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 Advances in Pure Mathematics, 2012, 2, 15-21 http://dx.doi.org/10.4236/apm.2012.21004 Published Online January 2012 (http://www.SciRP.org/journal/apm) Classification of Rational Homotopy Type for 8-Cohomological Dimension Elliptic Spaces Mohamed Rachid Hilal1, Hassan Lamane1, My Ismail Mamouni2* 1Faculté des Sciences Aïn Chock, Casablanca, Morocco 2Centre Pédagogique Régional, Rabat, Morocco Email: {rhilali, hlamanee}@hotmail.com, *mamouni.myismail@gmail.com Received September 21, 2011; revised November 8, 2011; accepted November 15, 2011 ABSTRACT The different methods used to classify rational homotopy types of manifolds are in general fascinating and various (see [1,7,8]). In this paper we are interested to a particular case, that of simply connected elliptic spaces, denoted X, by dis-cussing its cohomological dimension. Here we will the discuss the case when ; 8*HX0Xdim and . Keywords: Rational Homotopy Theory; Elliptic Spaces; Classification; Rational Homotopy Type; Minimal Model of Sullivan 1. Introduction Let us first recall some basic definitions of rational homo- topy theory. A simply connected space X is called elliptic, if both of and ;HX*πX dim;kkHX* are finite dimen- sion, and that its cohomological Euler-Poincar charac- teristic is given as 0k. We will fix this throughout this paper. The space is called rational if  :1Xc*πX is a -vector space. If it is not, by [4], we can associate a rational simply connected space, denoted X**;H, verifying , asalgebrasHX X**πasXX ,vectorspaces. The rational homotopy type of X is defined as the homotopy type of its rationalization X. Our purpose in this paper to give a complete classification this rational homotopy type when ;HX*mdi and 0cX. 2. Preliminaries The rational homotopy theory was founded in the the end of the sixties by Daniel Quillen and Denis Sullivan. One of the technical gadget of this theory is the minimal model of Sullivan, it is a free -commutative differen-tial graded algebra  associated to any simply connected CW complex X of finite type [3]. Here 2 is -graded vector space with ,Vd=dim .can aafdX even2iaiab  This is impossible. ,1ma aam dc312 4. In this case, 123 1cma  and 10dac, then 11>5 1>acafd X, and so 10ac. Let 1kV, such 1dac where iiV and 112 2m odd, in particular. But 13a is 0, because if not we will have 12 1odd147 3113 >21.iaiaaam aafdX  1c This is impossible. Proposition 6 If 123456, then X have one of the fo llowing r.h.t:  2nnn, where n is odd and 4fdX n.  2242, where n is odd and #nn nnfdX n. Proof. Let us recall that fdX and 4 are even, d thanat 1 and 6 are odd.  20First case: 4. Since 11 24 d 0 an1, then 140 and 240Hence . ,141 14241, ,,,,,2 2124  is a basis for *;HX, and therefore *222;,,,,HXabcabc , i.e., X has the r.h.t of 2nnn.  