Let $(M,\xi)$ be a contact manifold with contact form $\alpha$ and Reeb vector field $R$. The contact form can be rescaled by any positive smooth function $f$ to obtain a new contact form $\alpha_f:=f\alpha$.
How is the associated Reeb vector field $R_f$ affected?
I remember that it should be something of the form
$$R_f=\frac{1}{f}R+X_{\log f},$$
where $X_{\log f}$ should be the Hamiltonian flow of $\log f$. I'm not sure about this, especially because I don't know what is the symplectic space on which $\log f$ is defined, so I wouldn't know how to compute its Hamiltonian flow.
From the two conditions of the contact form ($\iota_R\alpha=1$ and $\iota_Rd\alpha=0$), I know that $R_f$ is something like $R_f=\frac{1}{f}R+X$ for some horizontal vector field $X$, and it also satisfies
$$\iota_{R_f}(df\wedge \alpha+fd\alpha)=\frac{1}{f}df(R)\alpha+df(X)\alpha-\frac{1}{f}df+f\iota_{X}d\alpha=0.$$
Can someone fill the gaps?
Best Answer
I guess that by horizontal you mean that $X\in \ker \alpha$.
Contracting both sides of your last equation with $R$ yields $df(X)=0$, so $$\iota_X d \alpha = \frac{1}{f^2} (df -df(R) \alpha).$$ This, together with the fact that $\iota_X \alpha =0$ uniquely determines $X$.
In other words, the contact form $\alpha$ defines the $C^\infty$-module isomorphism $$\begin{aligned} \flat_\alpha:\mathfrak{X}(M)&\to \Omega^1(M) \\ Y &\mapsto \alpha(Y) \alpha + \iota_Y d \alpha. \end{aligned}$$ Let $\sharp_\alpha$ denote the inverse of $\flat_\alpha$. Then, the vector field $X$ you want to determine is given by $$X=\sharp_\alpha\left(\frac{1}{f^2} (df -df(R) \alpha)\right).$$ Observe that $R=\sharp_\alpha(\alpha)$, and hence $$R_f=\sharp_\alpha\left(\frac{1}{f^2} (df -df(R) \alpha)+\frac{1}{f}\alpha\right).$$ The Hamiltonian vector field $X_f$ associated with $f\in C^\infty(M)$ (via the contact form $\alpha$) is given by $$X_f = \sharp_\alpha \left( df- (R(f)+f) \alpha\right),$$ so the vector field we are looking for is $R_f=X_{-\frac{1}{f}}$.