Consider in $\Bbb S^2$ the symplectic form $\omega \in \Omega^2(\Bbb S^2)$ given by $\omega_p(v,w) = \langle p, v \times w\rangle$ (that is, the usual area form). If $f \in \mathcal{C}^\infty(\Bbb S^2)$ is positive, then $\omega^f \doteq f\omega$ is also a symplectic form.

I would guess that there is a smart way to find coordinates $(\theta,\varphi)$ in $\Bbb S^2$ so that $\omega^f = {\rm d}\theta \wedge {\rm d}\varphi$, but I'm a bit at a loss of how to do that.

Attempt: (?) For $f = 1$, $\omega^f = \omega$ is the usual area form, and $x = \sqrt{1-z^2} \cos \theta$, $y = \sqrt{1-z^2}\sin \theta$ does the job. So I tried calling $$\begin{cases} x(\theta,\varphi) \doteq g(\varphi) \cos \theta \quad{\rm and} \\ y(\theta, \varphi) = g(\varphi) \sin \theta, \end{cases}$$with no restriction on $z = z(\theta,\varphi)$, a priori.

The condition $g(\varphi)^2 + z(\theta,\varphi)^2 = 1$ gives $\partial z/\partial \theta = 0$. With this I painstakingly computed the pull-back by expressing ${\rm d}x, {\rm d}y$ and ${\rm d}z$ in terms of ${\rm d}\theta$ and ${\rm d}\varphi$, and substituting in $$\omega^f = fx\,{\rm d}y \wedge {\rm d}z + fy\,{\rm d}z\wedge{\rm d}x + fz\,{\rm d}x \wedge {\rm d}y,$$and got the relation $$f g^2 \frac{\partial z}{\partial \varphi} - fz gg' = 1.$$Since the condition $g(\varphi)^2 + z(\theta,\varphi)^2 = 1$ also gives $gg' + z \partial z/\partial \varphi = 0$, that simplifies to $$f\frac{\partial z}{\partial \varphi} = 1.$$

This smells bad.

Is my attempt too wrong? What is the intelligent way to solve the problem?

It is exercise 5.1.1 here in case you're too curious. I'm in that stage of learning that you go around scavenging for nice exercises.


Let us consider the chart $$\mathbf x:(u,v)\mapsto \bigl(\cos (v) \cos (f(u)),\sin (v) \cos (f(u)),\sin (f(u))\bigr).$$ We evaluate the standard area form $\omega$ at the coordinate vectors corresponding to this chart: \begin{align} \omega(\partial_u,\partial_v)&=\det\begin{pmatrix}\cos (v) \cos (f(u)) & \cos (f(u)) \sin (v) & \sin (f(u)) \\ -\cos (v) \sin (f(u)) f'(u) & -\sin (v) \sin (f(u)) f'(u) & \cos (f(u)) f'(u) \\ -\cos (f(u)) \sin (v) & \cos (v) \cos (f(u)) & 0 \end{pmatrix} \\ &=-f'(u) \cos (f(u)). \end{align} It follows from this that $$\omega=-f'(u)\cos(f(u))du\wedge dv.$$ If we want this $\omega$ to be in canonical form in this chart, we need that $$(\sin(f(u)))'=f'(u)\cos(f(u))=-1.$$ Of course, for this we need that $\sin(f(u))=-u+c$ for some $c$ and then $$f(u)=\arcsin(u-c).$$ We this see that the chart $$(u,v)\mapsto\left(\sqrt{1-(c-u)^2} \cos (v),\sqrt{1-(c-u)^2} \sin (v),u-c\right)$$ turns the standard volume form into canonical form.

This worked because the standard volume form is rotationally invariant, so we knew that we could pick a chart with the same symmetry to take it to canonical form. Suppose instead that we consider the form $\omega^h=h\omega$, with $h$ any function on the sphere.

We pick the chart $$\mathbf y:(u,v)\mapsto\bigl(\cos (f(u)) \cos (g(v)),\cos (f(u)) \sin (g(v)),\sin (f(u))\bigr).$$ Then we have $$\omega^h(\partial_u,\partial_v)=-h(u,v)f'(u) \cos (f(u)) g'(v),$$ so that $$\omega^h=-h(u,v)f'(u) \cos (f(u)) g'(v)\,du\wedge dv.$$ Here we are writing $h(u,v)$ the value of the function $h$ on the point $\mathbf x(u,v)$, of course. If we want the chart to put $\omega^h$ into canonical form, we need that $$h(u,v)f'(u) \cos (f(u)) g'(v)=1.$$ The interesting thing is that now it is not true that there is always a way to pick $f$ and $g$ so that this holds! That means that no chart of the form that we picked works, and we have to try another candidate.

To deal with the general case, we need what is essentially the general chart: consider now $$\mathbf z:(u,v)\mapsto \bigl(\cos (f(u,v)) \cos (g(u,v)),\cos (f(u,v)) \sin (g(u,v)),\sin (f(u,v))\bigr).$$ A computation (I am using Mathematica to do all this, so it takes a few milliseconds!) shows that $$\omega^h(\partial_u,\partial_v)=h(u,v)\left(f_v(u,v) g_u(u,v)-f_u(u,v) g_v(u,v)\right)\cos (f(u,v)),$$ so to find a chart that canonicalizes $\omega^h$ we have to solve the PDE $$\cos(f)\begin{vmatrix}f_v&f_u\\g_v&g_u\end{vmatrix}=\frac{1}{h}.$$ At least locally, we can surely take $g(u,v)=u$, and then the equation becomes $$\frac{\partial}{\partial v}(\sin(f(u,v))=\cos(f(u,v))f_v(u,v)=\frac{1}{h(u,v)},$$ and one can solve this.

  • $\begingroup$ I proposed a coordinate chart of that form, and was lucky and found that for some choice of $f$ what we want actually happens. $\endgroup$ – Mariano Suárez-Álvarez Apr 3 '18 at 3:48
  • $\begingroup$ Thanks for taking the time to answer, but I'm not sure I follow. It seems that your $f$ is not the same as mine (I don't see why couldn't $f$ also depend on $v$). Or your $f$ is something to1be found, like the $g$ in my attempt? $\endgroup$ – Ivo Terek Apr 3 '18 at 3:48
  • $\begingroup$ No, your $f$ is not the same as mine. I trust you can translate from one notation to the other. Thi first part is just a way to find a chart that canonicalizes the usual volume form (it will work for other volume forms with rotational symmetry) The second part shows that that method does not necessarily work for all volume forms. $\endgroup$ – Mariano Suárez-Álvarez Apr 3 '18 at 3:49
  • $\begingroup$ I hadn't seen the edit. It's almost 1 a.m. local time and I'll get up early. I'll read it again in a few hours when I'm better rested and I'll get back to you. Thanks again! $\endgroup$ – Ivo Terek Apr 3 '18 at 3:51
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    $\begingroup$ I think your expectations were quite misguided! $\endgroup$ – Mariano Suárez-Álvarez Apr 4 '18 at 5:00

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