\section{What has to be computed}
\label{sec:anatomy}

We now carry the expansion of $Z[U]=\Tr(x^D\hat U)$ to second order and identify the pieces of $b$. This section contains no new number. Its purpose is to state precisely which objects are needed and which of them are known.

\subsection{The expansion to order \texorpdfstring{$g^4$}{g4}}

Insert \eqref{eq:Dexp} in \eqref{eq:ZU}. Since $D_2$ and $D_4$ commute with $D_0$ and with $\hat U$, the logarithm of the trace is a cumulant expansion,
\begin{equation}
-S_{\rm eff}[U]=-S_{\rm eff}^{(0)}[U]-G\,\vev{\cH}_U+\frac{G^2}{2}\,\vev{\cH^2}_{U,{\rm conn}}-\beta g^4\,\vev{D_4}_U+O(g^6).
\label{eq:cumulant}
\end{equation}
The connected average is $\vev{\cH^2}_{\rm conn}=\vev{\cH^2}-\vev{\cH}^2$. The dilatation operator also has a piece $g^3D_3$ that changes the number of letters, but it has no diagonal elements and first contributes as $D_3^2$ at order $g^6$.

All three averages are of order $N^2$. The quadratic parts were introduced above. The quartic parts are new, and we define them by
\begin{equation}
\begin{aligned}
\frac1{N^2}\vev{\cH^2}_{U,{\rm conn}}&=C_{41}|u_1|^2+Q_4(x)\,|u_1|^4+\ldots,\\
\frac1{N^2}\vev{D_4}_U&=d_4|u_1|^2+q_4(x)\,|u_1|^4+\ldots
\end{aligned}
\label{eq:Q4q4}
\end{equation}
Both $Q_4$ and $q_4$ are auxiliary functions of the temperature. Neither is physical by itself in general, as we show at the end of this section. The combination that enters $b$ is.

\subsection{The master formula}

Collecting the coefficient of $|u_1|^4$ in \eqref{eq:cumulant} gives $\tilde b=-\half G^2Q_4+\beta g^4q_4$. The coefficient of $|u_2|^2$ in $S_{\rm eff}^{(0)}$ is $m_2^2=\half A_2$ with
\begin{equation}
A_2(x)=1-z_B(x^2)+z_F(x^2),
\end{equation}
and the cubic coupling is $\kappa=Gc_3$ by \eqref{eq:H-avg}. Then \eqref{eq:b-from-btilde} gives
\begin{equation}
b=-G^2\Big[\half Q_4+\frac{2c_3^2}{A_2}\Big]+\beta g^4\,q_4 .
\label{eq:master}
\end{equation}
This is already of order $\lambda^2$. To this order every function can be evaluated at the free Hagedorn point $x_H$, where $\beta=\beta_0$ and $G=2g^2\beta_0$. There $A_2=0.9742786$. It is convenient to write the result as
\begin{equation}
b=g^4\big[-4\beta_0^2\,I+\beta_0\,q_4\big],\qquad I=\half Q_4+\frac{2c_3^2}{A_2},
\label{eq:master2}
\end{equation}
with everything at $x_H$. We need three numbers: $Q_4$, $c_3$ and $q_4$.

\begin{figure}[t]
\centering
\includegraphics[width=0.98\textwidth]{figs/anatomy.pdf}
\caption{The four kinds of contribution to the quartic coefficient $b$. Thick lines are letters that wind around the thermal circle and carry the Boltzmann factors. Thin lines are virtual states. Shaded circles are Polyakov loops $u_1$ or $u_{-1}$. (i) Two insertions of the one-loop operator $\cH$. (ii) Two insertions joined by the second moment $u_2$. (iii) The tree part of the two-loop operator $D_4$: three letters scatter into three. (iv) The loop part of $D_4$: two letters in different traces scatter forward through a one-loop diagram.}
\label{fig:anatomy}
\end{figure}

\subsection{Four kinds of contribution}

Figure~\ref{fig:anatomy} shows what the terms are. In each, four index loops carry the holonomies $U$, $U$, $U^{-1}$ and $U^{-1}$, and the question is what sits between them.

\emph{The square of the one-loop operator.} The term $Q_4$ is the connected average of two copies of $\cH$. Each copy acts on two letters. For the average to be connected and to produce four index loops, the two copies must share at least one letter, and the colour structures of \eqref{eq:Hfull} must be contracted in a non-planar way. It can be computed from the one-loop operator alone, and it needs no regularisation.

\emph{Exchange of $u_2$.} The cubic coupling $c_3$ is a one-loop quantity and the exchange is a tree diagram of the matrix model. It is the analogue, in this problem, of a contribution that Atick and Witten identified for strings near their Hagedorn temperature \cite{AtickWitten}. There the winding mode that becomes massless couples to the dilaton, and integrating out the dilaton gives the winding mode a negative quartic self-coupling. That is why they argued the string Hagedorn transition is first order. Here $u_1$ is the winding mode and $u_2$ plays the part of the dilaton. It turns out to be a small part of the answer.

