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I am simply looking for an argument proving the smoothness of the Reeb vector field of a given contact form.

If you don't know the relevant definitions, the problem is simply this: Let M be a manifold of odd dimension 2n+1 and let [itex]\alpha[/itex] be a 1-form on M such that

1) [itex]\alpha[/itex] is nowhere vanishing (Hence, for every p in M, [itex]H_p :=\ker\alpha_p[/itex] has dimension 2n.)

2) The (2n+1)-form [itex]\alpha\wedge d\alpha[/itex] never vanishes. Equivalently, for every p in M, the bilinear form [itex]d\alpha_p|_{H_p\times H_p}[/itex] is nondegenerate.

Since a nondegenerate bilinear form on an odd dimensional vector space does not exist, it must be that for every p in M, [itex]d\alpha_p[/itex] is degenerate. But by 2), the restriction of [itex]d\alpha_p[/itex] to H_p is nondegenerate. So it must be that [itex]\ker d\alpha_p=\{v\in T_pM : d\alpha_p(v,\cdot)=0\in T_p^*M\}[/itex] has dimension 1.

Define a vector field R of M axiomatically by:

(A1) [itex]R_p \in\ker d\alpha_p[/itex] for all p in M,

(A2) [itex]\alpha(R)=1[/itex].

Since [itex]\dim\ker d\alpha_p=1[/itex], axiom (A1) defines uniquely R up to scaling factor, and (A2) fixes that scaling factor. Hence R is well defined (and is called the Reeb vector field associated with the contact form [itex]\alpha[/itex]).

I am simply looking for a proof that R is smooth, given that M and [itex]\alpha[/itex] are too! Thanks!

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# Question about smoothness of a vector field (Reeb).

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