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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; kk HX * 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 * π X 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 0 cX . 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< i V i i VV and d a decomposable differential; that means (d does not have a linear part) and that . It is well known that the minimal model 2i V 1 i dV 2=0d ,Vd determines the rational homotopy type of X, in the sense that ** * ;, asalgebras πasvectorsp aces HXH Vd XV . 2 For example, the minimal model of an even sphere n is of the form x ,,yd with x 2n, y 41n, 0dx , 2 dy x*2 2 ; nxx 21n , while the H and minimal model of an odd sphere is of the form ,, x yd with 210dy V y n , . It will be utile for our proofs, to recall the reader this simple properties. For a homogeneous element x of , x denotes its degree, which verifies the following: 1xy x yy x; 1xxdy (Leibniz formula). dxy dxy 20x In particular x yyx , when x is odd and when x is even. :1dim kk 0k X V π π odd 0and 0 00 ,0 c c HVd is called the homotopic Euler-Poincar characteristic of X. In [5], S. Halperin have shown the following: (1) One other notion that we will use throughout this pa- per is the formal dimension of X, given as :max ,;0 n fdXn HX 1,, n aa . We know from [5] that, when are the elements of an homogene- *Corresponding author. C opyright © 2012 SciRes. APM M. R. HILAL ET AL. 16 ous basis of V, even 1. ii ii a oddaa fd Xa (2) Our proofs are essentialy based on this equality com- bined with an other equality established by J. Friedlander and S. Halperin in [2], that 21. afdX af dX *;HX ; n X even odd i i i a i a (3) Finally, let us recall that satisfies the Poincar duality, that means that the multiplication kn HX HH ;; k X X is a non degenerate bilinear form (here and nfd denotes the so called fundamental class of *;HX). For the reader interested by more details about the rational homotopy theory, we recommend the basic reference [3]. 3. The Main Theorem In all the remainder of this paper, X denotes a simply connected elliptic space with *;HXdim 80, c and will denotes it minimal model. Put 1 ,Vd , ,, 6 1, a basis for with the condi- tion that ; * HX 1ii and that i with aV ii a . The following table summarizes the classification of its rational homotopy type. Rational homotopy type of X Legend 3 21k 32 1X k fd 2nn 4 n Legend: 1) In [6], I. M. James has introduced the concept of reduced product when X is a based space. He put X X 1: and f dX n nn 4 and n is odd 22 2 # n n f dX n 21 21kk 1 2 and n is odd 2kp 22 f dXkk p nn 42 1nn p 22nn 42 1nn p 2 21 2np n 2nk 22kn 21nY 21 22 np n fdX 2 d 21np fdX k * 21 221 nk nk 21 2 nk nk d 21 3 nk 21 21213 12 k kk 21 E E : the total space of the fiber bundle with as base space 21 21pq 1111 :,,*,,*,,, . pp p XX Xxxxx n n 2) From this construction applied to an even sphere arises the James sphere p , satisfying *1 ; np Haa np. The use of the denotation p 1, ,5i means implicitly that n is supposed to be even. As the most of our proofs will be by contradiction, we will mark such proofs by (by contradiction) in its begin- ning and by (QED) when its end. In the spirit and desire to simplify the lecture of this paper, we will subdivide it on many propositions, lemmas and theorems. The first one is that: Lemma 1 There exists such that < i 1i. Proof. Suppose that 1ii n , then 11 510 fd Xn c . 