When talking about topological insulator and talking about bulk-edge correspondence, it seems to be widely accepted conclusion that the band Chern number (winding number) is equal to, when the boundary becomes open, the amount of edge states. But why? For example, SSH model or Graphene (treat $k_x$ as parameter of a 1D chain and $k_y$ as the wave vector along this 1D chain). When the 1D chain is closed, in the topologically non-trivial phase, Chern number can be calculated to be, say, one. And after cutting the 1D chain open, there is one pair of gapless state living on the two edges. Why is Chern number equal to the amount of gapless edge state pairs? Is there any proof of that?
[Physics] How to show that Chern number gives the amount of edge states
graphenequantum-hall-effecttopological-insulators
Related Solutions
I am still not sure what you precisely want to be a Klein Bottle, but let me make some comments that might help you clarify what exactly you want to know. (Warning: I am writing this while being very tired, people are invited to correct me.)
First of all one must be careful to distinguish band structure of the bulk from band structure of a semi-infinite strip (with edges). In your second figure, 1 og 2 are bulk band structures while 3 is band structure of a semi-infinite strip. I think you are mixing these.
- Bulk picture: Both momentum $k_x$ and $k_y$ are good quantum numbers. Assume you start with a trivial insulator (gap in the band structure, number 2 in your picure) and you continuously change the parameters of your system. As long as the bulk gap i open you have a trivial insulator. Suddenly the bulk gap closes, at one value of your parameters (critical point), and quickly opens up again, now you are in the topological phase BUT the bulk band structure still looks like figure 2 above. How to see the difference in these two phases? One way is looking for edge states.
- Edge picture: Make the system finite along $y$-direction, so now only $k_x$ is a good quantum number. Now you have alot of bands, depending on how many lattice sites you put along $y$. When you are in the trivial phase the system is still gapped and boring. While you change the system parameters and hit the critical point, the gap of all bands close and then most (but not all) bands open up again (picture 3 above). When one looks at the eigenvectors, one will see that the bands with a gap correspond to bulk states while the band(s) with no gap (with a Dirac point) correspond to states localized along the edge. Furthermore these edge states are robust.
This I mention, in case there were any misunderstanding here. Now to your question, how can one see the connection between band structure, Brillouin zone and topology? The interesting thing is that we only need to analyse the bulk in order to probe the topology of the system, although the interesting physics is along the edge (this is called bulk-edge correspondence in the litterature).
Assume we add a spin-orbit coupling to graphene that preserves $S_z$, meaning that although the spin $SU(2)$ symmetry is not preserved it is still well-defined to speak about spin-up/down (so a $U(1)$ subgroup is still a symmetry). I think that this term will open a gap at the Dirac points at $K$ and $K'$ and the system turns into an insulator. Since $K$ and $K'$ are not important anymore, we have a four-band model (two from spin up/down and two from sub lattice A/B). Since $S_z$ is conserved, the (bulk) Hamiltonian is block diagonal
$ H(\mathbf k) = \begin{pmatrix} H_{\uparrow}(\mathbf k) & \\ & H_{\downarrow}(\mathbf k) \end{pmatrix}$,
where $H_{\uparrow}$/$H_{\downarrow}$ are $2\times 2$ matrices. Since the system preserves time-reversal symmetry, the two Hamiltonians are related by a time-reversal transformation $H_{\downarrow} = \Theta H_{\uparrow}\Theta^{-1}$. Since the Brillouin zone is a torus, we need to classify maps $T^2\rightarrow \mathcal H$, where $\mathcal H$ is the appropriate space of $4\times 4$ matrices obeying some constraints (gapped and time-reversal invariance). For this model, however, we don't need to analyze $\mathcal H$. Since the model is block diagonal we can consider each block separately, any $2\times 2$ matrix can be written in the basis of Pauli matrices
$H_{\alpha}(\mathbf k) = d_0(\mathbf k) \mathbf I + \mathbf d(\mathbf k)\cdot\mathbf{\sigma}$, where $\alpha = \uparrow, \downarrow$ (the dependence of $d_0$ and $\mathbf d$ on $\alpha$ is suppressed for notational simplicity).
