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

+ 4 like - 0 dislike
357 views

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 $R^n$ 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 $\pi_1(M)\neq0$ case we can only have weak topological insulators.

asked May 3, 2014 in Theoretical Physics by Jia Yiyang (2,465 points) [ revision history ]
edited Jun 2, 2014 by Jia Yiyang

@L.Su, yes, and in fact I posted that link in the question

1 Answer

+ 5 like - 0 dislike

To elaborate on your last question: the torus \(T^2 \simeq S^1 \times S^1\) so any map \(f:T^2 \to M\) induces a pair of maps \(f_x:S^1 \to M\) and \(f_y :S^1 \to M\) each one giving a class in \(\pi_1(M).\) These maps are homotopic only if they give same class in \(\pi_1(M)\). Now the quotient of torus with respect the two generators is \(S^2\) (that is one has \(S^2 \simeq \frac{S^1 \times S^1}{S^1 \vee S^1}\)) so in conclusion a map with \(f_x \text{ and } f_y\)homotopic will descent to map \(S^2 \to M\). In your case as \(\pi_1(M)=0\) any map descends. You can find a detailed discussion in  Homotopy and quantization in condensed matter physics, Phys. Rev. Lett. 51 (1983), 51-53 by J.Avron, R.Seiler and B.Simon

answered Jun 1, 2014 by ahalanay (120 points) [ no revision ]

+1 and thanks for the reply. I roughly understood the argument you're offering, but I couldn't fill in one piece for myself, namely, the "?" map in following diagram

$$\require{AMScd} \begin{CD} T^2@>{f}>> M\\ @V{q}VV@VV{?} V\\ T^2/S^1\vee S^1@>{f'}>>M \end{CD},$$

where $f'$ will be the desired induced map from $f$. We need the "?" map(which should be a homotopy equivalence I suppose) to deform the image of two base circles under f to a point, because $f$ itself doesn't necessarily map the two base circles to a point, it only maps them to something homotopically trivial.  

EDIT: It seems we haven't installed packages that compile commutative diagrams, I upload it as a picture.

You may look at the torus as the subset on \(\mathbb{C} \times \mathbb{C}\)of the form \((e^{i\theta},e^{i\phi})\)so basically a point is uniquely determined by the two phases \(\theta \text{ and } \phi\). The map $q$ sends all the points with either $\phi=0$ or $\theta=0$ to the point $(1,1)$. Now the map $?$ equals $f$ for points which have both phases non-zero (not lying on the collapsed circles) and sends all the points on the images of the two circles to $y=f(1,1)$. By the simple-convexity of $M$ these circles both deform to $y$.

I'm a bit confused about the way you phrased it. "Two points of the torus will be equivalent if they have either the same $\theta$ or the same $\phi$", I'm not sure which part of my diagram it corresponds to, the only part I made use a equivalence relation is the map $q$, which identifies all the points on $ \theta=0$ and $\phi=0$. 

"Now your map ? will send any such loop to its base-point", is this supposed to be the definition of "?"? But then at this point it is not even clear if “?” is well-defined this way.

You are right. I edited my comment accordingly.

Ok, that was also my guess. Then how do we prove such "?" map is a continuous map? And how do we prove in such case the homotopy classification of $f$ coincide with the homotopy classification of $f'$? These are probably textbook materials, but I haven't found them yet. 

"Now the map ? equals f for points which have both phases non-zero (not lying on the collapsed circles) and sends all the points on the images of the two circles to y=f(1,1)"

On a second read, it seems you are defining $?$ as the identity map on $M-f(\text{base circles})$, and send the $f(\text{base circles})$ to $f(1,1)$, but this hardly looks continuous to me. What I guessed was one needs to deform $M$ in such a way that $f(\text{base circles})$ get deformed to the base point along the process.

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