Spectral methods and nonperturbative corrections

An approximation can agree with a partition function to four decimal places while missing an entire physical sector. Conversely, a numerical method can generate a discrepancy of precisely that size. How do we distinguish the two?

This reproduction and controlled training extension isolates a published correction to the ABJM sphere partition function at level four. Prepare with the research-entry method and the Airy formula and its omitted sectors. The latter supplies the exact density operator and canonical transform. Here the new task is to extract a coefficient, predict the next bias, and determine which numerical data can resolve it.

A fixed theory and an unfitted reference

Use the compact three-dimensional theory with gauge group $U(N)_4\times U(N)_{-4}$, no additional quotient, and the round-sphere contour and normalization inherited from the Airy lesson. The rank is $N$ for each factor. This is the generic N=6 theory, with twelve real Poincaré supercharges; the construction makes an N=2 subalgebra manifest and localization chooses a charge within it. We do not change the level to an exceptional enhancing value. The auxiliary Fermi gas is an ordinary ungraded statistical problem, not a supersymmetric index. All its coordinates and chemical potentials below are dimensionless, and $F=-\log Z$.

Let $Z_N$ denote the exact partition function and $p(N)$ its perturbative Airy approximation:

$$p(N)=e^A C^{-1/3}\operatorname{Ai}\!\left[C^{-1/3}(N-B)\right], \qquad C=\frac1{2\pi^2},\quad B=\frac14, \quad A=-\frac{\zeta(3)}{4\pi^2}-\frac{\log2}{2}. \tag{D12.1}$$

The constant is the level-four evaluation of the stated Hatsuda–Moriyama–Okuyama constant-map prescription (3.4). Numerically it is $-0.37702204733836592549$. It is supplied independently of the finite-rank data, with the conjectural/tested status of that prescription. We neither fit $A$ to $Z_1$ nor impose $p(0)=1$. The exact normalization is $Z_0=1$; those are different statements.

The modified grand potential used for the large-positive-$\mu$ expansion is a branch whose exponential is periodized to obtain the exact Fredholm determinant. It is not simply the globally periodic function $\log\det(1+e^\mu\rho)$. HMO section 3.2 explains the image contributions. Keep that distinction when interpreting the supplied expansion

$$\begin{aligned} J_{\mathrm{mod}}(\mu) &=\frac{C\mu^3}{3}+B\mu+A+e^{-\mu} +P(\mu)e^{-2\mu}+\frac{16}{3}e^{-3\mu}+\cdots,\\ P(\mu)&=-\frac{4\mu^2+2\mu+1}{2\pi^2}. \end{aligned} \tag{D12.2}$$

This is the $k=4$ entry of HMO (1.5). The source reconstructs these coefficients using exact data and other physical constraints. Our calculation tests their consequences; it does not claim to derive an instanton path integral or a complete nonperturbative theory. No uniform remainder bound is supplied by writing the dots.

Extract a coefficient rather than merely plotting a residual

The canonical transform contains $e^{-N\mu}$. Multiplication by $e^{-s\mu}$ therefore replaces $p(N)$ by $p(N+s)$, while multiplication by $\mu$ acts as $-\partial_N$. These derivatives refer to the analytic Airy expression, not to differentiation of an integer gauge rank.

The transform acts on $e^J$, not on $J$. Expanding its correction factor gives

$$e^{e^{-\mu}+P e^{-2\mu}+\cdots} =1+e^{-\mu}+\left(P+\frac12\right)e^{-2\mu}+\cdots. \tag{D12.3}$$

Consequently the first two canonical corrections are

$$\begin{aligned} \Delta_1(N)&=p(N+1),\\ \Delta_2(N)&=\left[\frac12- \frac{4\partial_N^2-2\partial_N+1}{2\pi^2}\right]p(N+2). \end{aligned} \tag{D12.4}$$

The factor $1/2$ is the two-copy contribution of the first sector. Omitting it is a change of statistical expansion, not a harmless choice of convention. The Airy equation supplies an efficient implementation: $p''(x)=(x-B)p(x)/C$, while $p'(x)$ is obtained from $\operatorname{Ai}'$ with one extra factor $C^{-1/3}$.

