††institutetext: Mathematical Institute, University of Oxford, Woodstock Road,
Oxford, OX2 6GG, UK

# AdS Virasoro-Shapiro from single-valued periods

Luis F. Alday, Tobias Hansen and Joao A. Silva<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract

We determine the full $1/\sqrt{\lambda}$ correction to the flat-space Wilson
coefficients which enter the AdS Virasoro-Shapiro amplitude in $\mathcal{N}=4$
SYM theory at strong coupling. The assumption that the Wilson coefficients are
in the ring of single-valued multiple zeta values, as expected for closed
string amplitudes, is surprisingly powerful and leads to a unique solution to
the dispersive sum rules relating Wilson coefficients and OPE data obtained in
Alday:2022uxp . The corresponding OPE data fully agrees with and extends the
results from integrability. The Wilson coefficients to order
$1/\sqrt{\lambda}$ can be summed into an expression whose structure of poles
and residues generalises that of the Virasoro-Shapiro amplitude in flat space.

##  1 Introduction

Closed string genus-0 four-point amplitudes are commonly referred to as
Virasoro-Shapiro amplitudes. The worldsheet theory to compute the
corresponding amplitude on AdS${}_{5}\times$ S5 space-time is still unknown,
which motivates us to explore alternative methods. In our favour we have that,
by the AdS/CFT duality, this amplitude is also a correlator in $\mathcal{N}=4$
SYM theory. The AdS/CFT dictionary maps the genus expansion in $g_{s}$ to the
expansion in inverse powers of the central charge $1/c$ while
$\alpha^{\prime}$ corresponds to inverse powers of the t’Hooft coupling
$1/\sqrt{\lambda}$ . We study the correlator of four stress-tensor multiplets
in Mellin space, to leading order in $1/c$, which in a $1/\lambda$ expansion
can be written as (see Alday:2022uxp  for further details)

$M(s_{1},s_{2})=\frac{8}{(s_{1}-\frac{2}{3})(s_{2}-\frac{2}{3})(s_{3}-\frac{2}{3})}+\sum\limits_{a,b=0}^{\infty}\frac{\Gamma(2a+3b+6)}{8^{a+b}\lambda^{\frac{3}{2}+a+\frac{3}{2}b}}\sigma_{2}^{a}\sigma_{3}^{b}\left(\alpha^{(0)}_{a,b}+\frac{\alpha^{(1)}_{a,b}}{\sqrt{\lambda}}+\frac{\alpha^{(2)}_{a,b}}{\lambda}+\cdots\right)$
(1)

where $s_{1}+s_{2}+s_{3}=0$, $\sigma_{2}=s_{1}^{2}+s_{2}^{2}+s_{3}^{2}$ and
$\sigma_{3}=s_{1}s_{2}s_{3}$, and we have suppressed an overall $1/c$.111These
Mellin variables are related to the ones in Alday:2022uxp  by
$s_{1}=s-\frac{4}{3}$, $s_{2}=t-\frac{4}{3}$, $s_{3}=u-\frac{4}{3}$. This
implies $\alpha^{(0)}_{a,b}=\alpha_{a,b}$ and
$\alpha^{(1)}_{a,b}=\beta_{a,b}-\frac{2}{3}(b+1)(6+2a+3b)\alpha^{(0)}_{a-1,b+1}$.
The leading coefficients $\alpha^{(0)}_{a,b}$ are known from the flat space
Virasoro-Shapiro amplitude via the flat space limit formula Penedones:2010ue ;
Fitzpatrick:2011hu .

In Alday:2022uxp  we used the bound on chaos Maldacena:2015waa ; Caron-
Huot:2021rmr  to derive dispersive sum rules that relate the Wilson
coefficients $\alpha^{(k)}_{a,b}$ to the OPE data of the exchanged heavy
single-trace operators with dimensions that grow as
$\Delta\sim\lambda^{\frac{1}{4}}$. In this paper, we present a solution to
these dispersive sum rules, determining the coefficients $\alpha^{(1)}_{a,b}$
and the corresponding $1/\sqrt{\lambda}$ corrections to the dimensions and
structure constants of said operators.

In Alday:2022uxp  it was assumed that $\alpha^{(1)}_{a,b}$ is in the ring of
multiple zeta values, has uniform transcendentality and that the sums over OPE
data related to $\alpha^{(1)}_{a,b}$ are given in terms of Euler-Zagier sums.
As discussed in Alday:2022uxp , these assumptions are not enough to fully fix
the $\alpha^{(1)}_{a,b}$. In the present paper we make the additional
assumption that $\alpha^{(1)}_{a,b}$ is in the ring of _single-valued_
multiple zeta values (i.e. $\alpha^{(1)}_{a,b}$ are single-valued periods).
This property is known to hold for tree-level closed string amplitudes in flat
space Stieberger:2013wea ; Stieberger:2014hba ; Schlotterer:2018zce ;
Brown:2019wna , and is expected from a world-sheet perspective. This
additional assumption turns out to be surprisingly powerful and leads to a
unique solution for $\alpha^{(1)}_{a,b}$. This also determines the
corresponding OPE data and we give the $1/\sqrt{\lambda}$ corrections to
conformal dimensions and OPE coefficients as analytic formulae for many Regge
trajectories. Our solution passes several checks. First, we reproduce the
dimensions of operators on the leading Regge trajectory (including the Konishi
operator) known from integrability and the two available Wilson coefficients
known from localisation. Second, we fix the solution for $\alpha^{(1)}_{a,b}$
by imposing single-valuedness for a few values of $a$. The resulting solutions
turn out to be single-valued for all values of $a$ that we are able to check,
in a non-trivial way. Third, our solution for $\alpha^{(1)}_{a,b}$ implies an
overconstrained linear system of equations for the OPE data. That this system
has a solution serves as a consistency check for $\alpha^{(1)}_{a,b}$.

Having found $\alpha^{(1)}_{a,b}$, the next step is to resum the low energy
expression in (1) to obtain a simpler expression that makes the analytic
structure of the amplitude manifest. As the sum over $a$ and $b$ in (1) is
divergent, we introduce the flat space transform, an integral transform
equivalent to the one in the flat space limit of Penedones:2010ue ;
Fitzpatrick:2011hu  but without sending the AdS radius to infinity. This is
equivalent to performing a Borel resummation. For the leading term this
reproduces the familiar Virasoro-Shapiro amplitude

$-\frac{\Gamma\left(-S\right)\Gamma\left(-T\right)\Gamma\left(-U\right)}{\Gamma\left(S+1\right)\Gamma\left(T+1\right)\Gamma\left(U+1\right)}=\frac{1}{STU}+2\sum_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(0)}_{a,b}\,,$
(2)

with $\hat{\sigma}_{2}=\frac{1}{2}(S^{2}+T^{2}+U^{2})$,
$\hat{\sigma}_{3}=STU$, for which Zagier:2019eus  found the representation

$2\sum\limits_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(0)}_{a,b}=\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{3}}\frac{y+2}{1-x-y}\binom{z+\delta-1}{\delta-1}^{2}\,,$
(3)

with $x=\hat{\sigma}_{2}/\delta^{2}$, $y=\hat{\sigma}_{3}/\delta^{3}$, and
$z=\delta\left(\sqrt{1-4y}-1\right)/2$. For the $1/\sqrt{\lambda}$ correction
we find

$2\sum\limits_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(1)}_{a,b}=\sum\limits_{\delta=1}^{\infty}\sum\limits_{n=0}^{\delta-1}\frac{1}{\delta^{4}}\mathcal{D}_{n}(\delta)\frac{y+2}{1-x-y}\binom{z+\delta-\frac{n}{2}-1}{\delta-n-1}^{2}\,,$
(4)

where $\mathcal{D}_{n}(\delta)$ is a third order differential operator in $x$,
$y$ and $z$ which produces a crossing-symmetric expression with poles up to
4th order in $S$, $T$ and $U$.

This paper is organised as follows. In section 2 we review the dispersive sum
rule for $\alpha^{(0)}_{a,b}$, its solution and other known data. Section 3
states the dispersive sum rule for $\alpha^{(1)}_{a,b}$, our precise
assumptions and, after a short primer on single-valued multiple zeta values,
constructs $\alpha^{(1)}_{a,b}$ for $0\leq b\leq 6$. In section 4 we
generalise the solutions to any value of $b$ by finding general expressions
for the sums over OPE data that appear in the dispersive sum rules. We use
this to determine OPE data for many Regge trajectories. In section 5 we apply
the flat space transform and resum the low energy expansion. We conclude in
section 6. Appendix A contains a derivation of general expressions and
recursion relations for the dispersive sum rules based on crossing-symmetric
dispersion relations. In appendix B we present an alternative representation
for the spin sums of section 4. Appendix C contains an analysis of extra
bootstrap constraints on the OPE data of heavy operators, other than the ones
in Alday:2022uxp . In appendix D we compute the residues of the highest order
poles of the amplitude at each order in $1/\lambda$ and resum them. In
appendix E we summarise the state of the art of the weak and strong coupling
expansions of the dimension of the Konishi operator.

##  2 Review of known data

The ‘stringy’ operators that enter the dispersive sum rules for
$\alpha^{(k)}_{a,b}$ are the single trace operators with twists
$\tau(r;\lambda)$ and OPE coefficients $\mathcal{C}^{2}(r;\lambda)$, which can
be parameterised by

$\displaystyle\tau(r;\lambda)$
$\displaystyle=\tau_{0}(r)\lambda^{\frac{1}{4}}+\tau_{1}(r)+\tau_{2}(r)\lambda^{-\frac{1}{4}}+\ldots\,,$
(5) $\displaystyle\mathcal{C}^{2}(r;\lambda)$
$\displaystyle=\frac{\pi^{3}}{2^{12}}\frac{2^{-2\tau(r;\lambda)}\tau(r;\lambda)^{6}}{\sin^{2}(\frac{\pi\tau(r;\lambda)}{2})}\frac{1}{2^{2\ell}(\ell+1)}f(r;\lambda)\,,$
(6) $\displaystyle f(r;\lambda)$
$\displaystyle=f_{0}(r)+f_{1}(r)\lambda^{-\frac{1}{4}}+f_{2}(r)\lambda^{-\frac{1}{2}}+\ldots\,,$
(7)

where $r$ labels collectively all quantum numbers characterising the
operators. The leading contribution to the twists is

$\displaystyle\tau_{0}(r)=2\sqrt{\delta}\,,\qquad\delta\in\mathbb{N}\,,$ (8)

and we will use $\delta$ along with the spin $\ell$ to label operators from
now on, i.e. $r=(\delta,\ell,\hat{r})$. As long as we are studying only a
single correlator we cannot access further quantum numbers $\hat{r}$ and will
denote the sum over them by

$\langle\ldots\rangle=\sum\limits_{\hat{r}}\ldots\,.$ (9)

The operators are organised into Regge trajectories by their dependence on
$\delta$ and $\ell$ as illustrated in figure 1.

$1$$2$$3$$4$$5$$\delta$$0$$2$$4$$6$$8$$\ell$$\ell=2(\delta-1)$$\ell=2(\delta-2)$$\ell=2(\delta-3)$$\ell=2(\delta-4)$
Figure 1: Chew-Frautschi plot of the stringy operators.

The dispersion relations imply the following expression for the first layer of
Wilson coefficients222 The derivation of the dispersive sum rules for
$\alpha^{(k)}_{a,b}$ for general $b$ is described in appendix A.1.

$\alpha^{(0)}_{a,b}=\sum\limits_{\delta=1}^{\infty}\sum\limits_{m=0}^{b}\frac{c^{(0)}_{a,b,m}}{\delta^{3+2a+3b}}F_{m}^{(0)}(\delta)\,,$
(10)

where

$\displaystyle c^{(0)}_{a,b,m}={}$
$\displaystyle\frac{(-1)^{m}(2a+3b-3m)\Gamma(a+b-m)}{2\Gamma(a+1)\Gamma(b-m+1)}$
(11)
$\displaystyle{}_{4}F_{3}\left(\tfrac{m+1}{2},\tfrac{m}{2},m-b,m+1-\tfrac{2}{3}a-b;m+1,m+1-a-b,m-\tfrac{2}{3}a-b;4\right)\,,$

and the leading contributions to the OPE coefficients appear in the sums

$F_{m}^{(0)}(\delta)=\frac{4^{m}}{\Gamma(2m+2)}\sum_{\ell=0,2,\ldots}^{2(\delta-1)}(\ell-m+1)_{m}(\ell+2)_{m}\langle
f_{0}(\delta,\ell)\rangle\,.$ (12)

The first few cases read explicitly

$\displaystyle\alpha^{(0)}_{a,0}={}$
$\displaystyle\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{3+2a}}F_{0}^{(0)}(\delta)\,,$
(13) $\displaystyle\alpha^{(0)}_{a,1}={}$
$\displaystyle\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{6+2a}}\left((a+\tfrac{3}{2})F_{0}^{(0)}(\delta)-F_{1}^{(0)}(\delta)\right)\,,$
$\displaystyle\alpha^{(0)}_{a,2}={}$
$\displaystyle\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{9+2a}}\left(\frac{1}{2}(a+1)(a+3)F_{0}^{(0)}(\delta)-(a+\tfrac{5}{2})F_{1}^{(0)}(\delta)+F_{2}^{(0)}(\delta)\right)\,,$
$\displaystyle\vdots$

All the coefficients $\alpha^{(0)}_{a,b}$ are known from the flat space limit,
for example

$\displaystyle\alpha^{(0)}_{a,0}={}$ $\displaystyle\zeta(3+2a)\,,$
$\displaystyle\alpha^{(0)}_{a,1}={}$
$\displaystyle\sum\limits_{\begin{subarray}{c}i_{1},i_{2}=0\\\
i_{1}+i_{2}=a\end{subarray}}^{a}\zeta(3+2i_{1})\zeta(3+2i_{2})=(a+\tfrac{3}{2})\zeta(2a+6)-2\zeta(2a+5,1)\,,$
$\displaystyle\alpha^{(0)}_{a,2}={}$
$\displaystyle\frac{2}{3}\sum\limits_{\begin{subarray}{c}i_{1},i_{2},i_{3}=0\\\
i_{1}+i_{2}+i_{3}=a\end{subarray}}^{a}\zeta(3+2i_{1})\zeta(3+2i_{2})\zeta(3+2i_{3})+\frac{1}{6}(a+1)(a+2)\zeta(9+2a)$
(14) $\displaystyle={}$
$\displaystyle\tfrac{1}{2}(a+1)(a+3)\zeta(2a+9)-2(a+\tfrac{5}{2})\zeta(2a+8,1)+\zeta(2a+7,2)+4\zeta(2a+7,1,1)\,,$
$\displaystyle\vdots\ \,$

written in terms of multiple zeta values of depth $k$ and weight
$s_{1}+\ldots+s_{k}$

$\zeta(s_{1},\ldots,s_{k})=\sum\limits_{n_{1}>\ldots>n_{k}>0}\frac{1}{n_{1}^{s_{1}}\cdots
n_{k}^{s_{k}}}\,.$ (15)

By comparing (LABEL:alpha_equations) with (14) it is apparent that
$F_{m}^{(0)}(\delta)$ are most naturally expressed in terms of Euler-Zagier
sums, defined by

$Z_{s_{1},\ldots,s_{k}}(N)=\sum\limits_{\begin{subarray}{c}n_{1},\ldots,n_{k}\\\
N\geq n_{1}>\ldots>n_{k}>0\end{subarray}}\frac{1}{n_{1}^{s_{1}}\cdots
n_{k}^{s_{k}}}\,,\qquad Z(N)=1\,,\qquad Z_{s_{1},\ldots,s_{k}}(0)=0\,,$ (16)

and which naturally lead to multiple zeta values when summed over $\delta$

$\zeta(s,s_{1},s_{2},\ldots)=\sum\limits_{\delta=1}^{\infty}\frac{Z_{s_{1},s_{2},\ldots}(\delta-1)}{\delta^{s}}\,.$
(17)

We see that

$F_{m}^{(0)}(\delta)=\sum\limits_{d=\lfloor\frac{m+1}{2}\rfloor}^{m}\sum\limits_{\begin{subarray}{c}s_{1},\ldots,s_{d}\in\\{1,2\\}\\\
s_{1}+\ldots+s_{d}=m\end{subarray}}2^{\sum_{i}\delta_{s_{i},1}}\delta^{m}Z_{s_{1},\ldots,s_{d}}(\delta-1)\,.$
(18)

Equating (18) and (12) fixes all the individual OPE coefficients $\langle
f_{0}(\delta,\ell)\rangle$. To see this, note that $F_{m}^{(0)}(\delta)$ is a
sum over $\delta$ different spins, so $F_{m}^{(0)}(\delta)$ for $0\leq m<n$
fixes all the OPE coefficients with $\delta\leq n$.

