quantum mechanics - Basic question on the Aharonov-Bohm effect

I have a very basic question on the Aharonov-Bohm effect.

Why is the curve integral $\oint_\Gamma {A}\cdot d{r}$ non-zero ? $\Gamma$ is the "difference" of both paths $P_1$ and $P_2$. If the magnetic field is limited to the interior of the solenoid $\operatorname{curl} {A}=0$ along the integral path $\Gamma$, so I can conclude that I can write ${A}=\nabla f$. A closed curve integral of a gradient function is zero.

I guess it is related with a possible singularity of $A$ in the very center of the solenoid.

Nevertheless if I travel around a point-like source of a gravitational field and compute the integral $\oint_\Gamma {F}\cdot d{r}$ where $F=-\nabla V(r)$ the closed curve integral over an conservative force field is certainly zero, whereas $V(r)$ even has a singularity and $F$ consequently too. I would be very grateful for an explanation.

Answer

From $$\operatorname{curl} A = 0 \tag{1}$$ in a region $U$, you can in general not conclude that $$A = \nabla f \tag{2}$$ for some function $f$ defined on all of $U$. Indeed this is related to the singularity, which removes a line through the origin. The degree to which (1) fails to imply (2) depends on the topology of $U$, more specifically it's de Rham cohomology. (1) implies (2) iff the de Rham cohomology is trivial. For $U = \mathbb{R} \times (\mathbb{R^2} \setminus \{ 0 \} )$, it turns out that the cohomology is not trivial, and hence your integral need not vanish.

To show that (1) does not imply (2) for 3-space with a line removed, we can assume that the line is the $z$-axis and, consider $$A = \begin{cases} \nabla \operatorname{arctan2} (y,x) & (x,y) \neq (0,-1) \\ (0, -1, 0 ) & (x,y) = (0,-1) \end{cases}$$ where also $(x,y) \neq (0,0)$. $A$ is smooth, and $\operatorname{curl} A = 0$, but you cannot find a function $f$, continuous away from the $z$-axis, such that $\nabla f = A$ everywhere.

It less trivial (or rather, one first needs to construct some sledgehammers, and then it's a 2-line proof) to show that for 3-space with a point (1) does indeed imply (2).

---

Update: If you want to understand why a point can be removed but not a line, one can think of it as needing to make a hole that is big enough. Suppose we have two curves $\gamma_1$ and $\gamma_2$ and $\int_{\gamma_i} A$, where $\operatorname{curl} A = 0$. If $\gamma_1$ can be deformed to $\gamma_2$ one can apply Stokes's theorem to argue that the integrals are equal. Suppose that the space is such that any curve can be deformed into any other curve. Then a standard prescription exists to find $f$: pick any point $x_0$ and let $$f(x) = \int_\gamma A$$ where $\gamma$ is any curve connecting $x_0$ and $x$. By the assumption the integral doesn't depend on the particular $\gamma$ chosen so $f$ is unambiguously defined. To make the argument fail you have to make a "hole" big enough that curves can't be deformed into each other. Removing a point isn't enough in three dimensions, but a line is: make a loop around the line singularity, you can't get it to not encircle the line without dragging it across the singularity. (In 2 dimensions a point is enough, because you can consider the projection of this example onto a plane.)