3.1. The Case Where 123456 Proposition 2 If 123456 3 21k , then X has the rational homotopy type (r.h.t) of , with 32 1kfdX * k and . Proof. Since 123 123 ,,aaa V, then . We distinguish two cases: 1) 1 is odd. Then . Let E be the vector space spanned by 222 123 0 132 ,, 12 3 . If dim 3E , we can take 12 1323 ,, as a basis of E. Let 1,, n bb an homogeneous basis of a complement of 123 ,,aaa in V with 1n and bb 123 ,,aaa, therefore 10db and 2 db * 123 1 ,, ,aaabHVd, what im- plies that * dim, 9HVd . So the minimal model of X is with da i = 0 and 123 ,, ,aaa d21 i ak 3 21k 1d . This is exactly the minimal model of . If im 2E , the there exist 12 , such that 2 30 11 2213 d a aa aa , and then 411221 23 3aa . According to the Poin- car duality, we have 123 , so 14 daa aaa and 123 123 0 aaa . This is impossible. 2) f dX and is even. Then are odd, be- 1 4 cause that 4 1131 0 fd X c . Therefore 222 1 10 12 32 323 i b V, and there exist tree generators of with even degrees such that 2 ii db a13i , . Then 414 1 even 333632 1 i aiaabaa fdX . Copyright © 2012 SciRes. APM M. R. HILAL ET AL. 17 This is impossible. re 3.2. The Case Whe 123456 Lemma 3 If 1234 65 , then f dX and 3 are even, 1 is odd, and 12 0 . Proof. First, of ther dualite have because Poincay, w 16 3 X2fd ev isen, and 1 , 6 hav e the same parity. Hence 13 241 1 c 0 2 , 1 is odd and 3 is eve (By contradiction) Suppo n. at 12 0se now th , sinc 2 herwise, the Po e 2 0 the ,1ai 12 n 4 iV. Ot incar duality let us to suppose that 16 25 and to conclude that 56 ,,14 i aa ai and that 56 ,aa are also generators of ,Vd. So 256 421.fd Xfd X 1 odd i aia This is impossible (QED). Lemma 4 If 123456 , getor b of V, * then there exists an homogeneousnera satis- fying 2 4124 dbaa aa , where . Proof. Since 0 12 , we can assume that3 a a 12 a and th at 4 aV, then 341 24 aa . Ot herwise 2 41 aa , then there exists an homogeneous generator , such 41 24 dbaa aa. Lemma 5 If 24 of V aa b2 123456 , then , ,,Vdx y 1 x * , 2 3 ,,x D with: 0Dy , 2 e 12 Dx Dx Dy y12 xxy wher 121 x xa et 4 y a. ** ,, Proof. We have H V Hd ,Wx y. We define the W D where algebra homo 123 ,,xx morph- :, ,WD Vd as 11 ism x a, 22 x a, 3 x b and 4 y a . is into because it trans- form 3 s the basis 12 ,,, x xxy ofW on linearly independent family 124 ,,,aaab a . Let the W and 01 VV V , since ** ,, 0 V 0 H Vd d then 1 \0dV V V . Assume tha and con- that H t 1 VV 0 10 1 cVsider such min 0x, then 12 dc b1,xV,cx db wh 12 12 ,,aa 22 124 0dc 0 er 12 e a 4 ,a. As n 1 4 aaa , the . We nis case have to discuss two cases: 11 dc a 2 24 ;2aan. In th 1 21 1cna and e therefor 11 1 421 92> . can a afdX even2 i aiab This is impossible. ,1 m a aam dc 312 4. In this case, 1 23 1cma and 10dac, then 11 >5 1>acafd X, and so 10ac. Let 1kV , such 1 dac where i iV and 11 2 2m odd, in particular. But 1 3a is 0 , because if not we will have 12 1 odd 1 47 3 113 >21. i aiaaam a afdX 