Quaternionic Structures in Mathematics and Physics SISSA - Trieste, 5-9 September 1994

ON A BRYLINSKI CONJECTURE FOR COMPACT SYMPLECTIC MANIFOLDS ^{∗}

Marisa Fernández, Raúl Ibáñez and Manuel de León

ABSTRACT

J. L. Brylinski in ([3], page 102) conjectured that for compact symplectic manifolds any de Rham cohomology class has a symplectically harmonic representative ν, that is, dν = δν = 0, for δ the Koszul differential. In this paper, a counterexample to this conjecture is constructed.

∗ Partially supported by DGICYT-SPAIN, Projects PB89-0571 and PB91-0142, and U.P.V., Project 127.310-EA 191/94.

1 Introduction

In this paper we shall consider symplectic manifolds as a particular type of Poisson manifolds; the Poisson bracket is given in a natural way by using the symplectic form as we explain bellow.

For a Poisson manifold M, J. L. Koszul in [12] defined a differential operator Δ : Λ^{k}(M) ⟶ Λ^{k−1}(M). This operator is given by Δ = [i(G), d], where G is the Poisson tensor defined by A. Lichnerowicz [13], i(G) denotes the contraction by G and d is the exterior derivative of M. We call it the Koszul differential; and using the notation of Brylinski [3] we shall write δ instead of Δ. Since δ^{2} = 0, it defines the so-called canonical homology of M.

For a symplectic manifold Brylinski [3] proved that δ is equal, up to a sign, to ⋆d⋆, where ⋆ is the symplectic star operator defined in a similar way to the Hodge star operator for Riemannian manifolds.

As in the case of Riemannian manifolds, one can define a symplectic Laplacian Δ = dδ+δd, which is identically zero. A k-form ν such that dν = δν = 0 is called symplectically harmonic. Brylinski in ([3]; page 102), inspired by the Hodge’s theorem for compact Riemannian manifolds, made the following

Conjecture: If M is a symplectic manifold which is compact, any de Rham cohomology class in H^{∗}(M) has a representative ν such that dν = δν = 0, or, in other words, ν is a symplectically harmonic representative.

For a compact manifold with (positive definite) Kähler metric Brylinski [3] proved that the conjecture is true. In this paper we shall give an example of a four dimensional compact symplectic manifold which does not satisfy the conjecture. Our example is the known compact symplectic manifold of Kodaira-Thurston (see [4, 6, 9, 11, 16]) which can be realized as the quotient of a nilpotent Lie group by a discrete subgroup.

The Kodaira-Thurston manifold KT is the simplest compact symplectic nilmanifold with no (positive definite) Kähler metric. KT cannot be Kählerian for topological reasons (see [4, 6, 7, 8, 10, 11, 16, 18]). However KT carries indefinite Kähler metrics ([4, 6, 7]), and hence symplectic structures.

2 The Koszul differential for symplectic manifolds

Let us briefly recall some results about the Koszul differential ([3, 12]). We also refer to [9] where a more detailed and more complete development may be found.

Let M be a symplectic manifold of dimension 2n, with symplectic form ω. Denote by X (M) the Lie algebra of C^{∞} vector fields on M and by F(M) the algebra of C^{∞} functions on M. For every f ∈ F(M) we shall denote by X_{f} the Hamiltonian vector field of f, that is, the vector field on M such that

i_{X_{f}} ω = df .

The Poisson bracket { , } on M is defined by

{f, g} = X_{f}(g) = −ω(X_{f}, X_{g}),  (1)

for any f, g ∈ F(M). A. Lichnerowicz [13] remarked that the Poisson bracket gives rise to a non-singular covariant skewsymmetric tensor field G of order 2 such that [G, G] = 0, where [ , ] is the Schouten-Nijenhuis bracket. G is called Poisson tensor, and it satisfies:

i(G)(df ∧ dg) = {f, g},  (2)

for f, g ∈ F(M), where i(G) denotes the contraction by G. Conversely, such a tensor field G gives rise to a Poisson bracket and, hence, to a symplectic form.

Next, we shall denote by Λ^{k}(M) the space of the differential k-forms on M. The Koszul differential δ : Λ^{k}(M) ⟶ Λ^{k−1}(M) is defined as the commutator of i(G) and the exterior derivative d, that is,

δα = [i(G), d](α) = i(G)(dα) − d(i(G)α) ,  (3)

for all α ∈ Λ^{k}(M). Brylinski [3], inspired by the differential of the Chevalley-Eilenberg complex C_{∗}(F(M), F(M)), has proved that δ can be alternatively defined by the formula:

δ(f_{0} df_{1} ∧ ⋯ ∧ df_{k}) = ∑_{1≤i≤k} (−1)^{i+1} {f_{0}, f_{i}} df_{1} ∧ ⋯ ∧ d̂f_{i} ∧ ⋯ ∧ df_{k}

+ ∑_{1≤i<j≤k} (−1)^{i+j} f_{0} d{f_{i}, f_{j}} ∧ df_{1} ∧ ⋯ ∧ d̂f_{i} ∧ ⋯ ∧ d̂f_{j} ∧ ⋯ ∧ df_{k} .  (4)

From (3) or (4) one can check the following identity (see [3, 12]):

δ^{2} = 0 .  (5)

Hence, we obtain the canonical complex of M:

⋯ ⟶ Λ^{k+1}(M) ⟶^{δ} Λ^{k}(M) ⟶^{δ} Λ^{k−1}(M) ⟶ ⋯

whose homology H_{∗}^{can}(M) is called the canonical homology of M.

