# 't Hooft limit of coupling fundamental fermions to Chern-Simons theory

This question is in reference to this paper: arXiv:1110.4386 [hep-th].

• I would like to know what is the derivation or a reference to the proof of their crucial equation 2.3 (page 12).

• In their analysis of fundamental fermions coupled to Chern-Simons theory, why have they been able to ignore the ghosts altogether? Does it have something to do with working in the light-cone gauge in which the self-interaction of the gauge field probably disappears and hence that removes the necessity to have ghosts?

• Their Lagrangian is 2.1 (page 11) is massless and they say that they can always tune the bare mass to be $0$ and they can always ignore the mass of the fermion. Can someone elaborate on this? Why was this possible? Isn't this tantamount to assumption of conformality which they want to prove in the 't Hooft limit? (Then isn't the argument becoming circular?)

Similarly if this were a scalar field theory then in the same strain one might want to say that the quadratic and the quartic scalar interactions can always be held at zero - but again wouldn't that be an assumption of conformality in the 't Hooft limit?

What would be a genuine proof of conformality for this theory or its scalar version?

• Is the scalar version of this theory somehow uninteresting or known?

• In this paper whatever is claimed as the higher-spin currents seem to have their conservation laws broken in in $1/k$ or $1/N$, then isn't the theory interesting only when $k$ and $N$ are both infinite and then isn't that a trivial theory?

What sense does it make to say the Fermion $2$-point function depends on the 't Hooft coupling (as in equation 2.23 page 16)? Isn't that a non-physical quantity to talk of?

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These are best asked as separate questions, but ok. – Ron Maimon Oct 19 '12 at 4:30

1. The identity is just the content of the diagram above it, as they say. The evaluation of the Feynman diagram is formula 2.3—the self-energy diagram is the sum of 1PI diagrams with one fermion line coming in and one line going out. I don't want to draw diagrams, but it's a sum of a line with a gluon wiggle in an arc, plus a line with two gluon wiggles in two arcs one above the other, then three wiggles, then four, and these are the "rainbow diagrams" they talk about. These are nested in a heirarchical way, every arc can include subarcs. The sum of the rainbow diagrams $R_1 + R_2 + R_3 +\cdots$ is the self-energy $\Sigma(p)$, with the term starting at $R_1$, which has one gluon. Notice that all terms of this sum share the outermost gluon propagator. So they all share the same outermost loop-momentum integral, so you can factor this integral out, and you can write out the inner electron line contribution for each value of the loop momentum, it's the sum of $P_0+ P_0 \Sigma P_0 + P_0 \Sigma P_0 \Sigma P_0 \cdots$ with free electron propagators between rainbow diagrams, which accounts for the little propagation from the outermost arc to the next less outermost arc. The inner propagator resums to the exact electron propagator again, because these are all the planar diagram contributions. So the self-energy is given by the one-loop gluon integral with the full propagator on the inner line. I don't think this explanation in words is more clear than the diagram they write in the paper. If you are confused on this, you probably just need to write down the diagrams, and review the derivation of the exact propagator from summing 1PI diagrams.
3. I don't understand this concern—they are fine tuning one parameter, the bare fermion mass, so that the exact $\sigma(p)$ ends up zero at zero momentum. You are allowed to do this—you are producing a conformal theory by going to a critical point. They aren't ignoring the bare mass, they are tuning it using the exact result for the Fermion propagator to ensure that the fermion self-energy is consistent with masslessness. For the scalar version, you would do the same thing—fine tune the mass to criticality.
5. It's not a trivial theory, because it has a gap equation that gives a nontrivial self-energy. Just because you are at large $N$ doesn't mean you are somehow in a free or trivial theory, you are just in a limit of enormous numbers of degrees of freedom at every point, so that only the planar diagrams are important. They don't have a zero coupling, just large $N$ at finite 't Hooft coupling. Their results are exact for large $N$ at any coupling, that's what makes it interesting.
6. The Fermion propagator is physical, although not as directly as a scalar propagator. It's the derivative of the $\log(Z)$ with respect to Fermion sources. If you aren't comfortable with Fermion sources yet, imagine introducing an external nearly free infinitely heavy Fermion field, which just has a small amplitude for changing into the Fermion in the theory we are talking about. Then if you put the heavy fermion at $x$ (it doesn't move) and asked how likely is it to propagate to $y$, this happens by changing to the light fermion and propagating according to the light propagator, so it's a way of giving meaning to a Fermion source.
If in a theory the bare and the interacting Fermionic propagator were to be denoted as $S_F$ and $S_F'$ and the self-energy were $\Sigma$ then this 1PI summing would give the relation, $S_F' = S_F ( 1 + \Sigma S_F')$ - but equation 2.3 is stronger than this - it gives a recursive determination of $\Sigma$ itself! do you have a reference which explains well this "rainbow diagram" summation thing - this seems to be an artefact of the large N limit. And is this somehow peculiar to the fermionic theory and not the scalar theory? – user6818 Oct 20 '12 at 21:58
The point about tuning the bare mass is this - that they can choose the bare mass such that $\Sigma (p=0)=0$ isn't that already an assumption of conformality? Aren't they trying to prove that in the limit of level and $N_c$ being infinite at finite 'thooft coupling the theory is a CFT? Doesn't that argument get circular then? And what does it mean for the propagator to depend on the 'tHooft coupling - shouldn't there have been a single unique theory sitting at level and colour being infinite rather than there being different theories depending on how you reaching the infinity? – user6818 Oct 20 '12 at 22:03