At the next order the requirement that (1) is an expansion in
$1/\sqrt{\lambda}$ leads to a sum rule for vanishing Wilson coefficients

$0=\sum\limits_{\delta=1}^{\infty}\sum\limits_{m=0}^{b}\frac{c^{(0)}_{a,b,m}}{\delta^{\frac{7}{2}+2a+3b}}\left(F_{m}^{(1)}(\delta)-(3+2a+3b)T_{m}^{(1)}(\delta)\right)\,,$
(19)

with

$\displaystyle T_{m}^{(1)}(\delta)={}$
$\displaystyle\frac{4^{m}}{\Gamma(2m+2)}\sum_{\ell=0,2,\ldots}^{2(\delta-1)}(\ell-m+1)_{m}(\ell+2)_{m}\langle
f_{0}(\delta,\ell)(\tau_{1}(\delta,\ell)+\ell+2)\rangle\,,$ (20)
$\displaystyle F_{m}^{(1)}(\delta)={}$
$\displaystyle\frac{4^{m}}{\Gamma(2m+2)}\sum_{\ell=0,2,\ldots}^{2(\delta-1)}(\ell-m+1)_{m}(\ell+2)_{m}\langle\sqrt{\delta}f_{1}(\delta,\ell)-f_{0}(\delta,\ell)(3\ell+\tfrac{23}{4})\rangle\,.$

This sum rule has the solution

$\displaystyle\tau_{1}(\delta,\ell)=-\ell-2\,,~{}~{}~{}~{}~{}\langle
f_{1}(\delta,\ell)\rangle=\langle
f_{0}(\delta,\ell)\rangle\frac{3\ell+\frac{23}{4}}{\sqrt{\delta}}\,.$ (21)

From a string theory perspective, we expect the corrections to energies of
string configurations to be spaced by half-integer powers of $\lambda$

$\hat{\Delta}(r;\lambda)=\sum\limits_{i=0}^{\infty}\hat{\Delta}_{2i}(r)\lambda^{\frac{1}{4}-\frac{i}{2}}\,.$
(22)

Our result for $\tau_{1}(\delta,\ell)$ suggests that the states we are
considering are dual to the energies from string theory by a shift of $2$ from
a supersymmetry transformation

$\hat{\Delta}(r;\lambda)=\Delta(r;\lambda)+2=\tau(r;\lambda)+\ell+2\,.$ (23)

For example, the dimension of the Konishi operator $\text{Tr }ZD^{2}Z$ is of
the form (22) at strong coupling and $\hat{\Delta}_{\text{classical}}=4$ in
the free theory (see appendix E). The superconformal primary of the Konishi
supermultiplet $\sum_{i=1}^{6}\text{Tr }\Phi_{i}^{2}$, which is exchanged in
the correlator (1), has classical dimension $\Delta_{\text{classical}}=2$, in
agreement with (23).333$\Phi_{i}$ are the six real scalar fields of
$\mathcal{N}=4$ SYM and $Z=\Phi_{1}+i\Phi_{2}$. For this reason we expect
$\tau_{1}(\delta,\ell)$ to be degenerate, i.e. the same for all species, so
that $\langle f_{0}(\delta,\ell)\tau_{1}(\delta,\ell)^{2}\rangle=\langle
f_{0}(\delta,\ell)\rangle(\ell+2)^{2}$ and so on.

Starting with the next layer of Wilson coefficients $\alpha^{(1)}_{a,b}$ and
the corresponding OPE data $\langle
f_{0}(\delta,\ell)\tau_{2}(\delta,\ell)\rangle$ and $\langle
f_{2}(\delta,\ell)\rangle$ we are truly starting to explore the Virasoro-
Shapiro amplitude in AdS. Of this data, the only pieces that were previously
known are, from integrability, the twists on the leading Regge trajectory
Gromov:2011de ; Basso:2011rs ; Gromov:2011bz

$\tau_{2}(\delta,2(\delta-1))=\frac{3\delta^{2}-\delta+2}{2\sqrt{\delta}}\,,$
(24)

and, from supersymmetric localisation, the Wilson coefficients444Localisation
also fixes $\alpha^{(2)}_{0,0}$ and $\alpha^{(3)}_{0,0}$ as in (92) below.
Binder:2019jwn ; Chester:2020dja

$\alpha^{(1)}_{0,0}=0\,,\qquad\alpha^{(1)}_{1,0}=-\frac{22}{3}\zeta(3)^{2}\,.$
(25)

In the remainder of the paper we will determine the rest of this data.

##  3 Solving the sum rules

The dispersive sum rule for the next layer of Wilson coefficients is

$\displaystyle\alpha^{(1)}_{a,b}=\sum\limits_{\delta=1}^{\infty}\sum\limits_{m=0}^{b}\frac{1}{\delta^{4+2a+3b}}\bigg{(}{}$
$\displaystyle
c^{(0)}_{a,b,m}\left(F^{(2)}_{m}(\delta)-(3+2a+3b)T^{(2)}_{m}(\delta)\right)$
(26)
$\displaystyle+c^{(2,0)}_{a,b,m}F^{(0)}_{m}(\delta)+c^{(2,1)}_{a,b,m}F^{(0)}_{m+1}(\delta)\bigg{)}\,,$

with new OPE data encoded in the sums

$\displaystyle T^{(2)}_{m}(\delta)={}$
$\displaystyle\frac{4^{m}}{\Gamma(2m+2)}\sum_{\ell=0,2,\ldots}^{2(\delta-1)}\sqrt{\delta}(\ell-m+1)_{m}(\ell+2)_{m}\langle
f_{0}(\delta,\ell)\tau_{2}(\delta,\ell)\rangle\,,$ (27) $\displaystyle
F^{(2)}_{m}(\delta)={}$
$\displaystyle\frac{4^{m}}{\Gamma(2m+2)}\sum_{\ell=0,2,\ldots}^{2(\delta-1)}(\ell-m+1)_{m}(\ell+2)_{m}\left(\delta\langle
f_{2}(\delta,\ell)\rangle-\frac{39}{4}\ell\langle
f_{0}(\delta,\ell)\rangle\right)\,.$ (28)

The coefficients $c^{(0)}_{a,b,m}$ are the ones given in (11) and the new ones
are given by

$\displaystyle c^{(2,0)}_{a,b,m}={}$
$\displaystyle\Big{(}-\left(\tfrac{1}{2}a+\tfrac{3}{4}b+\tfrac{27}{4}\right)m^{2}+\left(2a^{2}+6ab+\tfrac{26}{3}a+\tfrac{9}{2}b^{2}+13b-\tfrac{21}{8}\right)m$
(29)
$\displaystyle-\tfrac{4}{3}a^{3}-6a^{2}b-\tfrac{16}{3}a^{2}-9ab^{2}-16ab-\tfrac{1}{3}a-\tfrac{9}{2}b^{3}-12b^{2}-\tfrac{1}{2}b-\tfrac{277}{32}\Big{)}c^{(0)}_{a,b,m}\,,$
$\displaystyle c^{(2,1)}_{a,b,m}={}$
$\displaystyle(m+1)\Big{(}-\left(\tfrac{1}{2}a+\tfrac{3}{4}b+\tfrac{27}{4}\right)m+a^{2}+3ab+\tfrac{49}{12}a+\tfrac{9}{4}b^{2}+\tfrac{49}{8}b-\tfrac{75}{16}\Big{)}c^{(0)}_{a,b,m}\,.$

In contrast to $\alpha^{(0)}_{a,b}$, we now have unknown data on both sides of
the equations. There are some constraints arising from the fact that the sum
rule (26) is valid for $b=0,1,\ldots$ and $a=-b,-b+1,\ldots$ and that (1)
needs to be an expansion in positive powers of $\sigma_{2}$ and $\sigma_{3}$,
which implies

$\alpha^{(1)}_{a,b}=0\,,\quad\text{for }a=-b,-b+1,\ldots,-1\,.$ (30)

As discussed in Alday:2022uxp , this is not enough to fully fix
$\alpha^{(1)}_{a,b}$. However, we claim that all the data is uniquely fixed by
the following set of assumptions:

  * •

$\alpha^{(1)}_{a,b}$ is in the ring of single-valued multiple zeta values and
has uniform weight $4+2a+3b$.

  * •

$T^{(2)}_{m}(\delta)$ is a linear combination of Euler-Zagier sums (and
multiple zeta values) of maximal weight $m+2$ and maximal depth $m+1$.

  * •

$F^{(2)}_{m}(\delta)$ is a linear combination of Euler-Zagier sums (and
multiple zeta values) of maximal weight $m+3$ and maximal depth $m+1$.

Note that the last two assumptions imply that $\alpha^{(1)}_{a,b}$ will have
multiple zeta values of maximal depth $b+2$, which is the simplest possibility
compatible with the depth 2 localisation result for $\alpha^{(1)}_{1,0}$ (25).

It is known that the sphere integrals of tree-level closed string amplitudes
in flat space lead to Wilson coefficients in the ring of single-valued
multiple zeta values. In particular, all the $\alpha^{(0)}_{a,b}$’s can be
written in terms of single zeta values of odd arguments (which are single-
valued), due to the following representation for the flat space result

$\displaystyle\frac{f(S,T)}{STU}={}$
$\displaystyle-\frac{\Gamma\left(-S\right)\Gamma\left(-T\right)\Gamma\left(-U\right)}{\Gamma\left(S+1\right)\Gamma\left(T+1\right)\Gamma\left(U+1\right)}$
(31) $\displaystyle={}$
$\displaystyle\frac{1}{STU}\exp\left(2\sum\limits_{n=1}^{\infty}\frac{\zeta(2n+1)}{2n+1}\left(S^{2n+1}+T^{2n+1}+U^{2n+1}\right)\right)$
$\displaystyle={}$
$\displaystyle\frac{1}{STU}+2\sum_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(0)}_{a,b}\,,$

where $S,T,U$ are the dimensionless Mandelstam variables related to particle
momenta $p_{i}$

$\begin{gathered}S=-\frac{\alpha^{\prime}}{4}(p_{1}+p_{2})^{2}\,,\qquad
T=-\frac{\alpha^{\prime}}{4}(p_{1}+p_{3})^{2}\,,\qquad
U=-\frac{\alpha^{\prime}}{4}(p_{1}+p_{4})^{2}\,,\\\
\hat{\sigma}_{2}=\frac{1}{2}(S^{2}+T^{2}+U^{2})\,,\qquad\hat{\sigma}_{3}=STU\,,\end{gathered}$
(32)

satisfying $S+T+U=0$. $f(S,T)$ is the four-graviton amplitude of type IIb
superstring theory in flat space divided by the corresponding supergravity
amplitude. We are making the assumption that the (currently still unknown)
worldsheet description of closed strings in AdS also leads to single-valued
multiple zeta values.

We will concretise the assumptions on $T^{(2)}_{m}(\delta)$ and
$F^{(2)}_{m}(\delta)$ and construct the solutions after a short excursion on
single-valued multiple zeta values.

###  3.1 (Single-valued) multiple zeta values

In this section we give a practical introduction to working with (single-
valued) multiple zeta values (MZVs). There are many relations between MZVs of
the same weight, so in order to compare MZVs, we expand them in a basis for
the algebra $\mathcal{H}_{N}$ of MZVs of weight $N$. We denote by
$\mathcal{L}$ the space $\mathcal{H}$ modulo products of MZVs, i.e.
$\mathcal{H}$ is the polynomial algebra generated by the elements of
$\mathcal{L}$. We list some examples of possible basis elements and the
dimensions of both spaces and in tables 1 and 2.

$N$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---
$\mathcal{L}_{N}\text{ basis}$ | $\zeta(2)$ | $\zeta(3)$ |  | $\zeta(5)$ |  | $\zeta(7)$ | $\zeta(5,3)$ | $\zeta(9)$
$\mathcal{H}_{N}\text{ basis}$ | $\zeta(2)$ | $\zeta(3)$ | $\zeta(2)^{2}$ | $\zeta(5)$ | $\zeta(3)^{2}$ | $\zeta(7)$ | $\zeta(5,3)$ | $\zeta(9)$
|  |  |  | $\zeta(3)\zeta(2)$ | $\zeta(2)^{3}$ | $\zeta(5)\zeta(2)$ | $\zeta(5)\zeta(3)$ | $\zeta(7)\zeta(2)$
|  |  |  |  |  | $\zeta(3)\zeta(2)^{2}$ | $\zeta(3)^{2}\zeta(2)$ | $\zeta(5)\zeta(2)^{2}$
|  |  |  |  |  |  | $\zeta(2)^{4}$ | $\zeta(3)^{3}$
|  |  |  |  |  |  |  | $\zeta(3)\zeta(2)^{3}$
Table 1: Possible basis for $\mathcal{H}$ and $\mathcal{L}$. $N$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$\text{dim }\mathcal{L}_{N}$ | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 2 | 2 | 3 | 3 | 4 | 5 | 7 | 8 | 11
$\text{dim }\mathcal{H}_{N}$ | 1 | 1 | 1 | 2 | 2 | 3 | 4 | 5 | 7 | 9 | 12 | 16 | 21 | 28 | 37 | 49 | 65 | 86
Table 2: Dimensions of the space $\mathcal{H}$ ($\mathcal{L}$) of multiple
zeta values of weight $N$ (modulo products).

The task of rewriting MZVs in terms of a basis has been performed for weights
up to 30 in Blumlein:2009cf . In practice we use the program
HyperlogProcedures HyperlogProcedures  by Oliver Schnetz for this. An example
is

$\zeta(3,2,1)=3\zeta(3)^{2}-\frac{29}{30}\zeta(2)^{3}\,.$ (33)

Single-valued multiple zeta values were first studied by Brown Brown:2013gia
and are defined as single-valued multiple polylogarithms evaluated at unit
argument. There is a map on the ring of multiple zeta values $\mathcal{Z}$,
$\text{sv}:\mathcal{Z}\to\mathcal{Z}$ that sends multiple zeta values
$\zeta(s_{1},s_{2},\ldots)$ to single-valued multiple zeta values
$\zeta^{\text{sv}}(s_{1},s_{2},\ldots)$ which generate a smaller ring
$\mathcal{Z}^{\text{sv}}\subset\mathcal{Z}$. In particular we have

$\zeta^{\text{sv}}(2k)=0\,,\qquad\zeta^{\text{sv}}(2k+1)=2\zeta(2k+1)\,,\qquad
k\in\mathbb{N}\,.$ (34)

The space $\mathcal{H}^{\text{sv}}$ of single-valued MZVs is the polynomial
algebra generated by $\mathcal{L}^{\text{sv}}$, which is obtained by taking
the elements of odd weight of $\mathcal{L}$ and applying the sv map
Brown:2013gia . We show some basis generators and the dimensions of these
spaces in tables 3 and 4.

$N$ | 3 | 5 | 7 | 9 | 11 | 13 | 15
---|---|---|---|---|---|---|---
$\mathcal{L}^{\text{sv}}_{N}\text{ basis}$ | $\zeta^{\text{sv}}(3)$ | $\zeta^{\text{sv}}(5)$ | $\zeta^{\text{sv}}(7)$ | $\zeta^{\text{sv}}(9)$ | $\zeta^{\text{sv}}(11)$ | $\zeta^{\text{sv}}(13)$ | $\zeta^{\text{sv}}(15)$
|  |  |  |  | $\zeta^{\text{sv}}(5,3,3)$ | $\zeta^{\text{sv}}(7,3,3)$ | $\zeta^{\text{sv}}(9,3,3)$
|  |  |  |  |  | $\zeta^{\text{sv}}(5,5,3)$ | $\zeta^{\text{sv}}(7,3,5)$
|  |  |  |  |  |  | $\zeta^{\text{sv}}(6,4,3,1,1)$
Table 3: Possible basis for $\mathcal{L}^{\text{sv}}$. $N$ | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
$\text{dim }\mathcal{L}_{N}^{\text{sv}}$ | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 2 | 0 | 3 | 0 | 4 | 0 | 7 | 0 | 11
$\text{dim }\mathcal{H}_{N}^{\text{sv}}$ | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 2 | 2 | 3 | 3 | 5 | 5 | 8 | 8 | 13 | 14 | 21
Table 4: Dimensions of the space $\mathcal{H}^{\text{sv}}$
($\mathcal{L}^{\text{sv}}$) of single-valued multiple zeta values (modulo
products).

