Compare exact, planar, and gravitational predictions

An exact formula and a gravitational prediction can agree in their leading growth while differing by a large factor. Conversely, a small numerical residual can hide the wrong subleading coefficient. We will compare results at the level they actually determine, then design a calculation that resolves the first disagreement.

Start from the planar Wilson loop, the ABJM state-counting result, and the AdS source dictionary. Their derivations remain in those lessons. Here the new task is to decide what to hold fixed, what to subtract, and how accurately to evaluate the remaining quantity.

An exact answer still has a specified object

For the first comparison, use four-dimensional $SU(N)$ $\mathcal N=4$ Yang–Mills in its conformal vacuum, with sixteen real Poincaré supercharges and the chosen manifest $\mathcal N=2$ localization description. The observable is the normalized fundamental supersymmetric circle, $N^{-1}\operatorname{Tr}$, on the round four-sphere. Put $\lambda=g_{\rm YM}^2N>0$. Its half-BPS label fixes an operator, not every observable of the theory.

For the second, use three-dimensional compact $U(N)_k\times U(N)_{-k}$ ABJM, rank $N$ per node, positive integer $k$, and no extra center quotient. The round-sphere partition function has the absolute normalization and supersymmetric source prescription of the positive Fermi gas. Its full generic algebra has twelve real Poincaré charges, enhanced to sixteen at $k=1,2$; the chosen manifest subalgebra has four. We compare $F=-\log Z$ at fixed $k$, not a loop or a flat-space thermal ensemble.

These are different theories and different observables. A comparison within each row is meaningful:

Object	Exact information available	A subsequent approximation
Normalized $SU(N)$ circle	A finite Gaussian integral at any integer $N\geq2$ and positive $\lambda$	Planar, then strong coupling
ABJM sphere $Z_N$	An exact finite-rank matrix integral or exterior-power trace	Fixed-level large rank; perturbative Airy resummation
Pure $SU(2)$ Seiberg–Witten coupling	The local two-derivative prepotential and period data	A weak-coupling series, or a low-energy use of that action

The third row refers to four-dimensional $\mathcal N=2$ theory with eight real supercharges, a manifest $\mathcal N=1$ description, and a Coulomb-branch vacuum. One need not learn its curve to understand the scope: SW1 (2.7)–(2.9) explicitly specifies terms with at most two derivatives. Knowing their coupling exactly does not determine all higher-derivative interactions or all amplitudes at arbitrary energy. Close to a point where an omitted charged particle becomes massless, the action with only the photon multiplet must be enlarged. The effective-action lesson develops that distinction.

An exact integral also need not have an elementary closed form. Approximating its evaluation numerically introduces a computational error; replacing it by a saddle introduces a physical asymptotic truncation. Increasing arithmetic precision addresses the first of these.

Resolve a finite-rank correction

The circle’s exact finite-rank answer and planar limit are

$$\begin{aligned} W_N^{SU}(\lambda)&=e^{-\lambda/(8N^2)}\, \frac{e^{\lambda/(8N)}}{N}L_{N-1}^{1}\!\left(-\frac{\lambda}{4N}\right),\\ W_0(\lambda)&=\frac{2I_1(\sqrt\lambda)}{\sqrt\lambda}. \end{aligned} \tag{X04.1}$$

$L_{N-1}^{1}$ is the finite Laguerre polynomial defined in the prerequisite; $I_\nu$ is the modified Bessel function. Pestun (1.2) supplies the exact localization statement. Drukker–Gross (2.26) evaluates the auxiliary $U(N)$ integral. The exponential in front removes its free central Gaussian while keeping $\lambda$ fixed. It is part of the observable’s definition.

