The rational homotopy type of elliptic spaces up to cohomological dimension 8

DOI 10.1007/s13370-015-0377-9

The rational homotopy type of elliptic spaces up to cohomological dimension 8

Mohamed Rachid Hilali

· My Ismail Mamouni

· Hicham Yamoul

Received: 13 February 2014 / Accepted: 12 August 2015

© African Mathematical Union and Springer-Verlag Berlin Heidelberg 2015

Abstract Our goal in this paper is to give a full classification of the rational homotopy type of any elliptic and simply connected space when the sum of its Betti numbers is less or equal than 8.

Keywords Rational homotopy theory · Sullivan models · Elliptic spaces · Rational homotopy type

Mathematics Subject Classification Primary 55P62 · 55P15; Secondary 55P10 · 55Q05 · 55Q52

1 Introduction

Two continuous maps f, g : X −→ Y are called homotopic ( f g) if they are joined by a homotopy, that is a continuous map F : X × I −→ Y such that F ( x , 0 ) = f ( x ) and F ( x , 1 ) = g ( x ) for any x ∈ X. We say that X and Y have the same homotopy type, when there exist two continuous maps f : X −→ Y and g : Y −→ X , called homotopy equivalences, such that f g id

and g f id

, then we write X Y . This yields to an equivalence relation which may provide a classification of the homotopy type for any topological space.

Rational homotopy theory was interesting in a special homotopy type, the rational one. In fact, the founders D. Sullivan and D. Quillen were interesting in a kind class of topological spaces; the so called rational spaces, denoted in general X

and verifying the particular

B

[email protected] Mohamed Rachid Hilali [email protected] Hicham Yamoul [email protected]

1 Département de Mathématiques et d’Informatique, Faculté des Sciences Ain Chock, Km 8 Route d’El Jadida, BP: 5366, 20100 Maarif, Casablanca, Morocco

2 Département de Didactique des Mathématiques, Centre de Préparation à l’Agrégation, CRMEF Rabat, Avenue Allal Al Fassi, Madinat Al Irfane, BP: 6210, 10000 Rabat, Morocco

propriety that both π

( X

) and H

( X

; Z ) are Q -vector spaces. When X is simply connected we have π

( X ) is a Q -vector space if only if H

( X ; Z ) is also. Moreover [3, Theorem 9.11, page 111] X can be modelled, up to homotopy equivalence, by a rational CW complex X

as follow:

π

( X ) ⊗ Q ∼ = π

( X

) as vector spaces H

( X ; Z ) ∼ = H

( X

; Z ) as algebras

A continuous map f : X −→ Y is called a weak homotopy equivalence, when π

( f ) and all π

( f ) : π

( X ) −→ π

( Y ), n 1 are bijections. Two spaces X and Y have the same weak homotopy type if they are connected by a chain of weak homotopy equivalences

X ←− Y (0) −→ · · · ←− Y (n) −→ Y.

Proposition 9.8 [3, page 111] implies that all the rationalizations X

of X, have the same weak homotopy type, which depends only on the weak homotopy type of X, called the rational homotopy type of X. Then we write X

X

.

A simply connected topological space X, is said to be elliptic when both π

( X ) ⊗ Q and H

(X ; Q ) are of finite dimension. Our main purpose in this paper is to give a complete classification of the rational homotopy type of all simply connected and elliptic topological spaces X whose rational cohomological dimension is less than 8 (i.e., dim H (X; Q ) ≤ 8).

Our main theorem states that:

Theorem 1 If X is a simply connected and elliptic space, then its rational homotopy type is given by:

dimH^∗(X;Q) Rational homotopy type

1 ∗

2 _Sⁿ

3 _Sⁿ₍₂₎

4 _Sⁿ×S^m

Sⁿ₍3)

Y_λwith H^∗(Y_λ;Q)=Q[a,b]/(ab,a²−λb²) 5 _Sⁿ₍₄₎

Sⁿ₍₃₎#_S^m₍₂₎ 6andχπ=0 _Sⁿ₍₅₎

Sⁿ×S^m₍₂₎

P_λwith H^∗(P_λ;Q)=Q[a,b]/(ab,a²−λb⁴) Q_λwith H^∗(Q_λ;Q)=Q[a,b]/(b²+ab+λa²) R_λwith H^∗(R_λ;Q)=Q[a,b]/(a³,b²+λa²) 6andχπ=0 _Sⁿ×S^m₍2)

