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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.

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3 Answers 3

up vote 9 down vote accepted

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.

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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! –  klw1026 Apr 22 '11 at 15:52
    
Also, just to be completely clear, removing the ambiguity in $S$ is equivalent to finding $K$? Thanks again for the answer! –  klw1026 Apr 22 '11 at 20:29
    
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. –  Luboš Motl May 27 '12 at 5:27
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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.)

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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).

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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)? –  Heidar Apr 23 '11 at 12:27
    
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. –  Lawrence B. Crowell Apr 23 '11 at 12:57
    
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. –  Ron Maimon May 27 '12 at 10:05
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