I would greatly appreciate it if you could help me with the following:
I'm curious as to how to find the Reeb field $R_w$ associated to a
specific contact form $w$; does one actually find $R_w$ as the zero set of:
$$dw(R_w ,\cdot )=0,$$
where $R_w$ is known to exist by Linear Algebra, since manifold is odd-dimensional.
Or does one somehow use the fact that the Lie derivative of $ w$ with
respect to $R_w$ is zero (or something else)?
I know that a Reeb field is "stronger" than a plain contact field, in that a contact field preserves just the contact structure, while a Reeb field preserves the form itself, i.e., if $\phi$ is the flow of $R_w$ , then $\phi^*w$=$gw$ for contact fields (i.e., the kernel of the form $gw$ is the same as $\ker w$), but $\phi^*w=w$ for Reeb fields.
Hopefully, too, someone could help me figure out how to show that a contact vector field ( one whose flow preserves the contact structure) , that is transverse to the contact pages is a Reeb field. I'm kind of clueless here.
Thanks for any help, suggestions,
Gary K
Best Answer
Yes, the Reeb field $R_w$ associated to a contact form $w$ satisfies:
1) $w(R_w)=1$ , which is basically saying that $w(R_w) \neq 0$ (then you can rescale to 1)
2) $dw(R_w, .)==0$ , so that $R_w$ "kills" every vector vield under $dw$ . Notice that a Reeb field exists by linear algebra alone: your form is defined in an odd-dimensional manifold, so that it is an antisymmetric odd-dimensional quadratic form, meaning it is degenerate, so that $dw (R_w, .)$ has a solution.
You can then set, say for the standard form $w=xdy+dz$ in $\mathbb R^3 $ , then
$dw=dydx$ , so that you want $R_w:=R_x+R_y+R_z$ and, for any generic vector field $V:=V_x+V_y+V_z$, you want :
$dydx(R_x+R_y+R_z, V_x+V_y+V_z) $ so that,
$R_xV_y-R_yV_x=0$ , with $R_z, V_z$ free variables. But I don't see how to find
a general form (nor even a single solution ) for $R_w$ from this.