• 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.


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


(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 ...

  How to see the ground state degeneracy (GSD) from a $BF$ theory in $2+1$ $d$?

+ 1 like - 0 dislike

I have seen many times the $BF$ theory has non-trivial ground state degeneracy (typically on torus), but I can not see how the conclusion come out. Recently I found a paper by Hansson, Oganesyan and Sondhi,[Superconductors are topologically ordered] in which the superconductor is described by a Maxwell$-BF$ theory. They have a section of the GCD in a $BF$ theory in $2+1$ $d$. But actually I still have questions to understand it. 

The $BF$ theory in $2+1$ $d$ is given by the action
S = \frac{1}{\pi} \int d^3 x \epsilon^{\mu \nu \sigma} b_{\mu} \partial_{\nu} a_{\sigma}, \qquad (1)
where $a_{\mu}$ and $b_{\mu}$ are $U(1)$ gauge fields. 
$\mu,\nu,\sigma = 0,x,y$.

Working on $2-$torous, as in the section [IV.A] in Hansson's paper, the $BF$ theory can be written in the form 
S = \frac{1}{\pi}\int d^3x[\epsilon^{ij} \dot{a}_i b_j+
a_0 \epsilon^{ij} \partial_i b_j + b_0 \epsilon^{ij} \partial_i a_j],
where $\dot{a} = \partial_0 a$ and $i,j = x,y$.  They interpret $a_0$ and $b_0$ are multipliers for constraints 
$\epsilon^{ij} \partial_i b_j = 0$ and $\epsilon^{ij} \partial_i a_j = 0$. 
Upon inserting $a_i = \partial_i \Lambda_a + \bar{a}_i/L$
and $b_i = \partial_i \Lambda_b + \bar{b}_i/L$, 
where $\Lambda_{a/b}$ are periodic functions on the torus, $\bar{a_i}$ and $\bar{b_i}$ are spatially constant, $L$ denotes the size of the system, the above $BF$ theory reduces to
S = \frac{1}{\pi}\int d^3 x \epsilon^{ij} \dot{\bar{a}}_i \bar{b}_j. \qquad (2)

Then they say from the Eq.(2) one can obtain the commutation relation ( [Eq. (38)] in  their paper)
[\bar{a}_x, \frac{1}{\pi}\bar{b}_y] = i, \quad
[\bar{a}_y,-\frac{1}{\pi}\bar{b}_x] = i. \qquad (3)

Moreover, from the commutation relations Eq. (3), one can have ( [Eq. (39)] in  their paper)

A_x B_y + B_y A_x = 0, \quad
A_y B_x + B_x A_y = 0, \qquad (4)
where $A_i = e^{i\bar{a}_i}$ and $B_i = e^{i\bar{b}_i}$.
They claim that relations Eq. (4) indicates a $2\times2 = 4-$fold GCD and "$B_i$ can be interpreted either as measuring the $b$-flux or inserting an $a-$flux."

There are several points that I don't understand.

 1. How can I get communication relations Eq. (3) from the action Eq. (2)?
 2. Why relations Eq. (4) indicate a $4-$fold GCD?
 3. How should I understand the statement "$B_i$ can be interpreted either as measuring the $b$-flux or inserting an $a-$flux."?

I would be very appreciate if anyone can give me some hints or suggest me some relevant references.

asked Oct 21, 2014 in Theoretical Physics by hongchan (90 points) [ revision history ]
edited Oct 21, 2014 by hongchan

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:
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