because k > As a consequence,we get 2k-ld + (2k-1 - 1)k + i - 2k - k + i (mod 2k-1) for1 < i < k Since k > and d > we have 2k-ld + (2k-1 - l)k + i > (2k-1 - 1)2 + > 2-1 From this inequality it implies that, in the dyadic expansion of 2k-ld + (2k-1- )k + i, there is at least one nonzero term 2n with n > k- On the other hand, as 2k-1 - k + i < 2k-1 for < i < k, the dyadic expansion of 2k-1 - k + i is just a combination of the 2-powers 20, 21, , 2k-2 Therefore, in order to prove a(2k-ld+ (2k-1 )k i) > i for1 < i < k, we need onlyto showthat a(2 - k +i) >i This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions NGUYEN H V HUNG 4080 FromCorollary4.5, 2k-1 - (k- 1) is a (k- 1)-spike Then we have a(2k-1 - (k - 1) + j) > j for1 < j < k-1 Set i = j + 1; we get k +i) >i a(2k-1- for2 < i < k In addition,it is obviousthat > k+1) a(2k-1- In summary,we have shownthat a(2k-1 > i -k+i) for1 < i < k The lemmais proved O Remark6.4 (a) Lemma 6.3 cannot be improvedin the meaningthat the number 6~-2(d) = 2k-2d + (2k-2 - l)k is not a k-spike in general Indeed, takingd = 2t + - k witht big enoughso that d > 0, we have a(2k-2d + (2k-2 - l)k + (k - 1)) = a(2t+k-2 + (2k-2 - 1)) = k- By Lemma 3.3, 2k-2d + (2k-2 - 1)k is not a k-spike (b) However,a numbercould be a k-spikealthoughit is not of the form k-1(d) foranynonnegativeintegerd For instance,thisis thecase ofthe following numbers with k = 4: Stem(fl) = 40, Stem(e2) = 80, Stem(D3(2)) = 256, Stem(p) = 288, Stem(p2) = 144, wheree2,fl, P2,D3(2), p' are the usual elementsin Ext4 (F2,F2) This observation will be helpfulin the proofof Proposition7.2 below Proofof Theorem6.1 Accordingto Wood's theorem[32] (it was originallyPeter- son's conjecture), the primitive part PH,(BVk) is concentrated in the degrees d witha(d + k) < k This facttogetherwiththe equality + k) = a(2i(d + k)) = a(d + k) a(6'(d) show that, ifa(d + k) > k, then the domainand the targetof the homomorphism in the theoremare both zero If a(d + k) < k, then the theoremis an immediateconsequenceof Lemma 6.3 D and Proposition3.7 The theoremis proved Remark 6.5 Let k = and d = As 65-2(0) = 35, Theorem 6.1 claims that (Sq?)i-3 : PH (BV5)35 - PH*(BV5)5(2-_1) is an isomorphismof GL5-modulesfori > In the finalsectionwe will see that Sq? : F2 PH*(BV5)l5 GL5 - F2 PH*(BV5)35 GL5 is not a monomorphism.This showsthat Theorem6.1 cannotbe improvedin the meaningthat (k - 2) is, in general,the minimumtimesthat we must repeatedly apply SqO to get into "theisomorphism region"ofthe iteratedsquaringoperations A family {ail i > 0} of elements in ExtA(F,F2F2) is called an Sq?-family if = (Sq?)'(ao) foreveryi > An Sq?-familyin F2 PH*(BVk) is similarly GLk defined This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4081 Definition 6.6 Let ao cExtk(F2, F2) The root degreeof ao is the maximum nonnegativeintegerr such that Stem(ao) can be writtenin the form Stem(ao) = 5(d) = 2d + (2' - 1)k, forsome nonnegativeintegerd The followingtheoremis also numberedas Theorem1.6 in the Introduction Theorem 6.7 Let {ail i > 0} be an Sq?