1 c This is impossible. Proposition 6 If 123456 , then X have one of the fo llowing r.h.t: 2nnn , where n is odd and 4 f dX n. 22 4 2, where n is odd and # nn nn f dX n . Proof. Let us recall that f dX 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 1424 1, ,,,,, 2 2124 is a basis for *;HX, and therefore *222 ;,,,, H Xabcabc , i.e., X has the r.h.t of 2nnn . Second case: 2 40 . He14 re and 24 are both non null, because i an the opposite case we will have 14 a db or 24 aa db where b is a genera- tor of V , and in this cases 1256 odd >2 1 ii aaaaaa fdX . This is impossible. Recap 22 12 0 , 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 2 n . e 12 5634 Proposition 7 If 123456 and if dX is even, then X have the r.h.t of 21k f 2 21 kp k 212 with f dXkk p. of the parity of Proof. Because f dX, the dual incar an ity of Pod the fact that 0 c , then 1 and 3 ly oddare respective and even, so 22 12 0 . Assume that 22 0 and that 56 1234 ,,,, , then there exist 12 1234 ,, ii PP ,, such 1122 ii iPP that for 5,6i . This imp impossi- ble shat 1 lies the ituation t2 0 ii , but also th mption is false. Thus nea are h generators o at our second assucessarily 56 ,a f V bot and that 1256 odd 2>21 ii aaaaaa fdX fdX . This another impole situation issibmplies that our first assumption is also false. Put 13 0 , in this case ,,aa aa 34 12 , in particular j a are genera- tors of V for 3, 4j . The Poincar duality let us to write 1212jj aaa anconclude that 1j d to 0 and that 20 j and finally to write 51 , 26 j j . rity of the degree, then with da Recall that 20 j , because of the pa ,,,,,Vdabcx d 0dbdc 2 c. This is the mini- o 2 21 kp k and dx mal 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 o f f dX is odd, has on 2 1kp with 2 then Xe o f following r.h.t: 212kk 21 f dX k k p, 42 1n p, 21 22 np n with 2 d n 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 theven V 5, then fdX. This is 35 15i aaaaa even i a impossible. So 34 12 ,, with 120 and j a is an odd degree gator of Vener for 3 or 4j. A same jujstification as in the last proof let us to cat 51 onclude th j , 62 j and that X have the r.h.t of 2 21 21kp kk . Second case: 1 is even and 12 0 . Since 5 is odd, then 22 1212 dim, ,1 . Assme for example that 2 10 , the u n 2 1 j with j3 or 4 and j is odd. Therefore 10 j . Let sup- pose that 12 , j j ar and write 2= j are colline 1 j , then 2 221 ==0 jj , so odd 5 aV. Since that 12 =0 and that 22 12 = , then there exist two odd degree generators of V, b and c that 12 =dbaa 22 12 a, such and =dc a . We conclude that 551 odd 1 22 1 211>21 iij j aaaabcaa a fd Xafd X (impossible). Put 16 = j and 25 = j , then 22 12 == j j and 22 0 12 0, . The minimal model of X will be of the form 121 ,,Vd aad wi12 0a 23 ,,,,bth bb da d , 112 2 22 1 adba a, db 2 a , 30db and 12an, 2bn 22 nn 1p, i.e., X 21np 2 d d case: . Thir 1 is even and 120 s in th case, 2 , . Ae first we can write51 6 j j 22 12 , , 2 . Sin ce 121, then dim 11 12 and 2 2 212 . Suppose that 10 and write 3 1 ada, 2 3 10 ), then 11 aa 12 a ( 1 1 da ab , i.e., =0 1 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 12 da da12 0db, 2 11 db a, 2 22 db a, i.e., 21 22 np nn