Twist Accumulation in Conformal Field Theory: A Rigorous Approach to the Lightcone Bootstrap

Abstract

We prove that in any unitary CFT, a twist gap in the spectrum of operator product expansion (OPE) of identical scalar quasiprimary operators (i.e. $\phi \times \phi$) implies the existence of a family of quasiprimary operators ${\mathcal {O}}_{\tau , \ell }$ with spins $\ell \rightarrow \infty$ and twists $\tau \rightarrow 2 \Delta _{\phi }$ in the same OPE spectrum. A similar twist-accumulation result is proven for any two-dimensional Virasoro-invariant, modular-invariant, unitary CFT with a normalizable vacuum and central charge $c > 1$, where we show that a twist gap in the spectrum of Virasoro primaries implies the existence of a family of Virasoro primaries ${\mathcal {O}}_{h, {\bar{h}}}$ with $h \rightarrow \infty$ and ${\bar{h}} \rightarrow \frac{c - 1}{24}$ (the same is true with h and ${\bar{h}}$ interchanged). We summarize the similarity of the two problems and propose a general formulation of the lightcone bootstrap.

Appendices

Approximate Factorization of Conformal Blocks

In this appendix we will prove Lemma 2.7. We will find it convenient to prove a slightly different version of the lemma:

Lemma A.1

(Approximate factorization, alternative version) Let $a, b \in (0, 1)$, $a < b$. Let $d\geqslant 2$ and assume that $\Delta , \ell$ satisfy the unitarity bounds. If $\ell = 0$, assume in addition that $\Delta > \nu = \frac{d - 2}{2}.$ Then the conformal block $g_{\tau , \ell } (z, {\bar{z}})$ satisfies the following two-sided bound

$$\begin{aligned} 1 \leqslant \frac{g_{\tau , \ell } (z, {\bar{z}})}{\frac{(\nu )_{\ell }}{(2 \nu )_{\ell }} k_{\tau } ({\bar{z}}) k_{\tau + 2 \ell } (z)} \leqslant K_d (a, b) \left( 1 + \frac{\theta (d \geqslant 3)}{\Delta - \nu } \right) \qquad (\forall 0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1), \nonumber \\ \end{aligned}$$

(A.1)

where $K_d (a, b)$ is a finite constant independent of $\tau , \ell$ and of $z, {\bar{z}}$ in the shown range.

Let us quickly see why this version implies Lemma 2.7. We have an identity:

$$\begin{aligned} k_{\beta } (z)= & {} (4 \rho )^{\beta / 2} G_{\beta } (\rho ), \end{aligned}$$

(A.2)

$$\begin{aligned} G_{\beta } (\rho )&{:}{=}&{}_2 F_1 \left( \tfrac{1}{2}, \tfrac{\beta }{2}; \tfrac{\beta + 1}{2}; \rho ^2 \right) , \quad \rho = \frac{z}{\left( 1 + \sqrt{1 - z} \right) ^2}. \end{aligned}$$

(A.3)

We also have $1 \leqslant G_{\beta } (\rho ) \leqslant \frac{1}{\sqrt{1 - \rho ^2}}$ (Lemma A.4 below). Hence the denominator of (A.1) and the denominator of (2.29) are related to each other by

$$\begin{aligned} 1 \leqslant \frac{\frac{(\nu )_{\ell }}{(2 \nu )_{\ell }} k_{\tau } ({\bar{z}}) k_{\tau + 2 \ell } (z)}{{\bar{\rho }}^{\tau / 2} F_{\tau , \ell } (z)} \leqslant C (a), \end{aligned}$$

(A.4)

where C(a) does not depend on $\tau$ or $\ell$.

We will first prove the 2D case (App. A.1), then give an inductive proof in $d \geqslant 3$ dimensions (App. A.3), using Hogervorst’s dimensional reduction formula [14], which expands d-dimensional blocks in terms of $(d - 1)$-dimensional ones (App. A.2).

Remark A.2

Following the arguments below carefully, it’s easy to see that $K_d (a, b) = 1 + O (a)$ for $a \rightarrow 0$ and b fixed.

Remark A.3

There is another possible way of proving Lemma A.1 but we have not managed to realize it. In any dimension, one can expand the conformal block into collinear blocks:

$$\begin{aligned} g_{\Delta , \ell } (z, {\bar{z}}) = \underset{n = 0}{\overset{\infty }{\sum }} \underset{k = - n}{\overset{n}{\sum }} c_{n, k} {\bar{z}}^{\tau / 2 + n} k_{\Delta + \ell + 2 k} (z). \end{aligned}$$

(A.5)

Here the coefficients $c_{n, k}$ are positive for a group-theoretical reason: we decompose the $\text {SO} (2, d)$ representation into irreducible representations of the subgroup $\text {SL} (2; {\mathbb {R}})$ (which preserves the z-axis), and $c_{n, k}$ is the sum of squared inner products:

$$\begin{aligned} c_{n, k} = \sum \limits _{\psi } |\langle \psi |\phi (1) \phi (0) \rangle |^2, \end{aligned}$$

(A.6)

where the sum is over an orthonormal basis of collinear primary states $\psi$ with collinear twist $h = \frac{\Delta + \ell }{2} + k$ and scaling dimension $\Delta + n + k$.