-familyin Ext(F2,F2,) and let r be the rootdegreeofao If Trk detectsan forsome n > max{k-r-2, 0}, thenit detectsai for everyi > n and detectsaj moduloKer(Sq?)n-j formax{k - r - 2, 0} < j < n Proof.It is easy to see that ca(Stem(ai) + k) = a(2(Stem(ao) + k)) - a(Stem(ao) + k) Suppose a(Stem(ao) + k) > k; thenwe have a(Stem(ai) + k) > k foreveryi > By Wood's theorem[32] (it was originallyPeterson'sconjecture),PH*(BVk)t = in any degreet withac(t+ k) > k So, all elementsofthe family{ai i > 0} are not detectedby Trk Now we considerthe case wherea(Stem(ao) + k) < k We observethat a(Stem(ao) + k) = a(2r(d + k)) = a(d + k) < k Set q = max{k - r - 2, 0}, and we have +l (Stem(ao)) = 6++l(d) Stem(aq+l) -= Note that q+ r = max{k - r - 2,0}+ r+1 > (k-r-2) + r1 =k-1 So, by Lemmas 6.3 and 3.5, Stem(aq+l) is a k-spike Accordingto Theorem6.1, ifc = Stem(aq), then : PH*(BVk)c (Sq)i- -) PH*(BVk)2i-qc+(2i-q- )k is an isomorphismof GLk-modulesforeveryi > q Suppose Trk detects an with n > q, that is, an = Trk(an) for some an in As the squaring F2 PH*(BVk) If i > n, then we set = (SqO)i-n(a) GLk we have operationscommutewitheach otherthroughthe algebraictransfer, (an)= = (Sq)i-n = Trk(SqO)i- (Sq)i-nTrk(an) (an) = Trk(ai) Thus, is detectedby Trk foreveryi > n Next we considerj withmax{k - r- 2, 0} < j < n Then we set aj = [(Sq?)n-j]-l(n) This makes sense, as it is shownabove that (Sqo)n-j is isomorphicin the degree of aj Again, as the squaringoperationscommutewith each other throughthe we have algebraictransfer, (Sq?)n-iTrk((a j) = Trk(Sq?)n-j(aj) = Trk(an) = an = (Sq)n-j (aj) This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions 4082 NGUYEN H V HUNG As a consequence,we get Trk((aj) = aj (mod Ker(Sq?)n-j) This means that Trk detectsaj moduloKer(Sq)n~-j The theoremis proved D Remark6.8 (a) Underthe hypothesisof Theorem6.7, let a' = Trk(Sq?)'-'((a) foreveryi > max{k - r - 2, 0} no matterwhetheri > n or i < n Then we get a new Sq?-family{al i > maxk - r - 2,0}}, whose everyelementis detectedby Trk and ifi > n, ,_ ai, if i < n l (mod Ker(Sq?)n-i), The new Sq?-familyis called the adjustmentof the originalone (b) Theorem6.7 is still valid and can be shownby the same proofifwe replace max{k - r- 2, 0} by any numberq suchthat Stem(aq+i) is a k-spike.This remark will be usefulin the proofof Proposition7.2 forthe case k = Corollary 6.9 Let {ail i > 0} be an Sq?-familyin ExtA(F2,IF2) and let r be the rootdegreeof ao Suppose the classical Sq? is a monomorphism in the stems of the elements{ail i > max{k - r - 2,0}} If Trk detectsan for some n > max{k - r - 2, 0}, thenit detectsai for everyi > max{k - r - 2, 0} An Sq?-familyis calledfiniteifit has onlyfinitely manynonzeroelements,infinite if all of its elementsare nonzero The followingis also numberedas Corollary1.7 in the Introduction Corollary 6.10 (i) EveryfiniteSq?-familyin F2 PH,(BVk) has at most GLk (k - 2) nonzeroelements thenit does not detectany elementof a finite Trk is a monomorphism, If (ii) at least (k - 1) nonzeroelements in with Sq?-family ExtkA(F2, F2) Proof (i) Suppose that {ail i > 0} is an Sq?-familyin F2 PH,(BVk) withat GLk least (k - 1) nonzeroelements Then ao, a, ,ak-2 are its first(k - 1) nonzero elements Set d = deg(ao); then deg(ak-2) = 2k-2d + (2k-2 - 1)k Therefore,by Corollary6.2, (Sq)i-+2 : F2 0( PH,(BVk)2k-2d+(2k-2_l)k F2 GLk GLk PH,(BVk)2td+(2i-l)k is an isomorphismforeveryi > k- Therefore,fromak-2 7~ it impliesthat = (Sq0)i-k+2(ak-2) is nonzeroforeveryi > k - Thus, the Sq?