Tn,m,the total space of the fibration_S^n+m−1−→Tn,m−→Sⁿ×S^m 7 _Sⁿ₍₆₎

Sⁿ₍2)#_S^m₍₅₎ Sⁿ₍3)#_S^m₍₄₎

R with H^∗(R;Q)=Q[a,b]/(a²b,a³−b²) 8andχπ=0 _Sⁿ₍₇₎

Sⁿ₍₄₎×S^m₍₂₎ Sⁿ₍4)#_Sⁿ₍₄₎ Sⁿ₍5)#_S^m₍₃₎ Sⁿ₍₂₎×S^m₍₂₎×S^k₍₂₎

P_α,β,γwith H^∗(P_α,β,γ;Q)=Q[a,b,c]/(a²−αbc,b²−βac,c²−γab) 8andχπ=0 see[5]

Let us point out that other authors were also interesting in the classification of the rational homotopy type, namely [1,9,10,12,14]. Some other ones [8,13] considered broader con- cepts as C

or L

algebras. Note also that our classification generalises that of [11], since moreover we consider the both cases χ

= 0 and χ

= 0, we go furthermore and explicit in almost cases the Sullivan minimal model. In fact, our starting point to simplify our proofs is the use of the facts that X is rationally formal whenever dim H

( X; Q ) 10, and that such kind of spaces verify [6, Theorem 2]

dim π

( X ) ⊗ Q dim H

( X ; Q ).

The paper is organized as follows. In Sect. 2 we will recall basic definitions, denotations and theorems from rational homotopy theory that will be used in our proofs. Proof of Theorem 1 is detailed in Sect. 3.

2 Material

Given a graded Q -vector space V , one of the rational homotopy theory powerful gadget is the Sullivan minimal model, that is the free commutative differential graded algebra (V, d) defined by the tensor product

V := Exterior algebra V

⊗ Symmetric algebra V

and verifying the following: there exists some well ordered homogeneous basis (v

)

of V such that dv

∈

{v

, β < α}. The rational homotopy type of any simply connected space can be modelled, up to quasi isomorphism, by a Sullivan minimal model as follows:

π

( X ) ⊗ Q ∼ = Hom ( V , Q ) as vector spaces H

(X; Q ) ∼ = H

(V, d) as algebras

Example 2.1 The Sullivan minimal model of an odd sphere is of the form ({a}, 0), that of an even sphere is of the form ({a, x}, d ) with da = 0, d x = a

.

Throughout this paper X denotes a simply connected and elliptic topological space and (V, d) denotes its Sullivan minimal model. We define the following invariants:

(1) Formal dimension, fd( X) := max{k ∈ N , H

(X; Q ) = 0};

(2) Homotopical Euler–Poincaré characteristic, χ

( X) :=

k

(1)

dim V

; (3) Cohomological Euler–Poincaré characteristic, χ

( X ) :=

k

( 1 )

dim H

( X ; Q ) ; which verify:

Theorem 2 [4]

χ

(X) 0, χ

(X) 0 Furthermore, the following conditions are equivalent:

(1) χ

(X) = 0;

(2) χ

( X ) > 0;

(3) H

( X ; Q ) = H

( X ; Q ) .

Theorem 3 [2,4] There exists a homogeneous basis (a

) of V such that:

(1)

|ai|even

|a

| fd(X);

(2)

|aj|odd

| a

| 2fd ( X ) − 1;

(3)

|aj|odd

| a

| −

|ai|even

(| a

| − 1 )

= fd ( X ) .

Theorem 4 [4] H

( X ; Q ) satisfies the Poincaré duality property, which means that:

(1) dim H

(X; Q ) = 1 where n = fd( X) (i.e., H

(X; Q ) = Q μ. μ is then called the fundamental class of X);

(2) For any 0 k n, the cup-product

H

(X; Q ) × H

( X; Q ) −→ H

(X; Q ) ∼ = Q

is a non-degenerate bilinear form (in particular, H

(X; Q ) ∼ = H

(X; Q )).

Theorem 5 [6] If X is formal, then

dim V dim H

( X; Q ).

Let us recall that X is said to be formal, when its Sullivan minimal model is given by the quasi-isomorphism

( V , d ) ∼ = ( H

( X ; Q ), 0 ).

Roughly speaking, one can formally deduce the homotopy groups from the cohomology ones.

Remark 2.2 • From Theorem 2, one can deduce that dim V

dim V

.

The inequality is strict if and only if χ

( X )= 0 (i.e. dim H

( X ; Q ) = 2 dim H

( X ; Q ) ).