Second case: 240. He14re  and 24 are both non null, because ian the opposite case we will have 14a db or 24aa db where b is a genera-tor of V, and in this cases 1256odd >2 1iiaaaaaa fdX . This is impossible. Recap 22120, 12 0, 40, 14 0, 44 0, this leads us to conclude that X have the r.h.t of 22#nnn3.3. The Case Wher2n. e 12 5634   Proposition 7 If 123456 and if dX is even, then X have the r.h.t of 21kf 221 kpk 212with fdXkk p. of the parity of Proof. Because fdX, the dualincar anity of Pod the fact that 0c, then 1 and 3 ly oddare respective and even, so 22120. Assume that 22 0 and that 56 1234,,,, , then there exist 12 1234,,iiPP ,, such 1122iiiPPthat for 5,6i. This imp impossi-ble shat 1lies theituation t20ii , but also thmption is false. Thus nea are h generators oat our second assucessarily 56,af Vbot and that 1256odd 2>21iiaaaaaa fdX fdX . This another impole situation issibmplies that our first assumption is also false. Put 13 0, in this case ,,aa aa34 12, in particular ja are genera-tors of V for 3, 4j. The Poincar duality let us to write 1212jjaaa  anconclude that 1jd to 0 and that 20j and finally to write 51, 26jj. rity of the degree, then with daRecall that 20j, because of the pa ,,,,,Vdabcx d 0dbdc 2c. This is the mini- o221 kpk and dxmal modelf 21k. Copyright © 2012 SciRes. APM M. R. HILAL ET AL. 18 Proposition 8 If 1 62345  and ifdX is even, then x have one the r.h.tf  offdX is odd, has on21kp with  2 then Xe o f following r.h.t:  212kk 21fdX k k p, 42 1n p, 2122npn  with 2dnfd Xn22nn   21np with  42 1n pwill dses: First case: fd XnProof. We , iscuss three ca 1 is odd, then 5 is even. Suppose at thevenV 5, then fdX. This is 35 15iaaaaa eveniaimpossible. So 34 12,, with 120 and ja is an odd degree gator of Vener for 3 or 4j. A same jujstification as in the last proof let us to cat 51onclude thj, 62j and that X have the r.h.t of 221 21kpkk .  Second case: 1 is even and 12 0. Since 5 is odd, then 221212dim, ,1. Assme for example that 210, theun 21j with j3 or 4 and j is odd. Therefore 10j. Let sup-pose that 12,jjar and write 2=j are colline 1j, then 2221==0jj , so odd5aV. Since that 12=0 and that 2212=, then there exist two odd degree generators of V, b and c that 12=dbaa 2212a, suchand =dc a. We conclude that  551odd122 1211>21iij jaaaabcaa afd Xafd X   (impossible). Put 16=j and 25=j, then 2212==jj  and 220120,. The minimal model of X will be of the form 121,,Vd aad wi120a23,,,,bth bb da d, 112222 1adba a, db 2a , 30db and 12an, 2bn22nn1p, i.e., X21np2d d case: . Thir 1 is even and 120s in thcase, 2,. Ae first we can write51 6jj2212, , 2 . Sin ce121, thendim1112 and 22 212. Suppose that 10 and write 31ada, 2 310), then 11aa12a (11daab, i.e., =0112. That contradicts the main hypothesis in our third case. model of X wil,,,,bbb d with Hence the minimal l be of the form 12,,Vd aa12da da120db, 211db a, 222db a, i.e., 2122 npnnX . re3.4. The Case Whe 12 34 56 Proposition 9 If 1. f. As 2212npn ProofdX is even and =0c, then and 123456 and 2 are respecten and odd. Suppose (by contra-diction) that 12ively ev or13 is null (for example 12=0). Thf Poinat a V,e duality ocar insures th45,a then 2345odd >2 1iaiaaaaa fdX . This is impossible D). t 412=(QE Pu, 51=3n 23 0, the, because that 12 3=. Th us to 623is leadstake = and to conclude that 21=0. Hence  with ==dy d2=dyx 12,,,,,Vdxyyyd12=0dxy , and =2xn, 122 1np, i.e., ==yy21p. a 10 If 22nnXLemm 1234 56 and dX is odd, then 5,=a V. Let us suppose 4aV4af (for example) and discuss two ca ses: 44erator 7a of a 2=0 and th exists a gen-V is even, thenere such that 274=daa . If 1a is odd, then >2 1aaaaafdX 1237oddiai, impossible. Then 1a is even and necessary 2123dim ,1, i.e., there exists a generator 8a of V verifying 2=daaa a with 811223 1=0 or 2=0. Consequently 3787824 124221212121,aaaa aaa fdX 2odd23iaiaaaaaaa what is, once again, an impossible situation.  