ABJM small-N partition functions

An oscillatory integral can return many stable digits of the wrong limiting prescription. A positive integral can also mislead if its integration box grows faster than its numerical resolution. This lab gives each problem a separate diagnostic.

Begin with the regulated rank-one calculation. The default program needs only that lesson. The later Spectral checks section introduces an additional prerequisite and an optional program mode; you can complete the rank-one experiment and its first exercise independently.

Fix the observable before computing

We use the undeformed round-sphere partition function of three-dimensional compact $U(N)_k\times U(N)_{-k}$, with positive integer $k$, rank $N$ at each node, and no extra quotient by the center. The generic full theory has $\mathcal N=6$, or twelve real Poincaré supercharges, enhanced to sixteen at $k=1,2$. The calculation uses its $\mathcal N=2$ sphere subalgebra. Keep $H=-i/\ell$, matter scalar R-charge $1/2$, zero real masses and FI terms, and the original superpotential.

The dimensionless gauge scalars are $s=\ell\sigma$ and $t=\ell\widehat\sigma$. Our first node has phase $+i\pi k s^2$, and $F=-\log Z$. These are fixed finite-rank calculations, without a large-$N$ or semiclassical approximation.

At $N=1$, introduce positive Gaussian damping:

$$Z_\epsilon(1,k)= \int_{\mathbb R^2}ds\,dt\, \frac{\exp[i\pi k(s^2-t^2)-\epsilon(s^2+t^2)]} {4\cosh^2\!\pi(s-t)},\qquad \epsilon>0. \tag{LAB10.1}$$

The physical prescription is $\lim_{\epsilon\downarrow0}Z_\epsilon$, not an arbitrary rectangular oscillatory cutoff. The normalization agrees with FG (2.1) under $\mu=2\pi s$, $\nu=2\pi t$. KWY (3.68), (4.4) has the opposite first-node phase; exchange its nodes to compare. Its Weyl normalization is already included, with both factors trivial at rank one.

Give the Gaussian a fixed width

Set $x=s-t$ and $y=(s+t)/2$. The absolute Jacobian is one, while $s^2-t^2=2xy$ and $s^2+t^2=2y^2+x^2/2$. Absolute convergence at positive $\epsilon$ permits the Gaussian $y$ integration. Writing $C=\pi^2k^2+\epsilon^2$ gives

$$\begin{aligned} Z_\epsilon &=\sqrt{\frac{\pi}{2\epsilon}} \int_{\mathbb R}\frac{e^{-Cx^2/(2\epsilon)}}{4\cosh^2(\pi x)}\,dx\\ &=\frac{\pi}{\sqrt C}\, \mathbb E\!\left[ \frac1{4\cosh^2(\pi\sqrt{\epsilon/C}\,U)} \right],\qquad U\sim N(0,1). \end{aligned} \tag{LAB10.2}$$

The second line is the numerical representation: the integration variable $U$ always has unit width. A routine operating directly on the first line would have to resolve a peak narrowing as $\sqrt\epsilon$.

The limit is $1/(4k)$. For a useful finite-input guarantee, use $|\tanh v|\leq|v|$, $\mathbb E U^2=1$, and $1-(1+a)^{-1/2}\leq a/2$:

$$0\leq1-4kZ_\epsilon \leq\frac{\epsilon}{k^2} +\frac{\epsilon^2}{2\pi^2k^2}. \tag{LAB10.3}$$

This bounds the regulator bias. Raising numerical precision cannot remove it, and $\epsilon$ is not a physical coupling.

Run the rank-one experiment

Save these three files in one directory:

Download the complete lab (ZIP) with the program, pinned dependencies, reference output and run instructions.

• Portable Python calculation
• Pinned requirements
• Expected output

The tested environment is Python 3.13.7 with mpmath 1.3.0 in an isolated virtual environment. After activating such an environment, run:

python -m pip install -r requirements.txt
python small_n.py
python small_n.py --level 2

No notebook, repository imports, or network calls are required by the calculation itself. The program accepts levels one through four and 40–70 working decimal digits via --dps. The default is 50 digits, with a repeat at 70. The expected-output file contains two objects: entry is the default result, and spectral is the optional run below.