• In particular we conclude that dim V

= dim V

whenever dim H

( X; Q ) is odd and that H

( X; Q ) is a polynomial algebra truncated by a Borel ideal [4]. More precisely, we have

H

(X ; Q ) ∼ = Q [x

, . . . , x

]/( f

, . . . , f

)

where ( f

, . . . , f

) is a regular sequence of graded elements in the polynomial ring Q [x

, . . . , x

].

Moreover (see [11]), we have

| f

| 2 | x

| and dim H

( X ; Q ) 2

. The two previous inequalities and the following dimension formula

H

( X ; Q ) = | f

| · · · | f

|/| x

| · · · | x

| (1) can be viewed as powerful tools to check that our classification given here above is correct.

3 Classification

In our care of simplification, we divide the proof of our main result (Theorem 1) in many lemmas and propositions. We start by recalling the James construction [7] of the so called reduced product spaces from any based topological space (Y, ∗): We set

Y

= Y

Y

= Y × · · · × Y /(· · · , ∗, · · · ) (∗, · · · ) p 2

Applying this for even spheres, we get the James sphere S

whose rational cohomology is given by

H

( S

; Q ) ∼ = Q [ a ]/( a

).

The rational homotopy type of James spheres (when p 2) is fully described here below thanks to:

Lemma 3.1 X

S

if and only if χ

( X) = 0 and dim V = 2, where pn = fd(X).

Proof Suppose that X

S

and set x ∈ V such that H

(V, d) ∼ = Q [x ]/([x]

).

x is necessary of even degree |x | = n, because if not we will have x

= 0 (i.e. p = 1). Since [x]

= 0, there exists y ∈ X of odd degree such that d y = x

. Therefore, the Sullivan minimal model of X is of the form

({ x , y }, d ) with d x = 0 , d y = x

and is generated by two elements x and y whose degrees are of different parities (i.e. χ

( X ) = 0 and dim V = 2).

Conversely, suppose that χ

(X) = 0 and that dim V = 2. Let x ∈ V of even degree.

For ellipticity argument, there exists p ∈ N such that [x]

= 0. Let y ∈ V of odd degree such that d y = x

, then (V, d) ∼ = ({x , y}, d) with d x = 0, d y = x

. Therefore, X

S

.

The relation pn = fd ( X ) can be deduced obviously from the fact that H

( X ; Q ) ∼ = Q [ x ]/([ x ]

) with | x | = n.

Let us now begin by proving our classification.

Proposition 3.2 (Rational homotopy type when 1 dim H

(X ; Q ) 3) (1) If dim H

( X ; Q ) = 1, then X

∗ ;

(2) If dim H

(X; Q ) = 2, then X

S

with fd( X) = n;

(3) If dim H

(X; Q ) = 3, then X

S

with fd(X) = 2n.

Proof (1) Obvious from Theorem 5 and Remark 2.2.

(2) Can be obviously deduced from Theorem 5 (dim V = 1 , 2) and from Example 2.1.

(3) Can be obviously deduced from Theorem 5, Remark 2.2 (then dim V = 2), and from

Lemma 3.1.

Proposition 3.3 (Rational homotopy type when dim H

(X; Q ) = 4) If dim H

( X; Q ) = 4, then X has one of the following rational homotopy types:

(1) S

× S

with n + m = fd( X);

(2) S

with 3n = fd( X);

(3) Y

such that H

( Y

; Q ) = Q [ a , b ]/( ab , a

− λ b

) where λ ∈ Q

/( Q

)

and where a and b are both of even degrees.

Here and on the sequel ( Q

)

denotes the set of nonzero squares in Q

.

Proof As the case χ

( X ) = 0 was already well considered in [11], we will discuss only the opposite one when χ

(X) = 0. Hence by Remark 2.2, we have dim H

( X; Q ) = dim H

( X; Q ) = 2. Let B = {1, a, b, μ} be a well ordered basis of H

(V, d ) such that

|a| |b|. From the Poincaré duality property we have ab = μ. In particular fd(X) = |μ| =

|a| + |b|. Set a = [α], b = [β] and put n = |α|, m = |β|.

(1) If n and m are both odd, then α and β are generators of V with (V, d ) ∼ = ({α, β}, 0).

Hence

X

S

× S

.

(2) If n is odd and m is even, then α and β are generators of V with [β]

= 0. Thus V ∼ = ({α, β, x }, d )

with dα = dβ = 0, d x = β

and

X

S

× S

. (3) If n is even and m is odd (similar than the precedent case).

Proposition 3.4 (Rational homotopy type when dim H

( X; Q ) = 5)

If dim H

(X ; Q ) = 5, then X has one of the following rational homotopy types:

(1) S

with 4n = fd( X);

(2) S

# S

with n + m = fd(X).

