Consider the OPEs of a 2D CFT, with nondegenerate primary fields labeled by some collection $\mathtt{Prim}$ of indices. The data specifying the OPE can be taken as a collection $$\{C_{p,q}^{r;P}\}_{p,q,r\in \mathtt{Prim},P \in \mathrm{Part}}\subset \mathbb{C},$$ where $\mathrm{Part}$ is the collection of all multisets of positive integers (including the empty multiset). Hence, an element $P\in\operatorname{Part}$ can be uniquely identified with a partition of some nonnegative number into positive parts and vice versa. (To each such partition we can associate an element $L_{P}=L_{-n_1}\cdots L_{-n_K}$, where $n_1\geq\cdots\geq n_K\geq 1$ are the elements of $P$, with multiplicity.) The complex number $C^{r;P}_{p,q}$ is then the coefficient of $L_P$ applied to the $r$th primary field in the OPE between the $p$th and $q$th primary fields.

Given this setup, we can solve for $C^{r;P}_{p,q}$ in terms of $C^r_{p,q}=C^{r;\varnothing}_{p,q}$ using conformal invariance. See [1], Section 6.6.3. However, this algorithm seems to use non-degeneracy quite centrally, as solving for the coefficients $C^{r;P}_{p,q}$ at level $N$ requires inverting the Kac-Gram matrix at level $N$. Concretely, in order to solve Eq. 6.180 in [1], one must invert the matrix on the left-hand side, which is essentially the Kac-Gram matrix. Clearly, this fails when discussing the analogous problem in the presence of degenerate fields, e.g. when working with minimal models, since the Kac-Gram matrix is non-invertible. (The denominators in Eq. 6.181 can vanish, for instance.) This is discussed briefly in [1], Section 8.A, where the connection is made to the fusion rules of minimal models. However, I do not think that book contains a proof of something like the following, analogous to the result that applies when the Kac-Gram matrix is invertible at every level:

Letting $\{C^{r;P}_{p,q}\}_{p,q,r,P}$ denote a set of indended OPE coefficients for a theory with the primary field content specified by the minimal models, each $C^{r;P}_{p,q}$ is uniquely determined by the collection $\{C^{r;\varnothing}_{p,q}\}_{p,q,r}$ (up to the coefficients of null fields), and conversely given $\{C^{r;\varnothing}_{p,q}\}_{p,q,r} \subset \mathbb{R}$ (which may be subject to some specified finite set of constraints) there exists some collection $\{C^{r;P}_{p,q}\}_{p,q,r,P}$ of numbers such that the corresponding OPE is conformally invariant in the sense of Section 6.6 of [1].

Since this seems important to the construction of minimal models via the bootstrap, I assume it is true, but it does not appear straightforward to extract a proof from the content of [1]. Is there somewhere I can find a proof of this result? Requiring that the right-hand side of the $N$th level analogue of Eq. 6.180 is in the range of the matrix on the left-hand side for all $N$ naively imposes an infinite number of constraints on the conformal dimensions $h_p,h_q,h_r$ associated to $p,q,r$ when the $r$th primary field is degenerate, so from this perspective the proposition above would seem quite miraculous.

If one defines the term "fusion rules" to refer to the triples of $h_r,h_p,h_q$ such that the proposition holds, then the question becomes to show that the restrictions on $h_r,h_p,h_q$ described in Chapter 7 pf [1] are actually the fusion rules. (E.g., why does the discussion in Section 7.3 suffice?)

[1] Di Francesco, Mathieu, and Senechal, Conformal Field Theory

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    $\begingroup$ The idea is that the fusion rules (i.e. allowed triples $p,q,r$) are precisely such that what you want is true. Is [1] using a different definition of the fusion rules? (Personally, I always think about this or a trivial variation this as the definition.) $\endgroup$ Oct 31, 2021 at 3:04
  • $\begingroup$ One can define fusion rules in that way, sure. The problem then becomes to show that the finitely many restrictions on $h_p,h_q,h_r$ described in the yellow book are then (necessary and) sufficient to satisfy the a priori infinite number of restrictions (one for each $N$) described in the question. I believe the yellow book does the necessary part but not the sufficient part. $\endgroup$ Oct 31, 2021 at 10:54
  • $\begingroup$ To answer your question though, [1] does use a different definition of the fusion rules. See pg. 214: "The conditions under which a given conformal families occurs [in the OPE] are called fusion rules... we say, for instance, that the fusion of two conformal fields is possible if the three point function is not zero." I have edited the question to make clear what I'm getting at. $\endgroup$ Oct 31, 2021 at 11:14
  • $\begingroup$ That's the definition of a fusion rule in general. But to determine the fusion rules you're talking about (i.e. the set of non-zero three point functions in minimal models), [1] takes the standard approach of solving consistency conditions. $\endgroup$ Oct 31, 2021 at 17:15
  • $\begingroup$ Sure Connor, I'm not disagreeing with that; the point of the question was how to prove that [1] has really solved enough consistency conditions. That's why I said above that they cover necessity but not sufficiency. As my answer below gets at, the key is a non-trivial fact about generators of the null submodule. $\endgroup$ Oct 31, 2021 at 17:22

2 Answers 2


Since you are concerned with conformal invariance rather than unitarity, there is no need to consider a full set of Virasoro primaries which close under fusion. We can just show that one primary coefficient $C^{r,\varnothing}_{p,q}$ determines all of the descendant coefficients $C^{r,P}_{p,q}$ where the $P$'s are taken to span the Verma module that has null states modded out.

