\section{The free theory}
\label{sec:free}

We begin with the theory at $\lambda=0$. The results of this section are due to Sundborg \cite{Sundborg} and AMMPV \cite{AMMPV1}; AMMPV trace a special case to Hallin and Persson \cite{HallinPersson}, and the counting of single-trace words is also in Polyakov \cite{Polyakov}. A short review is \cite{AMMPVreview}. We review the free theory because every later section perturbs around it, and because the mechanism can be followed completely.

We put the theory on a three-sphere and set the radius to one. Energies are then measured in units of $1/R$. We write $\beta=1/T$ for the inverse temperature and
\begin{equation}
x=e^{-\beta},\qquad y=\sqrt{x}
\end{equation}
for the Boltzmann factor of a unit of energy and its square root. The square root appears because fermions on $S^3$ have half-integer energies. Both $x$ and $y$ are parametrisations of the temperature.

\subsection{Letters}

A free field on $S^3$ is a collection of harmonic oscillators, one for each spherical harmonic. The rotation group of the sphere is $SO(4)=SU(2)_L\times SU(2)_R$, and the harmonics are labelled by a pair of spins $(j_L,j_R)$. Table~\ref{tab:letters} lists the modes of the three kinds of field in $\cN=4$ super Yang--Mills theory. We call a single oscillator mode a \emph{letter}, since the states of the gauge theory will be words made from them.

\begin{table}[t]
\centering
\begin{tabular}{llll}
\toprule
field & harmonics $(j_L,j_R)$ & energy & single-letter partition function \\
\midrule
6 real scalars & $(j,j)$ & $2j+1$ & $6\times x(1+x)/(1-x)^3$ \\
4 Weyl fermions & $(j+\half,j)$ and $(j,j+\half)$ & $2j+\tfrac32$ & $4\times 4x^{3/2}/(1-x)^3$ \\
gauge field & $(j+1,j)$ and $(j,j+1)$ & $2j+2$ & $(6x^2-2x^3)/(1-x)^3$ \\
\bottomrule
\end{tabular}
\caption{The letters of $\cN=4$ super Yang--Mills theory on a unit $S^3$. In each row $j=0,\half,1,\ldots$ The scalars are conformally coupled, which gives the lowest mode energy one. Only the transverse modes of the gauge field are listed. In Coulomb gauge they are the only ones that propagate.}
\label{tab:letters}
\end{table}

The single-letter partition functions in the last column are the sums of $x^{E}$ over the modes in each row. For the scalars, for instance, the harmonic $(j,j)$ has $(2j+1)^2$ components of energy $2j+1$, and $\sum_{n\ge1}n^2x^n=x(1+x)/(1-x)^3$. Adding the rows gives the bosonic and fermionic totals
\begin{equation}
z_B(x)=\frac{6x+12x^2-2x^3}{(1-x)^3},\qquad
z_F(x)=\frac{16x^{3/2}}{(1-x)^3}.
\label{eq:zBzF}
\end{equation}
Their sum has a compact form in terms of $y=\sqrt x$,
\begin{equation}
z(x)=z_B+z_F=\frac{2y^2(3-y)}{(1-y)^3}.
\label{eq:z}
\end{equation}

There is a second way to read table~\ref{tab:letters}. The theory on $\mathbb{R}\times S^3$ is conformally equivalent to the theory on $\mathbb{R}^4$, and under that map the energy on the sphere becomes the scaling dimension. The letters are then the elementary fields at the origin of $\mathbb{R}^4$ with any number of derivatives acting on them, modulo the equations of motion: $\partial^k\phi$, $\partial^k\psi$, $\partial^kF$. The lowest scalar mode is $\phi$ itself with dimension one, and so on. We will use both languages.

\subsection{Gauss's law and the matrix model}

Every letter is an $N\times N$ matrix, because the fields are in the adjoint representation of $SU(N)$. If there were no further constraint, the partition function would be that of $N^2$ copies of table~\ref{tab:letters} and there would be no transition.

The sphere, however, is compact. Electric flux lines that leave a charge have nowhere to end, so the total charge on $S^3$ must vanish. For a non-abelian theory the statement is that physical states are singlets of the global $SU(N)$. This is the one effect of the gauge field that survives at zero coupling.

A projection onto singlets is an average over the group. If $\hat U$ denotes the operator that rotates every letter by the group element $U$, then
\begin{equation}
Z(x)=\int dU\;\Tr\big(x^{H}\hat U\big),
\label{eq:ZintU}
\end{equation}
where $dU$ is the Haar measure and the trace is over the unconstrained Fock space. In the path integral on $S^3\times S^1$ the same $U$ appears as the holonomy of the gauge field around the thermal circle, $U=P\exp(i\oint A_0)$. It is the one mode of the gauge field that has no quadratic action, which is why it must be integrated exactly. The two descriptions agree. We call $U$ the holonomy, or the Polyakov loop matrix.

The trace in \eqref{eq:ZintU} is that of free oscillators in a background $U$ and is elementary. An adjoint letter of energy $E$ contributes a factor $\det_{\rm adj}(1\mp x^EU)^{\mp1}$. Taking the logarithm, expanding it, and using $\tr_{\rm adj}U^n=|\tr\,U^n|^2$ at large $N$ gives
\begin{equation}
Z(x)=\int dU\,\exp\Big[\sum_{n=1}^\infty\frac{z_n(x)}{n}\,|\tr\,U^n|^2\Big],
\qquad
z_n(x)=z_B(x^n)+(-1)^{n+1}z_F(x^n).
\label{eq:Zfree}
\end{equation}
The sign $(-1)^{n+1}$ is the usual statistics factor for fermions.

