Fiber bundle over a symplectic manifold

Question

Let $(M,\omega)$ be a symplectic manifold. Does a vector fiber bundle $E$ exist, such that the Chern classes of $E$ are linked to the symplectic form as, for example, $\forall k$, $ch_k(E)=[\omega^k]$?

Answer 1

What you need is to construct principal circles bundles $E\to M$ over the symplectic manifold $(M^{2n}, \omega)$, whose Euler class is represented by the symplectic form. Recall that for principal circle bundles $E\to M$, the Euler class coincides with the first Chern class, which is an element in $H^{2}(M; \mathbb{Z})$. Thus, the symplectic form has to represent an integral class $[\omega]\in H^{2}(M; \mathbb{Z})$, which means that it has to lie in the image of the homomorphism $$ H^{2}(M; \mathbb{Z})\to H^{2}(M; \mathbb{R})\cong H^{2}_{DR}(M). $$ In fact, the existence of such a symplectic form has been described for example by R. E. Gompf (ftp://ftp.ma.utexas.edu/pub/papers/gompf/Gompf-symp.ps) as follows: Assume that $(M, \omega)$ is closed+symplectic. Then, for every Remannian metric on $M$ one can find a $\epsilon$-ball $B_{\epsilon}$ around the origin in the space of harmonic 2-forms on $M$, such that every element in $\omega+B_{\epsilon}$ is symplectic. However, the set of classes in $H^{2}(M; \mathbb{R})$ represented by these elements is open; hence there is a symplectic form which represents a rational cohomology class. Thus, multiplying with a suitable rational, one can always find a symplectic form representing an integral cohomology class.

Now, the classical Boothby-Wang construction (https://www.jstor.org/stable/1970165?seq=1#page_scan_tab_contents) associates to each symplectic manifold $(M, \omega)$ with an integral symplectic class $[\omega]$ a principal circle bundle ${\rm S}^{1}\to E\to M$ with Euler class equal to $[\omega]$. Moreover, $E$ carries a contact structure $\xi$.

In particular, fix a (closed+connected) symplectic manifold $(M^{2n}, \omega)$ with symplectic form $\omega$ representing an integral cohomology class, and write $[\omega]_{\mathbb{Z}}$ for the integral lift of $[\omega]\in H^{2}_{DR}(M)$. Let $\pi : E^{2n+1}\to M^{2n}$ be the principal circle principal bundle over $M$ with Euler class $e(E)=[\omega]_{\mathbb{Z}}$. By a result of Kobayashi, we can choose a ${\rm U}(1)$-connection, say $A$, on $E\to M$ with curvature form $\Omega=\frac{2\pi}{i}\omega$. Notice that $A$ is an invariant 1-form on $E$ with values in $\frak{u}(1)\cong i\mathbb{R}$, satisfying $$ dA=\pi^{*}\Omega \quad\text{and} \quad A(X)=2\pi i,$$ where $X$ is the fundamental vector field generated by the action of the element $2\pi i\in\frak{u}(1)$. Then we can prove that the real-valued 1-form $\phi:=\frac{1}{2\pi i}A$ on $E$ defines a contact structure on $E$ i.e. $\phi\wedge (d\phi)^{n}\neq 0$, with $\phi(X)=1$ and $d\phi=-\pi^{*}\omega$. Since $d\phi(X)=0$, the Reeb vector field $\xi$ associated to the contact form $\phi$ is given by the vector field $X$ along the fibres. Notice that such a structure defines a splitting $TE=\mathbb{R}\oplus\pi^{*}TM$ and hence an orientation on $E^{2n+1}$ (the $\mathbb{R}$-summand takes the orientation of $\xi=X$ and $TM$ is oriented by $\omega$).