Define an estimator for the leading coefficient,

$$\widehat c_N=\frac{Z_N-p(N)}{p(N+1)}, \qquad \widehat c_N^{\,(2)}= \frac{Z_N-p(N)-\Delta_2(N)}{p(N+1)}. \tag{D12.5}$$

The prediction is a coefficient tending to one, with a calculable finite-rank bias. This is more discriminating than observing that two positive functions become close. It checks the sign, exponent, shift, and normalization simultaneously.

For the comparison below, the exact finite-rank inputs come from HMO Appendix B, Figure 12. For example, $Z_1=1/16$ and $Z_2=(\pi^2-8)/(512\pi^2)$. Higher entries are exact rational expressions in $\pi$, evaluated before rounding. Set $r_0=(Z-p)/p$, $r_1=(Z-p-\Delta_1)/p$, and $r_2=(Z-p-\Delta_1-\Delta_2)/p$.

$N$	$r_0$	$r_1$	$r_2$	$\widehat c_N$
1	$5.86953\times10^{-3}$	$-7.80545\times10^{-5}$	$-2.23354\times10^{-8}$	$0.98687627$
2	$1.15487\times10^{-3}$	$-5.55986\times10^{-6}$	$-5.22700\times10^{-10}$	$0.99520880$
4	$9.97374\times10^{-5}$	$-8.56775\times10^{-8}$	$-1.19123\times10^{-12}$	$0.99914171$
6	$1.44594\times10^{-5}$	$-2.84124\times10^{-9}$	$-7.69747\times10^{-15}$	$0.99980354$

At $N=2$, $Z_2=0.00036998150558847232119$. Subtracting the first correction overshoots the exact result; the second, negative correction explains almost all that overshoot. The improved coefficient is $\widehat c_2^{\,(2)}=0.99999954956415$. At $N=6$ it is $0.99999999946775$. The remaining discrepancy is measured here against exact finite-rank inputs. It is not an error certificate for all higher ranks or for the full exponential series.

Check the finite-rank input through the spectrum

The independent object is the positive integral kernel on $L^2(\mathbb R,dx)$,

$$\rho(x,y)= \frac{1}{8\pi\sqrt{2\cosh(x/2)}\, 2\cosh[(x-y)/8]\,\sqrt{2\cosh(y/2)}}. \tag{D12.6}$$

Its normalization gives $\operatorname{Tr}\rho=1/16$, and fermionic statistics give $Z_2=[(\operatorname{Tr}\rho)^2-\operatorname{Tr}\rho^2]/2$. The center-coordinate integral in the two-cycle trace can be done exactly. With $u=x-y$ and $t=e^y$,

$$\int_{-\infty}^{\infty} \frac{dy}{2\cosh[(y+u)/2]\,2\cosh(y/2)} =\sqrt{e^u}\int_0^\infty\frac{dt}{(1+t)(1+e^ut)} =\frac{u}{2\sinh(u/2)}. \tag{D12.7}$$

The remaining even integral reduces to $\operatorname{Tr}\rho^2=I/(64\pi^2)$, where

$$I=\int_0^\infty\frac{x\,dx}{\sinh x\cosh^2(x/4)} =32\int_0^1\frac{(-\log q)q^2\,dq}{(1-q^4)(1+q)^2}=2. \tag{D12.8}$$

For a transparent evaluation, split the rational factor multiplying $-\log q$ into $8q/(1+q^2)-6/(1+q)+2/(1-q)$ and $d[4q/(1+q)^2]/dq$. The first three logarithmic integrals cancel; integration by parts leaves $\int_0^1 4\,dq/(1+q)^2=2$. Both endpoint terms vanish. Thus $\operatorname{Tr}\rho^2=1/(32\pi^2)$ and the exact $Z_2$ follows without using any Airy or instanton coefficient.

For higher ranks, a quadrature gives a finite positive matrix $K_{ij}=\sqrt{w_iw_j}\rho(x_i,x_j)$. If its eigenvalues are $\lambda_j$, calculate the coefficients of $\prod_j(1+z\lambda_j)$ recursively: start with $c_0=1$ and update $c_n\leftarrow c_n+\lambda_j c_{n-1}$ in descending order. Then $c_N$ approximates $Z_N$. This avoids alternating Newton sums of nearly equal traces.