At fixed $\lambda$, the known first correction gives

$$\frac{W_N^{SU}}{W_0}=1+\frac{g(\lambda)}{N^2}+O_\lambda(N^{-4}), \qquad g(\lambda)=\frac{\lambda^{3/2}}{96}\frac{I_2(\sqrt\lambda)}{I_1(\sqrt\lambda)}-\frac\lambda8. \tag{X04.2}$$

A plot of $W_N/W_0$ approaching one tests little about $g$: both a correct and an incorrect finite coefficient disappear. Instead form the scaled residual $r_N=N^2(W_N/W_0-1)$. The predicted limit is now $g$, not zero. If the expansion continues as $r_N=g+h/N^2+j/N^4+\cdots$, two ranks cancel its first bias:

$$\widehat g_N=\frac{4r_{2N}-r_N}{3} =g-\frac{j}{4N^4}+O_\lambda(N^{-6}). \tag{X04.3}$$

This is an extrapolation using a known power law. Derive it by inserting both series: the $h$ terms have equal coefficients before subtraction. It is not an error bound when the form or the remaining coefficients are unknown.

At $\lambda=16$, direct finite-polynomial evaluation gives:

$N$	$r_N$	$\widehat g_N$
$8$	$-1.542817660$	$-1.561266102$
$16$	$-1.556653991$	$-1.561299578$
$32$	$-1.560138181$	$-1.561301681$

The target is $g(16)=-1.561301822$. The extrapolated discrepancy falls by approximately sixteen under rank doubling, as an $N^{-4}$ term suggests. A separate positive-integral evaluation checks these numbers: with $T$ distributed with density $e^{-t}t^N/N!$ on $t>0$, DG (A.7) becomes $W_N^{SU}=e^{-\lambda/(8N)-\lambda/(8N^2)}\mathbb E[W_0(\lambda T/N)]$. It uses neither the finite polynomial nor a fit to the tabulated residual.

Subtraction amplifies numerical noise. If $W_N/W_0$ has absolute error at most $\eta$, the corresponding $r_N$ error can be $N^2\eta$. The extrapolation amplifies two independently bounded errors further. A smoother-looking extrapolated sequence can therefore be less trustworthy unless the input precision improves with rank. Exercise1 quantifies this cost.

Remove an unknown constant before testing the rank shift

The ABJM Airy lesson derives the perturbative resummation

$$Z_{\rm pert}(N,k)=e^{A(k)}C^{-1/3} \operatorname{Ai}\!\left(C^{-1/3}(N-B)\right), \qquad C=\frac2{\pi^2k},\quad B=\frac{k}{24}+\frac1{3k}. \tag{X04.4}$$

Here $\operatorname{Ai}$ is the decaying Airy solution and $A(k)$ is independent of rank. “Perturbative” describes the fixed-$k$ inverse-rank expansion being resummed; it does not mean weak field-theory coupling. FG (5.65)–(5.67) keeps an additional nonperturbative contribution. Equation (X04.4) is not the exact finite-rank partition function.

For $N>B$ large at fixed $k$, the Airy asymptotic form gives

$$F_{\rm pert}=a_k(N-B)^{3/2}+\frac14\log(N-B)+c_k +O_k(N^{-3/2}),\qquad a_k=\frac{\pi\sqrt{2k}}3. \tag{X04.5}$$

The constant is $c_k=-A(k)+\log(2\sqrt\pi)+\tfrac14\log C$. The leading coefficient $a_k$ is already fixed by state counting. We can test the next parameter without assuming $A(k)$. At fixed $k$, define $D_N=F(N+1,k)-F(N,k)=-\log[Z_{N+1}/Z_N]$ and $\Delta f(N)=f(N+1)-f(N)$. Expand $(N-B)^{3/2}=N^{3/2}-\tfrac32B\sqrt N+\tfrac38B^2N^{-1/2}+\cdots$. The rank-independent constant cancels from $D_N$, suggesting

$$\mathcal B_N=-\frac{2}{3a_k} \frac{D_N-a_k\Delta N^{3/2}-\tfrac14\log(1+1/N)} {\Delta\sqrt N}. \tag{X04.6}$$

For the perturbative Airy function this tends to $B$. More precisely, $\mathcal B_N=B+B^2/(4N)-B/(3a_kN^{3/2})+O_k(N^{-2})$. This is a residual diagnostic, not a new exact formula for $B$ at finite rank. Its division by $\Delta\sqrt N\sim1/(2\sqrt N)$ amplifies errors in the rank difference by order $\sqrt N$.