4a is odd, because of the Poincar duality, we must have 223349aVa be even and dim,= 1 . Let  such that 9124 234=daaaa a, then 9 24afdX23 24even221.iaiaaaa aaaa This is impossible. Lemma 11 If 12 34 56d anfdX is odd, then 0 and 13012. Pion) Assume, for example, that roof. (By contradict12=0. By theity of r, we have precedent lemma and the dualPoinca453 2512,aaa , ,a a and0, 34 0. Therefore 2252323,,, but fdX en 523is odd, th and 2252 3==0  . This is a contradictsition 12 If ion (QED). Propo 1 56234 and fdX is n x ha r.h.t: odd, theve one of the following 21nkk with 22kn, 21nY, where * and Y have a minimal Copyright © 2012 SciRes. APM M. R. HILAL ET AL. 19model of the form , ,uv d with ,,ab 0da db, du ab, 22dv ba . Proof. By the two last lemmas, we have 1 is odd and 1,,0,VdaVd ith 23,aa V w. *dim,=4HVd (case claBut tho ssified by the first au- r in his thesis), then 1nXY where 12nnY and 211nn or YY,,,d abuv wh,ere V d and ==0da db, =duab, 22=dv ba,. 3.5. Case Where 12345 6 Lemma 13 If 123456, then 412,aaa . 34,aa V, then there exist two of V satisfying 312,aaa or Proof. Suppose that generators 7a a25a and 8aa734=da a and 81da 13214=aa aa with 3V . We distinguish two cases:  First case: 3 is even, then1 is odd. As fdX and is even=0c, then 8 is even and conse- quently>fdX. 348eve iaiaaaan Second case: 3 is odd. As fdX is even an =0cd, then 8a is even and aaaafdX348even iaiThe two cases are both impossible. Lemma 14 If . 123456, then: 1) 2=0, 1) 2, 34 12,aaaa, 3) 34,aa V4) *514*624 and . of. 1Pro =0) suppose that 21, then 1 is even. Since fdX is even and=c0, then 2, 3 and 5 aroe bt 6=th odd. Pu21, then ,aaa V 2,,a344d an2345odd >1iaiaaaa dX  (con- tradiction). 2fa2) We have 222,,aa aaa34 1. If 2 is even, then 3 is odd and 3412aa,aa . If 2 is odd, thhat 2e result is evident diate consequencbecause te of 22=0. ). Hwe 3) It is an immeence can take 261=nd 2345,,aaa Since 222312==0, then there exis* aodd,aV. ts  4)such that 34=. So 140, 24 0 and *514, 6*24n 15 If . Propositio 123456 , then X havef 22nkn 121k the h.r.t o or that of 21nknk . 2122d Proof. Put 514=, 624= and 22124= , 2=42124X have one , then the mwing forms: inimal model of of the follo 123 21=dyx, ,Vd ith ,,,yyyd w0, and x112dx dy dy 1=2xn, 1y=2 1k, 2=2 1ykn, i.e., 21221nknkX. 12,,,Vd xydx dy 3,,yyd with 1120dy , 2312=dyxy y and 1=2xk, 1=21yn, 21n, i.e., 2nk =2yk2121nkX3.6. Case Where 2d. 123465  Lemma 16 If 123456, then 51234,,,aaaaa. Proof. Let 5aV and discuss many cases: 1) evenaV, then there exist two generators x and y 5of V such that 25 34=dxaaa a and 25=dyaa) 2a is odd, then >aaaa fdX12 69ievenaia. b)fdX is e=0cven and 2 is even. As a, then 1a odd isd, even1aV anodd >2 1iaiaxy fdX . aa2) odd5V. ) 2a is necessaodd, thenry odd2aV and 25= 34=. Hence there exists a generator x oVf  such that 2534=dxa aa a. even1aVthen i) , 3a and 2a are both odd since 0. Since 34 0 then jaV for j th 5=0j=codd =3 or=4j wi. Hence there exists a generat of Vor y su5jach that =dy a and so odd >1iaiaxy fdX2. od1aVoddii) d, then xV and 123odd 1iaia fdX ||>2aaa x. b) 2a is even. even2ecessar even and i) aV , then ny xV2even>iaiaaxfdX. 