Now, denote by Λ^{k}(G), (k ≥ 0), the associated pairing Λ^{k}(G) : Λ^{k}(T^{∗}M)×Λ^{k}(T^{∗}M) ⟶ F(M) which is (−1)^{k}-symmetric. We consider the volume form v_{M} on M given by

v_{M} = \frac{ω^{n}}{n!} . Then, imitating the Hodge star operator for Riemannian manifolds, Brylinski

[3] defined the symplectic star operator ⋆ : Λ^{k}(M) ⟶ Λ^{2n−k}(M) by

β ∧ (⋆α) = Λ^{k}G(β, α) · v_{M} ,  (6)

for all α, β ∈ Λ^{k}(M). Moreover, he proved that ⋆^{2} is the identity, and

δα = (−1)^{k+1} ⋆ d ⋆ (α),  (7)

for all α ∈ Λ^{k}(M).

The ⋆ operator permits us to relate the canonical homology with the de Rham cohomology of M.

Theorem 2.1 ([3]) The operator ⋆ establishes an isomorphism of the canonical homology group H_{i}^{can}(M) with the de Rham cohomology group H^{2n−i}(M).

In order to show, in the next section, that Brylinski’s conjecture is not true we shall need some results.

Definition 2.2 For any k ≥ 0, the map L : Λ^{k}(M) ⟶ Λ^{k+2}(M) is given by

L(α) = α ∧ ω ,  (8)

for all α ∈ Λ^{k}(M).

Because ω is closed, the map L commutes with the exterior derivative:

L ∘ d = d ∘ L .  (9)

The following lemma gives the relations among L, ⋆ and G and it will be useful in the next section.

Lemma 2.3 ([9]) We have

i(G) = − ⋆ L⋆ ,  (10)
[L, δ] = −d .  (11)

Moreover, any differential 1-form β on M satisfies

⋆ (β) = −(n − 1)! L^{n−1}(β) .  (12)

3 A counterexample to Brylinski conjecture

Consider the Heisenberg group H that is, the connected, simply-connected and nilpotent Lie group of dimension 3 of the form

H = \{( \begin{matrix}1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{matrix} ) / x, y, z ∈ R\} .

A standar computation shows that a basis for the left invariant 1-forms on H is given by {dx, dy, dz − xdy}. Now, to get a compact quotient, we take the space of right cosets Γ\H where Γ is the uniform subgroup of H with integer coordinates. Thus, Γ\H is a 3-dimensional compact nilmanifold. Hence the left invariant 1-forms dx, dy, dz − xdy all descend to 1-forms α, β, γ on Γ\H, and so we have

dα = dβ = 0, dγ = −α ∧ β .  (13)

Let us now consider the compact manifold KT = Γ\H × S^{1}. It is called the Kodaira-Thurston manifold.

Denote by η the canonical 1-form on S^{1}. Then, using the Nomizu’s theorem [15], we compute the de Rham cohomology groups of KT:

\begin{aligned}H^{0}(KT) & = & \{1\} , \\ H^{1}(KT) & = & \{[α], [β], [η]\} , \\ H^{2}(KT) & = & \{[α ∧ γ], [α ∧ η], [β ∧ γ], [β ∧ η]\} , \\ H^{3}(KT) & = & \{[α ∧ β ∧ γ], [α ∧ γ ∧ η], [β ∧ γ ∧ η]\} , \\ H^{4}(KT) & = & \{[α ∧ β ∧ γ ∧ η]\} . \end{aligned}

Furthermore, one can check that the 2-form ω given by

ω = α ∧ γ + β ∧ η ,  (14)

is a symplectic form on KT.

Let {X, Y, Z, T } be the dual basis to {α, β, γ, η}, and let G be the Poisson tensor on KT associated with the symplectic form ω and defined by (14). Then we have that G is given by

G = Z ∧ X + T ∧ Y .  (15)

Therefore, using (4), we have that the Koszul differential δ on the generators of Λ^{k}(KT), satisfies

δ(γ ∧ η) = α , δ(β ∧ γ ∧ η) = α ∧ β ,  (16)

and on the others is zero.

