Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

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

+ 4 like - 0 dislike
2667 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 (2,640 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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar\varnothing$sicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...