For $k=1$:

$\epsilon$	Computed $Z_\epsilon$	Relative bias $1-4Z_\epsilon$	Proven upper bound
$0.1$	$0.228848733109$	$0.08460507$	$0.10050661$
$0.03$	$0.242904986403$	$0.02838005$	$0.03004559$
$0.01$	$0.247547403082$	$0.00981039$	$0.01000507$

The precision repeat differs by about $1.7\times10^{-53}$ at the last input. That is evidence of numerical stability for this quadrature; the physical limit is still $0.25$, nearly one percent above the finite-$\epsilon$ result.

In the program, epsilon is the regulator, k the integer level, and z1_regulated implements (LAB10.2). Changing the level changes the physical theory. Changing the regulator or working precision changes how we approximate its answer.

Spectral checks

This section and --spectral require the Fermi-gas transformation. That lesson proves the relation to the regulated matrix integral and explains the auxiliary fermions. They are a representation of the sphere observable, not a replacement of ABJM by a microscopic free-fermion theory.

Keep the precisely normalized kernel on $L^2(\mathbb R,dx)$:

$$w(x)=\frac1{2\cosh(x/2)},\qquad \rho(x,y)= \frac{\sqrt{w(x)w(y)}}{4\pi k\cosh[(x-y)/(2k)]}. \tag{LAB10.4}$$

Here the auxiliary Planck constant is $2\pi k$. The operator has not been rescaled to trace one. Its normalization and permutation formula are FG (3.3)–(3.8).

Let $t_j=\operatorname{Tr}\rho^j$. The two permutations of two particles give

$$\begin{aligned} t_1&=\int_{\mathbb R}\rho(x,x)\,dx=\frac1{4k},\\ Z(2,k)&=\frac12(t_1^2-t_2)\\ &=\frac12\int_{\mathbb R^2} [\rho(x,x)\rho(y,y)-\rho(x,y)\rho(y,x)]\,dx\,dy. \end{aligned} \tag{LAB10.5}$$

The factor $1/2$ and the negative exchange term are both essential. A numerical kernel matrix can compute its trace and the trace of its square separately. We also want a comparison that avoids subtracting those two quantities.

Reduce the positive pair determinant

Substituting the kernel shows that the square bracket is $w(x)w(y)\tanh^2[(x-y)/(2k)]/(16\pi^2k^2)$. Use $d=x-y$ and $c=(x+y)/2$, again with unit Jacobian. Then

$$\begin{aligned} \int_{\mathbb R}w(c+d/2)w(c-d/2)\,dc &=\int_{\mathbb R}\frac{dc}{2[\cosh c+\cosh(d/2)]}\\ &=\frac{d}{2\sinh(d/2)},\\ Z(2,k)&=\frac1{32\pi^2k^2} \int_0^\infty \frac{d\,\tanh^2[d/(2k)]}{\sinh(d/2)}\,dd. \end{aligned} \tag{LAB10.6}$$

For the middle equality, substitute $t=\tanh(c/2)$; the integral becomes a rational integral on $-1<t<1$. Its value at $d=0$ is the continuous limit one. The last line uses evenness in $d$. Both (LAB10.5) and (LAB10.6) use the same $dx$ kernel measure; there is no additional $dx/(2\pi)$.

At $k=1$, put $d=2u$. The remaining integral is $4\int_0^\infty u\sinh u/\cosh^2u\,du$. Integration by parts reduces it to $4\int_0^\infty\operatorname{sech}u\,du=2\pi$. Thus

$$Z(2,1)=\frac1{16\pi}=0.0198943678864869\ldots,\qquad t_2=\frac1{16}-\frac1{8\pi}=0.0227112642270262\ldots. \tag{LAB10.7}$$

This is a complete finite-rank answer, not the leading term of a strong-coupling expansion.

