A standard Toda field theory action will be of the shape:

$$ S_{\text{TFT}} = \int d^2 x~ \Bigg( \frac{1}{2} \langle\partial_\mu \phi, \partial^\mu \phi \rangle - \frac{m^2}{\beta^2} \sum_{i=1}^r n_i e^{\beta \langle\alpha_i, \phi\rangle} \Bigg)$$

Where r is the rank of the algebra with simple roots $\alpha_i$. Adding the affine extension to the algebra corresponds to adding an extra root $\alpha_0$, and I read everywhere (e.g. Mussardo's stat. field theory book, and this PhD thesis: https://arxiv.org/abs/hep-th/0008200) that while normal Toda field theories are conformally invariant, this added root destroys the invariance.

My first question is: How is the non-affine case a CFT? When expanding the exponential to look at the quadratic terms, we get a mass-squared operator $M^2_{ij}= m^2 \sum_{k=1}^r n_i (\alpha_k)^i (\alpha_k)^j$. I don't see why this apparently is always zero in the non-affine case.

Second question: How can I see that the extra root finally makes these masses nonzero.



1 Answer 1


The simplest example of conformal Toda theory is Liouville theory. Conformal invariance of Liouville theory is explained in the Wikipedia article. The idea is that you should compute the dimension of the exponential interaction term using the appropriate energy-momentum tensor. (Looking at the action alone is not enough.)

In conformal Toda theory all the interaction terms have dimension $1$ and are therefore marginal. This does not yet show that the theory is conformal: a sum of marginal operators is not always exactly marginal. Intuitively, the idea here is that in a theory of $b$ bosons, a sum of $r$ independent marginal exponential interaction terms is still marginal if $r\leq b$. In conformal Toda theory you have $r=b$ so you do have conformal symmetry. In the affine case you have $r=b+1$ so you lose exact conformal symmetry.

The bound on the number of interaction terms comes from trying to perturbatively compute corrections to correlation functions. You are perturbing a free bosonic theory where momentum conservation constrains the contributions of interaction terms. There are $b$ momentum conservation rules, which completely determine the contributions of all interaction terms to a given correlator if $r\leq b$. This prevents conformal symmetry from being broken.

  • $\begingroup$ Do you maybe have a reference that goes into a bit more detail w.r.t. summing marginal exponential couplings? $\endgroup$
    – Abelaer
    Commented Jul 13, 2018 at 12:03
  • $\begingroup$ No idea, sorry. $\endgroup$ Commented Jul 13, 2018 at 12:37

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