-familyis infinite (ii) Let ao, al, , ak-2 be the last (k - 1) nonzeroelementsof the givenfinite Sq?-familyin Ext(F2, IF2) As ak-2 is the last nonzeroelementin the Sq?-family, we have Sq?(ak_2) = Set d = Stem(ao); then by Lemma 6.3, 2Stem(ak_2) + k 2k-ld + (2k-1 - l)k is a k-spike.So, ak-2 is critical Suppose to the contrarythat Trk detectssome (nonzero) elementin the Sq?family Then, as the squaringoperationscommutewith each otherthroughthe algebraictransfer,Trk also detectsthe criticalelementak-2 Accordingto Theorem5.9, thiscontradictsthehypothesisthatTrk is a monomorphism The corollary is proved D This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ON ALGEBRA BEHAVIOR AND GENERAL OF THE LINEAR ALGEBRAIC FOURTH GROUPS 4083 TRANSFER This sectionis an applicationof the previoussectionintothe studyof Tr4 We referto [30], [6], [17] foran explanationof the generatorsof ExtA (F2, F2) It has been known(see [17]) that the graded module ExtA(F2, F2) is generated by hihjhehm,hicj, di,ei, fi,gi+ pi, D3(i), p and subject to the relations = hihi+l O, hih+2 =0, hi = h2-hi+l, fori j-1,j,j +2, j 0, hicj = O hh+3 The followingis also numberedas Conjecture1.10 in the Introduction that detectsall elementsin Ext (F2, F2) Conjecture 7.1 Tr4 is a monomorphism in the the ones three Sq?-families except {gi\i > 1}, {D3(i)l i > 0} and {p[ i > 0} That Tr4 does not detectthe family{gij i > 1} is due to Bruner-Ha-Hurng[7] the authorabout his claim that Tr4 does not detect Recently,T N Nam informed the elementD3(0) The followingproposition,which is also numberedas Proposition 1.8 in the Introduction,is an attemptto preparefora proofof Conjecture7.1 Proposition 7.2 Let {biI i > 0} and {bi i > 0} betheSq?-familiesin Ext\(F2, F2) withbo one of the usual fiveelementsdo,eo,po,D3(0),p, and bo one of the usual two elementsfo,gi (i) If Tr4 detectsbnfor some n > 1, thenit detectsbi for everyi > (ii) If Tr4 detectsbnfor some n > O, thenit detectsbi for everyi > Proof Althoughthe stemsof b2 and bl cannot be writtenas 63(d) forsome nonnegativeintegerd (except forb2 = d2 and bl = 92), it is easy to check by using Lemma 3.3 that theyall are 4-spikes Followingpart (b) of Remark 6.8, we can show this propositionby the same as Sq? is a monomorargumentas givenin the proofofTheorem6.7 Furthermore, phism in positive stems of ExtA(F2,F2) (see e.g [17]), the propositionhas the as in Corollary6.9 strongformulation D The propositionis proved to show that: By means of Proposition7.2, to proveConjecture7.1 it suffices (1) Tr4 detectsdo,eo,fo,Po; (2) Tr4 does not detect gl, D3 (1),p; and (3) Tr4 is a monomorphism The followingtheoremis also numberedas Theorem1.9 in the Introduction Theorem 7.3 Tr4 does not detectany elementin the threeSq?-families{gi i > 1}, {D3(i)I i > 0} and {p'l i > 0} Outlineof theproof.First, we show that F2 PH,(BV4) is zero in degree 20 GL4 Therefore,Tr4 does not detect gl of stem 20 and, by Proposition7.2, does not detectany elementin the Sq?