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I am using the standard symbols of $V_\mu$ for the gauge field, $\lambda$ for its fermionic superpartner and $F$ and $D$ be scalar fields which make the whole thing a $\cal{N}=2$ vector/gauge superfield in $2+1$ dimensions.

Then the non-Abelian super-Chern-Simon's lagrangian density would be,

$$Tr[\epsilon^{\mu \nu \rho}(V_\mu \partial_\nu V_\rho - \frac{2}{3}V_\mu V_\nu V_\rho) +i\bar{\lambda_a}\lambda_a - 2FD]$$

Clearly this is classically scale invariant.

  • I would like to know of the argument as to why this is also quantum theoretically conformal (..may be there is some obvious symmetry argument which I am missing..)

  • Also is it true or obvious that if the above is perturbed by a $\lambda Tr[\Phi ^4]$ potential then this might flow to a fixed point which is $\cal{N}=3$ ? And then will it still be superconformal ?

I would like to know of a way of understanding this phenomenon of supersymmetry enhancement by renormalization flow. If someone could point me to some beginner friendly expository reference regarding this.

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The usual argument for why the Chern-Simons action is exactly conformal is that the action is gauge invariant only if the coupling constant, or Chern-Simons level $k$ (which you have not included in your Lagrangian) is integer valued. If the theory was not conformal there would be a non-zero beta-function which would make the coupling depend continuously on the renormalization scale $\Lambda$. But the integer $k$ can not be a continuos function of $\Lambda$, and hence the beta-function has to vanish. This argument doesn't depend on supersymmetry and hence holds equally well for the $\mathcal{N}=2$ case in your question.

As for the extension to $\mathcal{N}=3$: If I remember correctly you need to add two chiral multiplets $Q$ and $\tilde{Q}$ in conjugate representations of the gauge group and with a kinetic term plus a superpotential of the form you give (something like $W = \frac{1}{k} (\tilde{Q}T^aQ) (\tilde{Q}T^aQ)$). By $\mathcal{N}=3$ supersymmetry the coupling is the same as the Chern-Simons level, and hence the theory is again conformal. Unfortunately I don't know of any good reference which gives a general introduction to this theory. For some recent applications of it see eg papers by Gaiotto and Yin (arXiv:0704.3740) and Aharony, Bergman, Jafferis and Maldacena (arXiv:0806.1218). These references also contain discussions about how the $\mathcal{N}=3$ Chern-Simons theory appears as a conformal fixed point starting from $\mathcal{N}=2$ or $3$ Chern-Simons-matter theory or Yang-Mills-Chern-Simons.

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@Olof Thanks for your reply. Can you elaborate or give references to a source which will explain this point about why the Chern-Simon's level "k" needs to be integer valued for the action to be gauge invariant ? –  user6818 Jun 23 '11 at 9:51
    
@Olof I have seen this Gaiotto-Xi Yin paper. They too only seem to be referring to these results without proof or explanation. Can you give some more expository reference? –  user6818 Jun 23 '11 at 9:56
    
@Anirbit: I actually don't know of any good review article where the quantization of the Chern-Simons level is explained. I think the original reference would be: S. Deser, R. Jackiw and S. Templeton, Topologically massive gauge theories, Ann. Phys. 140, 372 (1982). The closely related question of instantons in 4D Yang-Mills is nicely discussed in Coleman's Aspects of symmetry. –  Olof Jun 23 '11 at 10:25
    
@Olof Can you elaborate on what you mean by "conjugate" representations ? Is it the same thing as what people also seem to call "bifundamental" like for $SU(N)$ theory it is denote as $(N,\bar{N})$ ? I haven't found a precise definition of these anywhere. Can you help with that? Also if you could write down explicitly what your $Q$ and $\tilde{Q}$ is. I guess you mean that they are $N \times N$ matrices ? Are they elements of some vector space supporting some specific (Adjoint?) representation of the gauge group ($SU(N)$?) ? –  user6818 Jun 24 '11 at 9:32
    
@Olof Also there is this issue of there being a 1-loop shift of $k$..although its not renormalized. Can you give more references to that ? –  user6818 Jun 24 '11 at 9:39
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