\section{Checks}
\label{sec:checks}

The internal checks of section~\ref{sec:computation} show that the computation is consistent with itself. This section compares it with the literature. There are three kinds of comparison: exact results in closed sectors, the pure Yang--Mills calculation of AMMPV, and the two-loop Hagedorn temperature of Harmark and Wilhelm. The last two were the most instructive.

\subsection{Exact results in closed sectors}

Table~\ref{tab:checks} lists the comparisons with exact results. All of them are passed exactly or to the numerical precision available.

\begin{table}[t]
\centering
\begin{tabular}{>{\raggedright\arraybackslash}p{0.66\textwidth}>{\raggedright\arraybackslash}p{0.26\textwidth}}
\toprule
check & outcome \\
\midrule
$|u_1|^2$ coefficient of $\vev{\cH}_U$ against Spradlin and Volovich \cite{SpradlinVolovich} & series identical; $d_1(x_H)=\tfrac32$ \\
confined phase at orders $g^2$ and $g^4$: Gaussian integral over the $u_n$ against direct traces over cyclic spin chains, including multi-traces & identical through $y^8$ \\
$su(2)$ sector: $Q_4=4x^3+8x^4$, $c_3=2x^3$; Hagedorn temperature of \cite{HO2} to second order in $\tilde G$ & exact \\
$su(2|3)$ sector at one loop: $\tilde T_H=(1+\tfrac38\tilde\lambda_{\rm HO})/\ln4$ \cite{HO1} & exact \\
non-planar two-loop $su(2)$ operator of \cite{BKS}: $d_4=-8x^2-8x^3$, hence the two-loop $su(2)$ Hagedorn temperature of \cite{GNS} & exact \\
trace of the two-loop $su(2|3)$ Hamiltonian of \cite{BeisertSU23} over three-letter states at $x=\tfrac14$: $-351/64$ & $-5.4843765$ \\
effective-Hamiltonian rules against exact diagonalisation of $U(2)$ and $U(3)$ matrix models & $10^{-10}$ \\
part of order $\beta^2$ of the three-loop diagrams against $\vev{\cH^2}$ from the harmonic action \cite{BeisertComplete} & $3.5\times10^{-10}$, every order \\
two-loop energies of non-BPS operators of lengths 3, 4 and 5 from the Bethe ansatz \cite{BeisertStaudacherLR}: $-24$, $-48$, $-42$, $-6$, $-34\pm10\sqrt5$ (in units of $g^4$) & see section~\ref{sec:cutoff} \\
\bottomrule
\end{tabular}
\caption{Comparisons with exact results.}
\label{tab:checks}
\end{table}

\subsection{Pure Yang--Mills theory}

If all lines are restricted to gauge bosons, our codes compute the quantities that AMMPV computed in \cite{AMMPV2}. The dictionary is $\lambda_A=\lambda/2$ for their coupling, $x_c=2-\sqrt3$ for the Hagedorn point of pure glue, and $C_{1,1,-2}=c_3/4\pi^2$, $C_{1,1,-1,-1}=(q_4-2\beta_cQ_4)/64\pi^4$ for their cubic and quartic coefficients, where $\beta_c=-\ln x_c$ and for pure glue $q_4$ includes the finite counterterm.

At two loops AMMPV give the three contributions to $C_{1,1,-2}$ from the quartic vertex, from two cubic vertices and from the Coulomb interaction. In units of $10^{-4}$ we find $6.53537$, $-22.87089$ and $9.16397$. They give $6.53536$, $-22.87088$ and $9.16396$. We also find the closed form $c_3(x)=-\tfrac34z_V(x^2)$ for pure glue, where $z_V$ is the last entry of table~\ref{tab:letters}. This fixes the normalisation of every type of vertex to six digits.