As the sv map is implemented in HyperlogProcedures, it is easy to obtain
explicit expressions for single-valued MZVs, for example

$\displaystyle\zeta^{\text{sv}}(5,3)$ $\displaystyle=-10\zeta(5)\zeta(3)\,,$
(35) $\displaystyle\zeta^{\text{sv}}(5,3,3)$
$\displaystyle=2\zeta(5,3,3)-\frac{8}{7}\zeta(5)\zeta(2)^{3}+\frac{12}{5}\zeta(7)\zeta(2)^{2}+90\zeta(9)\zeta(2)-5\zeta(3)^{2}\zeta(5)\,.$

###  3.2 Construction of the solutions

In order to solve the equation (26) subject to our assumptions, we will
construct an ansatz for $T^{(2)}_{m}(\delta)$ and $F^{(2)}_{m}(\delta)$ and
fix the coefficients by imposing the vanishing of Wilson coefficients for
negative values of $a$ (30) as well as single-valuedness for a finite number
of non-negative values of $a$

$\alpha^{(1)}_{a,b}\in\mathcal{Z}^{\text{sv}}\,,\quad\text{for
}a=0,1,\ldots\,.$ (36)

Let us begin with $\alpha^{(1)}_{a,0}$. We make the ansatz

$\displaystyle T^{(2)}_{0}(\delta)={}$ $\displaystyle
d_{0}Z(\delta-1)+d_{1}\delta
Z_{1}(\delta-1)+d_{2}\delta^{2}Z_{2}(\delta-1)\,,$ (37) $\displaystyle
F^{(2)}_{0}(\delta)={}$ $\displaystyle c_{0}Z(\delta-1)+c_{1}\delta
Z_{1}(\delta-1)+c_{2}\delta^{2}Z_{2}(\delta-1)+c_{3}\delta^{3}Z_{3}(\delta-1)+\widetilde{c}_{0}\delta^{3}\zeta(3)Z(\delta-1)\,,$

which we insert together with the solution for $F^{(0)}_{m}(\delta)$ (18) into
(26). As a first step we have to ensure that the sum over $\delta$ is
convergent. This is the case for $a>0$, but for $a=0$ we have the divergent
term

$\alpha^{(1)}_{0,0}=\sum\limits_{\delta=1}^{\infty}\left(\frac{\widetilde{c}_{0}\zeta(3)+c_{3}Z_{3}(\delta-1)}{\delta}+O(\delta^{-2})\right)\,.$
(38)

The asymptotic expansion for the Euler-Zagier sum is given by its relation to
the generalised harmonic numbers

$Z_{s}(\delta-1)=H^{(s)}_{\delta-1}=\zeta(s)-\frac{1}{(s-1)\delta^{s-1}}+O(\delta^{-s})\,,\quad
s=2,3,\ldots\,,$ (39)

so we impose convergence by setting

$\widetilde{c}_{0}=-c_{3}\,.$ (40)

Now the sums can be done using (17)

$\displaystyle\alpha^{(1)}_{a,0}={}$ $\displaystyle-
c_{3}\zeta_{\text{reg}}(2a+1)\zeta(3)+c_{3}\zeta_{\text{reg}}(2a+1,3)+\left(c_{2}-(2a+3)d_{2}\right)\zeta(2a+2,2)$
(41)
$\displaystyle+\left(2a^{2}-(2a+3)d_{1}+\tfrac{49}{6}a+c_{1}-\tfrac{75}{8}\right)\zeta(2a+3,1)$
$\displaystyle-\left(\tfrac{4}{3}a^{3}+\tfrac{16}{3}a^{2}+(2a+3)d_{0}+\tfrac{1}{3}a-c_{0}+\tfrac{277}{32}\right)\zeta(2a+4)\,.$

Here $\zeta_{\text{reg}}$ are shuffle-regularised multiple zeta values, as
described in section 2.1 of Broedel:2014vla . They are finite when the first
argument is 1 (for instance $\zeta_{\text{reg}}(1)=0$) and agree with the
usual multiple zeta values when the first argument is $2,3,\ldots$. For each
value of $a$ this expression can be rewritten in a basis of MZVs of weight
$4+2a$, for instance

$\displaystyle\alpha^{(1)}_{0,0}={}$
$\displaystyle\frac{1}{10}\left(4c_{0}+c_{1}+3c_{2}-5c_{3}-12d_{0}-3d_{1}-9d_{2}-44\right)\zeta(2)^{2}\,,$
(42) $\displaystyle\alpha^{(1)}_{1,0}={}$
$\displaystyle(\ldots)\zeta(2)^{3}+(\ldots)\zeta(3)^{2}\,,$
$\displaystyle\alpha^{(1)}_{2,0}={}$
$\displaystyle(\ldots)\zeta(2)^{4}+(\ldots)\zeta(3)\zeta(5)+(\ldots)\zeta(5,3)\,,$
$\displaystyle\alpha^{(1)}_{3,0}={}$
$\displaystyle(\ldots)\zeta(2)^{5}+(\ldots)\zeta(3)\zeta(7)+(\ldots)\zeta(5)^{2}+(\ldots)\zeta(7,3)\,.$

We now demand that each expression can be written in terms of a basis of
single-valued MZVs of the same weight

$\displaystyle\alpha^{(1)}_{0,0}={}$ $\displaystyle 0\,,$ (43)
$\displaystyle\alpha^{(1)}_{1,0}={}$ $\displaystyle\kappa_{1}\zeta(3)^{2}\,,$
$\displaystyle\alpha^{(1)}_{2,0}={}$
$\displaystyle\kappa_{2}\zeta(3)\zeta(5)\,,$
$\displaystyle\alpha^{(1)}_{3,0}={}$
$\displaystyle\kappa_{3}\zeta(3)\zeta(7)+\kappa_{4}\zeta(5)^{2}\,.$

Equating (42) and (43) fixes all the coefficients of the ansatz except for one
parameter. The remaining parameter (along with all new parameters in the
ansatz for $b=1,2$) is fixed by the constraints (30) and (36) once we impose
them up to $b=2$ as described below. The solution is consistent with the
localisation result (25), which we use here to immediately write the fully
fixed solution at $b=0$

$\displaystyle d_{0}$ $\displaystyle=2\,,\quad d_{1}=\frac{1}{4}\,,\quad
d_{2}=1\,,$ (44) $\displaystyle c_{0}$ $\displaystyle=\frac{405}{32}\,,\quad
c_{1}=\frac{89}{8}\,,\quad c_{2}=2\,,\quad c_{3}=-2\,.$

The result for $\alpha^{(1)}_{a,0}$ is given by

$\displaystyle\alpha^{(1)}_{a,0}={}$ $\displaystyle
2\zeta_{\text{reg}}(2a+1)\zeta(3)-2\zeta_{\text{reg}}(2a+1,3)-(2a+1)\zeta(2a+2,2)$
(45)
$\displaystyle+\frac{1}{3}\left(6a^{2}+23a+3\right)\zeta(2a+3,1)-\frac{1}{3}\left(4a^{3}+16a^{2}+13a+6\right)\zeta(2a+4)\,,$

and is in $\mathcal{Z}^{\text{sv}}$ for any value of $a$, which can be shown
by rewriting it in the form

$\alpha^{(1)}_{a,0}=-\left(a^{2}+\frac{35}{6}a+\frac{1}{2}\right)\sum\limits_{\begin{subarray}{c}i_{1},i_{2}=0\\\
i_{1}+i_{2}=a-1\end{subarray}}^{a-1}\zeta(3+2i_{1})\zeta(3+2i_{2})-2\sum\limits_{\begin{subarray}{c}i_{1},i_{2}=0\\\
i_{1}+i_{2}=a-1\end{subarray}}^{a-1}i_{1}i_{2}\zeta(3+2i_{1})\zeta(3+2i_{2})\,.$
(46)

The reason this rewriting in terms of single zeta values is possible is that
the expression has maximal depth two and the first non-trivial generators of
single-valued MZVs have depth three.

We can continue in the same way for $\alpha^{(1)}_{a,1}$ by inserting the
result for $T^{(2)}_{0}(\delta)$ and $F^{(2)}_{0}(\delta)$ and making an
ansatz for $T^{(2)}_{1}(\delta)$ and $F^{(2)}_{1}(\delta)$, and so on. For
general $m\geq 1$ we will use the following notation for an ansatz in terms of
Euler-Zagier sums with weights up to $w_{\text{max}}$ and depths up to
$d_{\text{max}}$

$A^{i}_{w_{\text{max}},d_{\text{max}}}=\sum\limits_{w=1}^{w_{\text{max}}}\sum\limits_{d=1}^{d_{\text{max}}}\sum\limits_{\begin{subarray}{c}s_{1},\ldots,s_{d}\in\mathbb{N}\\\
s_{1}+\ldots+s_{d}=w\end{subarray}}c^{i}_{s_{1},\ldots,s_{d}}\delta^{w}Z_{s_{1},\ldots,s_{d}}(\delta-1)\,.$
(47)

Our ansatz for $T^{(2)}_{m}(\delta)$ and $F^{(2)}_{m}(\delta)$ has the form

$\displaystyle T^{(2)}_{m\geq 1}(\delta)={}$ $\displaystyle
A^{1}_{m+2,m+1}+\delta^{3}\zeta(3)A^{3}_{m-1,m-1}+\delta^{5}\zeta(5)A^{5}_{m-3,m-3}+\delta^{6}\zeta(3)^{2}A^{7}_{m-4,m-4}+\ldots\,,$
$\displaystyle F^{(2)}_{m\geq 1}(\delta)={}$ $\displaystyle
A^{2}_{m+3,m+1}+\delta^{3}\zeta(3)A^{4}_{m,m}+\delta^{5}\zeta(5)A^{6}_{m-2,m-2}+\delta^{6}\zeta(3)^{2}A^{8}_{m-3,m-3}+\ldots\,,$
(48)

taking the explicit zeta values into account when determining the maximal
weights and depths of the terms. For the zeta values we include all basis
elements of $\mathcal{H}^{\text{sv}}$ that can be multiplied with
$A^{i}_{w_{\text{max}},d_{\text{max}}}$ with positive $w_{\text{max}}$ and
$d_{\text{max}}$. Once we fix the coefficients, only the terms with $A^{1}$,
$A^{2}$ and $A^{4}$ will survive. At higher orders in the expansion, similar
terms should produce the zeta values in the dimension of the Konishi operator
(154). Interestingly, non-trivial single-valued multiple zeta values were
observed to appear in the dimension of the Konishi operator at weak coupling
at 8 loops Leurent:2013mr , see (E). The sums in (47) start at $w=1$ to ensure
$F_{m\geq 1}(1)=T_{m\geq 1}(1)=0$, which follows from the definitions (27) and
(28). The number of coefficients in each ansatz is listed in table 5.

$m$ | 0 | 1 | 2 | 3 | 4 | 5
---|---|---|---|---|---|---
$T^{(2)}_{m}(\delta)$ | 3 | 6 | 15 | 33 | 70 | 145
$F^{(2)}_{m}(\delta)$ | 5 | 11 | 28 | 64 | 138 | 288
Table 5: Number of coefficients in the ansatz for $T^{(2)}_{m}(\delta)$ and
$F^{(2)}_{m}(\delta)$.

As it turns out, imposing (30) together with (36) for $a=0,1,2,3$ fixes all
the coefficients in each case, which we were able to show explicitly for
$b=1,\ldots,5$ and for $b=6$, where we used a smaller ansatz taking into
account some of the patterns we observed from the previous results. In each
case the resulting expression for $\alpha^{(1)}_{a,b}$ is in
$\mathcal{Z}^{\text{sv}}$ also for larger values of $a$, which we checked for
all cases with $b\leq 6$ and weights $4+2a+3b\leq 28$.

As in the case $b=0$, the expressions for $\alpha^{(1)}_{-b,b}$ always contain
$1/\delta$ terms that require cancellations in order for the sum over $\delta$
to be convergent. For $b>0$ it turns out that one can fix all coefficients by
imposing (30) for shuffle-regularised MZVs. The solutions always lead to
convergent sums in $\delta$ due to cancellations similar to (38). We checked
this by computing the asymptotic expansions of the Euler-Zagier sums using the
Mathematica package HarmonicSums Ablinger:2009ovq ; Ablinger:2012ufz ;
Ablinger:2014rba . For illustration we show the first few results for the sums
over OPE data (all Euler-Zagier sums are evaluated at $\delta-1$)

$\displaystyle T^{(2)}_{0}(\delta)={}$
$\displaystyle\delta^{2}Z_{2}+\tfrac{1}{4}\delta Z_{1}+2\,,$ $\displaystyle
T^{(2)}_{1}(\delta)={}$
$\displaystyle\delta^{3}\left(Z_{3}+2Z_{1,2}+3Z_{2,1}\right)+\delta^{2}\left(\tfrac{7}{4}Z_{2}+Z_{1,1}\right)+\tfrac{9}{2}\delta
Z_{1}\,,$ $\displaystyle T^{(2)}_{2}(\delta)={}$
$\displaystyle\delta^{4}\big{(}2Z_{1,3}+3Z_{3,1}+3Z_{2,2}+4Z_{1,1,2}+6Z_{1,2,1}+8Z_{2,1,1}\big{)}$
$\displaystyle+\delta^{3}\left(2Z_{3}+\tfrac{15}{4}Z_{1,2}+\tfrac{23}{4}Z_{2,1}+3Z_{1,1,1}\right)+\delta^{2}\left(2Z_{2}+9Z_{1,1}\right)\,,$
$\displaystyle T^{(2)}_{3}(\delta)={}$
$\displaystyle\delta^{5}\big{(}Z_{2,3}+2Z_{3,2}+4Z_{1,1,3}+6Z_{1,3,1}+8Z_{3,1,1}+6Z_{1,2,2}+7Z_{2,1,2}+8Z_{2,2,1}$
$\displaystyle+8Z_{1,1,1,2}+12Z_{1,1,2,1}+16Z_{1,2,1,1}+20Z_{2,1,1,1}\big{)}$
$\displaystyle+\delta^{4}\big{(}4Z_{1,3}+6Z_{3,1}+\tfrac{11}{2}Z_{2,2}+8Z_{1,1,2}+12Z_{1,2,1}+16Z_{2,1,1}+8Z_{1,1,1,1}\big{)}$
$\displaystyle+\delta^{3}\left(\tfrac{7}{2}Z_{1,2}+\tfrac{7}{2}Z_{2,1}+16Z_{1,1,1}\right)\,,$
(49)

and

$\displaystyle F^{(2)}_{0}(\delta)={}$
$\displaystyle-2\delta^{3}Z_{3}+2\delta^{2}Z_{2}+\tfrac{89}{8}\delta
Z_{1}+\tfrac{405}{32}+2\delta^{3}\zeta(3)F^{(0)}_{0}(\delta)\,,$
$\displaystyle F^{(2)}_{1}(\delta)={}$
$\displaystyle-2\delta^{4}\left(2Z_{1,3}+Z_{2,2}+3Z_{3,1}+Z_{4}\right)+\delta^{3}\left(6Z_{1,2}+11Z_{2,1}+Z_{3}\right)$
$\displaystyle+\tfrac{3}{8}\delta^{2}\left(204Z_{1,1}+59Z_{2}\right)+\tfrac{761}{16}\delta
Z_{1}+2\delta^{3}\zeta(3)F^{(0)}_{1}(\delta)\,,$ $\displaystyle
F^{(2)}_{2}(\delta)={}$
$\displaystyle-2\delta^{5}\big{(}2Z_{1,4}+2Z_{2,3}+3Z_{3,2}+3Z_{4,1}+4Z_{1,1,3}+2Z_{1,2,2}+6Z_{1,3,1}$
$\displaystyle+2Z_{2,1,2}+3Z_{2,2,1}+8Z_{3,1,1}\big{)}+4\delta^{4}\big{(}Z_{1,3}+3Z_{2,2}+2Z_{3,1}+4Z_{1,1,2}$
$\displaystyle+7Z_{1,2,1}+11Z_{2,1,1}-Z_{4}\big{)}+\tfrac{1}{8}\delta^{3}\left(707Z_{1,2}+763Z_{2,1}+2540Z_{1,1,1}+56Z_{3}\right)$
$\displaystyle+\tfrac{1}{32}\delta^{2}\left(5428Z_{1,1}+1341Z_{2}\right)+2\delta^{3}\zeta(3)F^{(0)}_{2}(\delta)\,.$
(50)

