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.
