The planar limit and strong coupling

A large number of colors and a large ‘t Hooft coupling are different limits. We will first take the color number to infinity at fixed coupling, obtaining a Wilson-loop function valid at both weak and strong coupling. Only then will we approximate that function at large coupling.

Use the finite-rank loop, color counting, and normalized eigenvalue densities. The observable remains the normalized fundamental half-BPS circle in massless four-dimensional N=4 SYM. The flat parent has sixteen real Poincaré supercharges, together with conformal supercharges; taking a limit does not change that field content. The physical target is $SU(N)$, whose gauge rank is $N-1$, with an explicitly labeled $U(N)$ Gaussian comparison.

Keep the inherited loop prescription and ordinary fundamental trace. With $g=g_{\rm YM}$,

$$\lambda=g^2N,\qquad M=2\pi\ell a,\qquad W_{U(N)}= \frac{\int dM\,e^{-2N\operatorname{Tr}M^2/\lambda} \,N^{-1}\operatorname{Tr}e^M} {\int dM\,e^{-2N\operatorname{Tr}M^2/\lambda}}. \tag{N16.1}$$

$M$ and $\lambda$ are dimensionless. Localization and the contour were established before this Gaussian calculation; color counting alone would not justify (N16.1).

Take the planar limit at fixed coupling

After diagonalization, the effective eigenvalue action is $2N\sum_i m_i^2/\lambda-2\sum_{i<j}\log|m_i-m_j|$. Apply the density method with $\int\rho(x)\,dx=1$. The force equation is

$$\operatorname{PV}\int\frac{\rho(y)}{x-y}\,dy=\frac{2x}{\lambda}.$$

The Gaussian solution from the prerequisite has width parameter $s=\lambda/4$:

$$\rho_0(x)=\frac{2}{\pi\lambda}\sqrt{\lambda-x^2}, \qquad |x|\leq\sqrt\lambda. \tag{N16.2}$$

It is zero outside this interval. Its integral is one because the area of the corresponding half-disc is $\pi\lambda/2$. Its resolvent $G(w)=2(w-\sqrt{w^2-\lambda})/\lambda$, with $\sqrt{w^2-\lambda}\sim w$ at infinity, has real boundary value $2x/\lambda$ on the cut. This verifies the force as well as normalization. The density is the large-$N$ limit, not the exact finite-rank density or a list of most probable eigenvalues.

Write $z=\sqrt\lambda>0$. The limiting loop is

$$W_0(\lambda)=\int_{-z}^z\rho_0(x)e^x\,dx =\frac2\pi\int_0^\pi e^{z\cos\theta}\sin^2\theta\,d\theta. \tag{N16.3}$$

The substitution is $x=z\cos\theta$; both the square root and the Jacobian supply a sine. To evaluate the integral without assuming special-function identities, put $x=zt$ and expand the exponential. Odd powers integrate to zero. The even moments follow from $u=t^2$:

$$\begin{aligned} \frac2\pi\int_{-1}^1 t^{2n}\sqrt{1-t^2}\,dt &=\frac{2\,\Gamma(n+\tfrac12)\Gamma(\tfrac32)} {\pi\,\Gamma(n+2)}\\ &=\frac{(2n)!}{4^n n!(n+1)!}. \end{aligned}$$

Here $\Gamma$ is the factorial extension, $\Gamma(n+1)=n!$. Dividing each moment by the exponential’s $(2n)!$ gives

$$\boxed{ W_0(\lambda)=\sum_{n=0}^\infty \frac{(\lambda/4)^n}{n!(n+1)!} =\frac{2I_1(\sqrt\lambda)}{\sqrt\lambda} =1+\frac{\lambda}{8}+\frac{\lambda^2}{192} +\frac{\lambda^3}{9216}+\cdots .} \tag{N16.4}$$

For integer $\nu\geq0$, the modified Bessel function used here is defined by $I_\nu(z)=\sum_{j\geq0}(z/2)^{2j+\nu}/[j!(j+\nu)!]$. Thus the last equality introduces a name for an already evaluated integral. The ratio of consecutive terms proves convergence for every finite $\lambda$. At $\lambda=0$, the continuous answer is one.

There is a useful check on taking this limit inside an unbounded exponential observable. The exact finite-rank sum can be written

$$W_{U(N)}= e^{\lambda/(8N)} \sum_{n=0}^{N-1} \frac{(\lambda/4)^n}{n!(n+1)!} \prod_{j=0}^{n}\left(1-\frac jN\right). \tag{N16.5}$$

For fixed $\lambda\geq0$, every product lies between zero and one and tends to one at fixed $n$. The convergent series in (N16.4) therefore bounds the sum, allowing its limit to be taken term by term. This independently establishes the result without assuming that weak convergence of densities controls every unbounded insertion.

