In real world there is always dissipation and therefore there is an
"arrow of time" and time-reversal symmetry does not hold (right?).
There are many phenomena in physics that we understand by invoking the time reversal symmetry of microscopic interactions. This is distinct from the "arrow of time" and the fact that an out-of-equilibrium system increases in entropy.
In terms of microscopic interactions, time reversal symmetry may be broken by an external magnetic field. This is the context in which we think of Quantum Hall states. For (most?) topological insulators, it is important that the microscopic interactions obey time reversal symmetry. This is very generally true: even if though you can see the "arrow of time" when a hot cup of coffee cools and increases the entropy of the room, the individual interactions (atoms bouncing off of each other) are time reversal symmetric. For topological insulator, the time reversal symmetry prevents edge spin currents from scattering.
If you have a solid foundation in undergraduate or graduate physics, the Review of Modern Physics articles by Hasan and Kane or Qi and Zhang are very good. For a more general audience, the Physics Today article by Qi and Zhang is intriguing.
The 'topological' in topological order means 'robust against ANY local perturbations'.
According to such a definition, topological insulator is not 'topological' since its properties are not robust against ANY local perturbations, such as the perturbation that break the U(1) and time reversal symmetry. So a more proper name for
topological insulator is 'U(1) and time-reversal symmetry protected insulator', which is one example of SPT order.
Some example of topologically ordered states (in the sense of 'robust against ANY local perturbations'):
1) $\nu=\frac{1}{3}$ FQH state
2) $Z_2$ spin liquid state
3) $\nu=1$ IQH state
4) $E_8$ bosonic QH state
The example 3) and 4) have no non-trivial topological quasi-particles (ie no non-trivial statistics, no non-trivial topological degeneracy), but have
gapless edge state that is 'robust against ANY local perturbations'.
-- Edit -- (I lifted some discussions below to here):
There are two kinds of topology in math. The "topology" in "topological order" is directly related to the first kind of topology in mathematics, as in algebraic topology, homology, cohomology, tensor category. The "topology"
in "topological order"
is different from the "topology"
in "topological insulator". The "topology"
in "topological insulator"
is related to the second kind of topology in mathematics, as in mapping class, homotopy, K-theory, etc. The first kind of topology is algebraic, while the second kind of topology is related to the continuous manifold of finite dimensions. We may also say that the first kind of topology is "quantum", while the second kind of topology is "classical".
The correct way to describe any gapped phases (such as topological orders and topological insulators) is to use the first kind of topology -- "quantum" topology, because the gapped phases are usually interacting. The second kind of topology -- the "classical" topology -- can be used to describe the one-body physics (include free fermion systems). The "classical" topology cannot be used to describe interacting many-body systems, which need "quantum topology".
One needs to go beyond "filling energy level" picture to understand topological order (the first kind of topology). Our education in traditional condensed matter physics (or traditional many-body physics) is almost all about "filling energy levels" (such as Landau Fermi liquid theory, band theory, etc), which is a trap that limit our imagination. The second kind of topology (the "topology"
in "topological insulator") can be understood within the framework of "filling energy level" picture.
To answer the question What are the geometric properties of states with topological order from which we could deduce the topological order with some kind of Chern number (but without starting from a Chern-Simons field theory and putting in the right one by hand ;) ). Is there anything like this? I like to say that topological order is algebraic, not geometric. So the topological invariants of topological order are very different from Chern numbers. The robust ground state degenercy and the robust non-Abelian geometric phases of the degenerate ground states are the topological invariants of topological order (which are the analogues of the Chern number).
Best Answer
The short answer: graphene is a counterexample.
The longer version: 1) You do not need to break the time reversal symmetry. 2) spin-orbit coupling does not break the time-reversal symmetry. 3) In graphene, there are two valleys and time inversion operator acting on the state from one valley transforms it into the sate in another valley. If you want to stay in one valley, you may think that there is no time-reversal symmetry there.
A bit more: It seems that time-reversal symmetry is not a good term here. Kramers theorem (which is based on time-reversal symmetry) says that state with spin up has the same energy as a state spin down with a reverse wavevector. It seems that in your question you use time-reversal symmetry for $E_{↑}({\bf k})=E_{↓}({\bf k})$ which is misleading and incorrect in absence of space-reversal symmetry.
Do you still need a citations or these directions will be enough?
UPD I looked through the papers I know. I would recommend a nice review Rev. Mod. Phys. 82, 3045 (2010). My answer is explained in details in Sec. II.B.II, Sec. II.C (note Eq. (8)) Sec. III.A, IV.A. The over papers are not that transparent. Sorry for the late update.