1db ff5 fsc fc0 sc0 ls0">
, 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
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
,,,,,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
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 fdX2
.
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.
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