\section{The mode cut-off and its counterterms}
\label{sec:cutoff}

The only regulator that is natural for a Hamiltonian on $S^3$ is a cut-off on the angular momentum of the modes. It is not a good regulator in the usual sense. This section describes what it breaks, the three finite counterterms that repair it, and why the quartic coefficient $b$ is not affected by any of them. The first counterterm was found by AMMPV. The second is a two-loop relative of theirs. The third has precedents as an operator: a renormalisation of the $A_0$ propagator, with a finite part, is among the quadratic counterterms of AMMPV \cite{AMMPV2}, and the term is listed for $\cN=4$ in \cite{ITT}. We have not seen its finite coefficient in $\cN=4$, or its effect on the spectrum, elsewhere.

\subsection{What the cut-off preserves}

The theory on $\mathbb{R}\times S^3$ is invariant under the superconformal group. A cut-off $J$ on the angular momentum of every mode preserves the rotations $SO(4)$ of the sphere, the R-symmetry $SU(4)$ and time translations. It preserves nothing else. The conformal boosts mix harmonics of neighbouring angular momentum and do not commute with the projection onto modes below the cut-off. Neither do the supercharges, once the interaction is switched on. If the cut-off is placed on the left spin $j_L$ only, as is convenient, parity is lost too.

The general principle is then that every local operator of dimension four or less that is allowed by the surviving symmetries will be induced, with a coefficient that is a power of $J$ fixed by the dimension of the operator, and a finite part that no surviving symmetry fixes. In a Lorentz-invariant regularisation most of these operators are forbidden. Here they must be found and removed by hand, by imposing the broken symmetries as renormalisation conditions. AMMPV met this at one loop \cite{AMMPV2}, with a quadratically divergent gluon mass and different wave-function renormalisations for the time and space components. Aharony, Marsano and Van Raamsdonk added a curvature-dependent gluon mass counterterm \cite{AMV}, and Mussel and Yacoby, who follow them with matter included \cite{MusselYacoby}, a scalar mass whose finite part is fixed by conformal invariance.

\subsection{The double-trace quartic term}

The first counterterm is the one of section~\ref{sec:computation}. It is a quartic, double-trace, dimension-four operator without derivatives, such as $(\tr A_iA_j)^2$ or $\tr(A_iA_i)\tr(X_aX_a)$, where $X_a$ are the scalars. Its divergent part cancels between diagrams and its finite part is proportional to $n_B-n_F$. For pure glue we find the coefficients $\tfrac1{60}$ and $\tfrac1{60}$ of AMMPV. For $\cN=4$ it vanishes, mode by mode.

This is the only one of the three counterterms that has the colour structure to contribute to $b$. It is absent. The argument for its absence assumes, however, that the cut-off is applied uniformly. If one cuts off different diagrams on different lines, or cuts off the total energy of the intermediate state, the diagrams see effectively different values of $J$ and the cancellation of logarithms leaves a finite piece. We apply the cut-off to the angular momentum of one specified line of the loop, the same in every diagram.

\subsection{The mass term}

\begin{figure}[t]
\centering
\includegraphics[width=0.62\textwidth]{figs/cutoff.pdf}
\caption{The two-loop self-energy $c_4(Z;J)$ of the lowest scalar mode on $S^3$, in units of $(\lambda/8\pi^2)^2$, as a function of the cut-off $J$ on the doubled left spin of all virtual modes. The curve is $-13.9J^2-15J$. The value required by supersymmetry is $+0.351$. The point at $J=0$ is the plane-wave matrix model.}
\label{fig:cutoff}
\end{figure}

The second counterterm appeared when we tried to compute the two-loop self-energy of a single letter directly. Figure~\ref{fig:cutoff} shows the result for the lowest scalar mode $Z$. In units of $(\lambda/8\pi^2)^2$ the values for $J=0,1,\ldots,6$ are
\begin{equation}
-1.594,\ -29.477,\ -84.509,\ -170.044,\ -281.395,\ -422.012,\ -588.717 .
\end{equation}
They are fitted by $-13.9(3)J^2-15J$. The correct answer, fixed by the vanishing anomalous dimension of $\tr\,Z^L$, is $(2\zeta_3-1)/4=0.351$.

An error in the code is excluded by a second code, which builds the Fock space explicitly for gauge group $SU(2)$ and normal-orders mechanically with no diagrammatic input. It reproduces every class of diagrams at $J=0$ and $J=1$ as exact rational numbers.

The point $J=0$ offers a test that needs no regulator. If only the lowest mode of each field is kept, the theory on $S^3$ reduces to the plane-wave matrix model \cite{BMN,KKP}. This is a consistent truncation with an exact $SU(2|4)$ symmetry of its own. Inside it our Hamiltonian, with no counterterm, gives exactly vanishing two-loop energies for the protected states, through cancellations between classes of diagrams of size several hundred. It gives $-90g^4$ for the state $\tr[Z,\phi]^2$, in agreement with \cite{FKP}, and it reproduces the one-body coefficient $-\tfrac{51}{32}$ and the singlet energy $-114$ of \cite{KKP}. The value $-1.594$ is $-\tfrac{51}{32}$. The Hamiltonian is therefore correct, and it is the cut-off theory at $J>0$ that is not supersymmetric.

The divergence is a mass term. For the scalar modes $Z$, $DZ$ and $D^2Z$ with energies $E=1,2,3$ the coefficient of $J^2$ in the self-energy, multiplied by $E$, is $-13.9$, $-16.2$ and $-14.8$. That is the pattern $\delta E=\delta m^2/2E$ with $\delta m^2\simeq-28J^2$. The $J^2$ comes from the region where all virtual lines are hard. For fermions the growth is linear in $J$, as befits an operator of dimension three. We also checked that more elaborate cut-offs, with different limits for inner and outer loops, do not help: the result then depends logarithmically on the ratio.