Lemma A.1 would follow if we had an upper bound on $c_{n, k}$ implying the inequality

$$\begin{aligned} \begin{aligned}&\underset{n = 0}{\overset{\infty }{\sum }} \underset{k = - n}{\overset{n}{\sum }} \frac{c_{n, k}}{c_{0, 0}} {\bar{z}}^n \frac{k_{\Delta + \ell + 2 k} (z)}{k_{\Delta + \ell } (z)} \leqslant K_d (a, b) \left( 1 + \frac{\theta (d \geqslant 3)}{\Delta - \nu } \right) \\&\quad \left( \forall 0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1,\quad \tau \leqslant \tau _{\max } \right) . \end{aligned} \end{aligned}$$

(A.7)

Such a bound can be easily derived in 2D, but we do not know how to derive it in $d \geqslant 3$. There have been papers deriving explicit expressions for the coefficients $c_{n, k}$ (or coefficients of similar expansions where z is replaced by u and ${\bar{z}}$ by $1 - v$) [27, 29, 30, 40, 41], but we did not manage to apply them. We leave realization of this strategy as an open problem.

1.1 $d = 2$

Recall that the ratio $(\nu )_{\ell } / (2 \nu )_{\ell }$ is defined for $\nu = 0$ as in footnote 7. In our normalization (2.26), the 2D conformal blocks are given by

$$\begin{aligned} g^{(2)}_{\tau , \ell } (z, {\bar{z}}) = {\left\{ \begin{array}{ll} k_{\tau } (z) k_{\tau } ({\bar{z}}) &{} (\ell = 0) \\ \frac{1}{2} [k_{\tau } ({\bar{z}}) k_{\tau + 2 \ell } (z) + k_{\tau } (z) k_{\tau + 2 \ell } ({\bar{z}})] &{} (\ell \geqslant 1). \end{array}\right. } \end{aligned}$$

(A.8)

For $\ell = 0$, the upper and lower bounds are obvious with constants 1 and in the full range of $z, {\bar{z}}$ (in $d = 2$ there is no pole $1 / (\Delta - \nu )$).

Consider then the case $\ell \geqslant 1$. Since $k_{\beta } (z) > 0$ for $z \in (0, 1)$, $\beta \geqslant 0$, the lower bound $g^{(2)}_{\tau , \ell } (z, {\bar{z}}) \geqslant {\frac{1}{2}} k_{\tau } ({\bar{z}}) k_{\tau + 2 \ell } (z)$ follows by dropping the second term in the block.

For the upper bound, let us write

$$\begin{aligned} g^{(2)}_{\tau , \ell } (z, {\bar{z}}) = \frac{1}{2} k_{\tau + 2 \ell } (z) k_{\tau } ({\bar{z}}) [1 + X], \quad X = \frac{k_{\tau + 2 \ell } ({\bar{z}}) k_{\tau } (z)}{k_{\tau + 2 \ell } (z) k_{\tau } ({\bar{z}})}. \end{aligned}$$

(A.9)

We will use identity (A.2). We have the following lemma about $G_{\beta } (\rho )$:

Lemma A.4

For any $0 \leqslant \beta _1 \leqslant \beta _2 < \infty$ we have

$$\begin{aligned} 1 \leqslant G_{\beta _1} (\rho ) \leqslant G_{\beta _2} (\rho ) \leqslant \frac{1}{\sqrt{1 - \rho ^2}} \,. \end{aligned}$$

(A.10)

Proof

$G_{\beta _1} (\rho ) \leqslant G_{\beta _2} (\rho )$ follows from the monotonicity of coefficients in the series expansion of $_2 F_1$. We also have $G_0 (\rho ) = 1$ and $G_{\infty } (\rho ) = \frac{1}{\sqrt{1 - \rho ^2}}$. $\square$

Using (A.2), the ratio X is rewritten as

$$\begin{aligned} X = \left( \frac{{\bar{\rho }}}{\rho } \right) ^{\ell } \frac{G_{\tau } (\rho )}{G_{\tau + 2 \ell } (\rho )} \frac{G_{\tau + 2 \ell } ({\bar{\rho }})}{G_{\tau } ({\bar{\rho }})} \leqslant 1 \cdot 1 \cdot \frac{1}{\sqrt{1 - \rho (a)^2}}, \end{aligned}$$

(A.11)

using ${\bar{\rho }} \leqslant \rho$, and Lemma A.4 for the other two factors. Hence X is uniformly bounded by C(a). This implies the upper bound with $C_2 (a, b) = 1 + C (a)$. The $d = 2$ case is complete.

1.2 Dimensional reduction formula

We will need Hogervorst’s dimensional reduction formula. We will state and use this result using blocks $g_{\Delta , \ell }$ labeled by the dimension and spin, not by the twist and spin as in the rest of the paper. Hopefully no confusion will arise.

Theorem A.5

(Hogervorst [14], Eq. (2.24)) Conformal blocks in d dimensions can be expanded in a series of conformal blocks in $d - 1$ dimensions, as

$$\begin{aligned} g_{\Delta , \ell }^{(d)} (z, {\bar{z}}) = \underset{n = 0}{\overset{\infty }{\sum }} \underset{p = 0}{\sum ^{[\ell / 2]}} {\mathcal {A}}_{n, \ell - 2 p}^{(d)} (\Delta , l) g^{(d - 1)}_{\Delta + 2 n, \ell - 2 p} (z, {\bar{z}}), \end{aligned}$$