\emph{The two-loop operator, tree part.} The operator $D_4$ contains a piece that acts on three letters at once and turns them into three others. It comes from tree-level scattering at order $g^4$, for instance two cubic vertices and a quartic one. When it acts across two traces it contributes to $q_4$. We call this part $q_4^{(a)}$. It has no loop and is finite.

\emph{The two-loop operator, loop part.} The operator $D_4$ also contains a piece that acts on two letters and returns them to themselves, through a one-loop diagram. The planar part of it renormalises $m_1^2$. The part that contributes to $q_4$ is of the double-trace form ${:}\tr(W_A\check W^A)\,\tr(W_B\check W^B){:}$, in which each letter is returned to its own trace while the loop connects the two. We call this part $q_4^{(b)}$. Each diagram in it is logarithmically divergent.

There is a third part of $D_4$, which acts on a single letter through a two-loop self-energy. It produces only two index loops and contributes to $m_1^2$, not to $b$. It will appear only in section~\ref{sec:checks}.

\subsection{What is known and what is not}

Two of the three traces of the one-loop operator that enter $c_3$ are in the work of Spradlin and Volovich, and the corresponding coefficient for general adjoint matter is given as a sum over angular momenta in \cite{MusselYacoby}. The function $Q_4$ has not appeared before, but it is a finite computation with a known operator.

The function $q_4$ requires the two-loop dilatation operator at the first non-planar order on all letters. It is known for the two holomorphic scalars of the $su(2)$ sector \cite{BKS}. The planar part is known on the larger $su(2|3)$ sector \cite{BeisertSU23}, for the pure scalar piece of the $so(6)$ sector, which is not closed at this order \cite{GGP}, and for twist-two operators through the anomalous dimensions of \cite{KLV,KLOV}. For the non-compact $su(1,1|2)$ sector Zwiebel found the two-loop generator and conjectured that its lift to finite $N$ holds for any gauge group \cite{Zwiebel}. None of this covers what we need. At the Hagedorn temperature all letters contribute. The scalars without derivatives, which make up the well-studied sectors, carry less than half of the weight in $z(x_H)=1$. We therefore construct the relevant parts of $D_4$ from the Hamiltonian.

\subsection{A freedom of scheme}

One subtlety must be addressed before numbers are quoted.

The effective action $S_{\rm eff}[U]$ is not unique. Only its integral over $U$ is physical. Consider adding to the dilatation operator a term proportional to the generator of gauge transformations, $\cG=\sum_A[W_A,\check W^A]$. Such a term vanishes on every gauge-invariant state and does not change the spectrum. The ensemble $\vev{\,\cdot\,}_U$ at fixed $U$ contains non-singlet states, however, and there the term does not vanish. One finds that it changes $Z[U]$ by a total derivative on the group manifold. In terms of the moments it is a change of variables, of the form $u_2\to u_2+g^2\gamma\,u_1^2$ and $u_1\to u_1+g^2\gamma'u_2u_{-1}$, with $\gamma$ and $\gamma'$ functions of the temperature.

This matters because two natural forms of the one-loop operator differ by exactly such a term. Beisert's form \eqref{eq:Hfull} has all of its content in a two-letter kernel. The operator that one obtains literally from second-order perturbation theory of the Hamiltonian has a different two-letter kernel, with no diagonal part, together with a one-letter self-energy. On singlets the two agree. At fixed $U$ they give different functions $c_3(x)$ and $Q_4(x)$. We call the two choices the commutator scheme and the literal scheme.

A change of variables cannot change the order of a phase transition. Indeed the redefinitions above shift $\tilde b$ and $\kappa^2/m_2^2$ in \eqref{eq:b-from-btilde} by equal and opposite amounts at the critical point, and shift $q_4(x)$ by an amount proportional to $1-z(x)$, which vanishes there. Hence $I$ and $q_4(x_H)$ are separately invariant. We have computed $I$ in both schemes as a check,
\begin{equation}
\begin{array}{lccc}
 & c_3(x_H) & Q_4(x_H) & I \\[2pt]
\mbox{literal scheme} & +0.16821 & 2.1222 & 1.11918(15) \\
\mbox{commutator scheme} & -0.4523 & 1.400 & 1.1200(16)
\end{array}
\label{eq:schemes}
\end{equation}
The individual entries differ, even in sign. The combination does not. The practical consequence is that $c_3$, $Q_4$ and $q_4$ must all be computed in one scheme. We use the literal scheme throughout, since that is what the Hamiltonian gives.