X . re3.4. The Case Whe 12 34 56 Proposition 9 If 1 . f. As 2 21 2np n Proo f dX is even and =0 c , then and 1 23456 and 2 are respecten and odd. Suppose (by contra- diction) that 12 ively ev or13 is null (for example 12 =0 ). Thf Poinat a V ,e duality ocar insures th45 ,a then 2345 odd >2 1 i aiaaaaa fdX . This is impossible D). t 412 = (QE Pu , 51 =3 n 23 0 , the , because that 12 3 = . Th us to 623 is leads take = and to conclude that 2 1= 0. Hence with ==dy d 2 =dyx 12 ,,,,,Vdxyyyd12 =0dxy , and =2 x n, 12 2 1np , i.e., ==yy 2 1p . a 10 If 2 2n n X Lemm 1234 56 and dX is odd, then 5 ,=a V. Let us suppose 4 aV 4 a f (for example) and discuss two ca ses: 44 erator 7 a of a 2=0 and th exists a gen- V is even, thenere such that 2 74 =daa . If 1 a is odd, then >2 1aaaaafdX 1237 oddi ai, impossible. Then 1 a is even and necessary 2 123 dim ,1 , i.e., there exists a generator 8 a of V verifying 2 =daaa a with 81 1223 1=0 or 2=0 . Consequently 378 78 24 1 24 22121 2121, aa aa a aa fdX 2 odd 23 i aiaaaaa aa what is, once again, an impossible situation. 4 a is odd, because of the Poincar duality, we must have 22334 9 aV a be even and dim,= 1 . Let such that 9124 234 =daaaa a , then 9 2 4 a fdX 23 24 even 2 21 . i aiaaaa aa aa This is impossible. Lemma 11 If 12 34 56 d an f dX is odd, then 0 and 130 12 . Pion) Assume, for example, that roof. (By contradict 12 =0 . By theity of r, we have precedent lemma and the dual Poinca 453 25 12 ,aaa , ,a a and0 , 34 0 . Therefore 22 52323 ,, , but f dX en 523 is odd, th and 2 252 3 ==0 . This is a contradict sition 12 If ion (QED). Propo 1 56 234 and f dX is n x ha r.h.t: odd, theve one of the following 21nkk with 22 kn , 21nY , where * and Y have a minimal Copyright © 2012 SciRes. APM M. R. HILAL ET AL. 19 model of the form , ,uv d with ,,ab 0 da db , du ab, 22 dv 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 1 n XY where 12 nn Y and 21 1nn or YY ,,,d abuv wh , ere V d and ==0 da 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 7 a a 25 a a nd 8 a a 734 =da a and 81 da 13214 =aa aa with 3V . We distinguish two cases: First cas e: 3 is even, then1 is odd. As f dX and is even=0 c , then 8 is even and conse- quently >fdX. 348 eve i aiaaaa n Second case: 3 is odd. As f dX is even an =0 c d , then 8 a is even and aaaafdX 348 even i ai The 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 2 1 , then 1 is even. Since f dX is even and= c0 , then 2 , 3 and 5 aroe bt 6 = th odd. Pu2 1 , then ,aaa V 2 ,,a344 d an 2345 odd >1 i aiaaaa dX (con- tradiction). 2fa 2) We have 22 2 ,,aa aaa 34 1 . If 2 is even, then 3 is odd and 34 12 aa,aa . If 2 is odd, thhat 2 e result is evident diate consequenc because t e of 2 2=0 . ). Hwe 3) It is an immeence can take 2 61 = nd 2345 ,,aaa Since 222 312 ==0 , then there exis* a odd ,aV. ts 4) such that 34 = . So 140 , 24 0 and * 514 , 6 * 2 4 n 15 If . Propositio 123456 , then X havef 2 2nk n 1 21k the h.r.t o or that of 21nk nk . 21 2 2 d Proof. Put 514 = , 624 = and 2 2124 = , 2= 42 124 X have one , then the m wing forms: i nimal model of of the follo 123 2 1 =dyx , ,Vd ith ,,,yyyd w 0, and x 112 dx dy dy 1=2 x n, 1 y =2 1k , 2=2 1ykn, i.e., 21 221nk nk X . 