At $k=1$, the target is $B=3/8$. The full positive-kernel calculation and the perturbative formula give:

$N$	$\mathcal B_N$ from the quantum kernel	$\mathcal B_N$ from the Airy formula
$8$	$0.37662137$	$0.37662137$
$16$	$0.37611071$	$0.37611071$
$32$	$0.37568869$	$0.37568869$

The kernel values use its actual elementary symmetric coefficients through rank 33. Changing the integration window from $[-64,64]$ to $[-80,80]$ and the quadrature from 768 to 1024 nodes changes these diagnostics by less than $1.2\times10^{-9}$. This is observed numerical stability, not a certified infinite-operator error bound. Eight displayed digits deliberately leave the tiny kernel/Airy difference unresolved. The common discrepancy from $3/8$ is instead a visible finite-rank bias. It is not evidence for a nonperturbative mismatch.

The Airy evaluation also has an independent positive-integral check, using $\operatorname{Ai}(x)=\sqrt{x}\,K_{1/3}(2x^{3/2}/3)/(\pi\sqrt3)$ for $x>0$, with $K_\nu(z)=\int_0^\infty e^{-z\cosh u}\cosh(\nu u)du$. Scaling out $e^{-z}$ avoids losing the small function before taking logarithms. Agreement between these evaluations tests numerical evaluation; it does not restore the sectors omitted from (X04.4).

Compare with gravity in its common regime

The four-dimensional circle and the ABJM sphere require different bulk calculations. For the circle, the leading classical string action yields $\log W\sim\sqrt\lambda$. The planar function instead obeys

$$W_0\sim\sqrt{\frac2\pi}\lambda^{-3/4}e^{\sqrt\lambda}, \qquad \log W_0=\sqrt\lambda-\frac34\log\lambda+O(1). \tag{X04.7}$$

Thus the classical action gets the leading logarithm; $e^{\sqrt\lambda}$ alone has a growing relative error as an approximation to $W_0$. The normalized prefactor contains further information. This is the comparison in DG §3, with the two meanings of agreement separated.

The D3 dictionary gives $\alpha'/L^2=\lambda^{-1/2}$ and $g_s=\lambda/(4\pi N)$. The near-horizon radius follows from Maldacena (2.2)–(2.4). In the weakly coupled original string description, the simple parametric overlap is $1\ll\lambda\ll N$. A small correction to one protected loop is not by itself a test of every bulk approximation.

For ABJM, the gravitational sphere action is instead $I_{\rm ren}=\pi L^2/(2G_4)=a_kN^{3/2}$ at leading order. It matches the same round-sphere free energy and normalization used above, as DMP (5.37)–(5.44) verifies with the opposite source sign $F_{\rm source}=\log Z$. At fixed level, the leading eleven-dimensional radius and circle grow with rank. In the separate IIA window, both $\lambda=N/k\gg1$ and $g_s^2=4\pi\sqrt2\,\lambda^{5/2}/N^2\ll1$ are needed. A fixed-$k$ error estimate cannot simply be reused while $k$ grows.

Matching $a_k$ does not yet match $B$, $A(k)$, or exponentially suppressed sectors. Nor does matching a Wilson loop establish an unrelated sphere free energy or an unprotected spectrum. A useful comparison identifies the same object, a common regime, a specified order, and a residual that the numerical calculation can resolve.

Exercises

1. What precision does extrapolation require?

Suppose $W_0$ is known exactly and each input ratio $W_N/W_0$, $W_{2N}/W_0$ has absolute numerical error at most $\eta$, with no assumption about their signs or correlation. Bound the numerical error in (X04.3). For $N=100$ and a desired error below $10^{-6}$, how small must $\eta$ be? Does the extrapolation itself bound the remaining physical truncation?