For the values of $\alpha^{(1)}_{a,1}$ and $\alpha^{(1)}_{a,2}$ we find

$\displaystyle\alpha^{(1)}_{a,1}={}$
$\displaystyle\zeta(3)(-4\zeta_{\text{reg}}(2a+3,1)+(2a+3)\zeta(2a+4))$
$\displaystyle+2\zeta_{\text{reg}}(2a+3,4)+4\zeta_{\text{reg}}(2a+3,1,3)+6\zeta_{\text{reg}}(2a+3,3,1)+2\zeta_{\text{reg}}(2a+3,2,2)$
$\displaystyle+2\zeta(2a+4,3)+2(2a+3)\zeta(2a+4,1,2)+(6a+7)\zeta(2a+4,2,1)$
$\displaystyle-\tfrac{2}{3}\left(6a^{2}+25a+15\right)\zeta(2a+5,2)-\tfrac{8}{3}\left(3a^{2}+19a+15\right)\zeta(2a+5,1,1)$
$\displaystyle+\tfrac{1}{6}\left(28a^{3}+212a^{2}+413a+249\right)\zeta(2a+6,1)$
$\displaystyle-\tfrac{1}{6}(2a+3)\left(4a^{3}+34a^{2}+88a+75\right)\zeta(2a+7)\,,$
(51)

and

$\displaystyle\alpha^{(1)}_{a,2}={}$
$\displaystyle\zeta(3)(2\zeta_{\text{reg}}(2a+5,2)+8\zeta_{\text{reg}}(2a+5,1,1)-2(2a+5)\zeta(2a+6,1)+(a+1)(a+3)\zeta(2a+7))$
$\displaystyle-4\zeta_{\text{reg}}(2a+5,1,4)-6\zeta_{\text{reg}}(2a+5,4,1)-4\zeta_{\text{reg}}(2a+5,2,3)-6\zeta_{\text{reg}}(2a+5,3,2)$
$\displaystyle-8\zeta_{\text{reg}}(2a+5,1,1,3)-12\zeta_{\text{reg}}(2a+5,1,3,1)-16\zeta_{\text{reg}}(2a+5,3,1,1)$
$\displaystyle-4\zeta_{\text{reg}}(2a+5,1,2,2)-4\zeta_{\text{reg}}(2a+5,2,1,2)-6\zeta_{\text{reg}}(2a+5,2,2,1)$
$\displaystyle+(2a+1)\zeta(2a+6,4)-4\zeta(2a+6,1,3)-4\zeta(2a+6,3,1)-2(2a+5)\zeta(2a+6,2,2)$
$\displaystyle-4(2a+5)\zeta(2a+6,1,1,2)-2(6a+13)\zeta(2a+6,1,2,1)-4(4a+7)\zeta(2a+6,2,1,1)$
$\displaystyle+(a+2)(a+3)\zeta(2a+7,3)+\left(10a^{2}+69a+85\right)\zeta(2a+7,1,2)$
$\displaystyle+2(2a+3)(3a+14)\zeta(2a+7,2,1)+4\left(6a^{2}+53a+73\right)\zeta(2a+7,1,1,1)$
$\displaystyle+\tfrac{1}{6}\left(-26a^{3}-249a^{2}-673a-528\right)\zeta(2a+8,2)$
$\displaystyle-\tfrac{2}{3}\left(20a^{3}+222a^{2}+682a+633\right)\zeta(2a+8,1,1)$
$\displaystyle+\tfrac{1}{6}\left(22a^{4}+307a^{3}+1472a^{2}+2945a+2088\right)\zeta(2a+9,1)$
$\displaystyle-\tfrac{1}{6}(a+1)(a+3)\left(4a^{3}+52a^{2}+217a+297\right)\zeta(2a+10)\,.$
(52)

One can check that these expressions cannot generally be expressed in terms of
single zeta values (as the $\alpha^{(0)}_{a,b}$ or $\alpha^{(1)}_{a,0}$ (46)),
for instance

$\alpha^{(1)}_{3,1}=-\frac{209279}{300}\zeta(13)-166\zeta(3)^{2}\zeta(7)-174\zeta(3)\zeta(5)^{2}+\frac{2}{25}\zeta^{\text{sv}}(5,5,3)\,,$
(53)

where

$\zeta^{\text{sv}}(5,5,3)=2\zeta(5,5,3)+10\zeta(5)\zeta(5,3)+20\zeta(9)\zeta(2)^{2}+275\zeta(11)\zeta(2)+50\zeta(3)\zeta(5)^{2}\,,$
(54)

is one of the basis elements that were chosen for
$\mathcal{L}^{\text{sv}}_{13}$, see table 3. In the next section we will lift
the results for $\alpha^{(1)}_{a,b}$, $b=0,\ldots,6$ to any value of $b$ by
finding general expressions for $T^{(2)}_{m}(\delta)$ and
$F^{(2)}_{m}(\delta)$.

##  4 CFT data

###  4.1 General structure

Our solution for $\alpha^{(1)}_{a,b}$ results in some corresponding CFT-data
to order $1/\sqrt{\lambda}$. At leading order a natural way to repackage the
CFT data is by considering the sums over spin $F^{(0)}_{m}(\delta)$ (18). At
this order it is natural to consider $T^{(2)}_{m}(\delta)$ and
$F^{(2)}_{m}(\delta)$. From the results for $m=0,\ldots,6$ we see that
$T^{(2)}_{m}(\delta)$ takes the form

$\displaystyle T^{(2)}_{m}(\delta)={}$
$\displaystyle\sum\limits_{w=m}^{m+2}\sum\limits_{d=\lfloor\frac{m+1}{2}\rfloor}^{m+1}\sum\limits_{\begin{subarray}{c}s_{1},\ldots,s_{d}\in\\{1,2,3\\}\\\
s_{1}+\ldots+s_{d}=w\end{subarray}}t^{m}_{\bm{s}}\delta^{w}Z_{\bm{s}}(\delta-1)\,,$
(55)

i.e. there are Euler-Zagier sums with words $\bm{s}=s_{1},\ldots,s_{d}$ of
letters from the alphabet $\\{1,2,3\\}$ with weights $m$ to $m+2$ and depths
$\lfloor\frac{m+1}{2}\rfloor$ to $m+1$. We define the following function
counting the number of occurrences of a given letter in a word

$n^{\bm{s}}_{k}=\sum\limits_{i=1}^{d}\delta_{s_{i},k}\,.$ (56)

We then find the following formula for the coefficients in (55)

$t^{m}_{\bm{s}}=\begin{cases}2^{n^{\bm{s}}_{1}}\left(Q^{m}_{w}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},n^{\bm{s}}_{3})+p^{m}_{w}P_{\bm{s}}\right),&n^{\bm{s}}_{3}\in\\{0,1\\}\,,\\\
0,&n^{\bm{s}}_{3}>1\,,\end{cases}$ (57)

with coefficients

$\displaystyle Q^{m}_{m+2}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},1)$
$\displaystyle=1\,,\quad$ $\displaystyle
Q^{m}_{m+2}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0)$
$\displaystyle=\frac{n^{\bm{s}}_{2}(n^{\bm{s}}_{2}+1)}{2}\,,$ (58)
$\displaystyle Q^{m}_{m+1}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},1)$
$\displaystyle=2\,,\quad$ $\displaystyle
Q^{m}_{m+1}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0)$
$\displaystyle=\frac{n^{\bm{s}}_{1}+n^{\bm{s}}_{2}(8n^{\bm{s}}_{2}+6)}{8}\,,$
$\displaystyle Q^{m}_{m}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},1)$
$\displaystyle=0\,,\quad$ $\displaystyle
Q^{m}_{m}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0)$
$\displaystyle=2-\frac{n^{\bm{s}}_{1}(n^{\bm{s}}_{1}+4n^{\bm{s}}_{2}-3)}{8}-\frac{n^{\bm{s}}_{2}(n^{\bm{s}}_{2}-1)}{2}\,,$

and

$p^{m}_{m+2}=\frac{1}{2}\,,\qquad p^{m}_{m+1}=1\,,\qquad p^{m}_{m}=0\,.$ (59)

The object $P_{\bm{s}}$ vanishes when the word $\bm{s}$ is lexicographically
ordered and is defined by

$P_{\bm{s}}=\sum\limits_{i=1}^{d}\delta_{s_{i},\max(\bm{s})}\sum\limits_{j=i+1}^{d}s_{j}(1-\delta_{s_{j},\max(\bm{s})})\,.$
(60)

Next we do the same for $F^{(2)}_{m}(\delta)$, which takes the form

$\displaystyle F^{(2)}_{m}(\delta)={}$ $\displaystyle
2\delta^{3}\zeta(3)F_{m}^{(0)}(\delta)+\sum\limits_{w=m}^{m+3}\sum\limits_{d=\lfloor\frac{m+1}{2}\rfloor}^{m+1}\sum\limits_{\begin{subarray}{c}s_{1},\ldots,s_{d}\in\\{1,2,3,4\\}\\\
s_{1}+\ldots+s_{d}=w\end{subarray}}f^{m}_{\bm{s}}\delta^{w}Z_{\bm{s}}(\delta-1)\,.$
(61)

We find the following expression for the coefficients

$f^{m}_{\bm{s}}=\begin{cases}2^{n^{\bm{s}}_{1}}\left(R^{m}_{w}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},n^{\bm{s}}_{3},n^{\bm{s}}_{4})+P^{n^{\bm{s}}_{3},n^{\bm{s}}_{4}}_{\bm{s},w}\right),&(n^{\bm{s}}_{3},n^{\bm{s}}_{4})\in\\{(0,1),(2,0),(1,0),(0,0)\\}\,,\\\
0,&\text{otherwise}\,.\end{cases}$ (62)

The coefficients $R^{m}_{w}$ determine the coefficients $f^{m}_{\bm{s}}$ for
lexicographically ordered words and are given by

$\displaystyle R^{m}_{m+3}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0,1)={}$
$\displaystyle R^{m}_{m+3}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},2,0)=-2\,,$
$\displaystyle R^{m}_{m+2}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0,1)={}$
$\displaystyle R^{m}_{m+2}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},2,0)=-4\,,$
$\displaystyle R^{m}_{m+1}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0,1)={}$
$\displaystyle R^{m}_{m+1}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},2,0)=0\,,$
$\displaystyle R^{m}_{m+3}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},1,0)={}$
$\displaystyle-2(n^{\bm{s}}_{2}+1)\,,$ $\displaystyle
R^{m}_{m+2}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},1,0)={}$ $\displaystyle
1+n^{\bm{s}}_{1}-2n^{\bm{s}}_{2}\,,$ (63) $\displaystyle
R^{m}_{m+1}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},1,0)={}$ $\displaystyle 2m+3\,,$
$\displaystyle R^{m}_{m+3}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0,0)={}$
$\displaystyle-\tfrac{1}{3}(n^{\bm{s}}_{2}-1)n^{\bm{s}}_{2}(n^{\bm{s}}_{2}+1)\,,$
$\displaystyle R^{m}_{m+2}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0,0)={}$
$\displaystyle\tfrac{1}{2}(m+2)n^{\bm{s}}_{2}(n^{\bm{s}}_{2}+1)\,,$
$\displaystyle R^{m}_{m+1}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},0,0)={}$
$\displaystyle-\tfrac{1}{6}m^{3}+4m^{2}+m\left((n^{\bm{s}}_{2})^{2}+\tfrac{1}{2}n^{\bm{s}}_{2}+\tfrac{467}{48}\right)+\tfrac{1}{3}(n^{\bm{s}}_{2})^{3}+(n^{\bm{s}}_{2})^{2}+\tfrac{1}{6}n^{\bm{s}}_{2}+\tfrac{89}{16}\,,$
$\displaystyle
R^{m}_{m}(n^{\bm{s}}_{1},n^{\bm{s}}_{2},n^{\bm{s}}_{3},n^{\bm{s}}_{4})={}$
$\displaystyle\tfrac{1}{96}\delta_{n^{\bm{s}}_{3},0}\delta_{n^{\bm{s}}_{4},0}\left(-20m^{3}+420m^{2}-24mn^{\bm{s}}_{2}+668m+1215\right)\,.$

The correction terms for unordered words are

$\displaystyle 2P^{0,1}_{\bm{s},m+3}={}$ $\displaystyle
P^{0,1}_{\bm{s},m+2}=-2P_{\bm{s}}\,,\qquad P^{0,1}_{\bm{s},m+1}=0\,,\qquad
P^{n^{\bm{s}}_{3},n^{\bm{s}}_{4}}_{\bm{s},m}=0\,,$ $\displaystyle
2P^{2,0}_{\bm{s},m+3}={}$ $\displaystyle
P^{2,0}_{\bm{s},m+2}=2\sum\limits_{i=1}^{d}\delta_{s_{i},3}(n_{3}^{>i}-1)(n_{1}^{>i}+2n_{2}^{>i})\,,\qquad
P^{2,0}_{\bm{s},m+1}=0\,,$ $\displaystyle P^{1,0}_{\bm{s},m+3}={}$
$\displaystyle\sum\limits_{i=1}^{d}\Big{(}\delta_{s_{i},3}(n^{>i}_{2}(-2n^{\bm{s}}_{2}+n^{>i}_{2}-1)+n^{>i}_{1}(-n^{\bm{s}}_{2}+n^{>i}_{2}-1))+\delta_{s_{i},2}n^{>i}_{1}(n^{>i}_{3}-1)\Big{)}\,,$
$\displaystyle P^{1,0}_{\bm{s},m+2}={}$
$\displaystyle\frac{1}{2}\sum\limits_{i=1}^{d}\Big{(}\delta_{s_{i},3}\left(n^{>i}_{1}(n^{\bm{s}}_{1}+n^{>i}_{1}-2n^{\bm{s}}_{2}+8n^{>i}_{2}+2)+2n^{>i}_{2}(n^{\bm{s}}_{1}-2n^{\bm{s}}_{2}+4n^{>i}_{2}+5)\right)$
$\displaystyle+4\delta_{s_{i},2}n^{>i}_{1}(n^{>i}_{3}-1)\Big{)}\,,$
$\displaystyle P^{1,0}_{\bm{s},m+1}={}$
$\displaystyle\frac{1}{4}\sum\limits_{i=1}^{d}\delta_{s_{i},3}\left(n^{>i}_{1}(4n^{\bm{s}}_{1}+n^{>i}_{1}+8n^{\bm{s}}_{2}+4n^{>i}_{2}+15)+4n^{>i}_{2}(2n^{\bm{s}}_{1}+4n^{\bm{s}}_{2}+n^{>i}_{2}+8)\right)\,,$
$\displaystyle P^{0,0}_{\bm{s},m+3}={}$
$\displaystyle\sum\limits_{i=1}^{d}\delta_{s_{i},2}n^{>i}_{1}(-n^{\bm{s}}_{2}+n^{>i}_{2}+1)\,,$
$\displaystyle P^{0,0}_{\bm{s},m+2}={}$
$\displaystyle\frac{1}{2}\sum\limits_{i=1}^{d}\delta_{s_{i},2}n^{>i}_{1}(n^{\bm{s}}_{1}-2n^{\bm{s}}_{2}+n^{>i}_{1}+8n^{>i}_{2}+5)\,,$
(64) $\displaystyle P^{0,0}_{\bm{s},m+1}={}$
$\displaystyle\frac{1}{4}\sum\limits_{i=1}^{d}\delta_{s_{i},2}n^{>i}_{1}(4n^{\bm{s}}_{1}+8n^{\bm{s}}_{2}+n^{>i}_{1}+4n^{>i}_{2}+1)\,,$

where $n_{k}^{>i}$ counts how many of the letters to the right of $s_{i}$
match $k$

$n_{k}^{>i}=\sum\limits_{j=i+1}^{d}\delta_{s_{j},k}\,.$ (65)

It would be interesting to study whether $T^{(2)}_{m}(\delta)$ and
$F^{(2)}_{m}(\delta)$ can be understood as coming from a specific sum, similar
to the expression for $F^{(0)}_{m}(\delta)$ that was considered in appendix
A.3.2 of Green:2008uj . In appendix B we derive an alternative representation
for $T^{(2)}_{m}(\delta)$ and $F^{(2)}_{m}(\delta)$ purely in terms of the
function $F^{(0)}_{m}(\delta)$.

###  4.2 Twists and OPE coefficients

By computing OPE data for many values for $\delta$ and $\ell$ we can determine
analytic formulae for the twists and OPE coefficients on different Regge
trajectories, which are illustrated in figure 1. Recall the results for the
leading OPE coefficients Alday:2022uxp

$\displaystyle\langle f_{0}(\delta,2(\delta-1))\rangle={}$
$\displaystyle\frac{r_{0}(\delta)}{\delta}\,,$ $\displaystyle\langle
f_{0}(\delta,2(\delta-2))\rangle={}$
$\displaystyle\frac{r_{1}(\delta)}{3}\left(2\delta^{2}+3\delta-8\right)\,,$
(66) $\displaystyle\langle f_{0}(\delta,2(\delta-3))\rangle={}$
$\displaystyle\frac{r_{2}(\delta)}{45}\left(10\delta^{4}+43\delta^{3}+8\delta^{2}-352\delta-192\right)\,,$

where

$r_{n}(\delta)=\frac{4^{2-2\delta}\delta^{2\delta-2n-1}(2\delta-2n-1)}{\Gamma(\delta)\Gamma\left(\delta-\left\lfloor\frac{n}{2}\right\rfloor\right)}\,.$
(67)

Similarly we can now find expressions for the corrections

$\displaystyle\langle f_{0}\tau_{2}\rangle(\delta,2(\delta-1))={}$
$\displaystyle\frac{r_{0}(\delta)}{2\delta^{3/2}}\left(3\delta^{2}-\delta+2\right)\,,$
$\displaystyle\langle f_{0}\tau_{2}\rangle(\delta,2(\delta-2))={}$
$\displaystyle\frac{r_{1}(\delta)}{18\sqrt{\delta}}\left(18\delta^{4}+25\delta^{3}-57\delta^{2}+50\delta-72\right)\,,$
(68) $\displaystyle\langle f_{0}\tau_{2}\rangle(\delta,2(\delta-3))={}$
$\displaystyle\frac{r_{2}(\delta)}{1350\sqrt{\delta}}\left(450\delta^{6}+1985\delta^{5}+1043\delta^{4}-12782\delta^{3}-2552\delta^{2}-35712\delta-11520\right)\,,$

as well as

$\displaystyle\langle f_{2}(\delta,2(\delta-1))\rangle={}$
$\displaystyle-\frac{r_{0}(\delta)}{96\delta^{2}}\left(112\delta^{3}-1872\delta^{2}+344\delta+201\right)+2\delta^{2}\zeta(3)\langle
f_{0}(\delta,2(\delta-1))\rangle\,,$ $\displaystyle\langle
f_{2}(\delta,2(\delta-2))\rangle={}$
$\displaystyle-\frac{r_{1}(\delta)}{864\delta}\left(672\delta^{5}-8272\delta^{4}+3072\delta^{3}+77038\delta^{2}-122559\delta+44136\right)$
$\displaystyle+2\delta^{2}\zeta(3)\langle f_{0}(\delta,2(\delta-2))\rangle\,,$
(69) $\displaystyle\langle f_{2}(\delta,2(\delta-3))\rangle={}$
$\displaystyle-\frac{r_{2}(\delta)}{324000\delta}\big{(}84000\delta^{7}-554800\delta^{6}+342368\delta^{5}+14918998\delta^{4}+28865953\delta^{3}$
$\displaystyle-197973672\delta^{2}+72891360\delta+108388800\big{)}+2\delta^{2}\zeta(3)\langle
f_{0}(\delta,2(\delta-3))\rangle\,.$

We include $\langle f_{0}(\delta,\ell)\rangle$, $\langle
f_{0}\tau_{2}\rangle(\delta,\ell)$ and $\langle f_{2}(\delta,\ell)\rangle$ for
the first seven Regge trajectories in a Mathematica notebook. For the leading
Regge trajectory with $\ell=2(\delta-1)$ the heavy operators are supposed to
be non-degenerate. Our result for $\tau_{2}(\delta,\ell)$ in this case agrees
exactly with the integrability results! Furthermore our procedure leads to a
wealth of CFT data, including structure constants of operators in the leading
Regge trajectory, which can hopefully be confronted with integrability results
in the near future.

