Q:

How can I infer the topology of a quantum state (or band) from its Chern number?

Whereas I can calculate the Chern number of a quantum state (or band) from the integration of the Berry curvature in all space.
How can I infer the topology of the quantum state from this result? What is the physical meaning of a quantum state with non-zero Chern number?

A:

I am in part trying to understand this myself.  The Berry phase is computed from differential forms, such as the one-forms $\omega$ constructed from states
$$
\omega~=~\langle\psi|d\psi\rangle
$$
and with the covariant differential $D~=~d~+~\omega$ the two-forms
$$
\Omega~=~D\omega~=~d\omega~+~\omega\wedge\omega
$$
The tensor components of the 2-form $F$ are elements of a self-adjoint principal bundle $P$.  The determinant of these elements
$$
det\Big|1~+~\frac{ixF}{2\pi}\Big|~=~\sum_nc_jx^n
$$
which is a characteristic polynomial which represents the Chern class.  Each $c_n(P)$ is an element of $H^{2n}(M)$.  So the curvature form for the Berry phase, or the Fubini-Study metric $\Omega=~dz\wedge d{\bar z}/(1~+~|z|^2)^2$ is evaluated $\int\Omega~=~2\pi i$ and gives $c_1~=~1$  So there is a nontrivial cocycle on the “2-level.  For this projective geometry there are alternating Betti numbers $1,~0$ for even and odd.
If you had some product of states $\prod_n |\psi_n\rangle$, say in an entangled state etc, you could apply the differential $d$ up to $n$ times and form and $n$-form.  For instance the product $|\psi_1,~\psi_2\rangle$ $=~|\psi_1\rangle|\psi_2\rangle$ defines the one-form
$$
\omega~=~d|\psi_1,~\psi_2\rangle~=~d|\psi_1\rangle|\psi_2\rangle~+~|\psi_1\rangle d|\psi_2\rangle
$$
and one could then build up a system of differential forms on various chains.  The analogue of the projective geometry for this is a $G_2(V)$ Grassmannian and this continues up for n-product spaces.