To achieve this, the Kac-Gram matrix doesn't need to be invertible... it just needs to be symmetric. In terms of the "partition function" from number theory, the $N$-th block has a size of $p(N) \times p(N)$ so we can diagonalize it level-by-level. Since the Verma module for $r$ is degenerate, some of the eigenvalues will be zero but this is where we mod out. We keep only the eigenvectors with non-zero eigenvalues and from these we get a collection of operators $$\mathcal{O}_{r,i} = \sum_{|P| = N} \alpha_{i,P} L_P \mathcal{O}_r$$ such that $$\left < \mathcal{O}_{r,i}(z_1) \mathcal{O}_{r,j}(z_2) \right > = \frac{\delta_{ij}}{z_{12}^{2h_r + 2N}}.$$

Now we just need to know the 3pt coefficients which appear in $\left < \mathcal{O}_p \mathcal{O}_q \mathcal{O}_{r,i} \right >$. It is tedious, but for all $L_P$ generators, we can always "take them out" of $\left < \mathcal{O}_p \mathcal{O}_q L_P \mathcal{O}_r \right >$ by repeatedly using $$L_{-n} \mapsto \mathcal{L}_{-n} = \sum_{i \neq 3} \frac{h_i(n - 1)}{(z_i - z_3)^n} - \frac{\partial_i}{(z_i - z_3)^{n-1}}$$ which the yellow book proves starting from the Virasoro Ward identity.

  • $\begingroup$ I don't believe this answers the question, since the non-trivial part is what restrictions on $h_p,h_q,h_r$ guarantee that one can iteratively solve for the higher coefficients. If the Kac-Gram matrix is not invertible, then one may not be able to solve the linear equation that arises at each step, hence the connection to the fusion rules. Thus, the heart of the question is to prove that the fusion rules which are stated in the yellow book actually suffice to iteratively solve for $C^{r;P}_{p,q}$. $\endgroup$ Oct 31, 2021 at 10:58
  • $\begingroup$ *intended fusion rules. $\endgroup$ Oct 31, 2021 at 11:04
  • $\begingroup$ The problem, when you have a non-invertible Kac-Gram matrix, is that the vector on the right-hand side of the equation to be solved may not lie in the range. $\endgroup$ Oct 31, 2021 at 11:21
  • $\begingroup$ I think the discussion in Section 7.3 is equivalent to writing down the constraints that come from the level N=2 discussion. Then, the claim should be that the differential equations under consideration are as a whole equivalent to the proposition in the original question, and moreover that the constraints coming from singular vectors at higher levels are not all independent. $\endgroup$ Oct 31, 2021 at 11:34
  • $\begingroup$ The important point missing from your answer is that when approaching the construction of the OPE in the way that you describe, while conformal invariance lets you solve for the coefficients of non-null descendants, each null descendant yields a constraint. Since there are infinitely many null descendants, you get infinitely many constraints. So, one could prove the proposition in the original question by (1) proving that it is equivalent to showing that the $h_p,h_q,h_r$ satisfy all the constraints, (2) proving that a small set of constraints on the $h$s suffice to satisfy all constraints. $\endgroup$ Oct 31, 2021 at 12:26

Here's a sketch of a proof, based partly on Connor's answer.

One first uses the symmetricness of the Kac-Gram matrix to find a basis $\{J_n\}_{n=0}^\infty \cup \{\Delta_n\}_{n=0}^\infty$ of the universal enveloping algebra of the negative part of the Virasoro algebra such that in the Verma module $V(c,h_r)$, the set $\{J_n|h_r\rangle\}_{n=0}^\infty$ is orthonormal and all $\Delta_n |h_r\rangle$ are null. Now, we can prove that the proposition in the original question will hold for $p,q,r \in \mathtt{Prim}$ if and only if $\mathcal{L}_{\Delta_n} \langle \mathcal{O}_p(w) \mathcal{O}_q(w')\mathcal{O}_r(z)\rangle=0$ holds for all $n$, where $\mathcal{L}_{\Delta_n}$ is the ordinary differential operator in $z$ associated to $\Delta_n$. (Here $\langle \mathcal{O}_p(w) \mathcal{O}_q(w')\mathcal{O}_r(z)\rangle$ denotes the 3-point function associated to $h_p,h_q,h_r$, with arbitrary nonzero coefficient.)

Now, we appeal to the claim that there exist two $\Delta,\Delta'$ (which I think we can take WLOG to be among the $\Delta_n$) such that all of the $\Delta_n$ are in the left-submodule of the Virasoro enveloping algebra generated by $\Delta,\Delta'$. Hence, the ODE $\mathcal{L}_{\Delta_n} \langle \mathcal{O}_p(w) \mathcal{O}_q(w')\mathcal{O}_r(z)\rangle=0$ holds for all $n$ if and only if the two ODEs $\mathcal{L}_{\Delta} \langle \mathcal{O}_p(w) \mathcal{O}_q(w')\mathcal{O}_r(z)\rangle=0$, $\mathcal{L}_{\Delta'} \langle \mathcal{O}_p(w) \mathcal{O}_q(w')\mathcal{O}_r(z)\rangle=0$ both hold. Each of these two is equivalent to a constraint on $h_p,h_q,h_r$, and this is discussed adequately in Section 8.3 of [1].


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