# Conformal QFTs for D > 2

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Which conformal QFTs do we know for spacetime dimension d > 2?

I know that for D = 4 we have N = 4 SYM and some N = 2 supersymmetric Yang-Mills + matter models.

What is the complete list of such QFTs? Is it plausible we know all of them? In particular, what examples do we have with D = 3?

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asked Dec 2, 2011
A lot of them in D=3 can be studied in $1/N$ expansions; e.g., for QED with $N_f$ flavors, a large-$N_f$ resummation strongly suggests that the theory is conformal in the IR for large enough $N_f$.

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Then there are the weakly-coupled Banks-Zaks theories for QCD or ${\cal N}=1$ SQCD, where you choose $N_f$ such that the one- and two-loop beta functions nearly cancel. These are the weakly-coupled end of a range of $N_f$ that are conformal.

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Those are just a few examples of the easier-to-study examples, but they're the tip of the iceberg. We certainly don't have a classification of all CFTs.

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@Matt Ree Thanks for the examples. Can you give some pedagogic reference for these examples - some canonical literature that explains these.

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## 1 Answer

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The class of 3d QFTs which are manifestly unitary and have classical on-shell lagrangians that are invariant under the rigid $N$-extended conformal superalgebra are known to admit a nice characterisation for $N>3$. This characterisation is due to Gaiotto and Witten (in section 3.2 of http://arxiv.org/pdf/0804.2907v1) and paved the way for a number of subsequent classification results for $N>3$ SCFTs in 3d over the past few years.

The generic on-shell $N=3$ lagrangian within this class involves a Chern--Simons term for a non-dynamical gauge field coupled to hypermultiplet matter fields in a real representation of the gauge group. Classical $N=3$ superconformal symmetry fixes the superpotential uniquely in terms of a canonical quartic function that is built from the matter representation. Non-renormalisation theorems suggest that any such theory enjoys exactly the same superconformal symmetry at the quantum level. A choice of gauge group, matter representation and Chern--Simons couplings is therefore sufficient to define any such lagrangian.

The rigid form of the $N=3$ superpotential can be deduced by looking at the possible structure of Yukawa couplings in the on-shell lagrangian. In particular, it follows from the requirement that they be invariant under the ${\mathfrak{so}} (3)$ R-symmetry in the $N=3$ superalgebra. The key insight of Gaiotto and Witten was to realise that enhancement to $N>3$ superconformal symmetry can only occur if these Yukawa couplings are invariant under the ${\mathfrak{so}} (4)$ R-symmetry in the $N=4$ superalgebra. In a nutshell, their characterisation for this to occur is that the gauge algebra and matter representation must collect into the even and odd parts of a certain auxiliary lie superalgebra.

For $N>4$, it turns out that any indecomposable lagrangian within this class is based on an irreducible matter representation characterised by an embedding into one of the classical simple lie superalgebras (classified by V.G. Kac in 1977). For $N=5$, each type of classical simple lie superalgebra encodes a superconformal lagrangian. The refinement to $N>5$ recovers the classification obtained by Schnabl and Tachikawa in http://arxiv.org/pdf/0807.1102v1, containing the celebrated ABJ(M) and BLG models as special cases. One can prove that $N=7$ does not occur, in that it automatically implies $N=8$.

For $N=4$, the general picture is a bit more complicated. Given $N>4$ superconformal symmetry, it follows that the target space for the hypermultiplet scalars must be flat. This is not necessarily the case for $N=4$ though and one can construct $N=4$ SCFTs where the hypermultiplet lagrangian is replaced by a gauged non-linear sigma model (with non-flat hyperKähler target). The indecomposable $N=4$ SCFTs with flat target have a rather elaborate classification (outlined at the end of section 3.1 in http://arxiv.org/pdf/0908.2125v2) in terms of certain chains of classical simple lie superalgebras whose linking rules are reminiscent of the game of dominoes!

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answered Dec 3, 2011 by (340 points)
Thx, that's a really nice answer!

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