What makes a system topological?

As I understand, if the Chern number which is obtained by integrating Berry curvature over a surface with a boundary is an integer, then the Chern number is a topological invariant. So when does Chern number become integer? If the berry phase is well defined along the closed curve will that be enough for the system to be topological?

• When you integrate the Berry curvature over the Brillouin zone (which is a torus topologically, so it is closed), you always get an integer. – Meng Cheng Sep 28 '15 at 15:13

Any system is topological. The question is, rather, whether it has non-trivial topology.

When considering translation invariant systems, we make a Bloch decomposition and so a system is specified by maps from the Brillouin zone into some space of Hamiltonians. Via the Bloch decomposition, this space of Hamiltonians will simply be the space of Hermitian $n\times n$ matrices which have a gap.

For simplicity let's assume we have no extra symmetries, and we are in two dimensions, and we are only considering two level systems. Then the Brillion zone is the $2$-torus, the space of Hamiltonians is $\mathbb{R}^3-{0}\cong S^2$ (think of the set of Hermitian invertible $2\times 2$ matrices, this space is spanned by the three Pauli matrices, with real coefficients, and such that not all three coefficients are zero), and a system is specified as a continuous map between these two spaces.

Then it turns out (after some study of topology) that these maps $$\mathbb{T}^2\to S^2$$ fall into classes indexed by $\mathbb{Z}$, where classes are defined by continuous deformations of the maps (i.e., homotopy).

For general $d$-dimensions and general $N$-level systems, the $2$-torus is replaced by a $d$-torus and the $2$-sphere is replaced by the Grassmannian manifold. Taking symmetries into account one have to then classify $G$-equivariant maps, rather than all continuous maps, which leads to a finer classification.

In this sense, a system (i.e. a map $\mathbb{T}^2\to S^2$) is topological when it is not a map homotopic to a constant map (the constant is irrelevant if your target space is path-connected).

Next, for non-translation invariant systems, one does not have a Brillouin zone anymore. So a Hamiltonian is just an operator in Hilbert space (with some conditions to make physical sense). One then classifies not a set of maps as before, but the set of possible Hamiltonians. The equivalence classes is now not homotopy as before, but path-connectedness.