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

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

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,853 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Group Cohomology and Topological Field Theories

+ 7 like - 0 dislike
54 views

I have a two-part question:

  1. First and foremost: I have been going through the paper by Dijkgraaf and Witten "Group Cohomology and Topological Field Theories". Here they give a general definition for the Chern-Simons action for a general $3$-manifold $M$. My question is if anyone knows of any follow-up to this, or notes about their paper?

  2. To those who know the paper: They say that they have no problem defining the action modulo $1/n$ (for a bundle of order $n$) as $n\cdot S = \int_B Tr(F\wedge F)$ $(mod 1)$, but that this has an $n$-fold ambiguity consisting of the ability to add a multiple of $1/n$ to the action - What do they mean here? Also, later on they re-define the action as $S = 1/n\left(\int_B Tr(F\wedge F) - \langle \gamma^\ast(\omega),B\rangle\right)$ $(mod 1)$ - How does this get rid of the so-called ambiguity?

Basically my question is if anyone can further explain the info between equations 3.4 and 3.5 in their paper. Thanks.

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user klw1026
asked Apr 21, 2011 in Theoretical Physics by klw1026 (120 points) [ no revision ]

3 Answers

+ 6 like - 0 dislike

First, the full paper is here:

http://citeseer.ist.psu.edu/viewdoc/download;jsessionid=807BE383780883ACB4CAB8BD48E8C90B?doi=10.1.1.128.1806&rep=rep1&type=pdf

Second, the paper has 150 citations. See all this information at INSPIRE (updated SPIRES):

http://inspirebeta.net/record/278923?ln=en

Third, the text between 3.4 and 3.5 looks totally comprehensible. At that point, they are able to define $n\cdot S$ modulo 1, which is equivalent to defining the action $S$ modulo $1/n$. The goal is to define the action $S$ itself modulo 1; I suppose that their normalization of the path integral has to have $\exp(2\pi i S)$ with the atypical $2\pi$ factor. Yes, confirmed, it's equation 1.2.

If you shift the action by 1 - or $2\pi$ in the ordinary conventions - it doesn't change the integrand of the path integral; it doesn't change the physics. So quite generally, if one is able to say that the action $S$ is equal to $S_0+n$ (or $2\pi n$ normally) for some integer $n$, he knows everything about the physics of the action he needs; shifting it by an integer doesn't change anything. That's why, in fact, the action is often defined modulo 1 only (up to the addition of an integer multiple of 1).

So it's enough to know the "fractional part" of the action; the integer part is irrelevant. However, at the point of the equation 3.4, their uncertainty is larger than that: they only know the action modulo $1/n$. For example, if the action is $9.37$ modulo $1/2$, it means that the fractional part may be $0.37$ but it may also be $0.87$. These two values of $S$ would change the physics because the contribution of the configuration to the path integral changes the sign if one changes $S$ by $1/2$ (in normal conventions, by $\pi$).

If one only knows $S$ modulo $1/n$, and if he thinks it's $S_0$ - in this case, the $F\wedge F$ expression - it means that the real action is $$ S = S_0 + K/n $$ and the integer $K$ has to be determined. Because the change of the action $S$ by an integer doesn't change physics, it doesn't matter if $K$ in the equation above is changed by a multiple of $n$. So the goal is to find the right $K$ to define the action - and $K$ is an unknown integer defined (or relevant) modulo $n$, i.e. up to the addition of an irrelevant and arbitrary multiple of $n$.

At some point, they find the right answer and it is $$ K = -\langle \gamma^*(\omega),B\rangle $$ which removes the ambiguity of $S$ - the missing knowledge whether $S$ should be the original $S$ or higher or smaller by a particular multiple of $1/n$. If you don't understand the text above, then apologies, I have no way to find out why, so I can't give you a better answer unless you improve your question.

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Luboš Motl
answered Apr 22, 2011 by Luboš Motl (10,248 points) [ no revision ]
No, I understand. So, do you have any ideas/motivation on how they came to adding $-\langle \gamma^\ast(\omega),B\rangle$? I agree that it works, just have no idea it would be that. Thanks for clearing things up!

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user klw1026
Also, just to be completely clear, removing the ambiguity in $S$ is equivalent to finding $K$? Thanks again for the answer!