(A.12)

where ($j = \ell - 2 p$)

$$\begin{aligned} {\mathcal {A}}^{(d)}_{n, j} (\Delta , \ell )= & {} Z_{\ell , j}^{(d)} \frac{\left( \frac{1}{2} \right) _n}{2^{4 n} n!} \frac{\left( \frac{\Delta + j}{2} \right) _n \left( \frac{\Delta - 2 \nu - j + 1}{2} \right) _n}{\left( \frac{\Delta + j - 1}{2} \right) _n \left( \frac{\Delta - 2 \nu - j}{2} \right) _n} \nonumber \\{} & {} \times \frac{(\Delta - 1)_{2 n} \left( \frac{\Delta + \ell }{2} \right) _n \left( \frac{\Delta - \ell - 2 \nu }{2} \right) _n}{(\Delta - \nu )_n \left( \Delta - \nu - \frac{1}{2} + n \right) _n \left( \frac{\Delta + \ell + 1}{2} \right) _n \left( \frac{\Delta - \ell - 2 \nu + 1}{2} \right) _n}, \end{aligned}$$

(A.13)

$$\begin{aligned} Z_{\ell , j}^{(d)}= & {} \frac{\left( \frac{1}{2} \right) _p \ell !}{p! j!} \frac{(\nu )_{j + p} (2 \nu - 1)_j}{\left( \nu - \frac{1}{2} \right) _{j + p + 1} (2 \nu )_{\ell }} (j + \nu - 1 / 2). \end{aligned}$$

(A.14)

Some remarks are due here. First, the existence of an expansion like (A.12) can be inferred very generally from decomposing a representation of the d-dimensional conformal block under the $(d - 1)$-dimensional conformal group. It also follows that the coefficients ${\mathcal {A}}^{(d)}_{n, j} (\Delta , \ell )$ have to be nonnegative if $(\Delta , \ell )$ satisfies the d-dimensional unitarity bounds. These condition are verified by the above expressions. Moreover, these coefficients are regular (no poles) for $\ell = 0, \Delta > \nu$ and $\ell \geqslant 1$, $\Delta \geqslant \ell + 2 \nu$.

Hogervorst [14] derived recursion relations satisfied by the coefficients ${\mathcal {A}}^{(d)}_{n, j} (\Delta , \ell )$. He then guessed (A.13) as the (unique) solution of those recursion relations. He then could check up to very high recursion order that (A.13) indeed remains compatible with the recursion. The complete proof that (A.13) is a solution of Hogervorst’s recursion relations is so far missing, but there is hardly any doubt that they are correct.

It may perhaps be possible to prove (A.13) by comparing with the pole structure of Zamolodchikov-like recursion relations [42] which have been established rigorously [43].

1.3 $d \geqslant 3$ lower bound

We will argue by induction using (A.12). The needed lower bound is obtained by keeping the term $n = 0, j = \ell$ and throwing out the rest (recall that (A.12) is a positive sum). We have

$$\begin{aligned} g_{\Delta , \ell }^{(d)} (z, {\bar{z}}) \geqslant {\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell ) g^{(d - 1)}_{\Delta , \ell } (z, {\bar{z}}). \end{aligned}$$

(A.15)

From (A.13), it can be seen that:

$$\begin{aligned} {\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell ) \frac{(\nu - 1 / 2)_{\ell }}{(2 \nu - 1)_{\ell }} = \frac{(\nu )_{\ell }}{(2 \nu )_{\ell }}. \end{aligned}$$

(A.16)

Hence by induction we get the lower bound

$$\begin{aligned} g_{\tau , \ell }^{(d)} (z, {\bar{z}}) \geqslant \frac{(\nu )_{\ell }}{(2 \nu )_{\ell }} k_{\tau } ({\bar{z}}) k_{\tau + 2 \ell } (z) . \end{aligned}$$

(A.17)

1.4 $d \geqslant 3$ upper bound

Here we will prove the upper bound in $d \geqslant 3$

$$\begin{aligned} \frac{g_{\tau , \ell } (z, {\bar{z}})}{\frac{(\nu )_{\ell }}{(2 \nu )_{\ell }} k_{\tau } ({\bar{z}}) k_{\tau + 2 \ell } (z)} \leqslant K_d (a, b) \left( 1 + \frac{1}{\Delta - \nu } \right) \qquad (\forall 0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1) .\qquad \end{aligned}$$

(A.18)

Note that the pole $\Delta = \nu$ can be approached only by the scalar blocks. Thus for $\ell \geqslant 1$ the bound says that the blocks are uniformly bounded.

As discussed in Sect. A.3, the first term in the reduction formula, ${\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell ) g^{(d - 1)}_{\Delta , \ell } (z, {\bar{z}})$, gives upon induction exactly the main term in the bound we want to obtain. Let Y be the sum of all the other terms, i.e.

$$\begin{aligned} Y = \underset{(n, p) \ne (0, 0)}{\overset{}{\sum }} {\mathcal {A}}_{n, \ell - 2 p}^{(d)} (\Delta , \ell ) g^{(d - 1)}_{\Delta + 2 n, \ell - 2 p} (z, {\bar{z}}). \end{aligned}$$

(A.19)

To show the induction step, we need to show a bound

$$\begin{aligned} \frac{Y}{{\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell ) g^{(d - 1)}_{\Delta , \ell } (z, {\bar{z}})} \leqslant C (a, b, d) \left( 1 + \frac{1}{\Delta - \nu } \right) \end{aligned}$$

(A.20)

with C(a, b, d) independent of $\ell , \Delta$.