Keep three numerical controls separate

Run:

python small_n.py --spectral
python small_n.py --spectral --level 2 --dps 60
python small_n.py --spectral --omit-exchange

The positive integral is evaluated directly. Its difference-coordinate cutoff D has an analytic omission bound:

$$0\leq Z(2,k)-Z_D(2,k)\leq \frac{(2D+4)e^{-D/2}}{16\pi^2k^2(1-e^{-D})}. \tag{LAB10.8}$$

Indeed $\tanh^2\leq1$ and $1/\sinh(d/2)\leq2e^{-d/2}/(1-e^{-D})$ for $d\geq D$. Integrating $d e^{-d/2}$ proves the bound. At $k=1$, $D=32$ omits about $4.85\times10^{-8}$; $D=64$ omits about $1.06\times10^{-14}$. These bounds concern the exact truncated integral, independently of quadrature accuracy.

The kernel check instead restricts both $x,y$ to $[-L,L]$. For Gauss–Legendre nodes $x_i$ and integration weights $\omega_i$ on that interval, form $R_{ij}=\sqrt{\omega_i\omega_j}\rho(x_i,x_j)$. Compute $\operatorname{Tr}R$ and $\operatorname{Tr}R^2$. The program also sums the positive pair determinants of this same matrix, checking their algebraic agreement; that last equality alone is not an independent quadrature test.

To test the actual quadrature, integrate the center variable analytically over its finite allowed range. For $0<d<2L$,

$$\begin{aligned} C_L(d)&= \frac{2\,\operatorname{atanh}\!\left[ \tanh(d/4)\tanh((L-d/2)/2)\right]}{\sinh(d/2)},\\ Z_{\mathrm{box}}&=\frac1{16\pi^2k^2} \int_0^{2L}C_L(d)\tanh^2[d/(2k)]\,dd. \end{aligned} \tag{LAB10.9}$$

Use $C_L(0)=\tanh(L/2)$. This one-dimensional reference has the same box as the kernel matrix. Its difference from the matrix answer isolates quadrature error. Separately, the diagonal mass outside the box is $\delta_L=\arctan(e^{-L/2})/(\pi k)$. Since the positive pair determinant is at most the diagonal product, $0\leq Z-Z_{\mathrm{box}}\leq t_1\delta_L-\delta_L^2/2$.

Representative $k=1$ results are:

Box half-width $L$	Nodes	Quadrature difference from the same-box integral	Bound on box omission
$16$	$32$	$2.57\times10^{-6}$	$2.67\times10^{-5}$
$16$	$64$	$1.81\times10^{-11}$	$2.67\times10^{-5}$
$16$	$128$	$4.32\times10^{-22}$	$2.67\times10^{-5}$
$32$	$32$	$7.23\times10^{-4}$	$8.96\times10^{-9}$
$32$	$128$	$2.08\times10^{-11}$	$8.96\times10^{-9}$

At fixed 32 nodes, a larger box is worse resolved. At fixed $L=16$, more nodes cannot remove the omitted tail. Increasing --dps cannot cure either error. The complete download reports all nine box/node combinations, while its precision repeat tests the positive integral at two arithmetic settings. Its top-level trace2 is reconstructed from the full positive integral; the separately evaluated kernel traces are in finite_box_kernel_checks.

The negative mode reports $t_1^2/2=1/32$ at $k=1$ instead of the correct (LAB10.7). It deliberately omits exchange, detects that discrepancy, and returns intentional_error_detected: true. A successful diagnostic exit does not endorse the wrong number. Likewise, checks_pass means the asserted identities, bounds and precision comparisons passed; it does not claim every coarse grid in the table is accurate.

Three experiments with solutions

1. Cancel the leading regulator bias

This exercise needs only the rank-one entrance. Expand $4kZ_\epsilon$ through order $\epsilon^2$, using $\mathbb E U^4=3$. Show that $2Z_{\epsilon/2}-Z_\epsilon$ cancels the linear bias. At $k=1$, compare its errors at $\epsilon=0.02$ and $0.01$. Does the extrapolation equal the exact answer at finite input?