The exact central-mode relation is $W_{SU(N)}=e^{-\lambda/(8N^2)}W_{U(N)}$. It tends to the same $W_0$ at fixed $\lambda$. The equality of planar answers does not remove this finite-rank distinction. These formulas reproduce Drukker–Gross (2.26)–(2.27).

Resolve the endpoint at strong coupling

Now take $\lambda\to+\infty$ in the planar answer. The insertion $e^x$ selects eigenvalues near the right endpoint $x=z$. Set $x=z-u$ in (N16.3):

$$W_0= \frac{2\sqrt2}{\pi}\frac{e^z}{z^{3/2}} \int_0^{2z}du\,e^{-u}u^{1/2} \sqrt{1-\frac{u}{2z}}. \tag{N16.6}$$

The important distance $u$ is of order one, while the support width is of order $z$. Expanding the last square root at fixed $u$ gives $1-u/(4z)-u^2/(32z^2)+\cdots$. Integrating powers uses $\int_0^\infty e^{-u}u^{j+1/2}du=\Gamma(j+3/2)$. Consequently,

$$\boxed{ W_0(\lambda)\sim \sqrt{\frac2\pi}\frac{e^{\sqrt\lambda}}{\lambda^{3/4}} \left[ 1-\frac{3}{8\sqrt\lambda} -\frac{15}{128\lambda} +O(\lambda^{-3/2}) \right].} \tag{N16.7}$$

This is an asymptotic expansion at fixed truncation order, unlike the convergent weak-coupling series. To justify the endpoint expansion, split the integral at $u=\varepsilon z$ for fixed $0<\varepsilon<2$. In the first region the square-root Taylor remainder is bounded by a constant times the next power of $u/z$; its integral has the indicated order. The remaining region is exponentially suppressed by $e^{-\varepsilon z}$. This explains both the approximation and its domain.

The power $\lambda^{-3/4}$ comes from the square-root edge density and the width of the endpoint region. Retaining only $e^{\sqrt\lambda}$ does not approximate $W_0$ with small relative error: its ratio to $W_0$ grows as $\sqrt{\pi/2}\lambda^{3/4}$. It does reproduce the leading term in the logarithm:

$$\log W_0=\sqrt\lambda-\frac34\log\lambda +\frac12\log\frac2\pi-\frac{3}{8\sqrt\lambda} +O(\lambda^{-1}). \tag{N16.8}$$

The endpoint calculation recovers the planar part of Drukker–Gross (3.5), including its first correction.

Derive the first finite-rank correction

The $N^{-2}$ topology from color counting predicts a power, not its coupling-dependent coefficient. We can obtain that coefficient from the exact answer. The finite polynomial $y(x)=L_{N-1}^{1}(x)$ obeys $xy''+(2-x)y'+(N-1)y=0$, verified by matching neighboring powers in its defining finite sum. Substitute $x=-\lambda/(4N)$ and $W=e^{\lambda/(8N)}y/N$ to get

$$\lambda W''+2W' =\left(\frac14+\frac{\lambda}{64N^2}\right)W, \qquad W(0)=1. \tag{N16.9}$$

Primes here mean $\lambda$ derivatives. A regular power series also has $W'(0)=1/8$. Write $W=\sum_{n\geq0}B_n(\lambda/4)^n/[n!(n+1)!]$. Matching coefficients in (N16.9) yields

$$B_0=B_1=1,\qquad B_{n+1}=B_n+\frac{n(n+1)}{4N^2}B_{n-1}.$$

At first order in $N^{-2}$, sum the increments: $B_n=1+n(n^2-1)/(12N^2)+O(N^{-4})$. Inserting this in the series and using $n(n^2-1)/[n!(n+1)!]=1/[(n-2)!n!]$ resums the correction:

$$\begin{aligned} W_{U(N)}&=W_0+\frac{\lambda}{48N^2}I_2(z) +O_\lambda(N^{-4}),\\ W_{SU(N)}&=W_0+\frac1{N^2} \left[\frac{\lambda}{48}I_2(z)-\frac{\lambda}{8}W_0\right] +O_\lambda(N^{-4}). \end{aligned} \tag{N16.10}$$

The subscript on $O_\lambda$ means $\lambda$ is fixed; it is not a bound uniform in arbitrarily growing coupling. This derives the first term of Drukker–Gross (A.10)–(A.11) and (B.2).