Using (A.13) we have

$$\begin{aligned} \frac{{\mathcal {A}}_{n, \ell - 2 p}^{(d)} (\Delta , \ell )}{{\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell )}= & {} \frac{(1 / 2)_p \ell !}{p! (\ell - 2 p) !} \frac{(\nu )_{\ell - p} (\nu - 1 / 2)_{\ell } (2 \nu - 1)_{\ell - 2 p} (j + \nu - 1 / 2)}{(\nu )_{\ell } (\nu - 1 / 2)_{\ell - p + 1} (2 \nu - 1)_{\ell }} \times \frac{(1 / 2)_n}{2^{4 n} n!} \nonumber \\{} & {} \times \frac{\left( \frac{\Delta + j}{2} \right) _n}{\left( \frac{\Delta + j - 1}{2} \right) _n } \frac{\left( \Delta - \nu - \frac{1}{2} \right) _n}{(\Delta - \nu )_n} \frac{(\Delta - 1)_{2 n}}{\left( \Delta - \nu - \frac{1}{2} \right) _{2 n}} \nonumber \\{} & {} \times \frac{\left( \frac{\Delta + \ell }{2} \right) _n}{\left( \frac{\Delta + \ell + 1}{2} \right) _n} \times \frac{\left( \frac{\Delta - 2 \nu - j + 1}{2} \right) _n}{\left( \frac{\Delta - j - 2 \nu }{2} \right) _n} \frac{\left( \frac{\Delta - \ell - 2 \nu }{2} \right) _n}{\left( \frac{\Delta - 2 \nu - \ell + 1}{2} \right) _n} \end{aligned}$$

(A.21)

In the third line we bound $\left( \frac{\Delta + \ell }{2} \right) _n \Big / \left( \frac{\Delta + \ell + 1}{2} \right) _n$ by 1 and we also use

$$\begin{aligned} \frac{\left( \frac{\Delta - 2 \nu - j + 1}{2} \right) _n}{\left( \frac{\Delta - j - 2 \nu }{2} \right) _n} \frac{\left( \frac{\Delta - \ell - 2 \nu }{2} \right) _n}{\left( \frac{\Delta - 2 \nu - \ell + 1}{2} \right) _n} \leqslant 1, \end{aligned}$$

(A.22)

(actually $= 1$ for $\ell = 0$, while to see this for $\ell \geqslant 1$ is easy using the unitarity bound $\Delta - \ell - 2 \nu \geqslant 0$).

We next study the second line in (A.21), denote it $F_n$. We need to pay attention to the pole at $\Delta = \nu$ and spurious poles at $\Delta = 1$ and $\Delta = \nu + 1 / 2$. For this reason we consider $n = 0, 1$ and $n \geqslant 2$ separately. We have

$$\begin{aligned} F_0&= 1, \end{aligned}$$

(A.23)

$$\begin{aligned} F_1&= \frac{(\Delta + j) (\Delta - 1) \Delta }{(\Delta + j - 1) (\Delta - \nu ) \left( \Delta - \nu + \frac{1}{2} \right) } . \end{aligned}$$

(A.24)

Considering separately the cases $\ell = 0$, $\ell = 1$ and $\ell \geqslant 2$ and using the unitarity bounds in each case, it’s easy to see that

$$\begin{aligned} F_1 \leqslant C \left( 1 + \frac{1}{\Delta - \nu } \right) \end{aligned}$$

(A.25)

with C uniform in $\Delta , \ell$, j.

Finally for $n \geqslant 2$ we can write

$$\begin{aligned} F_n = F_1 \frac{\left( \frac{\Delta + j}{2} + 1 \right) _{n - 1}}{\left( \frac{\Delta + j - 1}{2} + 1 \right) _{n - 1} } \frac{\left( \Delta - \nu + \frac{1}{2} \right) _{n - 1}}{(\Delta - \nu + 1)_{n - 1}} \frac{(\Delta )_{2 n - 2}}{\left( \Delta - \nu + \frac{1}{2} \right) _{2 n - 2}} . \end{aligned}$$

(A.26)

We can estimate the first factor by¹⁹

$$\begin{aligned} \frac{\left( \frac{\Delta + j}{2} + 1 \right) _{n - 1}}{\left( \frac{\Delta + j - 1}{2} + 1 \right) _{n - 1} } \leqslant 1 + 2 n \qquad (\Delta \geqslant 0), \end{aligned}$$

(A.27)

the second one by 1, and the last one by

$$\begin{aligned} \frac{(\Delta )_{2 n - 2}}{\left( \Delta - \nu + \frac{1}{2} \right) _{2 n - 2}} \leqslant (1 + 4 n)^{[\nu ]} \qquad (\Delta \geqslant \nu ). \end{aligned}$$

(A.28)

We conclude that for all n, $\Delta , \ell , j$ we have

$$\begin{aligned} F_n \leqslant C \left( 1 + \frac{1}{\Delta - \nu } \right) (1 + 4 n)^{[\nu ] + 1}. \end{aligned}$$

(A.29)

Using the discussed estimates for the second and third line in (A.21), we get an upper bound for the ratios of the ${\mathcal {A}}$ coefficients in (A.20):

$$\begin{aligned} \frac{{\mathcal {A}}_{n, \ell - 2 p}^{(d)} (\Delta , \ell )}{{\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell )}\leqslant & {} \frac{(1 / 2)_p \ell ! (\nu - 1 / 2)_{\ell }}{p! (\ell - 2 p) ! (2 \nu - 1)_{\ell }} \frac{(\nu )_{\ell - p} (2 \nu - 1)_{\ell - 2 p} (j + \nu - 1 / 2)}{(\nu )_{\ell } (\nu - 1 / 2)_{\ell - p + 1}} \nonumber \\{} & {} \times \frac{(1 / 2)_n}{2^{4 n} n!} C \left( 1 + \frac{1}{\Delta - \nu } \right) (1 + 4 n)^{[\nu ] + 1}. \end{aligned}$$

