Consider the "side of the cylinder" defined by ${x}^{2}+{y}^{2}=1$.
I want to make an 'incomplete derivative' (on $\mathbb{R}^{3}$ ) but having potential only on the side of the cylinder.
I probably make a closed 1-form only on the side of the cylinder (See (1-7)). Perhaps theoretically, if the $\omega $ is a 1-form on $\mathbb{R}^{3}$ , it will have potential $U$ on the side of cylinder if it is closed on the cylinder (Am I correct?) Here, the potential $U$ is a scalar-valued function on side of cylinder such that, $$dU = \omega . \tag {1-1}$$
【My Questions】
(1) Can I make an incomplete derivative (on $\mathbb{R}^{3}$)having potential only on the side of the cylinder? ( Can we make a one-form that can define the potential only on the side of the cylinder?)
(2) Following (1-7) might be the closed 1-form only on the "side of the cylinder" defined by ${x}^{2}+{y}^{2}=1$.
$$\omega = - \frac{1}{3} {y}^{3}dx - (\frac{1}{3}{x}^{3}-x) dy + dz \tag {1-7}$$ (3) If so, the (1-7) have a potential on the side of the cylinder? How can I express $U$ as a formula?
【Details】
Let ${f}_{1}(x,y)$ and ${f}_{2}(x,y)$ be a bivariate scalar-valued function and $\omega$ be a one-form defined by
$$\omega :={f}_{1}(x,y)dx + {f}_{2}(x,y)dy + dz \tag {1-2}$$
Then, $$d\omega := \frac{\partial {f}_{1}}{\partial y} - \frac{\partial {f}_{2}}{\partial x} dx\wedge dy \tag {1-3}$$
Therefore, if there are ${f}_{1}(x,y)$ and ${f}_{2}(x,y)$ satisfies the following equation, the $\omega $ should not be a closed-form on $\mathbb{R}^{3}$ but, closed on ${x}^{2} +{y}^{2}=1$ $$\frac{\partial {f}_{1}}{\partial y} - \frac{\partial {f}_{2}}{\partial x} = 0 \ \ , iff\ \ {x}^{2} +{y}^{2}=1 \tag {1-4}$$
For example, if there are ${f}_{1}(x,y)$ and ${f}_{2}(x,y)$ satisfies followings, then, these will satisfy (1-4). $$\frac{\partial {f}_{1}}{\partial y} = {y}^{2} \tag {1-5}$$ $$\frac{\partial {f}_{2}}{\partial x} = {x}^{2}+1 \tag {1-5'}$$
For example, if we define as follows, these seems to satisfy both (1-5) and(1-5'). $$ {f}_{1} := \frac{1}{3} {y}^{3} \tag {1-6}$$ $$ {f}_{2} := - \frac{1}{3}{x}^{3} + x \tag {1-6'}$$ Using (1-6) and(1-6)', $$\omega = - \frac{1}{3} {y}^{3}dx - (\frac{1}{3}{x}^{3}-x) dy + dz \tag {1-7}$$
and then, $$d\omega = ({x}^{2}+{y}^{2}-1) dx\wedge dy \tag {1-8}$$
This might be closed only on the ${x}^{2}+{y}^{2}=1$.
Is it correct?
P.S. I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions. I welcome any corrections and English review. (You can edit my question and description to improve them)
The cylinder is not simply connected, so even if the restriction of $\omega$ to the cylinder is closed, it need not be exact. There is no issue with $dz$ in your example, of course. But necessary and sufficient for $P(x,y)\,dx + Q(x,y)\,dy$ to be exact on the unit circle is that its integral over the circle evaluate to $0$. Yours does not. The form, restricted to the cylinder, cannot be exact.
EDIT: In fact, every $1$-form $\omega = P(x,y)\,dx + Q(x,y)\,dy + R(z)\,dz$ will be closed on the cylinder. The best way to convince yourself is to use coordinates $(\theta,z)$ on the cylinder. In these coordinates, any $1$-form $\omega = f(\theta)\,d\theta + g(z)\,dz$ is closed, and you can work much better in these coordinates to determine when it's exact.