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.


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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 $\langle{\alpha|\beta}\rangle = O(h^{{\rm min}\{n(\alpha), n(\beta)\}})$.

  • $\begingroup$ Yeah I agree it seems like the behavior you said, I'm not sure how one would even get to the result they have? But agreed that it doesn't affect the result of their analysis. $\endgroup$ Jul 4, 2020 at 15:04

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