# Question on Eq. 7.40 of "Conformal Field Theory" by Di Francesco et. al

I am trying to understand the second line of (7.40), which I've written below. \begin{align}\langle \alpha|\alpha\rangle &= c_\alpha h^{n(\alpha)}[1 + O(1/h)]\\ \langle \alpha | \beta \rangle &= O(h^{(n(\alpha) + n(\beta))/2 - 1}) + \cdots \end{align} where $$|\alpha\rangle$$ is a ''fixed length'' basis state of length $$n(\alpha)$$, i.e. is it a state of the form $$L_{-k_1}L_{-k_2}\cdots L_{-k_n}|{h}\rangle$$ where the number of operators appearing is the length $$n(\alpha)$$ and the $$k_i$$ appear in decreasing order. In particular it seems to me that the second line cannot be correct, with some straightforward counter examples. Take, for example $$|\alpha\rangle = L_{-1}^n|h\rangle$$ and $$\beta\rangle = L_{-n}|h\rangle$$. Then $$\langle \alpha |\beta\rangle = \langle h |L_1^n L_{-n}|h\rangle = h(n+1)!$$ which of course goes as $$O(h)$$ (not, as the formula would predict, as $$O(h^{(n-1)/2})$$). What am I doing wrong here? I don't see anything about this equation in the posted errata for the textbook.

I think you must be right, unless I'm also misunderstanding something, this doesn't even hold for the other off-diagonal elements in the Gram matrix at level 2 and 3, e.g. $$\langle{h}|L_1^2 L_{-2}|{h}\rangle = O(h)$$ at level 2, $$\langle{h}|L_2 L_1 L_{-1}^3|{h}\rangle = O(h^2)$$ and $$\langle{h}|L_3 L_{-1} L_{-2}|{h}\rangle = O(h)$$ at level 3, (your nice example differs even more though). I don't think this is important for the proof that the representations are unitarity for $$c\geq 1$$, and at a first glance I don't see anything like this discussion in the original CFT papers, but possibly the correct behavior might be something like $$\langle{\alpha|\beta}\rangle = O(h^{{\rm min}\{n(\alpha), n(\beta)\}})$$.