Shorter Notes: Some Simple Examples of Symplectic Manifolds

TLDR: In this paper, Liberman et al. gave a construction of closed symplectic manifolds with no Kaehler structure, and showed that such manifolds do not have even odd Betti numbers.

Abstract: This is a construction of closed symplectic manifolds with no Kaehler structure. A symplectic manifold is a manifold of dimension 2k with a closed 2-form a such that ak is nonsingular. If M2k is a closed symplectic manifold, then the cohomology class of a is nontrivial, and all its powers through k are nontrivial. M also has an almost complex structure associated with a, up-to homotopy. It has been asked whether every closed symplectic manifold has also a Kaehler structure (the converse is immediate). A Kaehler manifold has the property that its odd dimensional Betti numbers are even. H. Guggenheimer claimed [1], [2] that a symplectic manifold also has even odd Betti numbers. In the review [3] of [1], Liberman noted that the proof was incomplete. We produce elementary examples of symplectic manifolds which are not Kaehler by constructing counterexamples to Guggenheimer's assertion. There is a representation p of Z E Z in the group of diffeomorphisms of T2 defined by (1, 0) -P4 id, (0,1 I 0o 1l where [81 ]" denotes the transformation of T2 covered by the linear transformation of R2. This representation determines a bundle M4 over T with fiber T2: M4 = T2 XZ9Z T2, where Z E Z acts on T2 by covering transformations, and on T2 by p (M4 can also be seen as R4 modulo a group of affine transformations). Let Q1 be the standard volume form for T2. Since p preserves 21, this defines a closed 2-form i2 on M4 which is nonsingular on each fiber. Let p be projection to the base: then it can be checked that S21 + P*' 1 is a symplectic form. (It is, in general, true that "'j + Kp* 21 is a symplectic form, for any closed Q'1 which is a volume form for each fiber, and K sufficiently large.) But H1 (M4) = Z @ Z @ Z, so M4 is not a Kaehler manifold. Many more examples can be constructed. In the same vein, if M2k is a closed symplectic manifold, and if N2k+2 fibers over M2k with the fundamental class of the fiber not homologous to zero in N, then N is also a symplectic manifold. If, for instance, the Euler characteristic of the fiber is not zero, this Received by the editors July 31, 1974. AMS (MOS) subject classifications (1970). Primary 57D15, 58H05. C American Mathematical Society 1976