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Let $dw$ be a 2-form; $\alpha$ be a 1-form, $g$ a function (i.e., a 0-form), $X$ be a fixed vector field , all living in a 3-manifold, so that $$dw(X,.)=g\alpha .$$ Does this condition force $dw$ to be identically zero? I know if $g==0$ , the answer is no, since then a Reeb vector field $X_w$ for $w$ will do, by definition/construction. But what if $g$ is not a constant? I've been approaching this from the perspective of interior product, i.e., contracting , so that $g\alpha$ equals the contraction of $dw$, but this doesn't seem to line up. I'm thinking mostly of $w$ as a contact form, but I would like to know in general. Thanks.

EDIT: Here $w$ is a contact form, and $X$ is a transverse contact field to the structure $\xi$generated by the form. By definition, if $X$ is a contact field, we have $L_X w=g\alpha$ I'm trying to show that, if, in addition $X$ is transverse to $\xi$, we can conclude that $g=0$, so that $L_X w =0$ ,i.e., that $g$ must be identically zero. We start with $L_X w= g\alpha$, so that $ i _X dw +d(i _X w)=0$ . Now, $dw$ is a 2-form, $w$ is a 1-form, so we get:$dw(X,.)+d( w(X))=g\alpha$. From $X$ being transverse to $\xi$ we conclude $w(X)=1$ (so that d(w(X)=0) , since the structure is precisely the kernel of the form. So we're left with $L_X w=dw(X,.)=g\alpha$. I'm trying to see if I can conclude $g=0$ in this case.

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The equation $d\omega(X, . ) = g\alpha$ troubles me, for $\omega$ is a $2$-form, so $d\omega$ is a $3$-form, so $d\omega(X, . )$ is a $2$-form, but $g\alpha$ is a $1$- form. How so? – Robert Lewis Apr 1 '14 at 8:19
Unless of course by $d\omega(X)$ you mean $d(\omega(X, . ))$! Is that right? – Robert Lewis Apr 1 '14 at 8:21
That must be it! – Robert Lewis Apr 1 '14 at 8:22
@RobertLewis: Might be the OP mean "dw" is a two form? – John Ma Apr 1 '14 at 8:37
@John: of course! I misread it as $\omega$ is a $2$-form! – Robert Lewis Apr 1 '14 at 8:44

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