The integrand depends on $U$ only through its eigenvalues $e^{i\theta_a}$. At large $N$ they form a continuous density $\rho(\theta)$ on the circle, normalised to one, whose Fourier coefficients are the moments
\begin{equation}
u_n=\frac1N\tr\,U^n=\int d\theta\,\rho(\theta)e^{in\theta},
\qquad
\rho(\theta)=\frac1{2\pi}\Big[1+\sum_{n\ge1}\big(u_ne^{-in\theta}+{\rm c.c.}\big)\Big].
\label{eq:rho}
\end{equation}
The first moment $u_1$ is the Polyakov loop. The Haar measure, written in terms of the eigenvalues, is a repulsion between them, $\prod_{a<b}|e^{i\theta_a}-e^{i\theta_b}|^2$. In terms of moments it is $\exp(-N^2\sum_n|u_n|^2/n)$. Combining it with \eqref{eq:Zfree}, the partition function is $\int\prod du_n\,e^{-S_{\rm eff}}$ with
\begin{equation}
S_{\rm eff}^{(0)}=N^2\sum_{n=1}^\infty\frac{1-z_n(x)}{n}\,|u_n|^2 .
\label{eq:Seff0}
\end{equation}
This is the effective action of the free theory. The term $1$ is the repulsion of the eigenvalues, which favours a uniform distribution around the circle, $u_n=0$. The term $-z_n$ is an attraction produced by the matter, which favours a concentrated one. The repulsion is fixed and the attraction grows with temperature.

\subsection{The Hagedorn temperature}

At low temperature $z_n(x)<1$ for all $n$. Every $u_n$ has a positive mass and the saddle point is $u_n=0$. The action vanishes there, the free energy is of order one, and the Polyakov loop is zero. This is the confined phase. Performing the Gaussian integrals gives $Z=\prod_n(1-z_n(x))^{-1}$, which is the partition function of a gas of multi-trace operators. It could also be obtained by counting necklaces of letters with P\'olya's theorem \cite{Polya,Sundborg,Polyakov,BMS}.

Since $z_n(x)=z_1(x^n)$ up to the statistics sign and $x<1$, the first mode to go soft as the temperature is raised is $n=1$. It does so when $z(x)=1$. With \eqref{eq:z} this is a cubic equation for $y$,
\begin{equation}
y^3-3y^2-3y+1=(y+1)(y^2-4y+1)=0 .
\end{equation}
The root between zero and one is $y_H=2-\sqrt3$, and so
\begin{equation}
x_H=7-4\sqrt3=0.07180,\qquad
\beta_0\equiv-\ln x_H=2.6339158,\qquad
T_H^{(0)}=\frac1{\beta_0}=0.37966.
\label{eq:xH}
\end{equation}
The inverse Hagedorn temperature of the free theory, $\beta_0$, appears in almost every formula below.

This is a Hagedorn temperature in the usual sense. The confined partition function $\prod_n(1-z_n)^{-1}$ has a pole at $z(x)=1$. A pole at $x_H$ in a generating function $\sum_E\Omega(E)x^E$ means that the number of states grows like $\Omega(E)\sim x_H^{-E}=e^{\beta_0E}$. This exponential growth in the number of single-trace words is the same phenomenon as the Hagedorn growth in the number of states of a string \cite{Hagedorn,Sathiapalan,Kogan,AtickWitten}. Polyakov related such traces to the states of a closed string \cite{Polyakov}, and the free theory was read as a string theory of vanishing tension by Sundborg \cite{Sundborg,SundborgTensionless} and by Haggi-Mani and Sundborg \cite{HaggiManiSundborg}.

\subsection{Above the Hagedorn temperature}

For $x>x_H$ the coefficient of $|u_1|^2$ in \eqref{eq:Seff0} is negative. The action decreases as $|u_1|$ grows, and as a function of an unconstrained $u_1$ it is unbounded below. The run-away is stopped by the fact that the $u_n$ are not independent variables. They are Fourier coefficients of a density, and a density cannot be negative. With only $u_1$ turned on, $\rho=(1+2u_1\cos\theta)/2\pi$ is non-negative if and only if
\begin{equation}
|u_1|\le\half .
\label{eq:half}
\end{equation}
Beyond this point the density develops a gap, an interval of $\theta$ around $\pi$ in which there are no eigenvalues. The repulsion among the remaining eigenvalues is then stronger than the simple $|u_1|^2$ of \eqref{eq:Seff0}. We give the exact form in the next section.

The free theory therefore behaves as follows. For $T<T_H^{(0)}$, $u_1=0$. At $T=T_H^{(0)}$ the potential for $u_1$ is exactly flat all the way from $0$ to $\half$. For $T$ just above $T_H^{(0)}$ the minimum is just beyond $\half$, on the gapped branch. The Polyakov loop jumps from $0$ to $\half$ at a single temperature, and the free energy changes from $O(1)$ to $O(N^2)$.

This is a first-order transition in the sense that the order parameter jumps. It is not a conventional one. There is no barrier between the phases, no metastability and no coexistence. The potential is flat at the transition. That degeneracy is an accident of $\lambda=0$, where the action is quadratic in the $u_n$. Interactions add quartic and higher terms, and the smallest of them decides what the flat direction becomes.