I am studying analysis on manifolds, more specifically differential forms.
I have to show that form $\alpha = \cos(r)dz + r\sin(r)d\theta$ on $\mathbb{R}^3$ is well defined around the origin, where $(r, \theta, z)$ are cylindrical coordinates. I tried the following:
\begin{align}
\theta =& \arctan\Big(\frac{y}{x} \Big) \Rightarrow d\theta = -\frac{y}{x^2 + y^2} dx + \frac{x}{x^2 + y^2} dy \\
r =& \sqrt{x^2 + y^2} \Rightarrow dr = \frac{x}{\sqrt{x^2+y^2}}dx + \frac{y}{\sqrt{x^2+y^2}}dy \\
z =& z \qquad \quad\Rightarrow dz = dz
\end{align}
I inserted these into $\alpha$ I get:
\begin{align}
\alpha = \cos\big(\sqrt{x^2+y^2}\big) dz – \frac{y\sin(x^2 + y^2)}{\sqrt{x^2+y^2}} dx + \frac{x\sin(x^2 + y^2)}{\sqrt{x^2+y^2}} dy
\end{align}
Taking the limit $(x,y)\rightarrow (0,0)$ I get $\alpha(0,0)$ = dz.
Is this ok? I does not look ok, because then I could just set $r = \theta = 0$ and get the same?
And a seperate question, to determen the ker($\alpha$) do I have to solve:
\begin{align}
cos(r)\frac{df}{dz} = 0 \\
r\sin(r)\frac{df}{d\theta} = 0
\end{align}
I am just having some trubles understanding general forms on manifolds, so I would be grateful for any tips or an explanation, not necessarily a solution.
Best Answer
I will assume that you've agreed that $r^2d\theta$ extends continuously across $r=0$ (i.e., the $z$-axis), even though $\theta$ is not well-defined at $r=0$.
To find $\ker\alpha$, you need to find at each point $(r,\theta,z)$ the (tangent) vectors $v$ so that $\alpha(v)=0$. If $v=a\dfrac{\partial}{\partial r}+b\dfrac{\partial}{\partial \theta}+c\dfrac{\partial}{\partial z}$, this means we want all $(a,b,c)$ satisfying $$(r\sin r)b + (\cos r)c = 0.$$ Away from $\cos r = 0$, this is given by the $2$-plane $c=-(r\tan r)b$ in $(a,b,c)$-space.
The interesting question (which I am guessing you're not ready to tackle yet) is whether the differential equation $\alpha = 0$ is integrable. That is, are there surfaces whose tangent plane at $p\in\Bbb R^3$ are precisely given by $\alpha(p)=0$? The necessary and sufficient condition for (local) integrability is given by the Frobenius condition $$\alpha\wedge d\alpha = 0.$$ In this case, $\alpha\wedge d\alpha = (\sin r+\cos r(\sin r+r\cos r)) dr\wedge d\theta\wedge dz$, which is nonzero almost everywhere. Thus, this differential equation is not integrable.