Why do edge states in graphene exist between the valence and conduction band?

Question

I read in a review that there are two Dirac points in graphene, where the conduction band and valence band touch each other. Near these points electrons obey a linear dispersion relation. Breaking of time symmetry leads to a quantum hall state. The solution of the Schrödinger equation gives rise to a single band connecting the valence and conduction band, which represents the edge states, and the slope of that single band gives the chirality of edge states. Why do electronic states exist between valence and conduction band? Also can someone explain to me why Dirac points occur in graphene?

Topologically, a non-trivial state also occur when degeneracy is broken away from the Dirac points because of the spin orbit interaction. In that situation, when the band connecting the degenerate points cross the Fermi level an odd number of times, that leads to a topologically protected state and the even crossings don't. I didn't understand why.

Answer 1

There is a very nice review of this here.

The point is that graphene has (approximate) particle-hole symmetry, so any special has to happen right in the middle of the gap, since this symmetry exchanges the valence and conduction bands.

Answer 2

As I know there is a notion of "bulk boundary correspondence" if you terminate the infinite slab of graphene along an axis in an arbitarily direction then you get an edge and surface states. It will support chiral movement of electrons provided that its bulk(I.e unterminated graphene) be in a topological phase characterized by some topological quantity. This chiral movement is protected by some symmetry meaning that you can't get rid of that without breaking that symmetry.