Can a vector field be conservative if its domain is not a star domain? 
Can a vector field be conservative if its domain is not a star domain?

I was trying to figure out whether the vector field
$$\vec{f}(\vec{x}):=\frac{1}{\lvert \lvert \vec{x} \rvert \rvert} \begin{pmatrix}x-2y\\x+2y\end{pmatrix}$$
is conservative on $\Bbb R^2 / \text{{$\vec{0}$}}$.
I used the criterion:
$$\vec{f}\space \text{is conservative} \iff \frac{dP}{dy}=\frac{dQ}{dx} \space \text{and the domain is a star domain}$$
In this case $\frac{dP}{dy}\not=\frac{dQ}{dx}$ so the test fails already but I was wondering if I could have also argued that the domain is not a star domain so the field is not conservative.
Can a vector field ever be conservative if its domain is not a star domain?
What is so special about star domains that makes them so essential when dealing with conservative vector fields?
 A: Usually the definition of a conservative vector field is this:
Let $\vec{f}$ be a vector field in an open set $\Omega\subset\mathbb R^2$.
It is called conservative if there is a scalar function $\Phi:\Omega\to\mathbb R$ so that $\vec f=\vec\nabla\Phi$.
This definition makes no assumptions about the geometry.
Then we have these two theorems:


*

*If $f$ is conservative, then $\partial_1f_2=\partial_2f_1$.

*If $\Omega$ is simply connected, then $\partial_1f_2=\partial_2f_1$ implies that $f$ is conservative.


Star-shaped domains are simply connected, but the assumption is more general than that.
Using the first theorem we can see that since $f$ fails the conclusion, it must also fail the assumption; that is, $f$ is not conservative.
The special thing about simply connected domains is that they have no holes: you cannot take a path around a point outside $\Omega$.
It is the existence of such paths that causes theorem 2 to fail if $\Omega$ is not simply connected.
(There is also a measure for how badly it fails for a given domain, namely the first de Rham cohomology group.)
Star-shaped domains are nice because you can find a very simple path between any two points (through the center).
In more general simply connected domains you need more complicated paths but the basic idea is the same.
(Similar ideas work in higher dimensions as well, but I restricted myself to the plane for simplicity.)
A: The criterion would imply that the answer to your question is no, but the statement as written is incorrect. A correct modification of the statement is:

If a vector field ${\bf F} = (P, Q)$ defined on the open subset $U \subseteq \Bbb R^2$ satisfies $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ and the domain $U$ of $\bf F$ is star-shaped, then $\bf F$ is conservative.

As the question observes, neither of the hypotheses here hold, so we can't conclude anything from this statement alone.
On the other hand, at least when $\bf F$ is $C^2$, the equality $\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}$ is a necessary condition for conservativeness: If $\bf F$ is conservative, there is a function $f$ (called a potential) such that
$${\bf F} = \nabla f := \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right).$$
Then, by Clairaut's Theorem (the equality of mixed partials of sufficiently nice functions of two variables) we have
$$\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}\left(\frac{\partial f}{dx}\right) = \frac{\partial^2 f}{dy \, dx} = \frac{\partial^2 f}{dx \, dy} =\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right) = \frac{\partial Q}{\partial x}.$$ So, if $\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}$, then $\bf F$ cannot be conservative.
It is certainly possible for a vector field on a nonstarshaped domain to be conservative. To construct an example, simply pick any nonstarshaped (open) set $U \subseteq \Bbb R^2$ and any $C^1$ function $f : U \to \Bbb R^2$. Then, $\nabla f$ is conservative.
