Curl of Berry connection If $|n\rangle=|n( \textbf{R}(t) ) \rangle  $ satisfies the equation $$H(\textbf{R}(t))|n(\textbf{R}(t)) \rangle = E_{n}(\textbf{R}(t))|n(\textbf{R}(t))\rangle$$ then the berry phase $\gamma_{n}(t)$ for a closed path in the parameter space is given by: $$\gamma_{n}(t) = i\oint \langle n|\nabla_{\textbf{R}}|n\rangle d\textbf{R} =  i\iint(\nabla_{\textbf{R}} \times \langle n|\nabla_{\textbf{R}}|n\rangle) d\vec{S} = i \iint\sum\limits_{n\neq m}\frac{\langle n|(\nabla_{\textbf{R}} H)|m\rangle \times \langle m|(\nabla_{\textbf{R}}H)|n\rangle}{(E_{m}-E_{n})²}d\vec{S}$$
Im having trouble understanding the last equality. How does one get $$\nabla_{\textbf{R}} \times \langle n|\nabla_{\textbf{R}}|n\rangle = \sum\limits_{n\neq m}\frac{\langle n|(\nabla_{\textbf{R}} H)|m\rangle \times \langle m|(\nabla_{\textbf{R}}H)|n\rangle}{(E_{m}-E_{n})²}~?$$
 A: This identity is extremely useful - it's much more convenient to differentiate the Hamiltonian as a function of its parameters than it is to differentiate the corresponding eigenstates - but it is not trivial, so you shouldn't feel bad for not seeing it immediately.
We start by assuming that the Hamiltonian $H$ and its eigenstates $|n\rangle$ and eigenvalues $E_n$  depend on some set of parameters $\mathbf R$.  If this is the case, then
$$\nabla_\mathbf R(H|n\rangle) = (\nabla_\mathbf R H)|n\rangle + H\nabla_\mathbf R|n\rangle$$
$$ = \nabla_\mathbf R(E_n|n\rangle) = (\nabla_\mathbf RE_n)|n\rangle + E_n\nabla_\mathbf R |n\rangle$$
and so
$$ (H-E_n) \nabla_\mathbf R|n\rangle= \big[\nabla_\mathbf R(H-E_n)\big]|n\rangle$$
Note that the operator $(H-E_n)$ annihilates any component of $\nabla_\mathbf R|n\rangle$ which is proportional to $|n\rangle$.  To find the orthogonal component, we can apply the projection operator $P=\sum_{m\neq n} |m\rangle\langle m|$ to both sides, and then safely invert $(H-E_n)$ to obtain
$$\nabla_\mathbf R| n\rangle =\boldsymbol\alpha(\mathbf R)|n\rangle + (H-E_n)^{-1}\sum_{m\neq n} |m\rangle\langle m| \big[\nabla_\mathbf R(H - E_n)]|n\rangle$$
$$ =\boldsymbol \alpha(\mathbf R)|n\rangle + \sum_{m\neq n} |m\rangle \frac{\langle m|\big[\nabla_\mathbf R(H-E_n)\big] |n\rangle}{E_m-E_n}$$
$$ = \boldsymbol\alpha(\mathbf R) |n\rangle + \sum_{m\neq n} |m\rangle \frac{\langle m|(\nabla_\mathbf R H)|n\rangle}{E_m-E_n}$$
for some (vector-valued!) $\boldsymbol\alpha(\mathbf R)$.  Finally, we note that
$$\nabla_\mathbf R \times \langle n|\nabla_\mathbf R |n\rangle = \big(\nabla_\mathbf R \langle n|\big) \times \big(\nabla_\mathbf R |n\rangle\big) + \langle n| \nabla_\mathbf R \times \big(\nabla_\mathbf R |n\rangle\big)$$
$$=\sum_{m\neq n} \frac{\langle n|(\nabla_\mathbf R H)|m\rangle \times \langle m | (\nabla_\mathbf R H)|n\rangle}{(E_m-E_n)^2} + 0$$
I've skipped quite a large amount of algebra here, but that is an outline of the calculation.

Note also that we made several assumptions - most importantly that $|n\rangle$ is non-degenerate, so $E_n \neq E_m$ for all $n\neq m$.  If there is degeneracy, then we are going to have a problem, and indeed this is what happens when e.g. the band gap of some topological material closes.
As an example, the Chern number of a system is equal to the integral of the Berry curvature (summed over all occupied bands) and is proportional to the system's Hall conductivity. If we vary the parameters $\mathbf R$ slowly, then the Berry curvature (and therefore the Chern number) should change continuously.  However, since the Chern number can be shown to be an integer, the only way to change it continuously is to not change it at all, and so the Hall conductivity is topologically protected.
This goes out the window if, as we change the parameters $\mathbf R$, we close the band gap, thereby introducing energy degeneracy.  When the gap re-opens, it may have a different (but still topologically protected) Chern number and Hall conductivity; this is an example of a topological phase transition.