Since the spectrum is $E(\mathbf k) = d_0(\mathbf k) + \sqrt{\mathbf d\cdot\mathbf d}$, we can continously deform $d_0$ to zero and deform $\mathbf d\rightarrow \hat{\mathbf d}=\frac{\mathbf d}{|\mathbf d|}$ without closing the gap (and thus still be in the same topological class). Thus we can classify each block of the Hamiltonian, by the map $T^2\rightarrow S^2$, $\mathbf k\rightarrow \hat{\mathbf d}(\mathbf k)$. This is basically classified by the second homotopy group of the 2-sphere $S^2$ (winding number), $\pi_2(S^2) = \mathbb Z$ (well the map is from a torus $T^2$ and not $S^2$, so we need an argument for why we can use the second homotopy group but this I will not delve into right now). The winding number of $\hat{\mathbf d}(\mathbf k)$ is given by the degree of map formula
$C_1^{\alpha} = \frac 1{4\pi}\int_{T^2}d\mathbf k\;\hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y}\in\mathbb Z.$
Thus each block form a Integer Quantum Hall effect separately (see also this answer). Note that $C_1$ is basically the first Chern-number where $\epsilon^{\mu\nu}F_{\mu\nu} = \hat{\mathbf d}\cdot\frac{\partial \hat{\mathbf d}}{\partial k_x}\times\frac{\partial \hat{\mathbf d}}{\partial k_y}$, $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ where $A_{\mu}(\mathbf k) = -i\langle\psi(\mathbf k)|\partial_{\mu}\psi(\mathbf k)\rangle$ is the Berry phase for the filled states (to see how to rewrite the Berry phase in terms of $\hat{\mathbf d}$ see this paper. The reason I have avoided the Chern number approach is that, one needs to go into discussing vector (or principal) bundles which just would complicate everything even more).
Now we are almost done. We have found two Chern numbers $C_1^{\downarrow}$ and $C_1^{\uparrow}$, is there a Chern number for the combined system? Since $H_{\downarrow} = \Theta H_{\uparrow}\Theta^{-1}$ are related by time-reversal symmetry, one can show that $C_1^{\uparrow} = -C_1^{\downarrow}$. Thus the sum $C_1^{\uparrow} + C_1^{\downarrow} = 0$, but the difference $C_1^{\uparrow} - C_1^{\downarrow} = 2C_{spin}$ is well defined and sometimes called the spin Chern-number. Thus calculating $C_{spin}\in\mathbb Z$ will tell you if graphene with a spin-orbit coupling is topological or not.
However if one has a perturbation that breaks $S_z$, then our construction breaks down. But it turns out that $C_{spin}$ is still well defined but only modulo 2, $\nu = C_{spin} \text{mod} 2$. This is why people call this a $\mathbb Z_2$-topological insulator, see more details here (pdf-file).
There are of course many other ways of seeing the connection to topology and this is probably the most explicit and elementary way to do it. Is this what you wanted? I suspect that you are interested in the map $T^2\rightarrow\mathcal H$, $\mathbf k\rightarrow H(\mathbf k)$ and when you say "energy space" you mean $\mathcal H$. It turns out that this space is homotopy equivalent to $\mathcal H \approx U(8)/Sp(8)$ (called the symplectic class) by a band-flattening procedure. Sadly I am too tired to see if this is diffeomorphic to the Klein bottle. Is this what you are interested in?
This answer got much longer than originally planned.
The TKNN (bulk) and Büttiker (edge) explanations for the quantized Hall conductance correspond to different geometries.
In the TKNN theory, the "sample" consists of a torus closed on itself and therefore has no edges at all. In this case the electric potential is uniform, and the electric field is due to the time derivative of the vector potential (it lasts only as long as one varies the magnetic flux inside the torus). In this case, the Hall current is truly a bulk current.
Büttiker, on the other hand, considers a Hall bar with different electrochemical potentials on each side. If one (as does Büttiker) assumes that the electrostatic potential is uniform within the central region of the bar, and rises on the sides, then one finds that the current flows along the edges (in opposite directions), with more current flowing along one edge than along the other because of the different chemical potentials.
In a more realistic description, the electrostatic potential is not uniform in the bulk of the bar, so that the current flow takes place both along the edges and within the bulk of the bar. In any case, the net current is completely independent of the actual profile of the electrostatic potential accross the hall bar, and depends only upon the chemical potential difference. That is why why Büttuker and TKNN obtain the same answer for the (quantized) total current.
A nice discussion of this question is given by Yoshioka.
Best Answer
Yes, there is a proof of that. The first one appeared by a beautiful paper of Hatsugai in 1993 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.71.3697 for a particular special classes of models of the IQHE. More general proofs ensued, but it does turn out that the proofs rely on some non-trivial math, the most simple way to present it seems to be rooted in the context of some basic facts of complex analysis, as for example is presented in this paper: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.83.125109
There you see for example that essentially the Cauchy integral formula lies at the heart of the bulk-boundary correspondence. More general proofs rely on more sophisticated math, e.g., K-theory or Fredholm theory.