(A.30)

For the ratios of $(d - 1)$-dimensional conformal blocks in (A.20) we use the bounds from the previous induction step (the upper bound in the numerator and the lower bound in the denominator). We thus get

$$\begin{aligned} \frac{g^{(d - 1)}_{\Delta + 2 n, \ell - 2 p} (z, {\bar{z}})}{g^{(d - 1)}_{\Delta , \ell } (z, {\bar{z}})} \leqslant C' \frac{(2 \nu - 1)_{\ell } (\nu - 1 / 2)_{\ell - 2 p}}{(\nu - 1 / 2)_{\ell } (2 \nu - 1)_{\ell - 2 p}} \frac{k_{\tau + 2 n + 2 p} ({\bar{z}}) k_{\Delta + j + 2 n} (z)}{k_{\tau } ({\bar{z}}) k_{\Delta + \ell } (z)}. \end{aligned}$$

(A.31)

where $C' = K_{d - 1} (a, b) \left( 1 + \frac{\theta (d - 1 \geqslant 3) }{\Delta + 2 n - \nu _{d - 1}} \right)$ is unformly bounded in $\Delta , \ell$ since $\nu _{d - 1} = \nu - 1 / 2$.

For further simplification of (A.31) we use the identity (A.2). In addition to Lemma A.4, we will need another lemma about $G_{\beta }$:

Lemma A.6

For $\beta _1 \geqslant \beta _2 > 0$ we have

$$\begin{aligned} \frac{G_{\beta _1} (\rho )}{G_{\beta _2} (\rho )} \leqslant \frac{\Gamma \left( \frac{\beta _1 + 1}{2} \right) \Gamma \left( \frac{\beta _2}{2} \right) }{\Gamma \left( \frac{\beta _2 + 1}{2} \right) \Gamma \left( \frac{\beta _1}{2} \right) }\qquad (0 \leqslant \rho < 1). \end{aligned}$$

(A.32)

Proof

We use the integral representation of ${}_2 F_1$ valid for $\text {Re} (c)> \text {Re} (b) > 0$, $|\arg (1 - x) |< \pi$:

$$\begin{aligned} {}_2 F_1 (a, b; c; x) = \frac{\Gamma (c)}{\Gamma (b) \Gamma (c - b)} \int _0^1 \frac{t^{b - 1} (1 - t)^{c - b - 1}}{(1 - x t)^a} d t \end{aligned}$$

(A.33)

Taking $a = \frac{1}{2}, b = \frac{\beta }{2}, c = \frac{\beta + 1}{2}$ and $x = \rho ^2$ we get

$$\begin{aligned} G_{\beta _1} (\rho )= & {} \frac{\Gamma \left( \frac{\beta _1 + 1}{2} \right) }{\Gamma \left( \frac{1}{2} \right) \Gamma \left( \frac{\beta _1}{2} \right) } \int _0^1 \frac{t^{\frac{\beta _1}{2} - 1} (1 - t)^{- \frac{1}{2}}}{(1 - \rho ^2 t)^{\frac{1}{2}}} d t \,. \end{aligned}$$

(A.34)

We replace $t^{\beta _1 / 2}$ under the integral by a larger quantity $t^{\beta _2 / 2}$, and infer the stated bound. $\square$

By (A.2), we rewrite the ratio of k’s in (A.31) as

$$\begin{aligned} \frac{k_{\tau + 2 n + 2 p} ({\bar{z}}) k_{\Delta + j + 2 n} (z)}{k_{\tau } ({\bar{z}}) k_{\Delta + \ell } (z)} = (4 {\bar{\rho }})^{n + p} \frac{G_{\tau + 2 n + 2 p} ({\bar{\rho }})}{G_{\tau } ({\bar{\rho }})} (4 \rho )^{n - p} \frac{G_{\Delta + 2 n + j} (\rho )}{G_{\Delta + \ell } (\rho )}. \end{aligned}$$

(A.35)

By Lemma A.4, we have

$$\begin{aligned} \frac{G_{\tau + 2 n + 2 p} ({\bar{\rho }})}{G_{\tau } ({\bar{\rho }})} \leqslant G_{\tau + 2 n + 2 p} ({\bar{\rho }}) \leqslant \frac{1}{\sqrt{1 - {\bar{\rho }}^2}} \leqslant \frac{1}{\sqrt{1 - \rho (a)^2}}. \end{aligned}$$

(A.36)

By Lemmas A.4 and A.6 we have

$$\begin{aligned} \frac{G_{\Delta + 2 n + j} (\rho )}{G_{\Delta + \ell } (\rho )} \leqslant \frac{G_{\Delta + 2 n + j} (\rho )}{G_{\Delta + j} (\rho )} \leqslant \frac{\left( \frac{\Delta + j + 1}{2} \right) _n}{\left( \frac{\Delta + j}{2} \right) _n} \leqslant 1 + 4 n \qquad (\Delta \geqslant \nu ). \end{aligned}$$

(A.37)

Plugging these estimates into (A.31), we estimate it as

$$\begin{aligned} \frac{g^{(d - 1)}_{\Delta + 2 n, \ell - 2 p} (z, {\bar{z}})}{g^{(d - 1)}_{\Delta , \ell } (z, {\bar{z}})} \leqslant C'' (a, b, d) \frac{(2 \nu - 1)_{\ell } (\nu - 1 / 2)_{\ell - 2 p}}{(\nu - 1 / 2)_{\ell } (2 \nu - 1)_{\ell - 2 p}} (16 \rho {\bar{\rho }})^n \left( \frac{{\bar{\rho }}}{\rho } \right) ^p (1 + 4 n). \nonumber \\ \end{aligned}$$