The remedy is not to extrapolate but to renormalise. The mass counterterm is fixed by requiring that half-BPS operators and their descendants have the dimensions that the superconformal algebra dictates. Those conditions are the bootstrap equations used in section~\ref{sec:checks}, and the values \eqref{eq:c4} are the renormalised self-energies.

\subsection{The Coulomb term}

The bootstrap assumes that the two-body kernels need no finite correction. For letters without spin this can be tested in many ways and it holds. With letters that carry spin it fails, and that is the third counterterm.

It was found as follows. After the bootstrap the two-loop Hagedorn coefficient \eqref{eq:d4} came out as $-9.7$, short of the target by ten per cent. We then tested the two-loop Hamiltonian on the exact energies of short non-BPS operators, which are known from the Bethe ansatz \cite{BeisertStaudacherLR}. It reproduced them exactly in every sector built from scalars, fermions and field strengths. It failed in every sector that contains the letter $DZ$, a scalar with one derivative, by rational amounts such as $\tfrac{16}3$ and $\tfrac{80}9$. A dispersion test located the problem. For a single $DZ$ impurity moving in a background of $Z$ letters, the amplitude to hop by one site came out as $7$ where the algebra requires $9$. An independently written engine, with 5811 planar templates and explicit magnetic quantum numbers, gave $7.00$ as well, for every way of labelling the loop momentum. The discrepancy is therefore a property of the regulated theory and not of the code.

What is missing is a finite renormalisation of the instantaneous Coulomb interaction,
\begin{equation}
\Delta^{-1}\;\to\;\big(1-\tfrac43g^2\big)\,\Delta^{-1},\qquad\mbox{that is}\qquad\delta h_4=-\tfrac43\,h_2^{\rm Coul},
\label{eq:43}
\end{equation}
where $h_2^{\rm Coul}=PC_2P$ is the Coulomb part of the one-loop Hamiltonian, in units where $h_2$ is of order $g^2$. The tree-level Coulomb exchange between two letters vanishes identically when either letter has $j_L=0$ or $j_R=0$, since such a letter has no charge density in the relevant harmonic. This is why every earlier test, all of which involved letters of that kind, was blind to it.

The coefficient is heavily over-determined. We fitted three independent coefficients, for the Coulomb kernel in its two lowest multipoles and for a possible companion contact term, to fourteen exact two-loop energies in sectors of lengths three, four and five. The fit gives $-1.3332$, $-1.3296$ and $-0.0002$, with residuals below $10^{-3}$. With the values $-\tfrac43$, $-\tfrac43$ and $0$ imposed there is no free parameter and the residuals stay below $1.1\times10^{-3}$, which is the noise of the extrapolation in $J$. With the hop amplitude this makes fifteen exact conditions for one number.

We have not derived the coefficient $\tfrac43$. It is not an artefact of labelling: cut-offs on $j_L$, on $j_R$, on their maximum, minimum or mean, and smooth cut-offs on the energy, all give the same hop amplitude $7$. It does have a relative. The same factor $\tfrac43$ multiplies a finite kinetic counterterm for the fermions that we had found earlier, $\delta H=\tfrac43g^2H_0^{\rm ferm}+\ldots$ Both are what one would expect if the cut-off induced a finite relative renormalisation, by $\tfrac43g^2$, between the terms in the action with time derivatives and those with space derivatives. In a theory without boosts nothing forbids that. It is the two-loop analogue of the unequal wave-function renormalisations of \cite{AMMPV2,MusselYacoby}. A derivation along these lines should be possible, and would be valuable. A regulator that preserves sixteen supercharges does exist, in the form of the plane-wave matrix model expanded around fuzzy spheres \cite{IIST}, and in it none of these counterterms should arise.

\subsection{Why \texorpdfstring{$b$}{b} is not affected}

None of this affects the quartic coefficient, for two reasons.

The first is topological. The mass term and the Coulomb term are single-trace operators with two and four fields, and they are of order $g^4$. At the order we work they can only be inserted once, at tree level. The resulting diagrams have the topology of a one-loop or two-loop vacuum diagram, with at most three index loops. The coefficient of $|u_1|^4$ needs four. This is the argument of AMMPV \cite{AMMPV2}. These counterterms contribute to $m_1^2$, and to the cubic coupling at the next order, but they cannot contribute to $b$ at order $\lambda^2$.

The second is that the quantities that do enter $b$ are insensitive to how the cut-off is labelled, which is the practical symptom of a missing counterterm. Piece (a) has no loop. In piece (b) the double-trace kernel converges absolutely once the species are summed. As an illustration, the entry of the kernel between the lowest scalar and the first excited scalar multiplet comes out as $-606.0000$, $-605.9998$ and $-606.0002$ for three different choices of the line that carries the cut-off. The exact value is $-606$. The planar kernels that enter $d_4$ are different. Some of them converge only conditionally, and there the choice of line matters at the level of $\tfrac49$ to $\tfrac43$ in individual entries. The exact two-loop energies select one choice unambiguously.

The resulting picture is coherent. The quantity we set out to compute is the part of the three-loop free energy that is cleanest: double-trace, protected from counterterms by $n_B=n_F$, and absolutely convergent. The quantity we used to test the machinery, the two-loop Hagedorn temperature, is the part that is most sensitive to this regularisation. That it nevertheless comes out right, once the broken symmetries are restored, is the stronger test.