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Let us consider spherical coordinates $(r,\theta,\phi)$ on $\mathbb R^3$ and the manifold $M:=\mathbb R^3 \setminus \{0\}$.

Let us consider the 1-form on $M$ $$ \omega = zdz -r\sin^2{\theta}dr-\frac{1}{2}r^2\sin{2\theta}d\theta $$ where $z=r\cos{\theta}$. I think that we can write $\omega$ using only the spherical coordinates as $$ \omega = r\cos{2\theta}dr-r^2\sin{2\theta}d\theta $$ which is closed, $d\omega=0$. The ideal generated by $\omega$ is hence closed with respect to $d$, since $d(\eta \wedge \omega) = d\eta \wedge \omega$. This implies that the distribution $\Delta\subset TM$ (associated to $\omega$) is involutive, hence - by Frobenius theorem - integrable.

I have three questions:

1. I want to find the rank of the distribution and local generators of it: I think that the distribution is 2-dimensional and it is generated by $$ X = \frac{\partial}{\partial \phi} \qquad \text{and} \qquad Y=r^2\sin{2\theta}\frac{\partial}{\partial r}+r\cos{2\theta}\frac{\partial}{\partial\theta}. $$ Is it correct?

2. I want to find an integral variety of this distribution: I have found that the family of surfaces given by $$ \frac{r^2}{2}\cos{2\theta}-\phi+c=0, \qquad c \in \mathbb R. $$ are integral varieties of $\Delta$. Do you agree?

3. Finally, is $\Delta$ a trivial bundle over $M$? I do not know how I can do in order to answer this question. I have to establish if $\Delta \cong M \times \mathbb R$ (as vector bundles). Would you please help me?


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The two fields you gave are only defined on the open set where your coordinates are defined. What happens outside? – Mariano Suárez-Alvarez Feb 13 '13 at 22:25
@MarianoSuárez-Alvarez I'm sorry but I don't understand what you mean. Outside of what? Do I have to define the fields on all $\mathbb R^3$? I do not think so. Thanks for your comment. – Romeo Feb 13 '13 at 22:32
I mean outside of the open set on which your coordinates are defined (which is the complement of the $z$ axis) If the fields are not defined on all of your manifold, your claim that «they generate the distribution» does not mean anything, for what could it possibly mean at points where the fields are not defined? – Mariano Suárez-Alvarez Feb 13 '13 at 22:52
There also seems to be a problem defining $\omega$ near the $z$ axis. I think the form is singular there. – Harald Hanche-Olsen Feb 13 '13 at 23:10
I have edited the OP: now I have written the text exactly as it is in my paper. I am really puzzled, anyway; maybe we could consider an extension of the fields on all $M$ by a "continuity" (?) argument... – Romeo Feb 14 '13 at 10:42

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