I'm looking into the non-unitary minimal model $\mathcal M_{5,2}$ associated with the Yang-Lee edge singularity. I'm trying to justify which conformal dimensions we expect to appear (easy enough) but having a hard time completely justifying which solutions we don't choose. Let me lay out the theory as I understand it and a sketch of my current justifications and hopefully someone can reaffirm my theory, point out an error or provide an alternative way to achieve the same result.


The conformal dimensions that we expect to show up in this model are given by the Kac determinant (the determinant of the Gram matrix $G^{(n)}$ at level $n$) $$ \det G^{(n)} = A_n\prod_{\substack{r,s \geq 1 \\rs \leq n}} (h-h_{r,s})^{\mathcal P(n-rs)} $$ with $\mathcal P$ being the partition function from combinatorics and $h_{r,s}$ given by the coprime integers of the minimal model $\mathcal M_{p,q}$ as $$ h_{r,s} = \frac{(pr - qs)^2 - (p-q)^2}{4pq} $$ The integers $(p, q)$ clamp the Kac indices $r \in [1, q), s \in [1, p)$, hence we have $r = 1, s \in \{1,2,3,4\}$ for the Yang-Lee case. We care about the cases where the Kac determinant vanishes, so we care about the solutions $$ h_{1,1}, h_{1,2}, h_{1,3}, h_{1,4} $$ where $rs$ is the level at which the solution first occurs at. By a symmetry property present in the Kac solutions we identify \begin{align*} h_{1,1}&= h_{1,4}\\ h_{1,2}&= h_{1,3} \end{align*} and we expect to see these solutions appearing at levels 1 through 4 respectively. This is the Kac side of the theory, which gives a solution for the product of the eigenvalues (aka the determinant). On the other hand, we can work out the Gram matrix directly (which is easy enough for small $n$) and calculate its determinant. For $n=2$, the Gram matrix is \begin{align*} G^{(2)}(c,h) &= \begin{pmatrix} 4h(1+2h) & 6h \\ 6h & 4h+\frac{c}{2} \end{pmatrix} \end{align*} where for the Yang-Lee theory the central charge is $c=-22/5$. At level 3 the Gram matrix is \begin{align*} G^{(3)}(c,h) &= \left( \begin{array}{ccc} 24 (h+1) (2 h+1) & 12 h (3 h+1) & 24 h \\ 12 h (3 h+1) & h (c+8 h+8) & 10 h \\ 24 h & 10 h & 2 c+6 h \\ \end{array} \right) \end{align*}


Now, my confusion lies in the fact that the determinant of the Gram matrix at level 2 has three solutions, namely $h = 0, h = -1/5$ and $h = 11/8$ (for $c=-22/5$). The $h=0$ solution makes sense and this is $h_{1,1}=h_{1,4} = 0$. However, by symmetry we can only choose one of the latter two. The literature chooses $h=-1/5$, and indeed this value also features as a solution at level 3 as well (the other level 3 solutions I think are irrational—which is problematic—and 0) so this is reassuring. In the Big Yellow Book, an exercise in chapter 7 has the reader construct the OPE of the associated singular vector field with the stress-energy tensor to identify the conformal dimension that pops up in there, which again is $-1/5$.

Based on these two bits of data I can accept that $h=-1/5$ is the desired solution, but what exactly does the $h=11/8$ solution represent? I can disregard the irrational solutions at level 3 as this has implications regarding the finitely many primary operators in the theory, but this doesn't apply to $11/8$. The other thing that strikes me is that $11/8$ is positive yet the representation should be non-unitary. Does this mean we must choose the negative solution, or something? Non-unitarity should already be guaranteed by $c < 0$ so I think I might be barking up the wrong tree with this angle.


1 Answer 1


The $h = 11/8$ solution is $h_{2,1}$ which also obeys $r \geq 1$, $s \geq 1$ and $rs \leq 2$. The short answer for why people drop it is that it lies outside the Kac table. Are you asking for a proof of the fact that $r \in [1, q)$, $s \in [1, p)$ is necessary for getting finite fusion rules?

Proving that it's sufficient could be subtle because one needs to know that it's consistent to identify operators with the same conformal weight. But it should be easy to see that it's necessary. Fusing $\phi_{2,1}$ with itself produces $\phi_{3,1}$ which needs to be added since it has a conformal weight which has not been seen before. Then you can repeat the process.

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    $\begingroup$ Oh, this actually clears it up completely for me. Simply writing it as $h_{2,1}$ makes it obvious why we drop it. It's the same reason we don't consider, for example, $h_{4,2}$ in the $\mathcal M_{4,3}$ minimal model: it has $r \notin [1, q)$. Just another example of how just talking about a problem can lead you to see simple things that you've overlooked! $\endgroup$
    – Zxv
    Commented Aug 3, 2023 at 2:54

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