-family{gil i > 1} (Note again that this part of the theoremis due to Bruner-Ha-Hirng[7].) Second, as the stemsof D3(1) and pl are respectively126 and 142, we focusto the GL4-modulePH.(BV4) in degrees126 and 142 By routinecomputations,we showthat PH, (BV4) has dimension80 and 285 in degrees126 and 142,respectively, and furtherthat F2 PH.(BV4) is of dimension1 in these two degrees GL4 This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions NGUYEN H V HUNG 4084 of Note that,as Tr1 detectsthe family{hnl n > O} (see [28]),the homomorphism = > the n Tr detects the O} family{hnl algebras subalgebrageneratedby kTrk sends the two generatorsof its domainin degrees126 and 142 to So, Tr4 definitely the nonzeroelementsh2h2 and h2h4h7, respectively.Therefore,the two elements stems 126,142 are not detectedby Tr4 D3(1) and p' of,respectively, D The theoremis provedby combiningthis fact and Proposition7.2 AN OBSERVATION ON THE FIFTH ALGEBRAIC TRANSFER From Corollary5.8, the followingconjecturenaturallycomes up Conjecture 8.1 There are infinitely manydegreesin whichTr5 is not an isomorphism The factthat gn is not detectedby Tr4 and that Tr = )k Trk is a homomorphism of algebras not implythat hign is not detectedby Tr5 For instance, hog1= h2eo and h1gl = h2fo are presumablydetectedby Tr5, as eo and fo are expectedlydetectedby Tr4 The purposeof this sectionis to provethe following, whichis also numberedas Theorem1.11 in the Introduction Theorem 8.2 If hn+lgn is nonzero,thenit is not detectedbyTr5 of algebras, Outlineof theproof.We firstobservethat,as Sq? is a homomorphism > {hn+lgnl n 1} is an Sq?-family,that is, hn+lgn, (Sqo) n-(h2gl)= > foreveryn Next, using Lemma 3.3 we easily show that Stem(h2g1)= 23 is not a 5-spike, but 65(23) = 23 + = 51 is So, by Proposition3.7, (Sq ) : PH,(BV5)23 - PH,(BVk)2i.23+(2i-1)5 is an isomorphismof GL5-modulesforeveryi > In addition,a routinecomputationshowsthat PH,(BV5) is of dimension1245 in degree23, and furtherthat F2 PH*(BV5)23 = GL5 As a consequence,we get F2 ( GL5 PH*(BV5)2i.23+(2,-1)5 = 0, foreveryi > So, the domain of Tr5 is zero in the degreethat equals Stem(hn+lgn) = 2n-1 23 + (2n-1 - 1)5 foreveryn > Therefore,if hn+lgn is nonzero,thenit is not detectedby Tr5 The theoremis ? proved Corollary 8.3 If hn+lgn is nonzerofor everyn > 1, then thereare infinitely many degrees,namelythe degreesof hn+lgn for n > 1, in whichTr5 is not an epimorphism The corollary'shypothesisis claimedto be true by Lin [16] So, Conjecture8.1 is established This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4085 of algebras, Remark 8.4 As h3g2= h5gl (see [30]) and Sq? is a homomorphism Theorem8.2 also showsthat ifhn+4gnis nonzero,then it is not detectedby Tr5 Whichelementsin Ext(5 F2,F2) are detectedby Tr5? This questioncan be partiallyansweredby usingthe factthat Tr = Ok Trk is on elementsdetectedby Trk for and the information an algebra homomorphism k < For instance,h3D3(0) = hod2(see [6]) is presumablydetectedby Tr5, as ho is detectedby Tr1 and d2 is expectedlydetectedby Tr4 (see Conjecture7.1) Based on Theorem6.7 and concretecalculations,the followingconjecturepresentssome "new" families,whichare expectedlydetectedby Tr5 Conjecture 8.5 Tr5 detectseveryelementin the Sq?