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What is an example of a smooth vector field $V$ on an open set of the plane which is a geodesible vector field but there is no a conformal metric $g$ such that $V$ is geodesible vector field with respect to $g$.

A geodesible vector field is a non vanishing vector field for which there is a Riemannian metric $g$ such that all trajectories of $V$ are (unparametrized) geodesics with respect to $g$.

The motivation for this question is described here:

Limit cycles of quadratic systems and closed geodesics

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    $\begingroup$ By 'conformal metric $g$', do you mean that $g$ is conformal to the standard flat metric on the plane, i.e., that $g = e^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$ for some function $u$ on a domain in the plane? $\endgroup$ – Robert Bryant Jul 20 '17 at 11:25
  • $\begingroup$ @RobertBryant Yes. exactly. $\endgroup$ – Ali Taghavi Jul 20 '17 at 12:40
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Here is how one can construct an example: Consider the smooth, nonvanishing $1$-form $$ \omega = y^3(1{-}y)^2\,\mathrm{d}x + \big(y^3-2(1{-}y)^2\bigr)\,\mathrm{d}y. $$ Note: This $\omega$ came from Exercises 5 and 6 of Section 16 of Chapter XVIII of Volume IV of Dieudonné's Treatise on Analysis. These exercises show that $\omega$ cannot be written globally in the form $g\,\mathrm{d}f$ for two smooth functions on $\mathbb{R}^2$. (In other words, $\omega$ has no global 'integrating factor'.)

I am going to use Dieudonné's $1$-form to show that the nonvanishing vector field $$ V = y^3(1{-}y)^2\,\frac{\partial\ }{\partial x} + \big(y^3-2(1{-}y)^2\bigr)\,\frac{\partial\ }{\partial y}, $$ while geodesible, is not geodesible with respect to any conformal metric on the plane, i.e., a metric of the form $g = \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$.

First, I will show that $V$ is geodesible. To do this, it suffices to find a closed $1$-form $\phi$ such that $\phi(V)>0$. I construct $\phi$ as follows: Let $\rho \approx 0.639$ be the unique real root of $\rho^3-2(1{-}\rho)^2 = 0$. Now set $$ f(y) = \frac{\rho^3(1{-}\rho)^2-y^3(1{-}y)^2}{y^3-2(1{-}y)^2}, $$ and note that $f(y)$ is a rational function of $y$ with a quadratic denominator that never vanishes. Hence $f(y)$ is a smooth function of $y$. Now set $$ \phi = \mathrm{d}x + f(y)\,\mathrm{d}y. $$ Then $\phi$ is closed and computation yields $\phi(V) = \rho^3(1{-}\rho)^2>0$. Thus, $V$ is geodesible.

To see that $V$ not geodesible with respect to any metric of the form $g= \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2)$, write $$ V = h(y)\left(\cos\theta(y)\,\frac{\partial\ }{\partial x} + \sin\theta(y)\,\frac{\partial\ }{\partial y}\right) $$ where $h(y)>0$, and consider, for any function $u(x,y)$, the two $1$-forms $$ \eta_1 = \mathrm{e}^{u(x,y)}\bigl(\cos\theta(y)\,\mathrm{d}x{+}\sin\theta(y)\,\mathrm{d}y\bigr) \ \ \text{and}\ \ \eta_2 = \mathrm{e}^{u(x,y)}\bigl(-\sin\theta(y)\,\mathrm{d}x{+}\cos\theta(y)\,\mathrm{d}y\bigr). $$ If $V$ is to be geodesic with respect to the conformal metric $$ {\eta_1}^2+{\eta_2}^2 = \mathrm{e}^{2u(x,y)}(\mathrm{d}x^2+\mathrm{d}y^2), $$ then one must have $\mathrm{d}\eta_1 = 0$, which would imply, since $\mathbb{R}^2$ is simply connected, that $\eta_1 = \mathrm{d}v$ for some function $v=v(x,y)$ on $\mathbb{R}^2$. However, by construction, $$ \mathrm{d}v = \eta_1 = \mathrm{e}^{u(x,y)}\,\frac{\omega}{h(y)}, $$ so $\omega = \mathrm{e}^{-u(x,y)}h(y)\,\mathrm{d}v$, which is impossible by Dieudonné's Exercises. Thus, $V$ is not geodesible with respect to any conformal metric on $\mathbb{R}^2$.

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  • $\begingroup$ Thank you a lot for this very wonderful answer. $\endgroup$ – Ali Taghavi Aug 31 '17 at 13:23

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