###  4.3 Checks

We have performed several checks that back up both the assumptions made in
section 3 and the general expressions (55) and (61). By combining (26) with
(55) and (61) we can generate explicit expressions for $\alpha^{(1)}_{a,b}$
for any $b$ which satisfy (30) and (36), which we checked for all cases with
$b\leq 12$ and weights $4+2a+3b\leq 25$.

Another check is that the equations (27) = (55) for $T^{(2)}_{m}(\delta)$ or
(28) = (61) for $F^{(2)}_{m}(\delta)$ for a fixed value of $\delta$ depend on
$\delta$ unknown bits of CFT data (for $\ell=0,2,\ldots,2(\delta-1)$), but
there are $2\delta-1$ equations for these unknowns (for
$m=0,1,\ldots,2(\delta-1)$). That this overconstrained system of equation has
a solution is a consistency check between the solutions for
$T^{(2)}_{m}(\delta)$ or $F^{(2)}_{m}(\delta)$ for different values of $m$.

Finally, our results for $\tau_{2}(\delta,\ell)$ on the leading Regge
trajectory $\ell=2(\delta-1)$ agree precisely with the known result from
integrability (24), which we checked explicitly for $\delta=1,\ldots,28$, and
we reproduce the two Wilson coefficients that were previously determined using
localisation (25).

##  5 Summing the low energy expansion

Next we would like to do the sums over $a$, $b$ and $m$ to obtain an
expression for the amplitude with explicit poles. For the flat space amplitude
this was done in Zagier:2019eus  where they found essentially the formula (10)
by studying the Virasoro-Shapiro amplitude, without any reference to CFT
dispersion relations. In AdS we have the problem that the gamma function in
(1) makes the sums over $a$ and $b$ divergent. For this reason we will use the
flat space limit formula of Penedones:2010ue ; Fitzpatrick:2011hu  to regulate
the sums, not only for the flat space part of the amplitude but also for the
$1/\sqrt{\lambda}$ corrections. In this case we call it the flat space
transform. This essentially amounts to a Borel summation as it removes the
gamma function from the sums. The flat space transform is defined by

$\text{FS}(M(s_{1},s_{2}))=2\lambda^{\frac{3}{2}}c\int_{\kappa-i\infty}^{\kappa+i\infty}\frac{d\alpha}{2\pi
i}\,e^{\alpha}\alpha^{-6}M\left(\frac{2\sqrt{\lambda}S}{\alpha},\frac{2\sqrt{\lambda}T}{\alpha}\right)\,,$
(70)

where $S$, $T$ and $U$ are the Mandelstams introduced in (32) and
$M(s_{1},s_{2})$ is obtained by eliminating $s_{3}$ using
$s_{1}+s_{2}+s_{3}=0$.

Applying the flat space transform to our Mellin amplitude (1) we get

$\displaystyle\text{FS}(M(s_{1},s_{2}))={}$ $\displaystyle
A^{(0)}(S,T)+\frac{1}{\sqrt{\lambda}}A^{(1)}(S,T)+O(1/\lambda)\,,$ (71)
$\displaystyle A^{(0)}(S,T)={}$
$\displaystyle\frac{1}{STU}+2\sum\limits_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(0)}_{a,b}\,,$
$\displaystyle A^{(1)}(S,T)={}$
$\displaystyle-\frac{2}{3}\frac{\hat{\sigma}_{2}}{\hat{\sigma}_{3}^{2}}+2\sum\limits_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(1)}_{a,b}\,.$

We would like to do the sum in the expression for $A^{(0)}(S,T)$. To this end
we use the following representation of the coefficients $c^{(0)}_{a,b,m}$ from
(11)

$c^{(0)}_{a,b,0}=\frac{(2a+3b)(a+1)_{b-1}}{2\Gamma(b+1)}\,,\qquad
c^{(0)}_{a,b,m}=-\sum\limits_{k=0}^{\lfloor\frac{b-1}{2}\rfloor}\frac{\Gamma(3k+1)}{\Gamma(k+1)\Gamma(2k+2)}c^{(0)}_{a+1+3k,b-1-2k,m-1}\,,$
(72)

which we can use to write an expression for $c^{(0)}_{a,b,m}$ in terms of
$c^{(0)}_{a,b,0}$

$c^{(0)}_{a,b,m}=(-1)^{m}\sum\limits_{k_{1},\ldots,k_{m}=0}^{\infty}c^{(0)}_{a+m+3k,b-m-2k,0}\prod\limits_{i=1}^{m}\frac{\Gamma(3k_{i}+1)}{\Gamma(k_{i}+1)\Gamma(2k_{i}+2)}\,,\quad
k=k_{1}+\ldots+k_{m}\,.$ (73)

Next we sum the terms above over $a$ and $b$ for several fixed values of $k$
and $m$ and guess the general form

$\sum\limits_{a,b=0}^{\infty}c^{(0)}_{a+m+3k,b-m-2k,0}x^{a}y^{b}=\frac{1}{2}\frac{y+2}{1-x-y}\frac{y^{2k+m}}{(1-y)^{3k+m}}\,.$
(74)

Inserting this into (73), the sums over $k_{i}$ factorise and we find

$\sum\limits_{a,b=0}^{\infty}c^{(0)}_{a,b,m}x^{a}y^{b}=\frac{1}{2}\frac{y+2}{1-x-y}\left(\frac{\sqrt{1-4y}-1}{2}\right)^{m}\,,$
(75)

a formula that was already found in Zagier:2019eus . In order to perform the
sum over $m$, we consider the generating series found in Green:2008uj ;
Zagier:2019eus

$\sum\limits_{m=0}^{\infty}F^{(0)}_{m}(\delta)\left(\frac{z}{\delta}\right)^{m}=\binom{z+\delta-1}{\delta-1}^{2}\,.$
(76)

Combining everything, we can compute the sum

$\displaystyle
2\sum\limits_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(0)}_{a,b}={}$
$\displaystyle\sum\limits_{a,b,m=0}^{\infty}\sum\limits_{\delta=1}^{\infty}\frac{2}{\delta^{3}}x^{a}y^{b}c^{(0)}_{a,b,m}F^{(0)}_{m}(\delta)$
(77) $\displaystyle={}$
$\displaystyle\sum\limits_{m=0}^{\infty}\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{3}}\frac{y+2}{1-x-y}\left(\frac{\sqrt{1-4y}-1}{2}\right)^{m}F^{(0)}_{m}(\delta)$
$\displaystyle={}$
$\displaystyle\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{3}}\frac{y+2}{1-x-y}\binom{z+\delta-1}{\delta-1}^{2}\,,$

with

$x=\frac{\hat{\sigma}_{2}}{\delta^{2}}\,,\qquad
y=\frac{\hat{\sigma}_{3}}{\delta^{3}}\,,\qquad
z=\frac{\delta}{2}\left(\sqrt{1-4y}-1\right)\,.$ (78)

One can now use the formula

$\binom{a-1}{b-1}^{2}=\prod\limits_{j=1}^{b-1}\left(1-\frac{a(b-a)}{j(b-j)}\right)\,,$
(79)

and insert the definition of $\hat{\sigma}_{2}$ and $\hat{\sigma}_{3}$ to find

$A^{(0)}(S,T)=\frac{1}{STU}+\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{3}}\left(\frac{S}{\delta-S}+\frac{T}{\delta-T}+\frac{U}{\delta-U}+2\right)\prod\limits_{j=1}^{\delta-1}\left(1-\frac{STU}{j(\delta-j)\delta}\right)\,.$
(80)

A proof that this matches the familiar result (31)

$A^{(0)}(S,T)=-\frac{\Gamma\left(-S\right)\Gamma\left(-T\right)\Gamma\left(-U\right)}{\Gamma\left(S+1\right)\Gamma\left(T+1\right)\Gamma\left(U+1\right)}\,,$
(81)

was given in Zagier:2019eus .

In order to do the analogous sum for $A^{(1)}(S,T)$ we need to determine the
generating series for $T^{(2)}_{m}(\delta)$ and $F^{(2)}_{m}(\delta)$ which
appear in the expression for $\alpha^{(1)}_{a,b}$ in (26). By studying them
for fixed values of $m$ and $\delta$ we noticed that they can be written in
the form

$\displaystyle\sum\limits_{m=0}^{\infty}T^{(2)}_{m}(\delta)\left(\frac{z}{\delta}\right)^{m}={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\binom{z+\delta-\frac{n}{2}-1}{\delta-n-1}^{2}\,,$
(82)
$\displaystyle\sum\limits_{m=0}^{\infty}F^{(2)}_{m}(\delta)\left(\frac{z}{\delta}\right)^{m}={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}h_{n}(\delta)\binom{z+\delta-\frac{n}{2}-1}{\delta-n-1}^{2}\,,$

with

$\displaystyle g_{n}(\delta)={}$
$\displaystyle\frac{\delta}{(\delta-n)_{n}}\tilde{g}_{n}(\delta)\,,\qquad
h_{n}(\delta)=\frac{\delta}{(\delta-n)_{n}}\tilde{h}_{n}(\delta)\,,$ (83)
$\displaystyle\tilde{g}_{0}(\delta)={}$
$\displaystyle\frac{1}{2\delta}\left(3\delta^{2}-\delta+2\right)\,,\qquad\tilde{g}_{1}(\delta)=\frac{1}{72}\left(\delta^{2}+4\delta+6\right)\,,$
$\displaystyle\tilde{g}_{2}(\delta)={}$
$\displaystyle\frac{1}{7200}\left(-14\delta^{3}+3\delta^{2}+222\delta+210\right)\,,$
$\displaystyle\tilde{h}_{0}(\delta)={}$
$\displaystyle\frac{1}{96\delta}\left(-112\delta^{3}+1872\delta^{2}-2216\delta+1671\right)+2\delta^{2}\zeta(3)\,,$
$\displaystyle\tilde{h}_{1}(\delta)={}$
$\displaystyle\frac{1}{432}\left(-61\delta^{3}-703\delta^{2}-1221\delta+2934\right)\,,$
$\displaystyle\tilde{h}_{2}(\delta)={}$
$\displaystyle\frac{1}{216000}\left(4598\delta^{4}+33699\delta^{3}-103219\delta^{2}-297105\delta+581850\right)\,,$

and so on. We include further cases in a Mathematica notebook. Now the sums
can be done analogously to (77), with polynomials in $a,b,m$ turning into
differential operators acting on $x^{a}$, $y^{b}$ or $z^{m}$. In this way we
obtain

$\displaystyle
A^{(1)}(S,T)+\frac{2}{3}\frac{\hat{\sigma}_{2}}{\hat{\sigma}_{3}^{2}}=2\sum\limits_{a,b=0}^{\infty}\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha^{(1)}_{a,b}$
$\displaystyle={}$
$\displaystyle\sum\limits_{a,b,m=0}^{\infty}\sum\limits_{\delta=1}^{\infty}\frac{2x^{a}y^{b}}{\delta^{4}}\left(c^{(0)}_{a,b,m}\left(F^{(2)}_{m}(\delta)-(3+2a+3b)T^{(2)}_{m}(\delta)\right)+c^{(2,0)}_{a,b,m}F^{(0)}_{m}(\delta)+c^{(2,1)}_{a,b,m}F^{(0)}_{m+1}(\delta)\right)$
$\displaystyle={}$
$\displaystyle\sum\limits_{\delta=1}^{\infty}\sum\limits_{n=0}^{\delta-1}\frac{1}{\delta^{4}}\mathcal{D}_{n}(\delta)\frac{y+2}{1-x-y}\binom{z+\delta-\frac{n}{2}-1}{\delta-n-1}^{2}\,,$
(84)

with the differential operator $\mathcal{D}_{n}(\delta)$ given by (note that
$\partial_{z}$ only acts on $z$ but $\partial_{y}$ acts on $y$ and on $z$
through its definition)

$\displaystyle\mathcal{D}_{n}(\delta)=h_{n}(\delta)-g_{n}(\delta)\left(3+2x\partial_{x}+3y\partial_{y}\right)+\delta_{n,0}\big{(}-\left(\tfrac{1}{2}x\partial_{x}+\tfrac{3}{4}y\partial_{y}+\tfrac{27}{4}\right)(z+\delta)\partial_{z}z\partial_{z}$
$\displaystyle-6(x\partial_{x})^{2}y\partial_{y}-9x\partial_{x}(y\partial_{y})^{2}-16x\partial_{x}y\partial_{y}-\tfrac{4}{3}(x\partial_{x})^{3}-\tfrac{16}{3}(x\partial_{x})^{2}-\tfrac{1}{3}x\partial_{x}-\tfrac{9}{2}(y\partial_{y})^{3}$
$\displaystyle-12(y\partial_{y})^{2}-\tfrac{1}{2}y\partial_{y}-\tfrac{277}{32}+\left(6x\partial_{x}y\partial_{y}+2(x\partial_{x})^{2}+\tfrac{26}{3}x\partial_{x}+\tfrac{9}{2}(y\partial_{y})^{2}+13y\partial_{y}-\tfrac{21}{8}\right)z\partial_{z}$
$\displaystyle+\left(3x\partial_{x}y\partial_{y}+(x\partial_{x})^{2}+\tfrac{55}{12}x\partial_{x}+\tfrac{9}{4}(y\partial_{y})^{2}+\tfrac{55}{8}y\partial_{y}+\tfrac{33}{16}\right)\delta\partial_{z}\big{)}\,.$
(85)

Equation (84) has several nice properties. One can use (79) to check that

$\binom{z+\delta-\frac{n}{2}-1}{\delta-n-1}^{2}=\prod\limits_{j=1}^{\delta-n-1}\left(1-\frac{n^{2}-2n\delta+4\delta^{2}y}{4j(\delta-
n-j)}\right)\,,$ (86)

is a polynomial in $y$. By expanding around the location of the poles we can
also find general expressions for the residues

$A^{(1)}(S,T)+\frac{2}{3}\frac{\hat{\sigma}_{2}}{\hat{\sigma}_{3}^{2}}=\sum\limits_{k=1}^{4}\frac{R_{k}(T,\delta)}{(S-\delta)^{k}}+O((S-\delta)^{0})\,,\qquad\delta=1,2,\ldots\,.$
(87)

In terms of the function

$R(T,\delta)=-\frac{\Gamma(T+\delta)^{2}}{\Gamma(\delta)^{2}\Gamma(T+1)^{2}}\,,$
(88)

we find that the residues are the following polynomials in $T$

$\displaystyle R_{4}(T,\delta)={}$ $\displaystyle R(T,\delta)\,,$
$\displaystyle R_{3}(T,\delta)={}$
$\displaystyle\frac{1}{6\delta}\left(3\delta\partial_{T}-4\right)R(T,\delta)\,,$
$\displaystyle R_{2}(T,\delta)={}$
$\displaystyle\frac{1}{\delta^{2}}\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)R(T+\tfrac{n}{2},\delta-n)-\frac{3T(\delta+T)\partial_{T}^{2}+\tfrac{1}{2}(31\delta+46T)\partial_{T}+32}{12\delta^{2}}R(T,\delta)\,,$
$\displaystyle R_{1}(T,\delta)={}$
$\displaystyle\frac{1}{\delta^{3}}\sum\limits_{n=0}^{\delta-1}\left(h_{n}(\delta)-g_{n}(\delta)\delta^{2}(2+T\partial_{T})\right)R(T+\tfrac{n}{2},\delta-n)$
(89)
$\displaystyle+\frac{1}{96\delta^{3}}\left(8T^{3}\partial_{T}^{3}-4T(119\delta+90T)\partial_{T}^{2}-2(311\delta+402T)\partial_{T}-927\right)R(T,\delta)\,,$

where

$\displaystyle\partial_{T}R(T,\delta)={}$ $\displaystyle
2G_{1}(T,\delta)R(T,\delta)\,,$ (90)
$\displaystyle\partial_{T}^{2}R(T,\delta)={}$
$\displaystyle-2(G_{2}(T,\delta)-2G_{1}(T,\delta){}^{2})R(T,\delta)\,,$
$\displaystyle\partial_{T}^{3}R(T,\delta)={}$ $\displaystyle
4(G_{3}(T,\delta)-3G_{2}(T,\delta)G_{1}(T,\delta)+2G_{1}(T,\delta){}^{3})R(T,\delta)\,,$
$\displaystyle G_{k}(T,\delta)={}$ $\displaystyle
H^{(k)}(T+\delta-1)-H^{(k)}(T)=\sum\limits_{j=1}^{\delta-1}\frac{1}{(T+j)^{k}}\,,$

is given in terms of generalised harmonic numbers.