A useful second discretization makes the unitary change $t=\tanh(x/8)$. Its kernel is

$$\widetilde\rho(t,u)=\frac{\sqrt{w(t)w(u)}}{1-tu}, \qquad w(t)=\frac{(1-t^2)^2}{4\pi(1+6t^2+t^4)}, \qquad -1<t,u<1. \tag{D12.9}$$

The square roots of the Jacobians are essential: simply substituting coordinates in (D12.6) changes the spectrum. The two discretizations have different endpoint and resolution errors and can expose each other’s failures.

Change resolution while holding the theory fixed

First fix $N=4$, $k=4$, the source normalization and the Airy prescription. Vary only the quadrature resolution. On $x\in[-96,96]$, 128, 256 and 512 Gauss–Legendre nodes give relative errors in $Z_4$ of approximately $4.00\times10^{-5}$, $2.05\times10^{-12}$ and $8.0\times10^{-14}$ against the exact value. The coarse grid error is a substantial fraction of the physical $r_0$. The compact-coordinate calculation with 512 nodes differs by about $3.7\times10^{-13}$. These are observed numerical errors, not rigorous quadrature bounds; tiny negative matrix eigenvalues at rounding level are also monitored.

The real-line truncation does admit a separate analytic bound. The diagonal kernel obeys $\rho(x,x)\leq e^{-|x|/2}/(16\pi)$, hence its omitted trace satisfies $\delta\leq e^{-L/2}/(4\pi)$. Compression to $[-L,L]$ decreases each ordered eigenvalue. Telescoping the elementary symmetric products and using $e_j\leq(\operatorname{Tr}\rho)^j/j!$ gives

$$0\leq Z_N-Z_N^{\mathrm{box}} \leq\delta\frac{(1/16)^{N-1}}{(N-1)!}. \tag{D12.10}$$

At $L=96$, the relative bound is $3.47\times10^{-17}$ for $N=4$ but $1.87\times10^{-12}$ for $N=6$. It controls the box, not the grid. The finite-matrix checks support the leading correction and the next bias; they do not independently resolve $r_2$ at $N=6$. That table entry instead uses the exact published expression at high precision.

Next vary rank while keeping the desired coefficient accuracy fixed. Write $D=p(N+1)>0$. If the absolute errors in $Z,p,D$ are bounded by $e_Z,e_p,e_D$ with $e_D<D$, the ratio satisfies

$$|\widetilde c_N-\widehat c_N| \leq\frac{e_Z+e_p+|\widehat c_N|e_D}{D-e_D}. \tag{D12.11}$$

Subtracting two close numbers loses digits. At $N=6$, $D/p\simeq1.4462\times10^{-5}$: roughly five relative digits disappear before one studies the coefficient. Increasing rank improves the asymptotic hierarchy while making coefficient extraction harder. Use absolute error estimates before taking the ratio, repeat arithmetic at higher precision, and compare the final uncertainty with the specific sector being tested. Fitting $A$ would conceal part of this diagnostic.

A literature-grounded next experiment

Literature check: 20 October 2026. Mariño–Putrov gives the spectral/Airy starting point and explicitly retains an omitted term. HMO supplies the exact data, sector reconstruction and integer-level pole cancellation used here. These calculations are established reproductions, not unexplored effects.

The later Bootstrapping ABJM theory, 2602.10196v3, derives functional relations for spectral quantities and uses their consistency to establish several relations previously supported by conjectures and numerical evidence. Its conclusions qualify the scope; they should not be replaced by a claim that every nonperturbative observable is now proved. A useful next reproduction is to check one of its resolvent identities against the same positive operator before attempting a new deformation. This needs additional instruction in resolvents and analytic continuation, not just a larger matrix.

An immediately reachable extension is more modest: reproduce the third canonical sector below, then determine the smallest numerical resolution that can distinguish it from zero at each of $N=1,2,4$. Keep a record of absolute input errors and the unfitted normalization. A residual stable under every numerical improvement but inconsistent with the sourced expansion merits a convention and literature check first. No novelty follows from that residual alone.

Exercises

1. The next sector is not just another constant

Using (D12.2), derive the full canonical contribution at order $e^{-3\mu}$. Predict its sign at $N=2$, and calculate the residual after including it.