(A.38)

Combining (A.30) and (A.38), and using the inequality

$$\begin{aligned} \frac{\ell !}{(\ell - 2 p) !} \frac{(\nu )_{\ell - p} (\nu - 1 / 2)_{\ell - 2 p + 1}}{(\nu )_{\ell } (\nu - 1 / 2)_{\ell - p + 1}} \leqslant 2, \end{aligned}$$

(A.39)

we get

$$\begin{aligned} \frac{{\mathcal {A}}_{n, \ell - 2 p}^{(d)} (\Delta , \ell ) g^{(d - 1)}_{\Delta + 2 n, \ell - 2 p} (z, {\bar{z}})}{{\mathcal {A}}_{0, \ell }^{(d)} (\Delta , \ell ) g^{(d - 1)}_{\Delta , \ell } (z, {\bar{z}})}\leqslant & {} C''' (a, b, d) \left( 1 + \frac{1}{\Delta - \nu } \right) \frac{(1 / 2)_p }{p!} \left( \frac{{\bar{\rho }}}{\rho } \right) ^p \nonumber \\{} & {} \times \frac{(1 / 2)_n}{n!} (1 + 4 n)^{[\nu ] + 2} (\rho {\bar{\rho }})^n. \end{aligned}$$

(A.40)

The point of this bound is that it only depends on $n, p, \rho , {\bar{\rho }}$ but is uniform in $\Delta , \ell$. Also the dependence on n, p is factorized.

We now sum (A.40) over p, n to get a bound on the ratio (A.20). Although p in Hogervorst’s formula is bounded by $[\ell / 2]$ we will extend the sum over all p from 0 to $\infty$. The sum over p is convergent when ${\bar{\rho }} / \rho < 1$ while the sum over n is convergent when $\rho {\bar{\rho }} < 1$. Since we are assuming that $0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1$, we have

$$\begin{aligned} \quad \rho {\bar{\rho }} \leqslant \rho (a)< 1, \quad {\bar{\rho }} / \rho \leqslant \rho (a) / \rho (b) < 1. \end{aligned}$$

(A.41)

So the sums over n and over p are both bounded by a constant which only depends on a, b and d. We thus obtain the needed bound (A.20). This completes the induction step.

Proof of Lemma 2.14

Here we will prove Lemma 2.14, i.e. the improved logarithmic bound on the conformal blocks:

$$\begin{aligned} \frac{g_{\tau , \ell } (z, {\bar{z}})}{g_{\tau , \ell } (b, a)} \leqslant B {\bar{z}}^{\tau / 2} \log \left( \frac{1}{1 - z} \right) \qquad (0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1) \end{aligned}$$

(B.1)

for all $(\tau , \ell )$ satisfying $\tau \leqslant \tau _{\max }, \ell \leqslant \ell _{\max }$ and the unitarity bounds, i.e. $\tau \geqslant \nu$ for $\ell = 0$ and $\tau \geqslant 2 \nu$ for $\ell \geqslant 1$. We will show that the constant factor B in the upper bound is finite and depends only on a, b, $\tau _{\max }$ and $\ell _{\max }$.

We will divide the proof into two cases: (a) $1 \leqslant \ell \leqslant \ell _{\max }$, (b) $\ell = 0$. We will derive (B.1) with constants $B_1$ for (a) and $B_2$ for (b). In the end we take $B = \max \{ B_1, B_2 \}$.

1.1 $1 \leqslant \ell \leqslant \ell _{\max }$

In this case the range of $\tau$ we are concerned about is given by

$$\begin{aligned} 2 \nu \leqslant \tau \leqslant \tau _{\max }. \end{aligned}$$

(B.2)

As explained in Remark 2.8(c), Lemma 2.7 implies a two-sided bound for $2 \nu \leqslant \tau \leqslant \tau _{\max }$

$$\begin{aligned} C_1 \leqslant \frac{g_{\tau , \ell } (z, {\bar{z}})}{ {\bar{z}}^{\tau / 2} F_{\tau , \ell } (z)} \leqslant C_2 \qquad (0< {\bar{z}} \leqslant a< b \leqslant z < 1), \end{aligned}$$

(B.3)

where $F_{\tau , \ell }$ is given in (2.28), the constants $C_1$, $C_2$ are finite and they only depend on a, b and $\tau _{\max }$. Using the upper bound for $g_{\tau , \ell } (z, {\bar{z}})$ and the lower bound for $g_{\tau , \ell } (b, a)$, we get

$$\begin{aligned} \begin{aligned}&\frac{g_{\tau , \ell } (z, {\bar{z}})}{g_{\tau , \ell } (b, a)} \leqslant \frac{C_2 {\bar{z}}^{\tau / 2} F_{\tau , \ell } (z)}{C_1 a^{\tau / 2} F_{\tau , \ell } (b)} = \frac{C_2 {\bar{z}}^{\tau / 2} k_{\tau + 2 \ell } (z)}{C_1 a^{\tau / 2} k_{\tau + 2 \ell } (b)} \\&\left( 0< {\bar{z}} \leqslant a< b \leqslant z < 1,\quad 2 \nu \leqslant \tau \leqslant \tau _{\max } \right) , \end{aligned} \end{aligned}$$