From (87) we see that $F^{(2)}_{m}(\delta)$ contributes only to single poles
and $T^{(2)}_{m}(\delta)$ contributes also to double poles. The remaining
terms have poles up to fourth order and the whole expression (84) has no
simultaneous poles in different Mandelstams. The pole of fourth order might be
surprising if one expects the poles to arise from expanding a single pole in
$1/\lambda$. We show in appendix D that to any order in $1/\lambda$, the pole
of the highest order arises purely from the dispersive sum rule and does not
depend on corrections to the OPE data. We also resum the in $1/\lambda$
expansion for these poles.

We could now apply the inverse of the flat space transform (70)

$M(s_{1},s_{2})=\text{FS}^{-1}(A(S,T))=\frac{1}{2\lambda^{\frac{3}{2}}c}\int\limits_{0}^{\infty}d\beta
e^{-\beta}\beta^{5}A\left(\frac{\beta s_{1}}{2\sqrt{\lambda}},\frac{\beta
s_{2}}{2\sqrt{\lambda}}\right)\,,$ (91)

to obtain the summed Mellin amplitude. As discussed in Penedones:2010ue , the
poles of $A(S,T)$ will lead to exponential integrals and hence branch cuts.
These originate from many poles of the non-perturbative Mellin amplitude which
have vanishing separation at large $\lambda$, but become separated when
applying the flat space transform.

##  6 Conclusions

In this paper we determined the full $1/\sqrt{\lambda}$ contribution to the
Virasoro-Shapiro amplitude on AdS${}_{5}\times$ S5, by solving the dispersive
sum rules derived in Alday:2022uxp , using the crucial assumption that the
Wilson coefficients are single-valued periods. The resulting correction
possesses an analytic structure which naturally generalises that of the
Virasoro-Shapiro amplitude in flat space.

The natural next step is to determine the next layer of Wilson coefficients
$\alpha^{(2)}_{a,b}$ with similar arguments. A preliminary study shows that
single-valuedness is also powerful enough to determine the coefficients
$\alpha^{(2)}_{a,b}$ uniquely, once quantities like $\langle
f_{0}(\delta,\ell)\tau_{2}(\delta,\ell)^{2}\rangle$ are provided. This would
require solving a mixing problem to order $1/\sqrt{\lambda}$, considering more
general correlators. Single-valuedness is not powerful enough if we treat
these quantities as unknown, not surprisingly.

Certain universal parts of the answer, which do not depend on corrections to
the CFT data, can be studied to all orders in $1/\lambda$, see appendix D. It
would be interesting to explore this further.

Over the last few years there has been great progress in understanding how
supersymmetry, via localisation results, gives integrated constraints for the
correlator under consideration Binder:2019jwn ; Chester:2020dja ;
Chester:2019jas ; Chester:2020vyz ; Dorigoni:2021bvj ; Dorigoni:2021guq ;
Collier:2022emf . In the present context, this will lead to two linear
constraints for the Wilson coefficients at each order in $1/\lambda$. These
linear constrains are written in terms of single zeta values of odd arguments,
and it would be interesting to understand how this arises from our procedure.

Combining these two linear constraints and the flat space limit with our new
solution for $\alpha^{(1)}_{0,1}$, we can for the first time fully determine
the $D^{8}R^{4}$ term at planar order, which appears at
$O(1/\lambda^{\frac{7}{2}})$ and depends on four Wilson coefficients

$\displaystyle
M(s_{1},s_{2})=\frac{8}{(s_{1}-\frac{2}{3})(s_{2}-\frac{2}{3})(s_{3}-\frac{2}{3})}+\frac{120\zeta(3)}{\lambda^{3/2}}+\frac{210\left(3\sigma_{2}+7\right)\zeta(5)}{\lambda^{5/2}}+$
(92)
$\displaystyle\frac{140\left(108\sigma_{3}-99\sigma_{2}-320\right)\zeta(3)^{2}}{3\lambda^{3}}+\frac{35\left(2592\sigma_{2}^{2}-77328\sigma_{3}+73638\sigma_{2}+178909\right)\zeta(7)}{16\lambda^{7/2}}+O(\lambda^{-4})\,.$

Among the CFT data provided by our solution we reproduce the anomalous
dimensions of short Konishi-like operators, in full agreement with the results
from integrability, together with their structure constants. It would be
interesting to reproduce these structure constants from integrability methods,
along the lines of Basso:2022nny . A related direction is the interplay
between integrability and the conformal bootstrap, explored first in
Cavaglia:2021bnz ; Cavaglia:2022qpg  and in our context in Caron-Huot:2022sdy
. Now that our analytic methods allow to explore $1/\lambda$ corrections, it
would be interesting to feed this into the program of Caron-Huot:2022sdy .

Finally, we hope that our results can fuel progress towards determining the
worldsheet theory for strings on AdS${}_{5}\times$ S5. Recent progress on
determining the vertex operators has been made in Fleury:2021ieo . It would be
very interesting to see explicitly how the expression for the AdS Virasoro-
Shapiro amplitude, in a $1/\lambda$ expansion, arises from the worldsheet
theory.

## Acknowledgements

We thank Michael Green for useful discussions. The work of LFA and TH is
supported by the European Research Council (ERC) under the European Union’s
Horizon 2020 research and innovation programme (grant agreement No 787185).
LFA is also supported in part by the STFC grant ST/T000864/1. The work of JS
is supported by an early postdoc mobility fellowship (grant P2ELP2199748) by
the Swiss National Science Foundation (SNSF). The authors would like to
acknowledge the use of the University of Oxford Advanced Research Computing
(ARC) facility in carrying out this work.
http://dx.doi.org/10.5281/zenodo.22558

##  Appendix A Crossing-symmetric dispersion relations

Based on earlier work in Auberson:1972prg ; Sinha:2020win , a crossing
symmetric dispersion relation for Mellin amplitudes was derived in
Gopakumar:2021dvg . The idea is to use the variables

$a=\frac{y}{x}\,,\qquad\hat{s}_{k}(z,a)=a-\frac{a(z-z_{k})^{3}}{z^{3}-1}\,,\qquad
z_{k}=e^{\frac{2}{3}\pi i(k-1)}\,,\quad k=1,2,3\,,$ (93)

and in this appendix we used the rescaled variables

$\hat{s}_{k}=\frac{s_{k}}{2}\,,\qquad x=\frac{\sigma_{2}}{8}\,,\qquad
y=-\frac{\sigma_{3}}{8}\,.$ (94)

One can derive a fixed-$a$ dispersion relation by deforming an integration
contour in $z$. The derivation of the dispersion relation is described in
detail in Auberson:1972prg  and in the supplementary material of Sinha:2020win
. One starts with the expression

$M(z,a)=\frac{1}{2\pi
i}\frac{z^{3}}{1-z^{3}}\oint_{z}dz^{\prime}\frac{z^{\prime 3}-1}{z^{\prime
3}(z^{\prime}-z)}M(z^{\prime},a)+\text{const.}\,,$ (95)

and deforms the integration contour, picking up the poles corresponding to OPE
singularities, which lie on the unit circle (at least if $a$ lies in a certain
range). A constant contribution is not determined by this relation because of
poles at $z=0$ and $z=\infty$. The Regge limit $|s_{k}|\to\infty$ is mapped to
the three roots of unity $z=z_{k}$, and one assumes the following bound in the
Regge limit

$M(s_{1},s_{2})=o(s_{1}^{2})\ \text{for}\ |s_{1}|\to\infty\,,\quad s_{2}\
\text{fixed}\,,$ (96)

which amounts to

$M(z,a)=o\left(\frac{1}{(z-z_{k})^{2}}\right)\ \text{for}\ |z|\to
z_{k}\,,\quad a\ \text{fixed}\,.$ (97)

The factor $(z^{\prime 3}-1)/z^{\prime 3}$ in (95) is a subtraction that
ensures that these singularities do not contribute. In this way one finds the
following crossing symmetric expression for the Mellin amplitude in terms of
OPE data and Mack polynomials

$M(s_{1},s_{2})=\sum\limits_{\tau,\ell,m}\frac{c_{\tau,\ell}^{(m)}}{\tau_{m}}P_{\tau,\ell}(\tau_{m},\hat{s}_{2}^{\prime}(\tau_{m},a))H(\tau_{m},\hat{s}_{1},\hat{s}_{2},\hat{s}_{3})+\text{
const.}\,,$ (98)

where $\tau_{m}=\tau/2+m-2/3$,

$\displaystyle c^{(m)}_{\tau,\ell}={}$
$\displaystyle=\frac{2^{3\ell-7}\tau^{6}\Gamma\left(\ell+\frac{\tau}{2}+\frac{3}{2}\right)\Gamma\left(\ell+\frac{\tau}{2}+\frac{5}{2}\right)\Gamma\left(m+\frac{\tau}{2}-1\right)^{2}}{(\ell+1)\Gamma(m+1)\Gamma\left(\ell+\frac{\tau}{2}+2\right)^{2}\Gamma(\ell+m+\tau+3)}f(\tau,\ell)\,,$
(99)

is related to the OPE coefficient $f(\tau,\ell)$ from (7) and

$\displaystyle H(\tau_{m},\hat{s}_{1},\hat{s}_{2},\hat{s}_{3})$
$\displaystyle=\frac{\hat{s}_{1}}{\tau_{m}-\hat{s}_{1}}+\frac{\hat{s}_{2}}{\tau_{m}-\hat{s}_{2}}+\frac{\hat{s}_{3}}{\tau_{m}-\hat{s}_{3}}\,,$
(100) $\displaystyle\hat{s}_{2}^{\prime}\left(\tau_{m},a\right)$
$\displaystyle=-\frac{\tau_{m}}{2}\left(1-\left(\frac{\tau_{m}+3a}{\tau_{m}-a}\right)^{1/2}\right)\,,$
$\displaystyle P_{\tau,\ell}(\tau_{m},\hat{s}_{3})$
$\displaystyle=4^{-\ell}Q^{\tau+4,d}_{\ell,m}\left(2(\hat{s}_{3}+\tfrac{2}{3})-4\right)\,,$

where the Mack polynomial $Q_{\ell,m}^{\tau,d}(t)$ is defined as in
Alday:2022uxp .

###  A.1 Wilson coefficients

An expression for the Wilson coefficients in the expansion of the Mellin
amplitude

$M(s_{1},s_{2})=\sum\limits_{a,b=0}^{\infty}=\mathcal{M}_{a,b}x^{a}y^{b}\,,$
(101)

in terms of OPE data was derived in Gopakumar:2021dvg  by expanding (98) in
$x$ and $y$

$\mathcal{M}_{a-b,b}=\sum_{\begin{subarray}{c}\tau,\ell,m\end{subarray}}^{\infty}c_{\tau,\ell}^{(m)}\mathcal{B}_{a,b}^{(\tau,\ell,m)}\,,a\geq
1\,.$ (102)

Here $\mathcal{B}_{a,b}^{(\tau,\ell,m)}$ is given by

$\mathcal{B}_{a,b}^{(\tau,\ell,m)}=\sum_{q=0}^{b}\mathfrak{U}^{\left(\tau_{m}\right)}_{a,b,q}(-1)^{b+q}P_{\tau,\ell;q}\left(\tau_{m},0\right)\,,$
(103)

where
$P_{\tau,\ell;q}\left(\tau_{m},0\right)=\partial_{\hat{s}_{2}}^{q}P_{\tau,\ell}\left(\tau_{m},\hat{s}_{2}\right)|_{\hat{s}_{2}=0}$
and

$\begin{split}&\mathfrak{U}^{\left(\tau_{m}\right)}_{a,b,q}=\frac{(a-q-1)!(b+2a-3q)}{q!(a-b)!(b-q)!\left(\tau_{m}\right)^{b+2a-q+1}}\,_{4}F_{3}\left[\begin{array}[]{c}\frac{q}{2}+\frac{1}{2},\frac{q}{2},q-b,q+1-\frac{2a+b}{3}\\\
q+1,q+1-a,q-\frac{2a+b}{3}\end{array};4\right]\,.\end{split}$ (104)

By equating $M(s_{1},s_{2})$ to our reduced Mellin amplitude (1) (minus the
supergravity amplitude), the expansion

$\mathcal{M}_{a,b}=\sum\limits_{c=0}^{\infty}\frac{\mathcal{M}_{a,b}^{(c)}}{\lambda^{\frac{1}{2}(3+2a+3b+c/2)}}\,,$
(105)

is related to our conventions by

$\alpha^{(k)}_{a,b}=\frac{(-1)^{b}\mathcal{M}_{a,b}^{(2k)}}{\Gamma(6+2a+3b)}\,.$
(106)

Now we can insert the OPE data expansions (5) and (7), expand in large
$\lambda$ and sum over $m$ (as discussed in Alday:2022uxp ) to obtain the
expressions for $\alpha^{(0)}_{a,b}$ and $\alpha^{(1)}_{a,b}$ in (10) and
(26), generalising the expressions in Alday:2022uxp .

###  A.2 Recursion relations

For the reduced Mellin amplitude at hand, we can derive stronger crossing-
symmetric dispersion relations and indeed recursion relations for the Wilson
coefficients, because the reduced Mellin amplitude satisfies the stronger
Regge bound (we assume that for string theory the chaos bound is not
saturated)

$M(s_{1},s_{2})=o(s_{1}^{-2})\ \text{for}\ |s_{1}|\to\infty\,,\quad s_{2}\
\text{fixed}\,,$ (107)

or in terms of $z$ and $a$

$M(z,a)=o\left((z-z_{k})^{2}\right)\ \text{for}\ |z|\to z_{k}\,,\quad a\
\text{fixed}\,.$ (108)

This means we do not need the subtraction used above, and can even make use of
the fact that the Mellin amplitude vanishes at $z=z_{k}$. The trick to get a
nice relation is to use the combination

$\frac{z^{3}}{(z^{3}-1)^{2}}=-\frac{x}{27a^{2}}\,.$ (109)

We first consider the expression

$M(z,a)=\frac{1}{2\pi
i}\frac{(z^{3}-1)^{2}}{z^{3}}\frac{z^{3}}{1-z^{3}}\oint_{z}dz^{\prime}\frac{z^{\prime
3}}{(z^{\prime 3}-1)^{2}}\frac{z^{\prime 3}-1}{z^{\prime
3}(z^{\prime}-z)}M(z^{\prime},a)\,,$ (110)

which results in the following dispersion relation, that now also fixes the
constant part of the Mellin amplitude (that means (102) is also valid for
$a=0$)

$M(s_{1},s_{2})=\sum\limits_{\tau,\ell,m}\frac{c_{\tau,\ell}^{(m)}}{\tau_{m}}P_{\tau,\ell}(\tau_{m},\hat{s}_{2}^{\prime}(\tau_{m},a))\frac{\tau_{m}^{3}}{x(\tau_{m}-a)}H(\tau_{m},\hat{s}_{1},\hat{s}_{2},\hat{s}_{3})\,.$
(111)

The Regge bound is even strong enough to allow another ’addition’ so we can
start with

$M(z,a)=\frac{1}{2\pi
i}\frac{(z^{3}-1)^{4}}{z^{6}}\frac{z^{3}}{1-z^{3}}\oint_{z}dz^{\prime}\frac{z^{\prime
6}}{(z^{\prime 3}-1)^{4}}\frac{z^{\prime 3}-1}{z^{\prime
3}(z^{\prime}-z)}M(z^{\prime},a)\,,$ (112)

and find the relation

$M(s_{1},s_{2})=\sum\limits_{\tau,\ell,m}\frac{c_{\tau,\ell}^{(m)}}{\tau_{m}}P_{\tau,\ell}(\tau_{m},\hat{s}_{2}^{\prime}(\tau_{m},a))\left(\frac{\tau_{m}^{3}}{x(\tau_{m}-a)}\right)^{2}H(\tau_{m},\hat{s}_{1},\hat{s}_{2},\hat{s}_{3})\,.$
(113)

We can use the fact that we have two dispersion relations to obtain recursion
relations. To this end we can combine the expansion in $x$ and $y$ with the
one in OPE data and write

$\mathcal{M}_{a,b}=\sum\limits_{\tau,\ell,m}\mathcal{M}_{a,b}(\tau,\ell,m)\,.$
(114)

Equating (111) and (113) we find

$M(s_{1},s_{2})=\sum_{a,b}\sum\limits_{\tau,\ell,m}\mathcal{M}_{a,b}(\tau,\ell,m)x^{a}y^{b}=\sum_{a,b}\sum\limits_{\tau,\ell,m}\frac{\tau_{m}x-y}{\tau_{m}^{3}}\mathcal{M}_{a,b}(\tau,\ell,m)x^{a}y^{b}\,,$
(115)

which implies the recursion relation

$\sum\limits_{\tau,\ell,m}\mathcal{M}_{a,b}(\tau,\ell,m)=\sum\limits_{\tau,\ell,m}\frac{1}{\tau_{m}^{2}}\mathcal{M}_{a-1,b}(\tau,\ell,m)-\frac{1}{\tau_{m}^{3}}\mathcal{M}_{a,b-1}(\tau,\ell,m)\,.$
(116)

The next step is to expand in $1/\lambda$ and do the sums over $m$. To leading
order this implies the recursion relation

$\alpha^{(0)}_{a,b}(\delta)=\frac{1}{\delta^{2}}\alpha^{(0)}_{a-1,b}(\delta)+\frac{1}{\delta^{3}}\alpha^{(0)}_{a,b-1}(\delta)\,,$
(117)

where $\alpha^{(0)}_{a,b}(\delta)$ is the summand of $\alpha^{(0)}_{a,b}$

$\alpha^{(0)}_{a,b}=\sum\limits_{\delta=1}^{\infty}\alpha^{(0)}_{a,b}(\delta)\,.$
(118)

One checks that (10) is a solution of this recursion for the boundary
condition $\alpha^{(0)}_{-1,b}(\delta)=0$. As the boundary data
$\alpha^{(0)}_{-b,b}(\delta)$ is non-trivial ($\alpha^{(0)}_{-b,b}$ vanishes
only after summing over $\delta$), the recursion can be seen as a neat way to
encode the $a$ dependence of (10).