Theorem 3.1 The cohomology class [β∧γ∧η] in H^{3}(KT) does not admit a representative ν such that dν = δν = 0. Therefore, Brylinski’s conjecture is not true for compact symplectic no-Kähler manifolds.

Proof : Let us suppose that the cohomology class [β ∧ γ ∧ η] has a representative ν such that δν = 0. Then ν must be of the form

ν = β ∧ γ ∧ η + dθ ,  (17)

for some differential 2-form θ ∈ Λ^{2}(KT). Since δν = 0, from (5), (13), (16) and (17) we have

0 = δν = δ(β ∧ γ ∧ η + dθ)
= α ∧ β − d(δθ)
= d(−γ − δθ) ,

which implies that there is the cohomology class [γ + δθ] ∈ H^{1}(KT). Now, we distinguish the two cases: i) [γ + δθ] is zero in H^{1}(KT), and ii) [γ + δθ] is non-zero.

In the case i), i.e. [γ + δθ] = 0, we have

γ + δθ = df ,  (18)

for some function f ∈ F(KT ). Applying L to both sides of (18), from (9) and (14), we get

−β ∧ γ ∧ η + L(δθ) = d(Lf) ,

which implies

β ∧ γ ∧ η = L(δθ) − d(Lf) .  (19)

Now, we shall compute L(δθ). Firstly, from (11), we have

L(δθ) = δL(θ) − dθ .  (20)

Moreover, since θ is a differential 2-form on KT, we get L(θ) = gα ∧ β ∧ γ ∧ η, for some function g ∈ F(KT ). Now, from (7) and (12), we obtain

δ(Lθ) = − ⋆ d ⋆ (gα ∧ β ∧ γ ∧ η)  (21)
= − ⋆ (dg)
= L(dg) .

Substituting (20) and (21) in (19), it follows that

β ∧ γ ∧ η = d(Lg) − dθ − d(Lf)
= d(Lg − θ − Lf) ,

which is a contradiction because the cohomology class [β ∧ γ ∧ η] is non-zero in H^{3}(KT). Thus, the case i) is not possible.

Let us now consider the case ii), that is, [γ + δθ] is non-zero in H^{1}(KT). Then

[γ + δθ] = λ_{1}[α] + λ_{2}[β] + λ_{3}[η] ,

for some λ_{i} ∈ R, 1 ≤ i ≤ 3. This implies that

γ + δθ = λ_{1}α + λ_{2}β + λ_{3}η + dh ,  (22)

for some function h ∈ F(KT ). Applying L to both sides of (22) we obtain

−β ∧ γ ∧ η + L(δθ) = λ_{1}α ∧ β ∧ η − λ_{2}α ∧ β ∧ γ
+ λ_{3}α ∧ γ ∧ η + d(Lh) ,

from which we deduce

β ∧ γ ∧ η − λ_{2}α ∧ β ∧ γ + λ_{3}α ∧ γ ∧ η  (23)
= λ_{1}d(γ ∧ η) − d(Lh) + L(δθ) .

Proceeding as in the case i), we get L(δθ) = d(Lg) + dθ. Thus, from (23), it follows that the differential form β ∧ γ ∧ η − λ_{2}α ∧ β ∧ γ + λ_{3}α ∧ γ ∧ η defines the zero cohomology class in H^{3}(KT). But, it is not possible, because the cohomology classes [β ∧ γ ∧ η], [α ∧ β ∧ γ] and [α ∧ γ ∧ η] are the generators of the cohomology group H^{3}(KT). This

completes the proof.  QED

Furthermore, we consider the basis of vector fields {X, Y, Z, T } on KT; and define an almost complex structure J on KT by

JX = Y, JZ = T .  (24)

It is easy to check that J is integrable (i.e. the Nijenhuis tensor of J vanishes). Let # denote the symmetric product and let

ds^{2} = β#γ − α#η .  (25)

Then ds^{2} is an indefinite Kähler metric for J, and the corresponding Kähler form is the symplectic form ω.

Proposition 3.2 A Brylinski conjecture for the compact nilmanifold KT with the indefinite Kähler metric given by (25) is not true.

Proof : This follows from Theorem 3.1.  QED

These results shows that is not possible to develop a symplectic harmonic Hodge theory.

Remark 3.3 O. Mathieu has proved in [14] that a symplectic harmonic Hodge theory can be developped on a symplectic manifold M if and only if it satisfies the hard Lefschetz Theorem.

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Authors’ address: Marisa Fernández and Raúl Ibáñez Departamento de Matemáticas Facultad de Ciencias Universidad del País Vasco Apartado 644 Bilbao, SPAIN e-mail: mtpferol@lg.ehu.es and mtbibtor@lg.ehu.es

Manuel de León Instituto de Matemáticas y Física Fundamental Consejo Superior de Investigaciones Científicas Calle Serrano, 113-123 28006 Madrid, SPAIN e-mail: ceeml02@cc.csic.es