(B.4)

here we used the explicit form of $F_{\tau , \ell }$ in the last step. Then using the lower bounds $a^{\tau / 2} \geqslant a^{\tau _{\max } / 2}$, $k_{\tau + 2 \ell } (b) \geqslant b^{\tau _{\max } / 2 + \ell _{\max }}$ for the denominator, and the upper bounds $z^{\beta / 2} \leqslant 1$, $k_{\beta } (z) \leqslant {}_2 F_1 (\beta / 2, \beta / 2; \beta ; z) \leqslant {}_2 F_1 (\beta _{\max } / 2, \beta _{\max } / 2; \beta _{\max }; z)$ for the numerator, we get

$$\begin{aligned} \frac{g_{\tau , \ell } (z, {\bar{z}})}{g_{\tau , \ell } (b, a)} \leqslant C' {\bar{z}}^{\tau / 2} {}_2 F_1 \left( \frac{\tau _{\max } + 2 \ell _{\max }}{2}, \frac{\tau _{\max } + 2 \ell _{\max }}{2}; \tau _{\max } + 2 \ell _{\max }; z \right) \end{aligned}$$

(B.5)

in the regime where $0< {\bar{z}} \leqslant a< b \leqslant z < 1$, $2 \nu \leqslant \tau \leqslant \tau _{\max }$ and $1 \leqslant \ell \leqslant \ell _{\max }$. Here the coefficient $C' {:}{=}C_2 / (C_1 a^{\tau _{\max } / 2} b^{\tau _{\max } / 2 + \ell _{\max }})$ is finite and depends only on a, b, $\tau _{\max }$ and $\ell _{\max }$. For the hypergeometric function ${}_2 F_1$ we use the fact (mentioned in Remark 2.8(e)) that

$$\begin{aligned} {}_2 F_1 \left( \frac{\beta }{2}, \frac{\beta }{2}; \beta ; z \right) \sim \frac{\Gamma \left( \frac{\beta + 1}{2} \right) 2^{\beta - 1}}{\Gamma \left( \frac{\beta }{2} \right) \sqrt{\pi }} \log \left( \frac{1}{1 - z} \right) \qquad (z \rightarrow 1). \end{aligned}$$

(B.6)

This asymptotics together with the monotonicity of ${}_2 F_1 \left( \frac{\beta }{2}, \frac{\beta }{2}; \beta ; z \right)$ imply that

$$\begin{aligned} {}_2 F_1 \left( \frac{\beta }{2}, \frac{\beta }{2}; \beta ; z \right) \leqslant C'' \log \left( \frac{1}{1 - z} \right) \qquad \left( b \leqslant z < 1,\quad \forall \beta \geqslant 0 \right) , \end{aligned}$$

(B.7)

where $C''<\infty$ only depends on $0<\beta <\infty$ and $b>0$. Combining (B.5) and (B.7) we get

$$\begin{aligned} \frac{g_{\tau , \ell } (z, {\bar{z}})}{g_{\tau , \ell } (b, a)} \leqslant B_1 {\bar{z}}^{\tau / 2} \log \left( \frac{1}{1 - z} \right) \end{aligned}$$

(B.8)

in the regime $0< {\bar{z}} \leqslant a< b \leqslant z < 1$, $2 \nu \leqslant \tau \leqslant \tau _{\max }$ and $1 \leqslant \ell \leqslant \ell _{\max }$. Here $B_1: = C' C''$ is finite and depends only on a, b, $\tau _{\max }$ and $\ell _{\max }$. This finishes the proof of $\ell \geqslant 1$ case.

1.2 $\ell = 0$

The $d = 2$, $\ell = 0$ case can be shown exactly as $\ell \geqslant 1$ in the previous section, since in this case the $\Delta \rightarrow \nu$ pole is absent. In the rest of this section we consider $d \geqslant 3$.

By upper bound in Lemma 2.7 and ${\bar{\rho }} \leqslant {\bar{z}}$ we get the upper bound of $g_{\Delta , 0} (z, {\bar{z}})$ for $\Delta > \nu$:

$$\begin{aligned} g_{\Delta , 0} (z, {\bar{z}}) \leqslant K_d (a, b) \left( 1 + \frac{1}{\Delta - \nu } \right) {\bar{z}}^{\Delta / 2} F_{\Delta , 0} (z) \qquad (0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1),\nonumber \\ \end{aligned}$$

(B.9)

where $K_d (a, b)$ is a finite constant independent of $\Delta$. Then by the similar analysis as the $\ell \geqslant 1$ case, we get an upper bound on $g_{\Delta , 0} (z, {\bar{z}})$:

$$\begin{aligned} \begin{aligned}&g_{\Delta , 0} (z, {\bar{z}}) \leqslant D_1 \left( 1 + \frac{1}{\Delta - \nu } \right) {\bar{z}}^{\Delta / 2} \log \left( \frac{1}{1 - z} \right) \\&\left( 0 \leqslant {\bar{z}} \leqslant a< b \leqslant z< 1,\quad \nu < \Delta \leqslant \tau _{\max } \right) , \end{aligned} \end{aligned}$$

(B.10)

where $D_1$ is a finite constant, depending only on a, b and $\tau _{\max }$. The main subtlety here is the $(\Delta - \nu )^{- 1}$ term which blows up when $\Delta \rightarrow \nu$. To show that the ratio $g_{\Delta , 0} (z, {\bar{z}}) / g_{\Delta , 0} (b, a)$ has the ${\bar{z}}^{\tau / 2} \log \left( \frac{1}{1 - {\bar{z}}} \right)$ bound with the coefficient uniform in $\Delta$, we need a lower bound of $g_{\Delta , 0} (b, a)$ which also blows up as $(\Delta - \nu )^{- 1}$ when $\Delta \rightarrow \nu$. The lower bound in Lemma 2.7 does not show this behavior. So we need an improved lower bound of $g_{\Delta , 0} (b, a)$ which features the $(\Delta - \nu )^{- 1}$ singularity.