##  Appendix B Alternative representation for spin sums

The formulas (82) imply that there is an alternative representation for
$T^{(2)}_{m}(\delta)$ and $F^{(2)}_{m}(\delta)$. Adapting a computation from
Green:2008uj

$\displaystyle\sum\limits_{m=0}^{\infty}T^{(2)}_{m}(\delta)\left(\frac{z}{\delta}\right)^{m}={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\binom{z+\delta-\frac{n}{2}-1}{\delta-n-1}^{2}$
$\displaystyle={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\prod\limits_{k=1}^{\delta-n-1}\left(1+\frac{z+\frac{n}{2}}{k}\right)^{2}$
$\displaystyle={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\prod\limits_{k=1}^{\delta-n-1}\left(1+4\sum\limits_{s\in\\{1,2\\}}\left(\frac{z+\frac{n}{2}}{2k}\right)^{s}\right)$
$\displaystyle={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\sum\limits_{d=0}^{\infty}\sum\limits_{\begin{subarray}{c}\delta-n>k_{1}>\ldots>k_{d}>0\\\
s_{1},\ldots,s_{d}\in\\{1,2\\}\end{subarray}}\frac{2^{2d-s_{1}-\ldots-
s_{d}}(z+\frac{n}{2})^{s_{1}+\ldots+s_{d}}}{k_{1}^{s_{1}}\ldots
k_{d}^{s_{d}}}$ (119) $\displaystyle={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\sum\limits_{d=0}^{\infty}\sum\limits_{s_{1},\ldots,s_{d}\in\\{1,2\\}}2^{2d-s_{1}-\ldots-
s_{d}}\left(z+\frac{n}{2}\right)^{s_{1}+\ldots+s_{d}}Z_{s_{1},\ldots,s_{d}}(\delta-n-1)$
$\displaystyle={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\sum\limits_{w=0}^{\infty}\left(z+\frac{n}{2}\right)^{w}\frac{F^{(0)}_{w}(\delta-n)}{(\delta-n)^{w}}$
$\displaystyle={}$
$\displaystyle\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\sum\limits_{w=0}^{\infty}\sum\limits_{m=0}^{w}\binom{w}{m}\left(\frac{n}{2}\right)^{w-m}z^{m}\frac{F^{(0)}_{w}(\delta-n)}{(\delta-n)^{w}}\,,$

we can read off

$T^{(2)}_{m}(\delta)=\delta^{m}\sum\limits_{n=0}^{\delta-1}g_{n}(\delta)\sum\limits_{w=m}^{2(\delta-n-1)}\binom{w}{m}\left(\frac{n}{2}\right)^{w-m}\frac{F^{(0)}_{w}(\delta-n)}{(\delta-n)^{w}}\,,$
(120)

and similar for $F^{(2)}_{m}(\delta)$. We used in the final formula that
$F^{(0)}_{w}(\delta-n)$ vanishes for $w>2(\delta-n-1)$.

##  Appendix C More Bootstrap Constraints

We have explored two types of constraints that the bootstrap imposes on heavy
operators dual to short strings. One constraint is the fact that the
coefficients in the low energy expansion of the Mellin amplitude are related
to dimensions and OPE coefficients of heavy operators. Schematically,

$\displaystyle\xi_{a,b}=\sum_{\delta=1}^{\infty}\frac{p_{\xi,b}(a,\delta)}{\delta^{r+2a}},$
(121)

where $\xi_{a,b}$ is some generic coefficient in the low energy expansion of
the Mellin amplitude and the function $p_{\xi,b}(a,\delta)$ depends on the OPE
data through a sum over spins. It is expected that $\xi_{a,b}$ is a single
valued period, and so this constrains the CFT data. Another constraint that
the bootstrap imposes on the heavy operators is the equation

$\displaystyle\xi_{a,b}=0,~{}~{}~{}a=-b,~{}...~{},-1.$ (122)

In this section, we explore a third type of constraint. The essential idea is
that, at strong coupling, the Mellin amplitude should be similar to a tree
level string amplitude. Tree level string amplitudes obey stringent bootstrap
constraints, see Caron-Huot:2016icg ; Sever:2017ylk ; Figueroa:2022onw ;
Geiser:2022icl ; Coon:1969yw . We take advantage of this in the following way.
We write down a two sided dispersion relation, using the $s_{1}$ and $s_{2}$
channels. This dispersion relation stops converging for values of
$\mathop{\mathrm{Re}}(s_{3})$ sufficiently high, i.e. when we reach the first
pole in $s_{3}$. Furthermore, the two sided dispersion relation must diverge
in a precise way, such that it reproduces the residue in $s_{3}$. This gives
nontrivial constraints.

###  C.1 Flat Space

This idea is easier to express in flat space. The Virasoro-Shapiro amplitude
is given by

$\displaystyle
A^{(0)}(S,T)=-\frac{\Gamma\left(-S\right)\Gamma\left(-T\right)\Gamma\left(S+T\right)}{\Gamma\left(1+S\right)\Gamma\left(1+T\right)\Gamma\left(1-S-T\right)}\,.$
(123)

We ask the following question. Without knowing the form of the amplitude
(123), and assuming only the spectrum of particles exchanged (mass and spin)
and Regge boundedness, i.e.

$\displaystyle\lim_{S\rightarrow
i\infty}|A^{(0)}(S,T)|\leq\frac{1}{|S|^{2}},~{}~{}~{}\mathop{\mathrm{Re}}(T)<0,$
(124)

what can we say about particle couplings?

We write a two sided dispersion relation

$\displaystyle A^{(0)}(S,T)={}$
$\displaystyle-\frac{1}{U^{2}}\left(\frac{1}{S}+\frac{1}{T}\right)-\sum_{\delta=1}^{\infty}\sum_{\ell=0,2}^{2\delta-2}\frac{f_{0}(\delta,\ell)C_{\ell}^{(1)}(1+\frac{2U}{\delta})}{(1+\ell)\delta^{2}}\left(\frac{1}{S-\delta}+\frac{1}{T-\delta}\right)\,,$
(125)

where $U=-S-T$. By parametrising in this manner, the particle couplings
$f_{0}(\delta,\ell)$ turn out to be numerically equal to the OPE coefficients
that enter in (7). This is due to the flat space transform (70). Also, we
simply wrote down the part of the amplitude

$\frac{1}{STU}=-\frac{1}{U^{2}}\left(\frac{1}{S}+\frac{1}{T}\right)\,,$ (126)

because it won’t play any role. The dispersion relation converges for
$\mathop{\mathrm{Re}}(U)<1$. For $\mathop{\mathrm{Re}}(U)=1$, it must diverge,
in such a way as to give rise to the correct pole and residue in $U$. One very
natural way this can happen is if

$\displaystyle\lim_{\delta\rightarrow\infty}\sum_{\ell=0,2}^{2\delta-2}\frac{f_{0}(\delta,\ell)C_{\ell}^{(1)}(1+\frac{2U}{\delta})}{2(1+\ell)\delta^{3}}=f_{0}(\delta=1,\ell=0)\times\kappa\times\delta^{\kappa(U-1)-1}\,,$
(127)

where $\kappa$ is some number that is not fixed by this reasoning. The reason
for this asymptotic is that

$\displaystyle
f_{0}(\delta=1,\ell=0)\times\kappa\times\int_{\delta_{0}}^{\infty}d\delta\delta^{\kappa(U-1)-1}\underbrace{=}_{U\rightarrow
1}-f_{0}(\delta=1,\ell=0)\frac{1}{U-1}.$ (128)

This leads to the equation

$\displaystyle\boxed{\lim_{\delta\rightarrow\infty}\sum_{\ell=0,2}^{2\delta-2}\frac{f_{0}(\delta,\ell)C_{\ell}^{(1)}(1+\frac{2}{\delta})}{(1+\ell)\delta^{2}}=\kappa\,,}$
(129)

where $\kappa$ is some positive number, whose value is not fixed by this
reasoning. When we plug the numerical values of $f_{0}(\delta,\ell)$, it turns
out that (129) is precisely obeyed for every $\delta\in\mathbb{N}$, for
$\kappa=1$.

###  C.2 Flat Space from AdS

In this section we show how to derive (129) from: Mellin amplitudes in AdS,
CFT Regge boundedness and formulas (5) and (7) for the spectrum of exchanged
operators. Our discussion is similar to the one in Penedones:2010ue  (though
not exactly the same). The point of going through this is to later use the
same line of reasoning to derive a generalisation of (129) for the case
involving $f_{2}(\delta,\ell)$ and $\tau_{2}(\delta,\ell)$, see formula (145).
The starting point is the dispersion relation

$\displaystyle
M(s_{1},s_{2})=\sum_{\text{stringy}}C_{\tau,\ell}^{2}\omega_{\tau,\ell}(s_{1},s_{2}),$
(130)

where

$\displaystyle\omega_{\tau,\ell}(s_{1},s_{2})$
$\displaystyle\equiv\sum_{m=0}^{\infty}\mathcal{Q}_{\ell,m}^{\tau+4,d=4}\left(s_{3}-\frac{8}{3}\right)\left(\frac{1}{s_{1}+\frac{4}{3}-\tau-2m}+\frac{1}{s_{2}+\frac{4}{3}-\tau-2m}\right),$
(131) $\displaystyle\mathcal{Q}_{\ell,m}^{\tau,d}(t)$
$\displaystyle\equiv-\frac{2^{3\ell+2\tau-2}\Gamma\left(\ell+\frac{\tau}{2}-\frac{1}{2}\right)\Gamma\left(\ell+\frac{\tau}{2}+\frac{1}{2}\right)\Gamma\left(-\frac{d}{2}+\ell+\tau+1\right)Q_{\ell,m}^{\tau,d}(t)}{\pi\Gamma(m+1)\Gamma\left(\ell+\frac{\tau}{2}\right)^{2}\Gamma(\ell+\tau-1)\Gamma\left(-m-\frac{\tau}{2}+4\right)^{2}\Gamma\left(-\frac{d}{2}+\ell+m+\tau+1\right)}.$

$Q_{\ell,m}^{\tau,d}(t)$ is a Mack polynomial, see Alday:2022uxp  for our
conventions. This dispersion relation is valid at finite $\lambda$ and also
when the Mellin variables are finite. We now take the limit $s_{1}\rightarrow
s_{1}\lambda^{\frac{1}{2}},s_{2}\rightarrow s_{2}\lambda^{\frac{1}{2}}$ and
$\lambda\rightarrow\infty$. As explained in Penedones:2010ue , this is what
controls the flat space limit of AdS. In this regime, the sums in $m$ are
dominated by terms of order $\tau^{2}\sim\lambda^{\frac{1}{2}}$. We can thus
do the substitution $m=4\delta\lambda^{\frac{1}{2}}x$ and
$\sum_{m=0}^{\infty}\rightarrow
4\delta\lambda^{\frac{1}{2}}\int_{0}^{\infty}dx$ and we get

$\displaystyle\lim_{\lambda\rightarrow\infty}M(\lambda^{\frac{1}{2}}s_{1},\lambda^{\frac{1}{2}}s_{2})$
$\displaystyle=\frac{1}{\lambda^{\frac{3}{2}}}M^{(0)}(s_{1},s_{2})+O\left(\frac{1}{\lambda^{2}}\right),$
(132) $\displaystyle M^{(0)}(s_{1},s_{2})$
$\displaystyle\equiv\sum_{\delta=1}^{\infty}\sum_{\ell=0,2}^{\infty}f_{0}(\delta,\ell)\int_{0}^{\infty}dx\frac{e^{-\frac{1}{4x}}\left(-\frac{1}{\frac{s_{1}}{4\delta}-2x}-\frac{1}{\frac{s_{2}}{4\delta}-2x}\right)C_{\ell}^{(1)}\left(\frac{s_{3}}{4\delta
x}+1\right)}{4096(\ell+1)\delta^{3}x^{6}}\,.$

Note that Mack polynomials turn into Gegenbauer polynomials in this limit
Costa:2012cb .

At strong coupling, the Mellin amplitude develops cuts, as can be seen by
explicitly evaluating the above integrals for specific values of the spins.
This is intuitive, because the location of the poles is at
$s_{1},s_{2},s_{3}=\tau+2m+\frac{4}{3}$, $m\in\mathbb{N}_{0}$, so if
$\tau\sim\lambda^{\frac{1}{2}}$ is parametrically large, there is no way of
distinguishing between consecutive poles and the sequence of poles basically
condenses into a line. Because of this, it is useful to introduce the
transform

$\displaystyle
A^{(0)}(S,T)=2\int_{\kappa-i\infty}^{\kappa+i\infty}\frac{d\alpha}{2\pi
i}\alpha^{-6}e^{\alpha}M^{(0)}\left(\frac{2S}{\alpha},\frac{2T}{\alpha}\right).$
(133)

This the flat space transform (70) after having performed the large $\lambda$
expansion already in (132). From (133) we conclude that $A^{(0)}(S,T)$ is
bounded in the Regge limit like (124). It is crossing symmetric:
$A^{(0)}(S,T)=A^{(0)}(T,S)=A^{(0)}(S,U)$. Its poles are generated through the
following mechanism. Let us focus on the poles in $S$. Insert (132) into (133)
and commute the $x$ and $\alpha$ integrals. We can deform the contour in
$\alpha$ to the left due to the $e^{\alpha}$ in the integrand. The contour is
bent in such a way that we avoid the $\alpha=0$ singularity, but we pick up
the pole at $\alpha=\frac{S}{4\delta x}$. The $x$ integral then becomes, up to
a factor

$\displaystyle\int_{0}^{\infty}dx\frac{e^{-\frac{1}{4x}+\frac{S}{4\delta
x}}}{x^{2}}=-\frac{4\delta}{S-\delta}\,.$ (134)

We conclude that the poles and residues of $A^{(0)}(S,T)$ in $S$ are according
to

$\displaystyle
A^{(0)}(S,T)\approx-\frac{f_{0}(\delta,\ell)C_{\ell}^{(1)}(1+\frac{2U}{\delta})}{(1+\ell)\delta^{2}}\frac{1}{S-\delta}\,.$
(135)

So, we have established from (133) the necessary ingredients to write a two
sided dispersion relation for $T_{0}(S,T)$ and in this way obtain (129).

