Now I study very interesting lectures Superconformal symmetry and representations and I face some statements, which are unclear to me.

In unitary CFT there are unitary bounds for dimensions of operators. When the unitarity bounds are saturated, there are null-vectors (i.e. vectors with zero norm) in the representation. Hence representations with these null-vectors always have to saturate the bound and cannot acquire any anomalous dimensions in the quantum theory.

In exercise 4 there is nontrivial statement about recombination of such short representations:

Usually, conformal field theories depend on several parameters, such as the coupling constants of the theory. As these parameters are tuned, the scaling dimension of the multiplets typically changes, but the total number of states typically does not change. Thus, we have the phenomenon of recombination: as we move around in the parameter space of the theory, two or more short multiplets may join up to form one long multiplet, whose dimension is no longer protected by the unitarity bound. Argue the following recombination rule: $$[V ]_{∆=d−1} ⊕ [0]_{∆=d} → [V ]_{∆=d−1+ε} $$ where the multiplets on the right do not have a null-descendant anymore and can hence leave the unitarity bound, which we denoted by a shift of the scaling dimension by $ε > 0$.

Could somebody present some concrete model, which have such recombination phenomena?

  • 2
    $\begingroup$ See, e.g. arxiv.org/abs/1505.00963 $\endgroup$ Commented Jul 7, 2020 at 13:02
  • $\begingroup$ @PeterKravchuk this is really helpful, thank you! But what about concrete conformal multiplets, represented in question? $\endgroup$
    – Nikita
    Commented Jul 8, 2020 at 16:30
  • $\begingroup$ @PeterKravchuk In the paper, can you please say what's the unitary bound? Because it seems like. the recombination happens when $\Delta$ approaches 1 from below which is opposite to what I expected(from above). So I thought perhaps the unitary bound might not necessarily coincide with the critical recombination exponent. That's I ask about the critical bound to see if it's 1 and then wonder! $\endgroup$ Commented Aug 24, 2022 at 17:42

1 Answer 1


A concrete example of this recombination occurs in the compact free boson in two dimensions. Specifically, when you deform away from a point with enhanced symmetry called the self-dual radius. Consider the free boson CFT where the holomorphic and anti-holomorphic fundamental fields are allowed to take values in a circle: $X \sim X + 2\pi r$. An important set of operators in this theory is that of the exponentials $e^{i \alpha X(z)} e^{i \bar{\alpha} \bar{X}(\bar{z})}$. In my favourite conventions, the conformal weights are given by \begin{equation} h = \frac{\alpha^2}{2}, \bar{h} = \frac{\bar{\alpha}^2}{2}. \end{equation} Also, since $X$ and $\bar{X}$ are periodic, single-valuedness of these exponentials (integer spin) restricts the possible choices for $\alpha$ and $\bar{\alpha}$ to lie in a lattice which, in the same convention, is given by \begin{equation} (\alpha, \bar{\alpha}) \in \left \{ \left ( \frac{n}{r} + \frac{mr}{2}, \frac{n}{r} - \frac{mr}{2} \right ) \right \}, \end{equation} It is often said (e.g. in string theory where these are used to construct vertex operators) that the integer $n$ labels momentum while $m$ labels winding modes. In any case, this can all be found in Ginsparg's notes https://arxiv.org/abs/hep-th/9108028 or some other standard reference.

But now notice that if $r = \sqrt{2}$, we can take $m = n = \pm 1$ to get conserved currents other than $\partial X$. One of these is \begin{equation} e^{i \sqrt{2} X(z)} := [1]_1. \end{equation} The notation is justified because it is a primary under $SL(2; \mathbb{C})$ (also under larger algebras but never mind that) and the weights can be solved as $(h, \bar{h}) = (1, 0)$ which is the same as saying $(\Delta, J) = (1, 1)$. Next, this theory also contains the operator \begin{equation} e^{i \sqrt{2} X(z)} \bar{\partial} \bar{X}(\bar{z}) := [0]_2 \end{equation} where the weights again work out and we just multiplied a holomorphic primary with an anti-holomorphic primary. This estabishes the left hand side of the recombination rule.

To get the right hand side, consider the exact same $m = n = 1$ operator as before but now in a theory where $r = \sqrt{2} + \epsilon$. Calculating the charges to leading order in $\epsilon$, we find \begin{equation} e^{i\sqrt{2} X(z)} e^{-i \epsilon \bar{X}(\bar{z})} := [1]_{1 + \epsilon^2 + O(\epsilon^3)}. \end{equation} This is the multiplet we wanted to find. It is intuitively clear that it represents the "combination" of the two seen before because the conformal primary is infinitesimally close to what we called $[1]_1$ and its $\bar{\partial}$ descendant, now non-vanishing, is infinitesimally close to $[0]_2$ times a "coupling constant".


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