# Conservative field defined over a not simply-connected region

I am wondering if a field can be conservative if the region where it is defined is not simply-connected. By definition $$F$$ is conservative if there exists a differentiable function which satisfies $$F=grad(u)$$ If I find such function which is defined in the not simply connected region, am I OK?

• When you ask "am I OK?" are you asking "is the integral of this field along any closed path zero"? This is a common useful property of conservative fields on simply-connected domains. – Eric Towers Jan 31 at 14:55
• Yes I am asking this. And I think that a field is conservative and the integral on every closed path is zero are equivalent statements. – spyer Jan 31 at 15:02

• It can. Why not? In fact restriction of a conservative vector field on $\mathbb R^n$ to any open subset will still be a conservative. – Blazej Jan 31 at 15:13
Yes, for example $$X=\frac{\partial}{\partial x}$$ is $$\mathrm{grad}\,u$$ on $$\mathbb{R}^2-\{(0,0)\}$$, where $$u(x,y)=x$$.