###  C.3 AdS constraints

We can compute corrections to formula (132)

$\displaystyle\lim_{\lambda\rightarrow\infty}M(\lambda^{\frac{1}{2}}s_{1},\lambda^{\frac{1}{2}}s_{2})$
$\displaystyle=\frac{1}{\lambda^{\frac{3}{2}}}M^{(0)}(s_{1},s_{2})+\frac{1}{\lambda^{2}}M^{(1)}(s_{1},s_{2})+\ldots$
(136)

The basic ingredient is to understand how to expand Mack polynomials. The
formula we need is

$\displaystyle
Q_{\ell,m=x\tau^{2}}^{\tau,d=4}\left(s_{3}\lambda^{\frac{1}{2}}-\frac{8}{3}\right)=\lambda^{\frac{\ell}{2}}\left(\mathcal{C}(x,\delta,\ell;s_{3})+\mathcal{C}_{1}(x,\delta,\ell;s_{3})\frac{1}{\lambda^{\frac{1}{4}}}+\mathcal{C}_{2}(x,\delta,\ell;s_{3})\frac{1}{\lambda^{\frac{1}{2}}}+...\right),$
(137)

where
$\tau=2\sqrt{\delta}\lambda^{\frac{1}{4}}+\tau_{1}(\delta,\ell)+\tau_{2}(\delta,\ell)\lambda^{-\frac{1}{4}}+\ldots$
and

$\displaystyle\mathcal{C}(x,\delta,\ell;s_{3})=2^{\ell}x^{\ell}\delta^{\ell}C_{\ell}^{(1)}\left(1+\frac{s_{3}}{4x\delta}\right),$
(138)
$\displaystyle\mathcal{C}_{1}(x,\delta,\ell;s_{3})=\frac{1+4\tau_{1}(\delta,\ell)x}{4\sqrt{\delta}}\partial_{x}\mathcal{C}(x,\delta,\ell;s_{3}),$
$\displaystyle\mathcal{C}_{2}(x,\delta,\ell;s_{3})=\frac{3\ell+6x^{2}+20x}{48\delta+96\delta\ell}\partial^{2}_{x}\mathcal{C}(x,\delta,\ell;s_{3})+\frac{-12x^{2}-40x+3}{24\ell+12}\partial_{x}\partial_{s_{3}}\mathcal{C}(x,\delta,\ell;s_{3})$
$\displaystyle-\frac{1}{6\delta}\partial_{x}\mathcal{C}(x,\delta,\ell;s_{3})+\frac{x\tau_{2}(\delta,\ell)}{\sqrt{\delta}}\partial_{x}\mathcal{C}(x,\delta,\ell;s_{3})+\frac{x\tau^{2}_{1}(\delta,\ell)}{4\delta}\partial_{x}\mathcal{C}(x,\delta,\ell;s_{3})$
$\displaystyle+\frac{\tau^{2}_{1}(\delta,\ell)x^{2}}{2\delta}\partial^{2}_{x}\mathcal{C}(x,\delta,\ell;s_{3})+\frac{\tau_{1}(\delta,\ell)}{8\delta}\partial_{x}\mathcal{C}(x,\delta,\ell;s_{3})+\frac{x\tau_{1}(\delta,\ell)}{4\delta}\partial^{2}_{x}\mathcal{C}(x,\delta,\ell;s_{3}).$

The flat space transform (70) applied to (136) implies

$\displaystyle
A^{(i)}(S,T)=2\int_{\kappa-i\infty}^{\kappa+i\infty}\frac{d\alpha}{2\pi
i}\alpha^{-6}e^{\alpha}M^{(i)}\left(\frac{2S}{\alpha},\frac{2T}{\alpha}\right)\,.$
(139)

In formula (136), we assumed that there was no correction proportional to
$\frac{1}{\lambda^{\frac{7}{4}}}$. This is not automatic from (130) and
actually leads to the familiar conditions

$\displaystyle\tau_{1}(\delta,\ell)=-\ell-2\,,~{}~{}\langle
f_{1}(\delta,\ell)\rangle=\langle
f_{0}(\delta,\ell)\rangle\frac{3\ell+\frac{23}{4}}{\sqrt{\delta}}\,.$ (140)

$A^{(1)}(S,T)$ inherits a two sided dispersion relation from $M^{(1)}(S,T)$

$\displaystyle
A^{(1)}(S,T)=-\frac{2}{3}\frac{\hat{\sigma}_{2}}{\hat{\sigma}_{3}^{2}}+\sum_{\delta=1}^{\infty}\sum_{k=1}^{4}R_{k}(U,\delta)\left(\frac{1}{(S-\delta)^{k}}+\frac{1}{(T-\delta)^{k}}\right)\,,\qquad-
S-T<1\,.$ (141)

Note the appearance of poles of order $4$. The reason why we have higher poles
is because the expansion in $\frac{1}{\sqrt{\lambda}}$ generates extra powers
of $\frac{1}{x}$ which, upon doing an integral like (134), give rise to higher
order poles. The residues involve $f_{0}(\delta,\ell)$, $f_{2}(\delta,\ell)$,
$\tau_{2}(\delta,\ell)$, Gegenbauer polynomials and also derivatives of
Gegenbauer polynomials.

$\displaystyle R_{4}(U,\delta)$
$\displaystyle=-\sum_{\ell=0,2}^{2\delta-2}\frac{f_{0}(\delta,\ell)C_{\ell}^{(1)}\left(1+\tfrac{2U}{\delta}\right)}{(1+\ell)},$
(142) $\displaystyle R_{3}(U,\delta)$
$\displaystyle=-\sum_{\ell=0,2}^{2\delta-2}\Big{(}\frac{f_{0}(\delta,\ell)}{12\delta^{3}(\ell+1)(2\ell+1)}\big{(}4\delta^{2}\left(3\ell^{2}+2\ell-2\right)C_{\ell}^{(1)}\left(1+\tfrac{2U}{\delta}\right)$
$\displaystyle+6\delta(4\delta(\ell-1)-12U)C_{\ell}^{{}^{\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)-48U(\delta+U)C_{\ell}^{{}^{\prime\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)\big{)}\Big{)},$
$\displaystyle R_{2}(U,\delta)$
$\displaystyle=-\sum_{\ell=0,2}^{2\delta-2}\frac{f_{0}(\delta,\ell)}{24\delta^{5}(\ell+1)(2\ell+1)}\Big{(}2\delta^{3}(12\sqrt{\delta}(2\ell+1)\tau_{2}(\delta,\ell)+6\ell^{3}-25\ell^{2}-138\ell-32)$
$\displaystyle\times
C_{\ell}^{(1)}\left(1+\tfrac{2U}{\delta}\right)-4\delta^{2}\left(\left(3\ell^{2}+22\ell-31\right)4U+40\delta(\ell-1)\right)C_{\ell}^{{}^{\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)$
$\displaystyle+4U(\delta(8\delta(16-3\ell)+(47-6\ell)4U))C_{\ell}^{{}^{\prime\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)+192U^{2}(\delta+U)C_{\ell}^{{}^{\prime\prime\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)\Big{)},$
$\displaystyle R_{1}(U,\delta)$
$\displaystyle=-\sum_{\ell=0,2}^{2\delta-2}\frac{f_{2}(\delta,\ell)}{\delta^{2}(\ell+1)}C_{\ell}^{(1)}\left(1+\tfrac{2U}{\delta}\right)$
$\displaystyle+\sum_{\ell=0,2}^{2\delta-2}\frac{f_{0}(\delta,\ell)}{96\delta^{7}(\ell+1)(2\ell+1)}\Big{(}(192\sqrt{\delta}(2\ell+1)\tau_{2}(\delta,\ell)+1248\ell^{3}+4364\ell^{2}+2782\ell+927)$
$\displaystyle\times\delta^{4}C_{\ell}^{(1)}\left(1+\tfrac{2U}{\delta}\right)+4\delta^{3}(48\sqrt{\delta}(2\ell+1)U\tau_{2}(\delta,\ell)$
$\displaystyle+\left(6\ell^{3}-37\ell^{2}-383\ell+249\right)4U-314\delta(\ell-1))C_{\ell}^{{}^{\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)$
$\displaystyle+4U\delta^{2}\left(\left(-12\ell^{2}-216\ell+677\right)4U+4\delta(365-104\ell)\right)C_{\ell}^{{}^{\prime\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)$
$\displaystyle+32U^{2}\delta(4\delta(35-3\ell)+(45-4\ell)4U)C_{\ell}^{{}^{\prime\prime\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)$
$\displaystyle+768U^{3}(\delta+U)C_{\ell}^{{}^{\prime\prime\prime\prime}(1)}\left(1+\tfrac{2U}{\delta}\right)\Big{)}.$

We checked that these residues agree with the expressions (89) when plugging
in the OPE data from section 4.2.

Analogously, we can write a two channel dispersion relation in the $U$ and $T$
channels. When we equate it to the previous dispersion relation, and set
$U=0$, $T=-S$ and expand around $S=0$ we get

$\displaystyle
0=\approx\alpha^{(0)}_{-1,1}+\alpha^{(1)}_{-1,1}S+\alpha^{(1)}_{-2,2}S^{2}+\ldots\,.$
(143)

So, we recover the crossing equations $\alpha^{(0)}_{-1,1}=0$,
$\alpha^{(1)}_{-1,1}=0$, $\alpha^{(1)}_{-2,2}=0$, etc.

The next step is to set up the analogue of (129). At large $\delta$ the
summand in (141) must behave like $\frac{\log^{3}\delta}{\delta}$ when
$U\approx 1$ so as to give rise to a fourth order pole in $U$. The reason for
this is that

$\displaystyle\frac{1}{(U-1)^{4}}$
$\displaystyle=\frac{\kappa^{4}}{6}\lim_{\delta_{0}\rightarrow\infty}\int_{\delta_{0}}^{\infty}d\delta\delta^{\kappa(U-1)-1}\log^{3}(\delta)$
(144)

This reasoning leads to the equation

$\displaystyle\lim_{\delta\rightarrow\infty}\sum_{k=1}^{4}\frac{R_{k}(1,\delta)}{\delta^{k}}=\kappa\frac{\log^{3}\delta}{\delta}\,,$
(145)

where $\kappa$ is some constant, which we cannot determine by this type of
argument. (145) is a generalisation of (129). Equation (145) follows purely
from bootstrap, and so it can potentially act as a check on the assumption
that the $\alpha^{(1)}_{a,b}$’s are single-valued MZVs. In order to check
(145) we need to generate $f_{2}(\delta,\ell)$ and $\tau_{2}(\delta,\ell)$ for
large values of $\delta$. The amount of nested sums involved unfortunately
grows very rapidly, and so we were unable to generate a database of
$f_{2}(\delta,\ell)$ and $\tau_{2}(\delta,\ell)$ for sufficiently high values
of $\delta$ so as to check (145).

The explicitly known terms in the expression for the residues (89) fall off
faster than $\log^{3}\delta/\delta$, so (145) constrains the functions
$h_{n}(\delta)$ and $g_{n}(\delta)$

$\lim_{\delta\rightarrow\infty}\frac{1}{\delta^{4}}\sum\limits_{n=0}^{\delta-1}\left.\left(h_{n}(\delta)+g_{n}(\delta)\left(1-\delta^{2}(2+U\partial_{U})\right)\right)R(U+\tfrac{n}{2},\delta-n)\right|_{U=1}=\kappa\frac{\log^{3}\delta}{\delta}\,.$
(146)

##  Appendix D Leading poles

The poles of highest order in $A^{(k)}(S,T)$ are universal in the sense that
they do not depend on corrections to the OPE data. For $A^{(1)}(S,T)$ these
are the fourth order poles which originate from the terms in (29) that are
polynomials of degree three in $a$ and $b$.

To understand the structure of these terms better, we can compute the
analogous terms at higher orders in $1/\sqrt{\lambda}$ by expanding (102). We
compute the terms that produce the leading poles for $k\leq 5$ and find that
they have the following form, which we expect to hold for all $k$

$\alpha_{a,b}^{(k)}=\sum\limits_{\delta=1}^{\infty}\sum\limits_{m=0}^{b}\frac{1}{\delta^{3+2a+3b+k}}\frac{(-1)^{k}(2a+3b)^{3k}}{6^{k}\Gamma(k+1)}c^{(0)}_{a,b,m}F_{m}^{(0)}(\delta)+\text{
subleading poles in }A^{(k)}(S,T)\,.$ (147)

This can be summed over $k$ to give an exponential
$\text{exp}\left(-\frac{(2a+3b)^{3}}{6\delta\sqrt{\lambda}}\right)$. We can
apply the flat space transform and sum over $a$ and $b$ to get

$\displaystyle
2\sum\limits_{k,a,b=0}^{\infty}\frac{\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha_{a,b}^{(k)}}{\lambda^{\frac{k}{2}}}={}$
$\displaystyle\sum\limits_{\delta=1}^{\infty}\frac{1}{\delta^{3}}\text{exp}\left(-\frac{(2x\partial_{x}+3y\partial_{y})^{3}}{6\delta\sqrt{\lambda}}\right)\frac{y+2}{1-x-y}\binom{z+\delta-1}{\delta-1}^{2}$
(148) $\displaystyle+\text{ subleading poles}\,.$

The leading poles arise when the derivatives act on the denominator

$(2x\partial_{x}+3y\partial_{y})^{3k}\frac{y+2}{1-x-y}=\frac{(3k)!S^{3k+1}}{(\delta-S)^{3k+1}}+O\left(\frac{1}{(S-\delta)^{3k}}\right)\,,\qquad\delta=1,2,\ldots\,.$
(149)

This gives

$\displaystyle
2\sum\limits_{k,a,b=0}^{\infty}\frac{\hat{\sigma}_{2}^{a}\hat{\sigma}_{3}^{b}\alpha_{a,b}^{(k)}}{\lambda^{\frac{k}{2}}}={}$
$\displaystyle\sum\limits_{k=0}^{\infty}\frac{1}{\lambda^{\frac{k}{2}}}\left(\frac{R^{(k)}_{3k+1}(T,\delta)}{(S-\delta)^{3k+1}}+O\left(\frac{1}{(S-\delta)^{3k}}\right)\right)\,,$
(150)

with

$R^{(k)}_{3k+1}(T,\delta)=-\frac{(3k)!\delta^{2k-2}}{6^{k}k!}\frac{\Gamma(T+\delta)^{2}}{\Gamma(\delta)^{2}\Gamma(T+1)^{2}}\,.$
(151)

The leading order poles can be Borel resummed. We obtain

$\sum_{k=0}-\frac{(3k)!\delta^{2k-2}}{6^{k}k!}\frac{1}{\lambda^{k/2}(S-\delta)^{3k+1}}=-\int_{0}^{\infty}dt\frac{e^{\frac{\delta^{2}t^{3}}{6\sqrt{\lambda}(S-\delta)^{3}}-t}}{\delta^{2}(S-\delta)}\,.$
(152)

The integral can be written in terms of Airy and hypergeometric functions, if
desired.

##  Appendix E Dimension of the Konishi operator

At weak coupling, the dimension of the Konishi operator (which is related to
one of the operators exchanged in our correlator as discussed below (23)) was
calculated in Leurent:2013mr . It is given by

$\displaystyle\hat{\Delta}_{\text{weak}}=4+12g^{2}-48g^{4}+336g^{6}+96g^{8}(6\zeta(3)-15\zeta(5)-26)$
$\displaystyle-96g^{10}\left(54\zeta(3)^{2}-72\zeta(3)+90\zeta(5)-315\zeta(7)-158\right)$
$\displaystyle-48g^{12}\left(432\zeta(3)^{2}-3240\zeta(5)\zeta(3)+5472\zeta(3)-2340\zeta(5)-1575\zeta(7)+10206\zeta(9)+160\right)$
$\displaystyle+48g^{14}(2592\zeta(3)^{3}-8784\zeta(3)^{2}+8568\zeta(5)\zeta(3)-40320\zeta(7)\zeta(3)+108960\zeta(3)-20700\zeta(5)^{2}$
$\displaystyle-4776\zeta(5)-26145\zeta(7)-17406\zeta(9)+152460\zeta(11)-44480)$
$\displaystyle+96g^{16}\big{(}20736\zeta(3)^{3}+82656\zeta(3)^{2}-90(504\zeta(5)+721\zeta(7)-2688\zeta(9)+9664)\zeta(3)$
$\displaystyle+24840\zeta(5)^{2}+227799\zeta(7)+72\zeta(5)(3220\zeta(7)-2847)+97164\zeta(9)-1104246\zeta(13)$
$\displaystyle+566752-\tfrac{9}{5}\left(22800\zeta(3)^{2}\zeta(5)-194711\zeta(11)+792\zeta^{\text{sv}}(5,3,3)\right)\big{)}\,,$
(153)

where $g=\frac{\sqrt{\lambda}}{4\pi}$. At order $g^{16}$ we see that
$\zeta^{\text{sv}}(5,3,3)$ appears! This suggests that
$\hat{\Delta}_{\text{weak}}$ can be written in terms of single valued MZVs
only.

At strong coupling, the analytic prediction is Gromov:2011bz

$\displaystyle\hat{\Delta}_{\text{strong}}=2\lambda^{\frac{1}{4}}+2\lambda^{-\frac{1}{4}}+\left(\frac{1}{2}-3\zeta(3)\right)\lambda^{-\frac{3}{4}}+\left(6\zeta(3)+\frac{15\zeta(5)}{2}+\frac{1}{2}\right)\lambda^{-\frac{5}{4}}\,.$
(154)

Two more orders were computed numerically in Hegedus:2016eop

$\displaystyle\hat{\Delta}_{\text{strong}}\sim-91.976023725\lambda^{-\frac{7}{4}}+758.514613\lambda^{-\frac{9}{4}}\,.$
(155)

It would be interesting to find a function $\hat{\Delta}(\lambda)$ such that

  * •

at large $\lambda$ it can be expanded in powers of $\lambda^{-\frac{1}{4}}$,
matches (154) and (155), and all numerical coefficients are single valued
periods,

  * •

at small $\lambda$ it can be expanded in even powers of $g$, it matches (E),
and all numerical coefficients are single valued periods.

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