Proof Due to Theorem 2, Theorem 5 and Remark 2.2, necessary dim V

= dim V

= 1 , 2 and H

( X ; Q ) = H

( X ; Q ) . Let B = { 1 , a , b , c , μ} be a well ordered basis of H

( V , d ) with | a | | b | | c | . The Poincaré duality property imposes ac = b

= μ and forbids the two non-symmetric situations |a| = |b| < |c| and |a| < |b| = |c|. The case |a| = |b| = |c| is also impossible, because if not V

of dimension less than two, will have a, b, c as generators. The only remainder possibility is that |a| < |b| < |c| with a

= 0, because a

= 0 yields necessary to the impossible (for degree reasons) situation a

= μ = b

. Since a

= 0, the discussion will be around the nilpotency of a.

(1) If a

= 0, then H

( X ; Q ) ∼ = Q [ a ]/( a

) and X

S

where n = | a | ;

(2) If a

= 0, then necessary a and b are two generators of H

(X; Q ) with a

= c.

Therefore, a

= μ = b

and ab = 0. So H

( X; Q ) ∼ = Q [a, b]/(ab, a

− b

) and X

S

# _S

with n = |a| and m = |b|.

Lemma 3.5 Let P

such that H

( P

; Q ) ∼ = Q [a, b]/(ab, b

− λa

) with λ ∈ Q

/( Q

)

. Then

P

P

⇔ λ

/λ

∈ ( Q

)

. Proof Set

H

( P

; Q ) ∼ = ^Q [ a , b ]/( ab , b

− λ

a

) H

( P

; Q ) ∼ = Q [a

, b

]/(a

b

, b

− λ

a

)

Since all P

are formal, then P

P

if and only if H

( P

; Q ) ∼ = H

(P

; Q ). Let us search such isomorphism φ : H

( P

; Q ) −→ H

( P

; Q ) defined by

a

= φ(a) := pa

b

= φ( b ) := qb + sa

(2)

If λ

= λ

t

for some t ∈ Q

, take p = q = t and s = 0.

Conversely, suppose now that φ is an isomorphism. Since ab = 0

b

− λ

a

= 0

, a

b

= 0

b

− λ

a

= 0

and thanks to (2), we have s = 0 and 0 = b

− λ

a

= (λ

q

− λ

p

) a

. Thus λ

/λ

=

( p

/ q )

∈ ( Q

)

.

Lemma 3.6 Let R

such that H

(R

; Q ) ∼ = Q[a, b]/(a

, b

+ λa

) with λ ∈ Q

/( Q

)

. Then

R

R

⇔ λ

/λ

∈ ( Q

)

.

Proof Set H

( R

; Q ) ∼ = Q [ a , b ]/( a

, b

+ λ a

) and H

( R

; Q ) ∼ = Q [ a

, b

]/( a

, b

+ λ

a

). Since all R

are formal, then R

R

if and only H

(R

; Q ) ∼ = H

(R

; Q ).

Let us search such isomorphism φ : H

( R

; Q ) −→ H

(R

; Q ) defined by a

= φ(a) := la + hb

b

= φ(b) := pb + qb (3)

If λ

= λ

t

for some t ∈ Q

, then take φ(a) = ta and φ(b) = b.

Conversely, suppose now that φ is an isomorphism. Thanks to (3), we have b

+ λ

a

= ( p

+ λ

l

)a

+ 2( pb + λ

lh)ab + (q

+ λ

h

)b

. Due to b

+ λ

a

= b

+ λ

a

= 0, one may conclude that

λ

= − pq lh λ

= p

+ λ

l

q

+ λ

h

= p

q

· λ

+ λ

x

λ

+ λ

y

where x = l p , y = h

q

= λ

· p

x

q

since λ

y = − 1 x

This completes the proof.

Proposition 3.7 (Rational homotopy type when dim H

( X ; Q ) = 6 with χ

( X ) = 0) If dim H

(X; Q ) = 6, with χ

(X) = 0, then X has one of the following rational homotopy types:

(1) S

with 5n = f d( X);

(2) S

× S

with n + 2m = fd(X);

(3) P

such that H

( P

; Q ) ∼ = Q [ a , b ]/( ab , b

− λ a

) with λ ∈ Q

/( Q

)

; (4) Q

such that H

(Q

; Q ) ∼ = Q [a, b]/(a

b, b

+ ab + λa

) with λ ∈ Q

; (5) R

such that H

(R

; Q ) ∼ = Q [a, b]/(a

, b

+ λa

) with λ ∈ Q

/( Q

)

. Proof Thanks to Theorem 2 and Remark 2.2,

dim V

= dim V

= 1 , 2 , 3

and H

(X ; Q ) (as an algebra) admits at most two generators.

