I'm reading Geiges' notes. (https://arxiv.org/pdf/math/0307242.pdf) In the proof of Theorem 2.44 on page 17, the existence of the contact version Darboux coordinate is reduced to solving $H_t$ for each $t$, the PDE near the origin of $\mathbb{R}^{2n+1}$ $$\dot{\alpha}_t (R_{\alpha_t})+dH_t(R_{\alpha_t} )= 0$$ where $\alpha_t$ is a $1$-parameter family of contact forms and $R_{\alpha_t}$ is the corresponding reeb vector field. And he said that this equation always has a solution by integration if the neighborhood is small enough so that $R_{\alpha_t}$ has no closed orbit.

My question is why this is obvious? What I know is that this equation is a quasilinear first order PDE and can possibly be solved by the method of characteristics. But I can't find a reference that contains a clear statement when this kind of equation can be solved.

Thank you.


First, let us prove the following general result:

Proposition. Let $M$ be a manifold and $p\in M$, let $X$ be a vector field on $M$ and let $g\colon M\rightarrow\mathbb{R}$ smooth. If $X(p)\neq 0$, then there exists $U$ an open neighborhood of $p$ in $M$ and $f\colon U\rightarrow\mathbb{R}$ smooth such that: $$\mathrm{d}f(X)=g_{\vert U}.$$ Furthermore, if $g(p)=0$, one may assume that $f(p)=0$ and $\mathrm{d}f_p=0$.

Proof. Using the straightening theorem, there exists $(U,\phi)$ a chart of $M$ around $p$ such that $\phi_*X=\frac{\partial}{\partial x_1}$. Furthermore, one may assume that for $(x_1,\ldots,x_n)\in\phi(U)$, one has the following property: $$s\in[\min(0,x_1),\max(0,x_1)]\Rightarrow(s,x_2,\ldots,x_n)\in\phi(U)\tag{$\star$}.$$ Therefore, one can define a smooth map $F\colon\phi(U)\rightarrow\mathbb{R}$ by the following formula: $$F(x_1,\ldots,x_n):=\int_{0}^{x_1}g(\phi^{-1}(s,x_2,\ldots,x_n))\,\mathrm{d}s.$$ Notice that one has $F(0)=0$ and $\frac{\partial F}{\partial x_1}=g\circ\phi^{-1}$.

In this section, assume that $g(p)=0$, then there exists constants $a_2,\ldots,a_n$ such that $\mathrm{d}F_0=\sum\limits_{i=2}^na_i\mathrm{d}x_i$ and let us define $\overline{F}\colon\phi(U)\rightarrow\mathbb{R}$ in the following fashion: $$\widetilde{F}(x_1,\ldots,x_n):=F(x_1,\ldots,x_n)-\sum_{i=2}^na_ix_i.$$ Notice that $\widetilde{F}(0)=0$, $d\widetilde{F}_0=0$ and $\frac{\partial\widetilde{F}}{\partial x_1}=g\circ\phi^{-1}$, so that one can assume that $F(0)=0$ and $\mathrm{d}F_0=0$.

Finally, with $f=F\circ\phi$, using the chain rule, for all $x\in U$, one has: $$\mathrm{d}f_x(X(x))=(\mathrm{d}F_{\phi(x)}\circ T_x\phi)(X(x))=\mathrm{d}F_{\phi(x)}\left(\frac{\partial}{\partial x_1}_{\big\vert\phi(x)}\right)=\frac{\partial F}{\partial x_1}_{\big\vert\phi(x)}=g(x).$$ In addition, if $g(p)=0$, then one has $\mathrm{d}f_p=\mathrm{d}F_0\circ T_p\phi=0$. Whence the result. $\Box$

Remark. When I say that $(U,\phi)$ a chart aroud $p$, I mean $\phi(p)=0$.

Remark. Let $\|\cdot\|$ be the product norm on $\mathbb{R}^n$, since $0\in\phi(U)$ is open, there exists $\varepsilon>0$, s.t. $B(0,\varepsilon)\subset U$. Let $x\in B(0,\varepsilon)$ and $s\in[\min(0,x_1),\max(0,x_1)]$, then notice that $|s|\leqslant|x_1|$, so that one has: $$(s,x_2,\ldots,x_n)\in B(0,\varepsilon).$$ Therefore, shrinking $U$ to $\phi^{-1}(B(0,\varepsilon))$ establishes the technical assumption $(\star)$.

For all $t\in[0,1]$, since $R_{\alpha_t}(0)\neq 0$ (for example: $R_{\alpha_t}(0)\not\in\xi_0$) applying the result to the map $-\dot{\alpha_t}(R_{\alpha_t})$, there exists $U_t$ an open neighborhood of $0$ in $\mathbb{R}^{2n+1}$ and a map $H_t\colon U_t\rightarrow\mathbb{R}$ such that: $$\dot{\alpha_t}(R_{\alpha_t})+\mathrm{d}H_t(R_{\alpha_t})=0.$$ To conclude, one has to see that there exists a single neighborhood on which all $H_t$, $t\in[0,1]$ are defined. For all $t\in[0,1]$, let us define the following time of existence: $$\varepsilon_t:=\sup\{\varepsilon>0\textrm{ s.t. }B(0,\varepsilon)\subset U_t\},$$ then $t\mapsto\varepsilon_t$ is a lower semicontinuous function and by compacity of $[0,1]$, one has $\varepsilon:=\inf\limits_{t\in[0,1]}\varepsilon_t>0$. Finally, for all $t\in[0,1]$, $H_t$ is defined on $B(0,\varepsilon)$.

  • $\begingroup$ So the assumption of having no closed orbit isn't needed? $\endgroup$ – Chris Kuo Apr 26 '19 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.