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
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 210dy
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


*;HXdim 80, 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 d21
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 

.

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 Hd

,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 VV

0
10
1
cVsider 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

. 

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 jujstification 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 da12 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
,,,,,Vdxyyyd12 =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

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 that 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
aVthen 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 fdX2

.
od
1
aVodd ii) d, then
x
V and

123
odd 1
i
aia fdX 

||>2aaa x.
b) 2 a is even.
even
2ecessar 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

M. R. HILAL ET AL.
20
V such that da
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
aVhen
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

11

(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 Ve 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 Vuch 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
aaaahere

=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,, iimplies that
0
c
6, but this


.
2
2k
aaecessary 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, ,6o 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 a1 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.
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
=dyxyyce 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. Halperin, “An Arithmetic Charac-
terization of the Rational Homotopy Groups of Certain
Spaces,” Inventiones Mathematicae, Vol. 53, No. 2, 1979,
pp. 117
[3] Y. Felix, S. Halperin and J.-C. Thomas, “Rational Homo-
topy Theory,” Graduate Texts in Mathematics, Vol. 205,
Springer-Verlag, New York, 2001.
[4] P. Griffiths and J. Morgan, “Rational Homotopy Theory
and Differential Forms,” Progress in Mathematics, Birk-
häuser, Basel, 1981.
[5] S. Halperin, “Finitness in the Minimal Models of Sulli-
van,” Transactions of American Mathematical Society,
Vol. 230, 1977, pp. 173-199.
I. M. James, “Reduce[6] d Product Spaces,” Annals of Mathe-
matics, Vol. 62, No. 1, 1955, pp. 170-197.
doi:10.2307/2007107
[7] G. M. L. Powell, “Elliptic Spaces with the Rational
Homotopy Type of Spheres,” Bulletin of the Belgian
Mathematical Society—Simon Stevin, Vol. 4
pp. 251-263.
, No. 2, 1997,
[8] H. Shiga and T. Yamaguchi, “The Set of Rational Homo-
topy Types with Given Cohomology Algebra,” Homology,
Homotopy and Applications, Vol. 5, No. 1, 2003, pp. 423-
436.