# When can we take the Brillouin zone to be a sphere?

When reading some literatures on topological insulators, I've seen authors taking Brillouin zone(BZ) to be a sphere sometimes, especially when it comes to strong topological insulators. Also I've seen the usage of spherical BZ in these answers(1,2) by SE user Heidar. I can think of two possibilities:

(1)Some physical system has a spherical BZ. This is hard to imagine, since it seems to me that all lattice systems with translational symmetries will have a torodal BZ, by the periodicity of Bloch wavefunctions. The closest scenario I can imagine is a continuous system having $\mathbf{R}^n$ as BZ, and somehow(in a way I cannot think of) acquires an one-point compactification.

(2)A trick that makes certain questions easier to deal with, while the true BZ is still a torus.

Can someone elaborate the idea behind a spherical BZ for me?

Update: I recently came across these notes(pdf) by J.Moore. In the beginning of page 9 he mentioned

We need to use one somewhat deep fact: under some assumptions, if $π_1(M) = 0$ for some target space $M$, then maps from the torus $T^ 2\to M$ are contractible to maps from the sphere $S^2 \to M$

I think this is a special case of the general math theorem I want to know, but unfortunately Moore did not give any reference so I'm not sure where to look.

EDIT: The above math theorem is intuitively acceptable to me although I'm not able to prove it. I can take this theorem as a working hypothesis for now, what I'm more interested in is, granted such theorem, what makes a $\pi_1(M)=0$ physical system candidate for strong topological insulators(robust under local perturbations), and why in $\pi_1(M)\neq 0$ case we can only have weak topological insulators.

• I'm pretty sure that you can't have a spherical BZ. The BZ needs to be a shape that can be fit together to fill $\mathbf{R}^n$ without gaps. E.g. for graphene, which is 2D, the BZ is a hexagon, and hexagons can "tile a plane". Circles and spheres will always leave gaps between them, so they can't be a BZ. May 7, 2014 at 13:56
• @Inmaurer: Yes, I am also more inclined to option (2) May 7, 2014 at 16:50
• @JiaYiyang I have been very busy lately, but I will try to write an answer over the weekend. If I forget write a comment with @ to remind me. But the short answer is (2) but with some subtleties. The BZ is a torus, but if you instead think of it as a sphere you only get what people call strong topological insulators. If you take a torus, you get strong AND weak topological insulators, the reason for that is some complicated algebraic topology. But since the strong one is more interesting, we can for simplicity take the sphere. The weak ones are not interesting since they are not really robust. May 9, 2014 at 6:40
• Hi @Heidar, this @ is just in case you forgot, and if you are still busy please take your time. May 12, 2014 at 2:17
• @Heidar: I plead you not to forsake this quetion :) Jun 27, 2014 at 3:03

The answer to your question is covered in these two papers: Homotopy and Quantization in Condensed Matter Physics

J. E. Avron, R. Seiler, and B. Simon http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.51.51

and

Homotopy Theory of Strong and Weak Topological Insulators

Ricardo Kennedy, Charles Guggenheim

http://arxiv.org/abs/1409.2529

The first one explains how to use maps from the sphere in the case of the quantum Hall effect, whereas the second one discusses a more general context of all symmetry classes and dimensions.

Let $X$ be some topological space and $\mathbb{T}^2$ be the $2$-torus. Consider the task of classifying the set $[\mathbb{T}^2\to X]$, the set of homotopy classes the maps $\mathbb{T}^2\to X$.

Then as Avron, Seiler and Simon argue the following:

As you should know, $\pi_1(\mathbb{T}^2)\cong\mathbb{Z}\times \mathbb{Z}$, that is, there are two basic loops $S^1\to \mathbb{T}^2$ which are not homotopic to each other. Call them $\gamma_1$ (blue) and $\gamma_2$ (red).

Then any map $f:\mathbb{T}^2\to X$ induces two maps $S^1\to X$ via $$f\circ\gamma_i:S^1\to X$$

Then one can prove that for two maps $f_1:\mathbb{T}^2\to X$ and $f_2:\mathbb{T}^2\to X$, if the either $f_1\circ\gamma_1$ is not homotopic to $f_2\circ\gamma_1$ OR $f_1\circ\gamma_2$ is not homotopic to $f_2\circ\gamma_2$, then $f_1$ is not homotopic to $f_2$.

However, "even if the two induced maps $S^1\to X$ are pair-wise homotopic, there could still be a left over piece, which is a map $S^2\to X$.". In this way, you could classify maps $\mathbb{T}^2\to X$ by two pieces of $\pi_1(X)$, one piece of $\pi_2(X)$, etc...

For our case, $X$ is the set of gapped Hamiltonians, which topologically is the Grassmannian ($G_m(\mathbb{C}^n)$ for an $n$ level system with the Fermi energy above $m$ levels).

For instance, for a two-level system with 1 occupied level, we have $$G_1(\mathbb{C}^2)\cong S^2$$. But the the $2$-sphere is simply connected (that is, $\pi_1(S^2)=0$), so the two aforementioned loops $\gamma_1$ and $\gamma_2$ will not contribute to the classification, and we would only have the "left over" piece $$S^2\to S^2$$ And it is well known that $$[S^2\to S^2]\equiv\pi_2(S^2)\cong\mathbb{Z}$$ The integer being the degree of the map $\mathbb{T}^2\to S^2$ which would be the same degree as the induced map $S^2\to S^2$ which in a more general setting (of a complex line bundles over $\mathbb{T^2}$) would be the Chern number of that line bundle. Indeed, isomorphism classes of line bundles over $\mathbb{T^2}$ are isomorphic to homotopy classes of maps $\mathbb{T^2}\to Gr_1\left(\mathbb{C}^2\right)\cong S^2$.

• It would be very helpful if you could give a rough sketch of the relevant arguments in the papers. Mar 5, 2016 at 12:56
• @Martin, I hope this helps.
– PPR
Mar 5, 2016 at 23:33