At three loops AMMPV list fourteen diagrams, which they label 3a to 3n, with a finite part for each after subtraction of the logarithmic divergence. The diagrams can be grouped by vertex content into seven classes that are separately meaningful in both calculations. Both calculations use a sharp cut-off on angular momentum, so the finite parts can be compared without a translation constant. The logarithmic coefficients agree exactly, diagram by diagram. Table~\ref{tab:ym} compares the finite parts.

\begin{table}[t]
\centering
\begin{tabular}{llrr}
\toprule
class & AMMPV diagrams & this work & AMMPV \\
\midrule
four cubic vertices & 3i, 3l & $-7.84$ & $-7.6(5)$ \\
two cubic, one quartic & 3g, 3j & $3.574$ & $3.564$ \\
two cubic, one Coulomb & 3h, 3k & $3.326$ & $3.33$ \\
two quartic & 3a, 3d & $0.217$ & $0.217$ \\
quartic and Coulomb & 3b, 3e & $-1.065$ & $-1.066$ \\
cubic and quintic & 3m & $0.462$ & $0.46$ \\
two Coulomb, and sextic & 3c, 3f, 3n & $-2.095$ & $-3.388$ \\
\midrule
sum & & $-3.42$ & $-4.5(3)$ \\
\bottomrule
\end{tabular}
\caption{Finite parts of the three-loop contributions to $C_{1,1,-1,-1}$ in pure Yang--Mills theory, in units of $10^{-4}$, grouped by vertex content. The AMMPV column gives the sums of their printed entries. The error on the first row is ours, from the single digit of their entry 3l; the error on the sum is theirs.}
\label{tab:ym}
\end{table}

Five classes agree with AMMPV within the rounding of their entries, and the class of 3g and 3j, whose entries sum to $3.564$, agrees to within one unit in the last digit of 3g. The single diagrams 3d and 3e, which can be isolated, agree to three digits, and 3m to the two that AMMPV quote. We also reproduce their finite counterterm exactly, as described in section~\ref{sec:computation}.

The seventh class does not agree. It contains two Coulomb interactions $C_2$, together with the sextic vertex $C_4$ whose self-contraction is of the same form. We find $-2.095$, and the entries of AMMPV sum to $-3.388$. This class accounts for the difference between our pure-glue result $b=-4.3\times10^{-4}\lambda_A^2$ and their $-5.7\times10^{-4}\lambda_A^2$. In the units of table~\ref{tab:ym} it differs by $+1.29$ and the sums by about $+1.1$. The first row, where their entry 3l has a single digit, gives back $-0.24$.

We have not resolved this. The following can be said. On our side the part of this class proportional to $\beta^2$ is obtained from three codes. The logarithmic coefficient is exact. The first two finite coefficients of exactly this class, at $x^4$ and $x^5$, were computed by two engines with independently written ordering remainders, with results $36.9553(5)$ against $36.957(6)$ and $65.93(5)$ against $65.87(6)$. The sextic vertex is the one in \eqref{eq:C234}, from the Weyl-ordered Hamiltonian. On the side of AMMPV, in the Lagrangian formalism this class involves contact terms $\delta(t)$. They recomputed 3c and 3n, though not 3f, with a second regulating function, and the sum of each diagram and its counterterms did not change. They print the expression for every diagram and point to a Mathematica file for the integrals, so the comparison can be carried further, diagram by diagram, than we have taken it here. One of the two calculations has an error in this class.

For the purpose of this paper the important point is its size. In units of $q_4$ the disagreement is $0.8$. It is in the direction of a more negative $b$ if AMMPV are right. The margin for the sign of $b$ in $\cN=4$ is fourteen. The disagreement therefore does not affect our conclusion. It should be resolved in the future, and section~\ref{sec:discussion} says what would be needed.

\subsection{The two-loop Hagedorn temperature}

The second external check is of a different nature. It does not test $b$. It tests the Hamiltonian, the perturbation theory and the engines on a quantity that is known exactly, through the coefficient of $|u_1|^2$ at order $g^4$.