Hint 1

At cubic order, the exponential contains the new grand-potential term, the product of the first two terms, and three copies of the first.

Hint 2

The coefficient is $16/3+P+1/6$. Apply the same signed derivative rule as in (D12.4), now at $N+3$.

Solution

Expanding before transforming gives

$$\Delta_3(N)=\left[\frac{11}{2}- \frac{4\partial_N^2-2\partial_N+1}{2\pi^2}\right]p(N+3).$$

At $N=2$, the derivative polynomial is large enough that $\Delta_3$ is negative. Its relative contribution is approximately $-5.22867\times10^{-10}$, accounting for the negative $r_2$. Direct substitution gives $(Z_2-p-\Delta_1-\Delta_2-\Delta_3)/p\simeq1.66716\times10^{-13}$. This small positive residual is consistent with a further sector; its size alone supplies neither a complete remainder bound nor an exact completion. Keeping only $(16/3)p(N+3)$ would give the wrong sign because it omits the product with $P$.

2. A regulator cannot remove the whole second correction

HMO (1.3)–(1.4) separates the second worldsheet coefficient $d_2(k)$ and the first membrane polynomial $M_k(\mu)$. At $k=4+\epsilon$, their expansions are

$$\begin{aligned} d_2(k)e^{(2-8/k)\mu} &=-\frac8{\pi^2\epsilon^2} -\frac{4(\mu+1)}{\pi^2\epsilon} -\frac{\mu^2+\mu+1/2}{\pi^2}-\frac76+O(\epsilon),\\ M_k(\mu) &=\frac8{\pi^2\epsilon^2} +\frac{4(\mu+1)}{\pi^2\epsilon} -\frac{\mu^2}{\pi^2}+\frac76+O(\epsilon). \end{aligned}$$

Add the sectors before taking the limit. Explain why setting $k=4$ separately in each term fails, and why replacing $e^{-8\mu/k}$ by $e^{-2\mu}$ before expanding also fails. Is $\epsilon$ a continuous family of compact Chern–Simons levels here?

Hint 1

The pole coefficients cancel, but their finite parts still depend on the expansion of the exponential.

Hint 2

A term of order $\epsilon^2$ in an exponential can multiply a double pole to give a finite quadratic polynomial in $\mu$.

Solution

The sum of the finite terms is $-(2\mu^2+\mu+1/2)/\pi^2=P(\mu)$, exactly (D12.2). Each isolated term is undefined at the integer level. Freezing the exponential discards the finite contributions produced by its first and second derivatives; in particular it misses part of the $\mu^2$ term. The regulator analytically continues the coefficient formulas to define their combined limit. It is not an assertion that arbitrary noninteger levels define the same compact gauge theory. Finally the canonical second sector is still $P+1/2$, since cancellation of poles occurs in $J$ before exponentiation.

3. Specify the precision for the quantity actually measured

At $N=6$, suppose $Z,p,D$ each have relative error at most $\eta=10^{-8}$. Use (D12.11) to bound the absolute uncertainty of $\widehat c_6$. Can these inputs reliably test the departure of $\widehat c_6^{\,(2)}$ from one? State what must change.

Hint 1

Use $Z/p=1+r_0$, $D/p=r_0-r_1$ and $\widehat c_6\simeq0.999804$.

Hint 2

For the second estimator, include the uncertainty of $\Delta_2$ as an additional absolute numerator error. Its remaining physical deviation is only about $5.32\times10^{-10}$.

Solution

Substitution gives the conservative bound

$$|\widetilde c_6-\widehat c_6| \leq\frac{\eta[Z+p+|\widehat c_6|D]}{D(1-\eta)} \simeq1.383\times10^{-3}.$$

Even the $1.96\times10^{-4}$ finite-rank departure of the first estimator from one is below this bound. The second estimator’s departure is much smaller. To test it at a ten-percent relative level, the combined absolute numerator/denominator error must be below roughly $5\times10^{-11}D$, about $7\times10^{-16}p$. This asks for more than accurate printed decimals: the box, quadrature, arithmetic, and Airy/derivative evaluations must all meet an allocated error budget. The $L=96$ box bound and the reported double-precision grids do not certify that target at $N=6$. Using the exact published expression at higher precision is a different, explicitly stated source of data.