If H

( X ; Q ) is generated by only one, then H

( X ; Q ) ∼ = Q [ a ]/( a

) and X

S

with n = |a|.

Suppose now that H

(X; Q ) is generated by two cohomological classes a and [b]. Let B = {1, a, b, c, d , μ} be a basis of H

(V, d ) ordered in increasing degree with ad = bc = μ with ad = bc = μ . We have to distinguish two possibilities.

First case: |a| < |b|. Since a

= 0, we discuss on the nilpotency of a.

(1) If a

= 0, then B = { 1 , a , a

, b , a

, μ} with b

= λ a

and ab = λ

a

. If ab = 0, then

X

P

.

If not (i.e. λ

= 0), then set a

= λ

a and b

= λ

b−λ

a

to get a

b

= 0 and b

−λ

a

= 0, where λ

= λ

λ

+ 1. In other words,

X

P

.

(2) If a

= 0 but a

= 0, then B = {1, a, a

, b, ab, μ} with a

b = μ and b

= λa

b.

If one set a

= b − (λ/2)a

and b

= a he gets a

= b

= 0. Thus H

(X; Q ) ∼ = Q [ a

, b

]/( a

, b

) and

X

S

× S

, where n = |a| and m = |b|.

(3) Finally, suppose that a

= 0. Since ad = μ, then necessarily ab = 0 and one can set c = ab. On the other hands, ab

= bc = 0, then take d = b

. In particular b

(for degree reasons). So H

( X ; Q ) ∼ = Q [ a , b ]/( a

, b

) and therefore

X

S

× S

, where n = |a| and m = |b|.

Second case: |a| = |b|. By the Poincaré duality property, we have a non trivial monomial relation of the type:

λ

a

+ λ

ab + λ

b

= 0 . (4)

If λ

= 0 (or equivalently λ

= 0), then (4) can be reduced to

b

+ νab + λa

= 0, (5)

and therefore dim Q {a

, b

, a

b, ab

} = 1.

(1) If a

= 0 (or equivalently b

= 0), then we put a

= a, b

= b + (ν/2)a to get a

= 0 , b

+ (λ − ν

/ 4 ) a

= 0. Thus

X

R

.

(2) If a

= 0 and b

= 0, then we set a

b = ρa

and put a

= a, b

= b − ρa to get a

b

= 0 , b

+ (ν − 2 ρ) a

b

+ (ρ

+ νρ + λ) a

= 0. If ν − 2 ρ = 0, then

X

Q

. If ν − 2ρ = 0, then b

= 0 and thus

X

R

.

Suppose now λ

= λ

= 0, then ab = 0. Moreover, a

and b

are both non null and linearly dependent. Thus we set a

= λ b

(with λ = 0) and put a

= a + b , b

= a − λ b to get a

b

= 0 and b

+

a

b

−

a

= 0.

If λ = ± 1, then

X

R

. If λ = ±1, then

X

R

.

Remark 3.1 • Unfortunately, this don’t achieve entirely the classification since we still

dont know under which conditions Q

Q

.

• Our classification given here above in Proposition 3.7 seems not totally identical to that given by Mimura and Shiga in [11]. The main reason is that our context is rational while that of Mimura-Shiga is complex. However, it is worth to point out that our classification leads to that of Mimura-Shiga thanks to the following remarks:

– P

P

S

# S

for any λ ∈ C

;

– Q

S

# S

for any λ =

, after considering the variables change a

= μ

(b − λ

a ) and b

= μ

( b −λ

a ) . Here λ

denotes a given solution of the equation x

+ x +λ and μ

denotes one of its arbitrary 3-roots.

– R

S

# _S

after considering the variables change a

= b + i μ a and a

= b − i μa where μ is a complex square root of λ.

Lemma 3.8 Suppose dim H

( X ; Q ) = 6 with χ

( X ) = 0 and let _B = { 1 , [α], [β], [α

], [β

], μ} be a basis of H

(V, d) ordered in increasing degrees. Then,

(1) dim V ∈ { 3 , 5 } ; (2) fd ( X ) is odd;

(3) β

∈ / V and {β, α

} ⊂ V .

Proof (1) From Theorem 2, Theorem 5 and Remark 2.2, we conclude that dim V ∈ { 1 , 3 , 5 } . The situation dim V = 1 is impossible, because if it holds, we obtain V = {α}

with |α| even and [α]

= 0 (i.e H

( X; Q ) = 0). This is impossible since dim H

(X ; Q ) = dim H

( X; Q ) = 3.