Hint 1

Expand $\operatorname{sech}^2v=1-v^2+2v^4/3+\cdots$ as well as the prefactor $\pi k/\sqrt C$.

Hint 2

A term $b\epsilon^2$ becomes $-b\epsilon^2/2$ in the extrapolated combination.

Solution

The Gaussian moments in (LAB10.2) give

$$4kZ_\epsilon= 1-\frac{\epsilon}{k^2} +\left(\frac2{k^4}-\frac1{2\pi^2k^2}\right)\epsilon^2 +O(\epsilon^3).$$

Consequently the combination has no term linear in $\epsilon$, but retains minus one-half the displayed quadratic coefficient. Direct evaluation gives absolute errors $8.98396\times10^{-5}$ and $2.33748\times10^{-5}$ at the two requested inputs. The ratio is about $3.84$, approaching four as $\epsilon$ decreases. This is quadratic improvement, not exact removal of the regulator. It also does not automatically inherit the one-sided bound (LAB10.3).

2. Change the physical level and budget the error

After the spectral preparation, use $k=2$. Derive $Z(2,2)=1/(32\pi^2)$ from (LAB10.6), and find $t_2$. Can the bound (LAB10.8) certify a difference-cutoff omission below $10^{-10}$ at $D=32$? At $D=64$? Explain why that is not yet a total numerical error certificate.

Hint 1

Set $d=4u$ and use $\sinh(2u)=2\sinh u\cosh u$.

Hint 2

The remaining integral involves $u\sinh u/\cosh^3u$. Differentiate $\operatorname{sech}^2u$.

Solution

The unnormalized integral in (LAB10.6) becomes $8\int_0^\infty u\sinh u/\cosh^3u\,du=4\int_0^\infty\operatorname{sech}^2u\,du=4$. Including its prefactor gives $Z(2,2)=1/(32\pi^2)\approx0.003166286989$. Because $t_1=1/8$, the trace is $t_2=1/64-1/(16\pi^2)$.

The bounds are approximately $1.21\times10^{-8}$ at $D=32$ and $2.65\times10^{-15}$ at $D=64$. Only the latter certifies the requested tail budget. A quadrature comparison and a precision comparison still address separate errors in evaluating the truncated integral. A box cutoff $L$ also has its own domain and bound; it cannot be substituted for $D$.

3. Diagnose two normalization mistakes

After the spectral preparation, consider two modifications at $k=1$: dropping $t_2$ from (LAB10.5), or replacing $\rho$ by $\rho/t_1$ to make it a trace-one density operator. Find both incorrect two-particle answers, and identify a rank-one check that catches the second error. What physical approximation, if any, do these modifications implement?

Hint 1

Every term in the two-particle determinant is quadratic in $\rho$.

Hint 2

The rank-one sphere integral is $1/4$, whereas the trace of the rescaled operator is one.

Solution

Dropping exchange gives $1/32=0.03125$, an overestimate by $1/32-1/(16\pi)\approx0.01135563$. Rescaling $\rho$ by four gives $16Z(2,1)=1/\pi$ and changes $Z(1,1)$ to one. In general $\rho\mapsto c\rho$ changes $Z(N,k)$ by $c^N$, so this is an observable normalization change.

Neither modification is a controlled approximation to the fixed observable here. The first discards the antisymmetric occupation rule of its exact auxiliary representation; the second discards the absolute normalization inherited from localization. The three-dimensional theory remains interacting. Correct auxiliary fermion statistics and the rank-one normalization are both needed to reproduce its sphere partition function.

The successful comparison is between the specified Abel limit, the normalized kernel traces, and a positive two-particle integral with explicit cutoffs. It supports these finite-rank computations. It does not establish arbitrary contour deformations, a large-$N$ expansion, or a microscopic free-fermion description.