Gauge transformations and observables
The phase assigned to an eigenvector is conventional; measurable holonomy and curvature are not.
$$\mathcal A\mapsto\mathcal A-\nabla_k\phi$$
What this means
The phase assigned to an eigenvector is conventional; measurable holonomy and curvature are not.
Deep dive: mathematical and physical mechanism
With $\mathcal A=i\langle u|\nabla_k u\rangle$, the gauge change $|u\rangle\mapsto e^{i\phi}|u\rangle$ gives $\mathcal A\mapsto\mathcal A-\nabla_k\phi$. The curvature $\Omega=\nabla_k\times\mathcal A$ is invariant. Berry phases around closed loops are gauge invariant modulo $2\pi$.
How it appears in quantum phenomena
Gauge-invariant Berry curvature enters anomalous electron velocity.
Assumptions, caveats and limitations
A gauge choice can be singular without implying a physical singularity.
Connect the ideas
For context, see the Foundations of Topology topic guide and the reading references.