(2) Assume that fd ( X ) is even. From the Poincaré duality property, we have [αβ

] = [βα

] = μ . In particular |α|, |β

| (and also |α

|, |β| ) are of the parity. By not- ing that H

( X ; Q ) ⊕ H

( X ; Q ) ⊂ H

( X ; Q ) , this yields to the contradiction dim H

(X ; Q ) ∈ {2, 4, 6}.

(3) Suppose that β

∈ V or that {β, α

} ⊂ V. Since [αβ

] = [βα

] = μ), there exists a generator c ∈ V of even degree, such that dγ = αβ

− βα

. Since |α| and |β

| are of different parities, then (for example when |α| is even)

|ai|even

> |α| + |γ | =

|α| + fd ( X ) − 1 > f d ( X ) . This contradicts Theorem 3.

Proposition 3.9 (Rational homotopy type when dim H

( X ; Q ) = 6 with χ

( X ) = 0) If dim H

( X ; Q ) = 6 with χ

( X ) = 0, then X has one of the following rational homotopy types:

(1) S

× S

with n odd and n + 2m = fd(X );

(2) T

, the total space of the fibration

S

−→ T

−→ S

× S

, where n, m are of the same parity.

Proof Let B = { 1 , [α], [β], [α

], [β

], μ} be a basis of H

( V , d ) as in Lemma 3.8 and consider P the set describing all possible parities of the degrees of the elements of _B . Set a = [α], b = [β].

(1) If P = { 0 , odd , odd , even , even , odd } , then necessary α, β ∈ V and α

, β

∈ / V . There- fore dim V = 3 with [αβ] = 0. So

( V , d ) ∼ = ({α, β, x ), d ) with d α = d β = 0 , d x = ab and B = {1, [α], [β], [α x], [βx ], [αβx ]}. Hence

X

T

.

(2) If P = {0, odd, even, odd, even, odd}, then α, β ∈ V and α

, β

∈ / V. So necessary ab = [α

], ab

= μ and a

= b

= 0, i.e., H (X; Q ) ∼ = Q [a, b]/(b

) and

X

S

× S

, where n = | a | and m = | b | .

(3) If P = {0, even, even, odd, odd, odd}, then (for degree arguments) we should have ab = b

= 0.

If a

= 0, then (V, d) ∼ = {a, b, x, y, z}, d) with d x = a

, d y = ab and d z = b

. In other words

X

T

, where n = |a| and m = |b|.

If a

= 0, then b = a

and b

= aa

where a

= [α

] and b

= [β

]. Thus a

= 0 and H

( X; Q ) ∼ = Q [a, a

]/(a

). In other words,

X

S

× S

,

where n = | a

| and m = | b | verifying n + 2m = fd ( X ) since a

a

= μ .

(4) If P = {0, even, odd, even, odd, odd}, then α, β ∈ V (for degree reasons) and α

, β

∈ / V (from Lemma 3.8). Hence a

= [α

] and ab = [β

]. In other words, H

(X; Q ) ∼ = Q [a, b]/(a

) and

X

S

× S

,

where n = |a| and m = |b|.

Proposition 3.10 (Rational homotopy type when dim H

( X ; Q ) = 7) If dim H

( X ; Q ) = 7, then X has one of the following rational homotopy types:

(1) R with H

(R, Q ) ∼ = Q [a, b]/(a

b, a

− b

);

(2) S

# S

with 3n = 4m = fd(X );

(3) S

# S

with 2n = 5m = fd ( X ) ;

(4) _S

with 6n = fd ( X ) .

Proof We know from Theorem 5 and Remark 2.2 that dim V

= dim V

∈ { 1 , 2 , 3 } and that

H

(V, d) = H

(V, d).

Let B = { 1 , [α

], [α

], [α

], [α

], [α

], μ} be a basis of H

( V , d ) ordered in increasing degree with all n

= |α

| are even. Thanks to the Poincaré duality property we should have

[α

α

] = [α

α

] = [α

] = μ.

The both situations n

= n

= n

= n

= n

and n

< n

= n

= n

< n

are forbidden;

because if not, V of dimension 3 will have at least 4 generators {α

, i = 1 , .., 4 } .

The case n

= n

< n

< n

= n

is also impossible. Indeed, assume that it holds, then necessary {α

, α

} ⊂ {α

, α

, α

}. Because if not (for example if α

∈ V ), then

|ai|even

| a

| |α

| + |α

| + |α

| = fd ( X ) + |α

| > fd ( X ) and this contradicts Theorem 3. Therefore

V ⊂ {α

, α

, α

}.