-familiesinitiatedby the classes n, x, hog2,D1, H1, hlD3 (0), h2D3(0), Q3, h4D3(0), h6gl,hog3ofstems31, 37, 44, 52, 62, 62, 64, 67, 76, 83, 92, respectively Conjectures8.5 and 7.1, togetherwiththe factthat Tr = Dk Trk is an algebra homomorphism, predictthat Tr5 detects all Sq?-familiesinitiatedby the classes of stems < 125, exceptpossiblythe threefamilies,whichare respectivelyinitiated by Phl, Ph2 and hop' Since Sq?(Phl) = h2gl, everyelementof the Sq?-family initiatedby Phl is not detected by Tr5 (see [28] for Phl and Theorem 8.2 for hn+lg,) It has been knownthat Tr5 does not detectthe Sq?-familyofexactlyone nonzeroelement{Ph2} (see Remark5.7) We have no predictionon whetherthe Sq?-familyinitiatedby hop' of stem 69 is detectedor not FINAL REMARKS or Remark 9.1 We still not know whetherTrk fails to be a monomorphism fails to be an epimorphismfork > If Singer's Conjecture 1.4 that Trk is a foreveryk is true,thenby Theorem1.5 the algebraictransferdoes monomorphism not detectthe kernelof Sq? in k-spikedegrees This leads us to the studyof the kernelof Sq? in F2 ? PH*(BVk) The map GLk Sq : PH*(BVk) PH*(BVk) is obviouslyinjective Taking this observationtogetherwith Corollary3.8 into account,one would expect that the Kameko map Sq? = GLk 0-~0 Sq: F2 ? PH,(BVk) GLk -+ F2 GLk PH*(BVk) is also a monomorphism However,this is false Indeed, PH,(BV5) has dimension whileF2 PH (BV5) has dimen432 and 1117 in degrees15 and 35, respectively, GL5 sion and in degrees15 and 35, respectively.We obtainedthis claim by usinga computerprogramof S Shpectorovwrittenin GAP As in theproofofTheorem5.9, let hnalso denotetheelementin F2 PH, (BV), GL1 Tr1 is the usual hn C Ext4(F2, F2) In the folwhoseimageunderthe isomorphism lowingremark,we willuse the productof ODk(F2 PH,(BVk)) definedby Singer in [28] and his result that Tr = is a homomorphism of algebras Dk Trk k(F2 GLk PH*(BVk)) GLk - Ext(F2, This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions 2) NGUYEN H V HUNG 4086 Remark 9.2 (a) Let t5 = Tr5 (t) - h4h4 (b) If tk E F2 PH,(BVk) GLk h4h4 (F2 PH,(BV5))15 GL5 Then Sq?(t5) = and is a positivedegreeelementwith Sq?(tk) = and Trk(tk) - 0, then Sq?(hntk) = and Trk+l(hntk) 2n > 4(deg(tk))2 $/0 for every n with "Proofof Remark9.2 " Part (a) of this proofproceedsundertwo hypotheses: (i) F2 PH,(BV5) has respectivelydimension2 and in degrees15 and 35 GL5 (This is knownby a computercalculationas writtenabove.) (ii) E Im(Tr4) (This is a part of Conjecture1.10 When this paper was beingrevised,Le M Ha privatelyinformedthe authorthat he provedthis claim.) It would be betterto writea directproofin the framework of invarianttheory forthe factSq0(h4h4) = in F2 PH*(BV5) GL5 (a) As is well known,Ext 5+15(F2,JF2) = Span{h4h4,h1do} Combiningthe fact that Tr = (k Trk is an algebra homomorphism with the one that hn is in the image of Tr1 foreveryn, and is in the image of Tr4, we conclude that hoh4 and h1do are both in the image of Tr5 On the otherhand, the domainof Tr5 has dimension2 in degree 15 So, Tr5 is an isomorphismin degree 15 As F2 PH*(BV5) has respectivelydimension2 and in degrees15 and 35, there GL5 existsa nonzeroelementt5 C F2 ( PH*(BV5) in degree15 such that Sq?