2a20ii) 21=a, then 15 (because of the Poincar duality) and 150. Put 61athen ja5=aa, oddV for =3j or =4j (=0c) and 50j= (Poincar duality). Let x be a generator of V such t at 5jdxa , theh=an odd52121aijaaaxaa fdX 5>.ijLemma 17 If 123456, then 61234,,,aaaaa. Proof. Let 6aV and discuss many cases: 1) even6aV. Then there exist generators 7a, 8a of Copyright © 2012 SciRes. APM M. R. HILAL ET AL. 20 Copyright © 2012 SciRes. APM V such that daa) 71=a and 286=daa . 634aa a1a is even, then  odd 121 21.iaifd Xafd X 78 66121>aafdX a a b) 1a is odd, then 67even >iaiaaa fdX. odd6aV2a2) . a) evenV, then 29>a fdXodd2aVhen even iaiaa. b) , t9>2 1fdX. 126oddiaiaaa aaLemma 18 If 123456, then 1232,,, 3aaa . 4a VProof. Put 1234=,,,NaaaaV. iiaa for all =1, ,6im01) If =1N, th, this cenplies the contradiction i. f16==2) I =4N. We have 34. a) 3a and 4a are bot th h even,enfdX is even an =0 and rators ofd 24et 7a some gen V. L8a be e with 71634=daa aaa an2=d da84a , therefore. 78odd >2iaiafdXb)1aa  3a and 4a are both odd, then fdX is and odd 12=2 1=0aac , so 11 (for example) is odd and 7iaiX 134odd >2 1aaaaa fd . c) 3a is even and 4a is od exd (forample), then 37even >iaiaaafdX. Lemma 19 If 123456d anif 1234,,aaaa V then 1a and , 2,ja have dif-ferent4,aaV. Proof. 1 parities where jaa Suppose that a23=, and ja have the par-itye sam. 1) If 1a and ja are both even, then all ia are even for =1, ,6i and 0c. 2) If 1a and ja are odd, then necessary 2=jaa and 3,4 12 (because that 2,,aa ), but pos 34aa 1sible. ProposIf this is imition 20 123456 and if 1234,,, 2aa V, then 13nmXa a21= ,. 34,,,Proof. Put 12jahe fact that aaa Ve the duality of Poincar, taa . Becaus 1a, ja have different anijaa for all parities 1,a=1,id the fact that . Let 1,,6jaaa such that a is odd and1,kjaaaa, it is evident that ka is even. As *,=, ,=,1mnkHVdn mthen 7=kk0,  7= . This allows us to take 1, ,ppaa aabec77kkk w=aith =3p, ause if not 8d*dim ,HV. Hence 4=0k and 0nk. Con-clude that 0,a Wd , that evendim =1V ,,Vdand that m,= 4HWd. In [? ,Wd dl of n, then (3)X*di ], is theminimal moe(3)21nm where =kna and 2m 21 If 1=a. Lemma 123456 and if 12,,aa34, 3aa V, theongat n only one am 3a or 4a is in V. Proof. Assume th34,aa V, then 1a and 2a at narause thecessary 221aa. Te both even, bechere- fore 33 1=0a , i.e., =1 1fd XcfdX is d, and thereodnerator 7a, shat exists a ge of Vuch t 73416=daaaa a with 37even iai>aaafdX. Lemma 22 If 123456 and if 1234,,, 3aaaa V, then aaa ere 12=jwh34=,jaaaV. Proof. Suppose 12=0, we know, froof Poi16 25==m the duality ncar, that  and by the Lemmas 16 and 17 that 56 2,,,1jaa ae deduce that aa . W52,jaaa , 616,aaa and that 2ja a w, then 61=22121,,kjaaaahere =3,4kj. 1) If 21kaa,ajaaa and 2522,jaaaa. As 56aa, then 252aa, and so ia is even for all =1,,iimplies that 0c6, but this . 22kaaecessary 2 and 22=jkj2) If , then n1=0= . Hence 2=2jfdXa and 2a5jaaa, 26jaa. Since ja and 2a are both even, then16=fdXaa and 6=2jaa. So ia is even ifo =1, ,6o the contradiction cr all, but this leads t0. 3) If 1kjaaa, 0then 21= and =j21. Sup-pose that2d discusso case. a) 21 2=0 an tws2jaa n a, the1a, 2a and 3a are even and c0. , theb) 25aa n 22=3fdX a is even, but also 161==2jfdXa a, then a a1a There- fore is even. iis evenca for all =1, ,6i and 0f 1kjaa . 4) Ia, then ==0 an22212d 2==jjk  (P). Hence 5oincar duality2jaa, 36jaa and 2=12=3jjfdXa aa. So a12=