• If [α

]

= [α

]

= 0, then necessary [α

.α

] = 0. Because the opposite requires [α

] = [α

.α

] and leads to the impossible result μ = [α]

= 0. With the same arguments, one may conclude that [α

.α

] = [α

.α

] = 0. Consequently, V should have five generators β

such that d β

= α

, d β

= α

and d β

= α

α

, d β

= α

α

, d β

= α

α

. This contradicts dim V 3.

• Suppose now [α

]

= 0 (for example).

If α

∈ {α

, α

}, then one can set α

= α

and μ = [α

]

. Since {α

, α

} ⊂ {α

, α

}, then one can set also α

= α

α

and α

= α

α

with i + j = k + = 3. But noting that dim Q {α

, α

, α

α

} = 1 leads to the contradiction that α

and α

are linearly dependent.

If α

∈ / {α

, α

} , then necessary α

∈ {α

, α

} and α

= μ . Thus

|ai|even

| a

| |α

| + |α

| + |α

| > 3 |α

| = fd ( X ).

This contradicts Theorem 3.

Hence, the only remainder possible case is that when n

< n

< n

< n

< n

(the other non-symmetric ones are forbidden by the Poincaré duality property). Thanks to the useful remark of Mimura and Shiga in [11], that the cohomology algebra of X is isomorphic, as a graded algebra, to either Q [ a ]/( a

) or Q [ a , b ]/( f

, f

) , where ( f

, f

) is a regular sequence.

In particular [α

]

= 0, because it is not possible to find one or two generators of H ( X ; Q ) such that is isomorphic to either Q [a]/(a

) or Q [a, b]/( f

, f

) whenever [α

]

= 0.

The following discussion will focus on the nilpotency of a = [α

].

(1) If a

= 0, then put b = [α

] and let us distinguish two cases:

If ab = 0, then _B = { 1 , a , b , a

, ab , a

, a

} . Thanks to the Poincaré duality property we should have b

= a

, and hence (for degree arguments) a

b = b

= 0. In other words, H

(X ; Q ) ∼ = Q [a, b]/(a

− b

, a

b), i.e.,

X

R.

If ab = 0, then B = { 1 , a , b , a

, b

, a

, a

} with a

= b

(thanks to the Poincaré duality property). Thus H ( X ; Q ) ∼ = Q [ a , b ]/( ab , a

− b

) and

X

S

# S

.

(2) If a

= 0 but a

= 0, then B = {1, a, a

, b, a

, a

, a

} with b

= a

where b = [α

] . Since 2 | a | < | b | < 3 | a | , then ab = 0 and therefore H

( X ; Q ) ∼ = Q [ a , b ]/( ab , b

− a

) . In other words,

X

S

# S

.

(3) If a

= 0 but a

= 0, then B = {1, a, a

, a

, a

, a

, a

} and H

( X; Q ) ∼ = Q [ a ]/( a

) . In other words,

X

S

.

Proposition 3.11 (Rational homotopy type when dim H

(X ; Q ) = 8 with χ

(X) = 0) If dim H

(X; Q ) = 8 with χ

( X) = 0, then X has one of the following rational homotopy types:

(1) S

with 7n = fd(X);

(2) S

× S

with 2n + 4m = fd ( X ) ; (3) _S

# _S

with 4n = fd ( X ) ; (4) S

# S

with 3n = 5m = fd(X );

(5) S

× S

× S

with 2 ( n + m + k ) = fd ( X ) ;

(6) P

such that H

( P

; Q ) ∼ = Q [a, b, c]/(a

− αbc, b

− βac, c

− γ ab).

Proof Thanks to Remark 2.2, especially the dimension formula (1), H

( X; Q ) (as algebra) admits at most three generators and is fully described by

H (X; Q ) ∼ = Q [x

, . . . , x

]/( f

, . . . , f

)

where ( f

, . . . , f

) is a regular sequence of graded elements in the polynomial ring Q [x

, . . . , x

]. We will discuss on the number of its generators.

First case: H

( X ; Q ) is monogenic (i.e., H

( X ; Q ) ∼ = Q [ a ]/( a

) ), then X

S

with 7n = fd( X).

Second case: H

(X; Q ) is bigenic (i.e., H

( X; Q ) ∼ = Q [a, b]/( f

, f

)). Let us compare the degrees of |a| and |b|.

(1) If |a| < |b| and | f

| = k|b| for some k ≥ 1, then 2|b| ≤ | f

| =

|a| <

|b|. In particular k = 1, 2.