(t5) = GL5 Since Tr5 is an isomorphism in degree15, Tr5(t5) $ Next, we show that Tr5(ts) = h4h4 Indeed, we suppose to the contrarythat Tr5(t5) = Ah4h4 + hldo, for some A IF2 Then, as Sq?(hO) - we have Ext, (F2, F2) and Sq? is an algebra homomorphism, Sq?Tr(ts) = ASq?(h4h4) + Sq?(hdo) h4 in = Sq?(hldo) = h2d Since Tr5 commuteswiththe squaringoperations,we get Tr5Sq?(t5) = Sq?Tr5(t5) = h2dl : This contradictsthe above conclusionthat Sq?(t5) = O Therefore, Tr (ts) + Ah4h4+ h1do forany A C F2 Combiningthiswiththe factthat Trs(ts) / in Span{h h4,hldo}, we get Trs(t5) hh4 With ambiguity of notation, we also have Tr5(h4h4) is an isomorphismin degree15, we obtain t5 = h4h4 we have (b) As Sq? is an algebrahomomorphism, Sqo(hntk) = Sqo(hn)Sqo(tk) = h4h4 = Tr5(t5) As Tr5 - On the otherhand, as Tr = (k Trk is also an algebrahomomorphism, we get Trk+l (htk) Trl (hn)Trk(tk) = hnTrk(tk) This content downloaded from 128.235.251.160 on Sat, 07 Mar 2015 06:03:45 UTC All use subject to JSTOR Terms and Conditions STEENROD ALGEBRA AND GENERAL LINEAR GROUPS 4087 As shownin the proofof Lemma 5.3, a consequenceof Davis' Theorem5.1 claims that,ifTrk(tk) 0, then hnTrk(tk) / foreveryn with 2n > 4(Stem(Trk(tk)))2 = 4(deg(tk))2 [2 The remarkis proved As an immediateconsequence,we have Corollary 9.3 (i) Ker(Sq?) n (F2 PH,(BVk)) GLk is nonzero for k = and has an infinitedimension for k > k = 5, Trk detects a nonzero element in the kernel of Sq?, and for each For (ii) k > 5, it detectsinfinitely manynonzeroelementsin thiskernel It has been known(see [28], [4]) that Sq? is injectiveon F2 PH*(BVk) for GLk k < in positivedegreesofF2 PH* (BV4) Conjecture 9.4 Sq? is a monomorphism GL4 In otherwords,Sq? is a monomorphism in positivedegreesof F2 PH*(BVk) if GLk and only if k < The followingis an analogue of Corollary6.2 and is relatedto Corollary6.10 Conjecture 9.5 (Sq? is eventuallyisomorphicon the Ext groups) Let Im(Sq0)i denotethe image of (Sq?)i on Ext(lF2, F2) There is a numbert dependingon k such that Im(SqO)t (Sq)i-t - Im(Sq?)i is an isomorphismforeveryi > t In otherwords,Ker(Sq0)i = Ker(Sq?)t on Ext(lF2,F2) foreveryi > t As a consequence,any finiteSq?-familyin Ext (F2, F2) has at mostt nonzeroelements Is the conjecturetruefort = k- 2? An observationon the knowngeneratorsof the Ext groupssupportsthe above conjecturewitht muchsmallerthan k - It also leads us to the questionon whetherSqO is an isomorphismon Im(Sq?)t C F2 PH*(BVk) GLk forsome t < k- (This questionhas an affirmative answergivenby Corollary6.2 fort = k - 2.) 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[32],Singer[28],Priddy[26],Kuhn [15 ]and others.) In particular, one of the most directattemptsin studyingthe cohomologyof the Steenrod of the generallineargroupswas the algebra by means of modularrepresentations the so-called... REPRESENTATIONS ALGEBRA AND OF THE STEENROD OF THE GENERAL LINEAR GROUPS NGUYEN H V HUNG Dedicated to Professor Nguye^nHtruAnh on the occasion of his sixtiethbirthday ABSTRACT Let Trk be the algebraic... exploitthe relationshipbetweenthe algebraictransferand the squaringoperationSq? It is well known(see [18]) that thereare squaringoperationsSqi (i > 0) actingon the cohomologyofthe Steenrod algebra