atomic-physicsatomsgraphenesolid-state-physics
Graphene is always in the news now a days and its key features are that it is; very strong, conductive and transparent. It is so transparent that each layer of graphene will only absorb 2% of Light passing through it.
But what is it about the structure of Graphene which makes it (almost) transparent?
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.
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.
In a perfect graphene sheet, all carbon atoms are sp$^2$-hybridized, with three in-plane $\sigma$-orbitals and two out-of-plan $\pi$-orbitals. This means that each carbon atom can form equivalent s-bonds to each of its three neighboring atoms. The bonding energy of one C-C bond in graphene amounts to 4.93 eV [doi:10.1088/0953-8984/14/4/312]. These strong (covalent) $\sigma$-bonds are responsible for the extraordinary mechanical properties of graphene.
The out-of-plane orbitals are responsible for the $\pi$ bands (Dirac cones) and lead to the nice electric properties of graphene, but they do not significantly contribute to the mechanical "strongness" of graphene.
Best Answer
Graphene is only transparent because it is very thin (one atom thick). If it absorbs 2% per layer then just a few hundred layers would absorb almost all light and that would still be a very thin sheet of graphite.
The question should be why does graphene absorb so much light compared to diamond which really is transparent? A simplified answer is that graphene is a very good conductor because it has only three covalent bonds per atom compared to the full four in diamond. This makes it possible for electrons to move freely over a sheet of graphene to conduct electricity. Like metals this means it will absorb or reflect light because the free electrons can absorb the small amount of energy in the photon. In diamond the photons would need to have enough energy to release an electron from the covalent bonds. For visible light this is not possible so the photons pass right through the diamond and are only stopped or deflected by impurities.