12 ,,,Vd xy dx dy 3 ,,yyd with 112 0dy , 2 312 =dyxy y and 1=2 x k, 1=21 y n , 21n, i.e., 2 nk =2yk 21 21nk X 3.6. Case Where 2 d. 12346 5 Lemma 16 If 1 23456 , then 51234 ,,,aaaaa. Proof. Let 5 aV and discuss many cases: 1) even aV, then there exist two generators x and y 5 of V such that 25 34 =dxaaa a and 2 5 =dya a) 2 a is odd, then >aaaa fdX 12 69i evenaia. b) f dX is e=0 c ven and 2 is even. As a, then 1 a odd isd, even 1 aV an odd >2 1 i aiaxy fdX . a a 2) odd 5V. ) 2 a is necessaodd, thenry odd 2 aV and 25 = 34 = . Hence there exists a generator x oVf such that 2534 =dxa aa a . even 1 aVthen i) , 3 a and 2 a are both odd since 0. Since 34 0 then j aV for j th 5=0 j = c odd =3 or =4j wi . Hence there exists a generat of Vor y su5ja ch that = dy a and so odd >1 i aiaxy fdX2 . od 1 aVodd ii) d, then x V and 123 odd 1 i aia fdX ||>2aaa x. b) 2 a is even. even 2ecessar even and i) aV , then ny xV 2 even> i aiaaxfdX . 2 a20 ii) 2 1 =a, then 15 (because of the Poincar duality) and 150 . Put 61 athen j a 5 =aa, odd V for =3j or =4j (=0 c) and 50 j = (Poincar duality). Let x be a generator of V such t at 5j dxa , the h =an odd 5 2121 ai j aaax aa fdX 5 >. ij Lemma 17 If 123456 , then 61234 ,,,aaaaa. Proof. Let 6 aV and discuss many cases: 1) even 6 aV. Then there exist generators 7 a, 8 a of Copyright © 2012 SciRes. APM M. R. HILAL ET AL. 20 Copyright © 2012 SciRes. APM V such that da a) 71 =a and 2 86 =daa . 634 aa a 1 a is even, then odd 1 21 21. i ai fd Xafd X 78 6 6 12 1> aafdX a a b) 1 a is odd, then 67 even > i aiaaa fdX . odd 6 aV 2 a 2) . a) even V, then 29 >a fdX odd 2 aVhen even i aiaa . b) , t 9 >2 1fdX. 126 oddi aiaaa aa Lemma 18 If 123456 , then 123 2,,, 3aaa . 4 a V Proof. Put 1234 =,,,NaaaaV. ii aa for all =1, ,6i m0 1) If =1N, th, this c en plies the contradiction i . f16 == 2) I =4N. We have 34 . a) 3 a and 4 a are bot th h even,en f dX is even an =0 and rators ofd 2 4 et 7 a some gen V. L8 a be e with 71634 =daa aaa an2 = d da84 a , therefore . 78 odd >2 i aiafdX b) 1aa 3 a and 4 a are both odd, then f dX is and odd 12 =2 1=0 aa c , so 11 (for example) is odd and 7i aiX 134 odd >2 1aaaaa fd . c) 3 a is even and 4 a is od exd (forample), then 37 even > i aiaaafdX . Lemma 19 If 123456 d an if 1234 ,,aaaa V then 1 a and , 2, j a have dif- ferent 4 ,aaV. Proof. 1 parities where j aa Suppose that a23 =, and j a have the par- ity e sam . 1) If 1 a and j a are both even, then all i a are even for =1, ,6i and 0 c . 2) If 1 a and j a are odd, then necessary 2 = j aa and 3 ,4 12 (because that 2 ,,aa ), but pos 34 aa 1 sible. ProposIf this is imition 20 123456 and if 1234 ,,, 2aa V, then 1 3 nm X a a2 1 = , . 34 ,,,Proof. Put 12 j a he fact that aaa Ve the duality of Poincar, t aa . Becaus 1 a, j a have different an ij aa for all parities 1,a =1, i d the fact that . Let 1, ,6 j aa a such that a is odd and 1, kj aaaa , it is evident that k a is even. As *,=, ,=,1 mn k HVdn m then 7= kk 0, 7= . This allows us to take 1 , , pp aa aa bec 77kk k w=aith =3p, ause if not 8d * dim ,HV . Hence 4= 0 k and 0 n k . Con- clude that 0,a Wd , that even dim =1V ,,Vd and that m,= 4HWd. In [? ,Wd dl of n , then (3) X * di ], is the minimal moe(3) 21nm where =k na and 2m 21 If 1=a . Lemma 123456 and if 12 ,,aa34 , 3aa V , theong at n only one am 3 a or 4 a is in V. Proof. Assume th 34 ,aa V, then 1 a and 2 a at narause thecessary 2 21 aa. T e both even, bechere- fore 3 3 1=0 a , i.e., =1 1fd X c f dX is d, and thereodnerator 7 a, shat exists a ge of Vuch t 73416 =daaaa a with 37 even i ai>aaafdX . Lemma 22 If 123456 and if 1234 ,,, 3aaaa V , then aaa ere 12 = jwh 34 =, j aaaV . Proof. Suppose 12 =0 , we know, fro of Poi16 25 == m the duality ncar, that and by the Lemmas 16 and 17 that 56 2 ,,, 1 j aa ae deduce that aa . W 52 , j aaa , 616 ,aaa and that 2 j a a w , then 61 = 22 121 ,, kj aaaahere =3,4kj. 1) If 2 1k aa ,a j aaa and 2 522 , j aaaa. As 56 aa, then 2 52 aa, and so i a is even for all =1,, iimplies that 0 c 6, but this . 2 2k aaecessary 2 and 2 2= jkj 2) If , then n1=0 = . H ence 2 =2 j f dXa and 2 a 5 j aaa , 2 6 j aa. Since j a and 2 a are both even, then 16 = f dXaa and 6=2 j aa . So i a is even i fo =1, ,6o the contradiction c r all, but this leads t 0 . 3) If 1kj aaa , 0then 2 1= and = j 2 1 . Sup- pose that2 d discusso case. a) 21 2=0 an tws 2 j aa n a, the1 a, 2 a and 3 a are even and c0 . , theb) 25 aa n 2 2 =3 f dX a is even, but also 161 ==2 j f dXa a , then a a1 a There- fore is even. iis evenc a for all =1, ,6i and 0 f 1kj aa . 4) Ia , then ==0 an 2 22 12 d 2 == jjk (P). Hence 5 oincar duality2 j aa, 3 6 j aa and 2 =1 2=3 j j f dXa aa . So a 12 =< j j a aa. a tion 23 ProposiIf 123456 and if 12 3V 34 ,,,aaaa , then X have on r.h.t: 21p e of the following E: the total space of the fiber bundle with M. R. HILAL ET AL. Copyright © 2012 SciRes. APM 21 21q as base sp nn 1, 2 n and 3 n are both even. ye pre = ace, 3 12 wher Proof. cedent lemmas and by the Poincar duality 12 = j n We know b e n th that kj . Put 51 =k aaa dan 62 = j aaa , then 1=0 and 21 = 22 j . W distinguish two cases: 1) 0 e , then 1 a and j a are both odd because that 0 c . Replace 1 a by 1 a and put =1 , then the minimal model of X is of the form , , 12 ,, x yyyd with 12 2 1=2=x y 21q and 2 12 =dyxyyce X ~2n 21 21 = <<yp n . Henere E wh p q E is a fibratiof the KS-complex n o 121 ,,0 , , . yyyy xy =0 2, ,, d xy d 2) , thehe minimal model n X have t 12 ,, , , x yyyd 3 12 n nn with 12 ==0dy dy and 2 =dyx , i.e., X with 1, 3 n are both eve 4. Acknowledgements Th uthors reviewers for their constructive comments and an earlier draft of r is also a Paul Goerss and Kathryn Hess for their interest and en- grateful to Hiroo Shiga and y Type of Nilmanifolds Up to Dimension 6,” arXiv: 1001.3860v1, 2010. -133. doi:10.1007/BF01390029 n2 n and n. e a would like to think the anonymous suitable advices on this pape. Itpleasure to thank couragement. The authors are Toshihiro Yamaguchi for the several email discussion exchanged before the submission of this paper. REFERENCES [1] G. Bazzoni and V. Muõz, “Rational Homotop [2] J. B. Friedlander and S. 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Yamaguchi, “The Set of Rational Homo- topy Types with Given Cohomology Algebra,” Homology, Homotopy and Applications, Vol. 5, No. 1, 2003, pp. 423- 436. |