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Let $X$ be the Hopf vector field on the three-sphere. Is there a smooth nowhere zero function $f$ so that the modified vector field $fX$ is not the Reeb vector field of any contact form on the three-sphere?

This question is a follow-up on aglearner's interesting question on how to construct volume-preserving flows on the three-sphere that are not Reeb vector fields.

I'm also looking for evidence to support the following guess: If $X$ is a vector field on the three-sphere, there exists a nowhere zero smooth function $f$ so that the vector field $fX$ is not a Reeb vector field for any contact form on the three-sphere.

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  • $\begingroup$ If you can make sure that $fX$ has at most one closed orbit, you violate Cristofaro-Gardiner--Hutching's refinement of Taubes' theorems (a.k.a. the Weinstein conjecture). $\endgroup$ – Marco Golla Jan 17 '17 at 0:59
  • $\begingroup$ @MarcoGolla, I think Reeb flows are really special and I'm looking for easier arguments. $\endgroup$ – alvarezpaiva Jan 17 '17 at 16:59
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The answer is yes.

Proposition. Let $\alpha$ be the standard contact form on the three-sphere (for which the Reeb vector field is the Hopf vector field $X$). If $f$ is a strictly positive function on $S^3$ such there exist two Hopf circles $\gamma_1$ and $\gamma_2$ for which $$ \int_{\gamma_1} f^{-1} \alpha \neq \int_{\gamma_2} f^{-1} \alpha \ , $$ then the vector field $fX$ is not the Reeb vector field of any contact form on $S^3$.

The key idea is that all solutions of $\dot{x}(t) = f(x(t))X(x(t))$ are periodic and that if $fX$ were a Reeb vector field, all these solutions would have the same period.

Indeed, assume $fX$ is the Reeb vector field of a contact form $\beta$. If $\gamma_1$ and $\gamma_2$ are integral curves of $fX$ with periods $T_1$ and $T_2$, then $$ T_2 - T_1 = \int_{\gamma_2} \beta - \int_{\gamma_1} \beta = \int_\Sigma d\beta \ , $$ where $\Sigma$ is a cylinder with boundary $\gamma_2 - \gamma_1$ and foliated by Hopf circles (I'm being a bit sloppy with the word foliated). It follows that this last integral is zero and so $T_1 = T_2$.

Now we just have to notice that the period of $\gamma_i$ ($i = 1, 2$) equals the integral of $f^{-1}\alpha$ over $\gamma_i$: $$ \int_{\gamma_i} f^{-1} \alpha = \int_0^{T_i} f^{-1}(\gamma_i(t))\alpha(\dot{\gamma_i}(t)) \, dt = \int_0^{T_i} f^{-1}(\gamma_i(t))\alpha(f(\gamma_i(t)) X(\gamma_i(t))) \, dt = T_i $$

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