For this we use the dimensional reduction formula (A.12), restricting to $\ell = 0$:

$$\begin{aligned} g_{\Delta , 0}^{(d)} (z, {\bar{z}}) = \underset{n = 0}{\overset{\infty }{\sum }} {\mathcal {A}}_{n, 0}^{(d)} (\Delta , 0) g^{(d - 1)}_{\Delta + 2 n, 0} (z, {\bar{z}}), \end{aligned}$$

(B.11)

where the reduction coefficients are given by

$$\begin{aligned} {\mathcal {A}}^{(d)}_{n, 0} (\Delta , 0) = \frac{\left( \frac{1}{2} \right) _n \left( \left( \frac{\Delta }{2} \right) _n \right) ^3}{4^n n! (\Delta - \nu )_n \left( \Delta - \nu - \frac{1}{2} + n \right) _n \left( \frac{\Delta + 1}{2} \right) _n}. \end{aligned}$$

(B.12)

When $\Delta > \nu$, (B.11) is a positive sum, so $g_{\Delta , 0}^{(d)}$ is bounded from below by the $n = 1$ term:

$$\begin{aligned} g_{\Delta , 0}^{(d)} (z, {\bar{z}}) \geqslant \frac{\left( \frac{1}{2} \right) \left( \frac{\Delta }{2} \right) ^3}{4 (\Delta - \nu ) \left( \Delta - \nu + \frac{1}{2} \right) \left( \frac{\Delta + 1}{2} \right) } g^{(d - 1)}_{\Delta + 2, 0} (z, {\bar{z}}). \end{aligned}$$

(B.13)

This estimate gives us the needed lower bound of $g_{\Delta , 0}^{(d)} (b, a)$ since it has the $(\Delta - \nu )^{- 1}$ factor. The whole prefactor in front of $g^{(d - 1)}_{\Delta + 2, 0} (z, {\bar{z}})$ is bounded from below by

$$\begin{aligned} \frac{\left( \frac{1}{2} \right) \left( \frac{\Delta }{2} \right) ^3}{4 (\Delta - \nu ) \left( \Delta - \nu + \frac{1}{2} \right) \left( \frac{\Delta + 1}{2} \right) } \geqslant \frac{D_2}{\Delta - \nu } \qquad (\nu < \Delta \leqslant \tau _{\max }), \end{aligned}$$

(B.14)

where $D_2 {:}{=}\frac{\nu ^3}{32 \left( \tau _{\max } - \nu + \frac{1}{2} \right) (\tau _{\max } + 1)} > 0$. Now take $z = b$ and ${\bar{z}} = a$, the estimate of $g^{(d - 1)}_{\Delta + 2, 0} (b, a)$ is similar to the previous section:

$$\begin{aligned} g^{(d - 1)}_{\Delta + 2, 0} (b, a)\geqslant & {} 2^{- \tau _{\max } - 2} a^{(\Delta + 2) / 2} F_{\Delta + 2, 0} (b)\nonumber \\= & {} 2^{- \tau _{\max } - 2} a^{(\Delta + 2) / 2} 4^{(\Delta + 2) / 2} b^{\Delta + 2} {}_2 F_1 (\Delta + 2, \Delta + 2; 2 \Delta + 4; b) \nonumber \\\geqslant & {} 2^{- \tau _{\max } - 2} a^{(\tau _{\max } + 2) / 2} 4^{(\nu + 2) / 2} b^{\tau _{\max } + 2} {}_2 F_1 (\nu + 2, \nu + 2; 2 \nu + 4; b)=: D_3\nonumber \\ \end{aligned}$$

(B.15)

Here in the first line we used the lower bound of Lemma 2.7 and the fact that ${\bar{\rho }}^{(\Delta + 2) / 2} \geqslant ({\bar{z}} / 4)^{(\Delta + 2) / 2} \geqslant 2^{- \tau _{\max } - 2} {\bar{z}}^{(\Delta + 2) / 2}$ for $\Delta \leqslant \tau _{\max }$, in the second line we used the explicit form of $F_{\Delta + 2, 0} (b)$, in the third line we bounded $a^{(\Delta + 2) / 2}$, $4^{(\Delta + 2) / 2}$, $b^{\Delta + 2}$ and ${}_2 F_1$ by their minimal values in the range $\nu \leqslant \Delta \leqslant \tau _{\max }$. $D_3>0$ because $a, b \in (0, 1)$ and $\tau _{\max } < \infty$.

Putting these together, we have for $0 \leqslant {\bar{z}} \leqslant a< b \leqslant z < 1$ and $\nu < \Delta \leqslant \tau _{\max }$:

$$\begin{aligned} \frac{g_{\Delta , 0} (z, {\bar{z}})}{g_{\Delta , 0} (b, a)}\leqslant & {} \frac{D_1}{D_2 D_3} (\Delta - \nu ) \left( 1 + \frac{1}{\Delta - \nu } \right) {\bar{z}}^{\Delta / 2} \log \left( \frac{1}{1 - z} \right) \nonumber \\\leqslant & {} B_2 {\bar{z}}^{\Delta / 2} \log \left( \frac{1}{1 - z} \right) , \end{aligned}$$

(B.16)

where $B_2 {:}{=}\frac{D_1}{D_2 D_3} (1 + \tau _{\max } - \nu )$ is finite and depends only on a, b and $\tau _{\max }$.