Hint 1

The residual at $2N$ multiplies its input error by $(2N)^2$.

Hint 2

Use the triangle inequality on the weighted difference; cancellation of an asymptotic coefficient does not imply cancellation of arbitrary numerical errors.

Solution

The worst-case propagated error is $[4(2N)^2+N^2]\eta/3=17N^2\eta/3$. At $N=100$, requiring this to be below $10^{-6}$ gives $\eta<3\times10^{-6}/170000\simeq1.765\times10^{-11}$. This bounds only errors satisfying the stated input assumptions. It says nothing about the unknown $-j/(4N^4)$ bias, a wrong assumed power law, or an inaccurate shared $W_0$. A common error in the denominator must be propagated separately. Numerical precision and asymptotic control answer different questions.

2. Which change can a rank difference detect?

At fixed $k$, multiply every $Z_N$ by a positive rank-independent factor $e^\alpha$. Separately consider $Z_N\mapsto e^{\beta N}Z_N$, with real $\beta$. Calculate how $F$, $D_N$ and $\mathcal B_N$ change. Which modification is invisible to (X04.6), and does that make it irrelevant to the absolute sphere partition function?

Hint 1

Take the logarithm before the finite difference. Keep the minus sign in $F=-\log Z$.

Hint 2

A constant shift in $D_N$ is divided by $\sqrt{N+1}-\sqrt N$ in the diagnostic.

Solution

The first change gives $F\mapsto F-\alpha$, while $D_N$ and $\mathcal B_N$ remain unchanged. It changes the absolute quantity and its constant term, even though this diagnostic cannot see it. In the second case $F\mapsto F-\beta N$ and $D_N\mapsto D_N-\beta$. Therefore

$$\delta\mathcal B_N=\frac{2\beta}{3a_k(\sqrt{N+1}-\sqrt N)} \sim\frac{4\beta}{3a_k}\sqrt N.$$

A rank-dependent normalization of this kind destroys the inferred finite limit. The two changes are diagnostic models; neither is an authorized redefinition of the fixed ABJM sphere prescription. To test $A(k)$, one must also compare an absolutely normalized quantity, not only adjacent-rank ratios.

3. Can an all-orders inverse-rank test establish equality?

For a positive model $Z_N^{(0)}$, define $Z_N^{(1)}=Z_N^{(0)}[1+\epsilon e^{-c\sqrt N}]$ with $c>0$ and $0<\epsilon<1$. Show that the two free energies have the same expansion in every inverse power of $\sqrt N$, yet are unequal at every finite positive $N$. Bound their free-energy difference. How does that difference affect the rank-shift diagnostic at leading exponential order?

Hint 1

For $x>0$, $x/(1+x)\leq\log(1+x)\leq x$. An exponential beats every fixed power.

Hint 2

Write $\delta F\simeq-\epsilon e^{-c\sqrt N}$ and take its adjacent-rank difference before dividing by $\Delta\sqrt N$.

Solution

With $x_N=\epsilon e^{-c\sqrt N}$, $\delta F=-\log(1+x_N)<0$ and $x_N/(1+x_N)\leq|\delta F|\leq x_N$. For every fixed $p$, $N^p x_N\to0$, so no coefficient in an inverse-rank expansion distinguishes the two functions. At finite rank the logarithm is nonzero.

Their adjacent-rank difference is $\delta D_N=\epsilon c e^{-c\sqrt N}\Delta\sqrt N\,[1+O(N^{-1/2})+O(e^{-c\sqrt N})]$. Consequently $\delta\mathcal B_N=-2\epsilon c e^{-c\sqrt N}/(3a_k)$ at leading exponential order. It still vanishes faster than every power. The positive toy multiplier is not a claimed ABJM instanton coefficient; it isolates why agreement to all perturbative orders cannot establish exact equality. Resolving such a sector requires its own observable, accuracy and controlled remainder analysis.