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Consider a Symplectic manifold D (with $H^1(D)=0$) with symplectic form $w$. Let V be the total space of a circle bundle over D with non-trivial Euler class $e\in H^2(D)$. You may think of V as the set of unit vectors in a complex line L bundle over D with chern class e. Then we can construct a symplectic form ,denote it by same symbol $w$, on total space of L whose restriction to D is the original w. The question is whether V has a contact structure with a contact form $\alpha$ such that $d\alpha=w\mid_V$.

Looking the the Gysinn sequence : $0\rightarrow H^0(D) \rightarrow H^2(D) \rightarrow H^2(V) \rightarrow 0$ it seems that the answer is:

Yes iff $w$ is multiple of e.

I Just wanted to make sure that my conclusion is correct!

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  • $\begingroup$ How do you construct the symplectic form on $L$? $\endgroup$
    – Tim Perutz
    Commented Apr 30, 2010 at 1:36
  • $\begingroup$ Lets fix a hermitian metric on L and let r be the distance function. for $\pi:L\rightarrow D$ consider the two form : $\tilde{w} = π^*w+ r \pi^*e + dr ∧ \beta$ Where $\beta$ is the connection one-form on L\{zerosection} satisfying $\beta{\partial{\theta}}=1$. $\tilde{w}$ extends across D and its restriction to D is $w$. See page 15 of annals paper:" The symplectic sum formula for Gromov-Witten invariants" $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Apr 30, 2010 at 5:41
  • $\begingroup$ This, presumably, is symplectic near the zero-section only, in general? Anyway, I concur that if $e$ isn't a multiple of $[w]$, this form is not exact on circle-bundles in $L$, and that conversely, a multiple of a connection form with curvature $const. w$ is contact. $\endgroup$
    – Tim Perutz
    Commented Apr 30, 2010 at 12:06
  • $\begingroup$ My question is local, so I am interested just in a small neighborhood of D in L. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Apr 30, 2010 at 19:19

2 Answers 2

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If the Euler class is integral and the bundle is principal, the answer is yes. Moreover, the contact structure you obtain is regular. This is Theorem 3 in Boothby,W. M. and Wang, H. C.: On contact manifolds. Ann. of Math. (2) 68 1958 721–734. L.O.

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  • $\begingroup$ The Euler class is integral because it is first chern class of a line bundle and the bundle is principal because it is by definition! The question was something else! $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Apr 29, 2010 at 21:03
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Let me explain my reason:

Let's $w$ be the symplectic structure on D and $\tilde{w}$ be the one on total space of line bundle . Then we have Gysin exact sequence above which gives $H^2(V)=H^2(D)/ \left\langle e\right\rangle$ and so for $\tilde{w}$ to be of the form $[d\alpha]=[0] \in H^2(V)$ it has to be a multiple of e, when restricted to D.

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