# Berry's phase and solid angle

I've hit a wall with the Berry's phase.

So here's the thing:

• I know Berry's phase is defined as $$\gamma_C = \int_S \mathcal F d^2\mathbf k$$
• I'm considering the 2D $H$ of $$H = \left(\begin{matrix}d_z & d_x - id_y \\ d_x + id_y & -d_z\end{matrix}\right) = \hat {\mathbf d}(\mathbf k) \cdot \vec\sigma$$ where $\hat{\mathbf d}$ is essentially a point on $S^2$.

What I don't understand is why it now automatically follows that $$\mathcal F = \frac12 \epsilon_{ij}\hat{\mathbf d} \cdot (\partial_{i}\hat{\mathbf d} \times \partial_j\hat{\mathbf d}).$$

Hopefully somebody can clarify, in particular a geometric argument would be nice. The method I'm reviewing now states,

The Berry curvature is given by the solid angle per unit area in $\mathbf k$ space, which is simply half the solid angle element for the mapping $\mathbf {\hat d}(\mathbf k)$.

Edit: Maybe this additional information is useful: $\mathcal F = \nabla \times \mathbf A$ with $$\mathbf A = -i\left< u(\mathbf k)\right| \nabla \left| u(\mathbf k)\right>.$$ with $\left|u(\mathbf k)\right>$ the cell periodic eigenstate of the Bloch Hamiltonian

So in the original Berry paper, the following identity is derived: $$\mathcal F = \operatorname{Im} \frac{\left<+\right|\nabla\hat H\left|-\right> \times \left<-\right|\nabla\hat H\left|+\right>}{(E_+ - E_-)^2}$$ for the two state system. (Please refer to the paper from '84 for a more detailed overview hereof.)
The original $\hat H$ had a factor $\frac12$ in front of it, such that $\nabla \hat H = \frac12 \vec \sigma$. Using the identity $\left<\pm\right|\vec\sigma\left|\mp\right> = \hat{\mathbf i} \mp i\hat{\mathbf j}$ after rotation of axes rotated alongside $\mathbf d$, the resulting outer product is $2i\hat{\mathbf k}$, of which the imaginary part is obviously $2\hat{\mathbf k}$. This leaves a factor $\frac14$ from the previous expression for $\nabla\hat H$ to be plugged in, and the fact that $E_+ = -E_- = \frac12\lvert\mathbf d\rvert$ (just calculate when the determinant of $\hat H$ above is zero). Apparently cancelling the axis rotation adds in a factor $\hat{\mathbf N} = \frac{\mathbf d}{d}$ (this is the part I can't quite figure out). The result is $$\mathcal F = \frac{2\mathbf {\hat k}}{d^2} \cdot \frac14 \cdot \frac{\mathbf d}{d} = \frac{\mathbf d}{2d^3}$$
Note that the general expression for $\gamma_C = \frac12 \Omega(\mathbf d)$ quickly follows.
Lastly, why $\gamma_C$ must be a multiple of $2\pi$ from this, is unclear to me...