, 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. 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.