\emph{The target.} The Hagedorn temperature is the zero of \eqref{eq:m1-twoloop}. Write $w(\beta)=1-z(e^{-\beta})$ and $\beta_H=\beta_0+g^2\beta_1+g^4\beta_2$. At order $g^2$ one finds $\beta_1=-2\beta_0$, as in section~\ref{sec:oneloop}. At order $g^4$,
\begin{equation}
\beta_0\,d_4=2\beta_0^2\,C_{41}-\Big[w'\beta_2+\half w''\beta_1^2+2\beta_1d_1+2\beta_0\beta_1\,\partial_\beta d_1\Big],
\label{eq:target}
\end{equation}
with everything evaluated at $\beta_0$. The ingredients are $w'=\tfrac32$ and $w''=-\tfrac32\sqrt3$ from \eqref{eq:z}; $d_1=\tfrac32$ and $\partial_\beta d_1=-(8\sqrt3-9)$ from the closed form \eqref{eq:d1}; $\beta_2=\beta_0(4+11.50399)$ from the result \eqref{eq:HW} of Harmark and Wilhelm; and $C_{41}=8.353829$, which we computed from the one-loop operator as $\vev{H_{12}^2}$ on two letters plus $\vev{H_{12}H_{23}+H_{23}H_{12}}$ on three. The four terms in the bracket are $61.25$, $-36.05$, $-15.80$ and $+134.77$, and the first term on the right is $115.91$. The result is
\begin{equation}
d_4(x_H)=-10.7287 .
\label{eq:d4target}
\end{equation}
This is the number that our two-loop operator must reproduce. We validated the conversion formula \eqref{eq:target} by brute force on $su(2)$ spin chains, where everything is known.

\emph{The direct computation.} The coefficient $d_4$ receives contributions from all three parts of $h_4$: the tree part acting on three letters, the planar loop part acting on two, and the two-loop self-energy acting on one. The first is finite and gives $-0.156$. The second is logarithmically divergent diagram by diagram and finite in the sum, like piece (b) above, and gives $-5.65(2)(6)$, where the second error is an estimate of the extrapolation in the cut-off. The third is problematic. The two-loop self-energy of a letter, computed with a cut-off $J$ on the angular momenta of the virtual states, does not converge. It grows like $J^2$. We explain why in the next section. The remedy is to fix the one-body terms by symmetry instead. The anomalous dimensions of half-BPS operators must vanish, and those of their descendants are determined. Requiring this of the full $h_4$, acting on states with one impurity in a background of $Z$ letters, determines the one-body energies $c_4$ in terms of two-body kernels that do converge. In units of $g^4$ per letter we find
\begin{equation}
c_4(Z)=2\zeta_3-1,\qquad c_4(\psi)=2\zeta_3-9,\qquad c_4(DZ)=2\zeta_3-\tfrac{105}8,
\label{eq:c4}
\end{equation}
and numerical values for the next three letters. The six lightest letters carry 98 per cent of the thermal weight at $x_H$, and their one-body terms sum to $-4.476$. For the remaining letters we extrapolate the measured trend, which gives $-0.41(2)$.

There is one more ingredient. The exact two-loop energies in the last line of table~\ref{tab:checks} revealed that the cut-off leaves a finite renormalisation of the Coulomb interaction, which we describe in the next section. It shifts the two-body term by $-0.017$ and enters the one-body terms through the bootstrap. The value of $c_4(DZ)$ in \eqref{eq:c4} includes it.

Adding the pieces,
\begin{equation}
d_4(x_H)=-0.156-5.65-0.017-4.476-0.41=-10.71(2)(6),
\label{eq:d4}
\end{equation}
against the target $-10.7287$. The agreement is at the level of half a per cent, within the errors.

The value of this check should be stated precisely. It is not a blind test. The first version of our number was $-9.7$, short by ten per cent. The discrepancy led to the discovery of the Coulomb counterterm, whose coefficient was then fixed by other data and not by the target. After that the number agreed. The check therefore shows that the regulated Hamiltonian, supplemented by finite counterterms that are fixed independently, reproduces a non-trivial two-loop result of integrability. It does not show that the correct value would have been obtained had the target not been known.