I want to prove the fact that there are infinite number of primary operators in CFT by Conformal bootstrap.

However, for that I need to show that the crossed conformal blocks $g_{\Delta,\ell}(1-z,1-\bar{z})$ behave like $\log z$ in the limit $z \to 0$ and $z = \bar{z}$.

In 2D and 4D, rather in any even dimensions, using the explicit expression due to Dolan and Osborn I managed to prove this.

However I am clueless about how to do it for general dimensions.

It would be extremely helpful if someone could shed some light into this matter.


2 Answers 2


You don't need to set $z=\bar{z}$ for the $\log z$ to appear, you can leave $\bar{z}$ arbitrary. Then, take the limit $\bar{z} \to 0$ and the block in any dimension becomes $$ g_{\Delta,\ell}(z,\bar{z}) \to \bar{z}^{\frac{\Delta-\ell}2} k_{\Delta+\ell}(z) $$ where $$ k_\beta(z) = z^{\beta/2} {}_2F_1(\beta/2,\beta/2,\beta,z) $$ is the $SL_2(\mathbb{R})$-block. You can see this from the Casimir operator, $$ C_2 = D_z + D_{\bar{z}} + (d-2) \frac{z\bar{z}}{z-\bar{z}}[(1-z)\partial_z-(1-\bar{z})\partial_{\bar{z}})] $$ where $D$ is some differential operator. The third term vanishes like $\mathcal{O}(\bar{z})$ and hence the conformal block factorizes. Now take the limit $z \to 1$ and the $\log 1-z$ ($\log z$ in the t-channel) appears.

  • 2
    $\begingroup$ See also arxiv.org/abs/1509.00014 $\endgroup$ Jul 12, 2021 at 15:33
  • $\begingroup$ Thank you very much. I think I understand your answer. In the quadratic casimir if we take the term $2(d-2)\frac{z\bar{z}}{z-\bar{z}}(.....)$ to be 0 I get the answer. Please correct me if I am wrong. $\endgroup$ Jul 13, 2021 at 10:03
  • $\begingroup$ Yes, I've edited the answer for clarification. $\endgroup$
    – and008
    Jul 13, 2021 at 10:37
  • $\begingroup$ I would like to ask another question. The hypergeometric functions are known to be analytic everywhere(if I am not wrong) then how does the logz term appear which is not analytic near z→0. I mean I understand there is a derivation. But if we look at the equation broadly, we have an analytic function in the L.H.S. and a non-analytic function in the R.H.S. $\endgroup$ Jul 17, 2021 at 9:59
  • $\begingroup$ The hypergeometric function (for generic arguments) has a branch cut from $1$ to $\infty$, so in particular it's not analytic at $z=1$. $\endgroup$
    – and008
    Jul 17, 2021 at 17:07

I am hereby writing an answer to my own question. Please actively comment on your opinions and let me know if my answer works.

From DSD's notes, we have the following expression for conformal blocks using the "rho-configuration" of the 4-pt functions:

$g_{\Delta_{O},\ell_{O}}(z,\bar{z}) = \sum_{n = 0,2,\cdots,j}B_{n,j}r^{\Delta + n}C_{j}^{\frac{d-2}{2}}(\cos\Theta)$

Now, restricting to $z = \bar{z} \implies \Theta = 0$.

So that means we are looking at $C_{j}^{\frac{d-2}{2}}(1)$. But lets, do it in a different way.

Lets, look at $C_{j}^{\frac{d-2}{2}}(1-x)$ where $x \to 0$ also let us look at only the $n=0$ order term. From, wikipedia: Any n-th order coefficient can be written as:

$C_{n}^{(\alpha)}(x) = \frac{(2\alpha)_{n}}{n!}{}_{2}F_{1}(-n,2\alpha + n;\alpha +1/2;\frac{1-x}{2})$

So, we get:

$ C_{0}^{(\alpha)}(1-x) \propto \log x$ when $x \to 0$.

Thus, we get a logarithmic dependence in the crossed blocks in any dimensions.

  • $\begingroup$ No, this expansion is valid only if $j$ is an integer. Hence, the Gegenbauer polynomial is indeed a polynomial and a log never appears. Moreover, this log would correspond to the limit $\theta \to 0$ which doesn't make sense physically. You are interested in the limit $r \to 1$. $\endgroup$
    – and008
    Jul 13, 2021 at 10:45
  • $\begingroup$ Thank you very much for the comment. I understand the mistake here. $\endgroup$ Jul 13, 2021 at 10:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.