Why the integral of Berry curvature over a closed surface is not zero?

I read [1,2] that for a spin-1/2 particle under magnetic field, the Berry curvature is a monopole, $$\mathbf F_{\pm} = \mp\frac{\mathbf B}{2B^3},$$ of which the integral over a closed surface is $$2\pi$$.

Because the Berry curvature is defined by the curl of Berry connection $$\mathbf F = \nabla_B\times\mathbf A$$, the divergence of a curl $$\mathbf A\Rightarrow\nabla_B\cdot \mathbf F =0$$ must be zero, and so is the aforementioned integral $$\oint_{S_B}\mathbf F\cdot d\mathbf S_B=\int_{V_B}\nabla_B\cdot\mathbf F =0.$$

Is there anything missing here?

Edit: Both daniel and Sebastian answer my problem. Another discussion is also helpful to me.

• Hint: Remember what you learnt about point charges. – Sebastian Riese Sep 25 '18 at 17:23
• @SebastianRiese Well, I learned that the point-charge solution in electromagnetism can be expressed as the divergence of a scalar potential, but never as the curl of a vector potential. But this rule seems to fail here. I edited my problem for clarity. – Toto Sep 26 '18 at 8:20
• Note (from Wikipedia) that the divergence theorem applies to continuously differentiable functions, i.e. functions whose derivative is continuous. Is ∂F continuous in your integration region? – danielsmw Sep 26 '18 at 12:31
• Also, the relation $\vec F = \nabla \times A$ breaks down at $B = 0$ (it only holds in the non-degenerate) – Sebastian Riese Sep 26 '18 at 16:21
• (.. case). My hint was supposed to mean, that $\nabla \cdot \vec F \propto \delta(\vec B)$, since the field is the same as that of a point charge. And $\nabla \cdot (\nabla \times \vec A) = 0$ simply does not hold in general at a singularity. – Sebastian Riese Sep 26 '18 at 16:29