Assume k = 1. Then f

= a

and | b | = m | a | with m ≥ 2. By the dimension formula (1), we have | f

| =

| b | . Since f

∈ / ( a ) , then | f

| is an integer (i.e., m = 2 or 4).

If m = 2 then ( f

, f

) = ( a

, b

) . If m = 4 then ( f

, f

) = (a

, b

).

Assume now that k = 2. Then | f

| = k|b| and | f

| = 4|a|. Therefore |b| ≤ 2|a|.

If |b| < 2|a|, then necessarily ( f

, f

) = (b

, a

) since |a|+|b| < 2|b| = | f

| < 4|a| =

| f

|.

If | b | = 2 | a | , then ( f

, f

) = ( a

, b

) since | f

| = 2 | b | .

In short, all the discussed cases lead to the same situation, that is:

H

( X ; Q ) ∼ = Q [ a , b ]/( a

, b

).

In other word,

S

× S

.

(2) If | a | < | b | and | f

| = k | a | for some k ≥ 1, then necessarily k = 2 or 4

Assume k = 2. Then | f

| = 2|a| and | f

| = 4|b|. In particular ( f

, f

) = (a

, b

) Assume now that k = 4. Then | f

| = 4|a| and | f

| = 2|b|.

If |b| = 2|a|, then ( f

, f

) = (a

, b

).

If |b| = 2|a|, then ( f

, f

) = (αa

+ βb

, γ b

+ δa

b) for some |α|

, |β|

, |γ |

,

|δ|

∈ Q

. Put c = γ b + δ a

, then

H

( X ; Q ) ∼ = Q [ b , c ]/( bc , λ c

+ λ

b

)

where λ = α

δ

and λ

= αγ

δ

+ β.

(3) If |a| < |b| and| f

| is not a multiple neither of |a| nor of |b|, then by regularity | f

| = k|b|

for some integer k ≥ 2. Therefore, the dimension formula (1) leads to the impossible situation that | f

| =

| a | with k = 2 or 4.

(4) If | a | = | b | , then f

and f

are homogeneous polynomials of respective degrees 2 and 4. As in Lemma 3.1 [11], we may set

H

( X ; Q ) ∼ = Q [ a , b ]/( a

− α b

, a

+ β a

b ) with α = λ

∈ ( Q

)

. Two situations arise for the basis of H

( X ; Q ) . The first one is that B = {1, a, a

, a

, a

, b, ab, μ} with

a

= αb

a

= β a

b (6)

By setting a

= a + λb and b

= a − λb, the system (6) turns into a

b

= 0

a

− b

= 0

Then H

( X ; Q ) ∼ = Q [ a

, b

]/( a

b

, a

− b

) and consequently X

S

# _S

.

The second possibility for the cohomology basis of H

( X ; Q ) is that B = { 1 , a , b , a

, a

, b

, a

, μ} with μ = a

= b

(thanks to the Poincaré duality property) and a

= μ ab with μ > 0 and b

= 0 (for degree reasons). Then set the variables change a

= b − μa and b

= b to get a

b

= a

− b

= 0. Therefore H

( X; Q ) ∼ = Q [a

, b

]/(a

b

, a

− b

) and

X

S

# S

.

Third case: H

( X; Q ) is trigenic (i.e., H

(X; Q ) ∼ = Q [a, b, c]/( f

, f

, f

)). The only cases

permitted by the dimension formula (1) are the following:

(1) | a | < | b | < | c | . Then a basis of H

( X ; Q ) can be given by B = { 1 , a , b , c , ab , ac , bc , μ}

with μ = abc (thanks to the Poincaré duality property) and a

= b

= c

= 0 (for degree reasons). Hence H

( X ; Q ) = Q [ a , b , c ]/( a

, b

, c

) and

X

S

× S

× S

with 2 ( n + m + k ) = fd ( X ) .

(2) | a | = | b | = | c | . Then f

, f

and f

are homogeneous polynomials of degree 2. Hence, put f

= a

− αbc, f

= b

− βac, f

= c

− γ ab with α, β, γ ∈ ( Q

)

. Thus H

(X; Q ) = Q [a, b, c]/(a

− αbc, b

− βac, c

− γ ab) and

X

P

.

Acknowledgments The authors would like to thank all members of the research group Moroccan Area in Algebraic Topology (MAAT) for the beautiful ambiance and atmosphere of work that prevailed within the group. This work is based on several discussions exchanged during their monthly seminar. Finally, we are grateful to our anonymous referee(s) for the suitable advice and assistance during the refereeing process.

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