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user klw1026
Yup, removing the ambiguity of $S$ is equivalent to finding $K$, more precisely finding $K$ mod $n$. But in the full quantum theory, $S$ only matters mod $1$ (in normal normalizations of physics, $2\pi$), because it appears in $\exp(2\pi i S)$ in the path integral only.

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Luboš Motl
+ 5 like - 0 dislike

Dijkgraaf and Witten used $\mathcal H^3[G,U(1)]$ to define CS theory for gauge group $G$. Recently, group cohomology has found applications in condensed matter physics. It may classify the so called "symmetry protected topological phases" of interacting bosons:

The $d$-dimensional symmetry protected topological phases of interacting bosons with symmetry group $G$ can be one-to-one labeled by elements in $\mathcal H^{d+1}[G,U(1)]$. ($d$ is the space dimensions.)

(The symmetry protected topological phases are for interacting systems, which are similar to the topological insulators of non-interacting fermions. They are short-range entangled states with symmetry.)

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Xiao-Gang Wen
answered May 27, 2012 by Xiao-Gang Wen (3,319 points) [ no revision ]
+ 0 like - 0 dislike

The integer part of this is a cocycle condition, which is a measure of the winding number for a gauge transformation. The Chern-Simons (CS) theory is a $2~+~1$ dimensional quantum field theory for a non-dynamical gauge field$A_\mu$. The action for such a theory is $$ S_{CS}~=~\frac{k}{4\pi}\int A\wedge dA~+~\frac{2}{3}A\wedge A\wedge A $$ Where $A_\mu$ is a component of the one form ${\underline A}~=~A_\mu{\underline e}^\mu$ for a non-abelian gauge field transforming in the adjoint representation of the gauge group $U(N)$.

The theory to make sense must be well behaved under gauge transformations. While it is relatively easy to show invariance in the abelian case, the non-abelian case is a little more subtle. In this case $$ S_{CS}~\rightarrow~S_{CS}~+~2\pi kN $$ Where $N$ is a integer for the winding number of the gauge transformation performed. Quantization of the theory using Feynman’s path integral formalism requires that$e^{iS_{CS}}$ be gauge invariant. This leads to the condition that $k~\in~{\mathbb Z}$. The integer $k$ is the Chern-Simons level $A_\mu$. Typically every gauge group in the Chern-Simons theory has a level associated to it.

This form of the Chern-Simons theory is not supersymmetric. However it is possible to make the gauge field $A_\mu$ a component of an ${\cal N}~=~2$ vector multiplet. This necessarily introduces two scalar fields $A_\mu$ $F$, an auxiliary field, and a 2-component Dirac spinor $\psi$ to the theory in a superfield $$ \Psi~=~\psi~+~\theta \sigma^\mu A_\mu~+~H.C.~+~{\bar\theta}\theta F. $$ It is possible to extend this theory to admit the full ${\cal N}~=~8$ SUSY (16 supercharges).

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Lawrence B. Crowell
answered Apr 23, 2011 by Lawrence B. Crowell (590 points) [ no revision ]
Since I know very little about supersymmetry, I have a few simple question. How does the $\cal N = 2$ SUSY Chern-Simons action look like in terms of the superfield $\Psi$? Is it possible to write this down without the superspace/superfield notation? Finally, does this SUSY extension add new mathematical features to the CS theory (relation to knot theory, quantum groups, modular tensor categories, CFT and so on)?

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Heidar
The generic form of the CS is the same as the above. The CS Lagrangian has that cubic form, which means that in $2~+~1$ spacetime a fermion can have bosonic statistics and visa versa. The exchange statistics which apply in $3$ space gets "pushed" into the time part --- so to speak. This gives anyonic behavior. In string theory this describes the $M_2$ brane, and in condensed matter physics graphene.

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Lawrence B. Crowell
This is not an answer to the question, which is not about the more elementary relation which restricts k to an integer, but about fixing up the action when the group gauge group is a quotient or discrete. You are answering a different (much simpler) question.

This post imported from StackExchange Physics at 2014-04-05 04:26 (UCT), posted by SE-user Ron Maimon

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$\hbar$y$\varnothing$icsOverflow
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).
To avoid this verification in future, please log in or register.




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

Your rights
...