For fixed integer $\nu$, the defining series also gives $I_\nu(z)=\pi^{-1}\int_0^\pi e^{z\cos\theta}\cos(\nu\theta)\,d\theta$. Near its dominant endpoint $\theta=0$, every fixed $\cos(\nu\theta)$ tends to one. The common Gaussian endpoint therefore gives $I_2(z)/I_1(z)\to1$ as $z\to\infty$. The leading relative correction for the $U(N)$ benchmark is

$$\frac{W_{U(N)}-W_0}{W_0} =\frac{\lambda^{3/2}}{96N^2}\frac{I_2(z)}{I_1(z)} +O_\lambda(N^{-4}). \tag{N16.11}$$

For $SU(N)$ subtract $\lambda/(8N^2)$ at this order. The strong-coupling expansion and the genus expansion thus test different small quantities:

Approximation	Leading diagnostic
Keep only the leading endpoint prefactor	$1/\sqrt\lambda$
Keep only the planar expectation value	$\lambda^{3/2}/N^2$ at large $\lambda$
Drop the central Gaussian	$\lambda/N^2$

A useful parametric window for the first two is $1\ll\lambda\ll N^{4/3}$. This diagnoses the hierarchy of asymptotic terms; it is not a rigorous uniform error theorem obtained from one coefficient. The exact finite-rank answer remains available to test a chosen pair $(N,\lambda)$.

Worked example: separate two error budgets

At $\lambda=100$, the exact planar value is $W_0\approx534.1976607403$. Define a relative error as “approximation divided by the stated exact answer, minus one.” The leading expression in (N16.7) has a $4.0358\%$ error relative to $W_0$. Including $-3/(8\sqrt\lambda)$ reduces it to $0.13448\%$; including the displayed $-15/(128\lambda)$ reduces it to $0.012561\%$.

Finite-rank effects are a separate comparison:

$N$	$W_{U(N)}/W_0-1$	$W_{SU(N)}/W_0-1$
10	$0.09242484$	$-0.03593846$
50	$0.00356451$	$-0.00144079$
100	$0.00089011$	$-0.00036022$

These numbers use the finite polynomial, not a discretized saddle. At $N=50$, (N16.10) predicts $0.00355911$ and $-0.00144089$, respectively. Increasing the rank improves this comparison without improving the strong-coupling truncation at fixed $\lambda$.

An independent numerical check represents the semicircle by a half-line tridiagonal operator with adjacent entries $z/2$. Its vacuum return walks count the same even moments. A finite section of size $D$ misses no return walk of length below $2D$; its exponential’s missing contribution is bounded by $\sum_{j\geq2D}z^j/j!$. Diagonalizing that finite operator reproduces (N16.3) with an independently controlled cutoff. Here $D$ is a numerical truncation, not the gauge color number $N$.

Keep the order of limits explicit

The statement “$\log W$ has leading term $\sqrt\lambda$” asks for less than a small relative error in $W$. The correction to the logarithm suggested by the large-coupling genus expansion is $\lambda^{3/2}/(96N^2)$. Comparing this with $\sqrt\lambda$ gives $\lambda/N^2\ll1$, the weaker scale appearing in Drukker–Gross (3.6) and its following discussion. Even a subleading correction to a logarithm can exponentiate to a large multiplicative correction. The two criteria address different questions; neither licenses unrestricted exchange of the limits.

For a direct countercheck, hold an integer $N\geq2$ fixed and take $\lambda\to\infty$. The highest term of the Laguerre polynomial gives

$$\begin{aligned} W_{U(N)}&\sim \frac{(\lambda/4N)^{N-1}}{N!}\,e^{\lambda/(8N)},\\ W_{SU(N)}&\sim \frac{(\lambda/4N)^{N-1}}{N!}\, e^{\lambda(N-1)/(8N^2)}. \end{aligned} \tag{N16.12}$$

The leading logarithms now grow linearly in $\lambda$, not as $\sqrt\lambda$. This is a different limiting sequence. Likewise, a planar limit at fixed $\lambda$ sends $g^2=\lambda/N$ to zero, but it need not make the collective parameter $\lambda$ small. “Planar,” “weak coupling,” and “strong coupling” are separate statements.

Exercises

1. Bound the endpoint approximation

Define $A(z)=W_0(z^2)/[\sqrt{2/\pi}\,e^z z^{-3/2}]$. Use an inequality for $\sqrt{1-v}$ on $0\leq v\leq1$ to give finite-$z$ lower and upper bounds on $A(z)$ as one-dimensional integrals. Derive the next term in (N16.8). Does knowing $\log W_0/z\to1$ imply $W_0/e^z\to1$?

Hint 1

The chord and tangent bounds are $1-v\leq\sqrt{1-v}\leq1-v/2$. Keep the original endpoint $u=2z$ when integrating these inequalities.

Hint 2

For the logarithm, expand $\log(1+a/z+b/z^2)$ through order $z^{-2}$, including the square of the first correction.

Solution

Equation (N16.6) gives $A(z)=\Gamma(3/2)^{-1}\int_0^{2z}e^{-u}u^{1/2} \sqrt{1-u/(2z)}\,du$. Writing $\gamma(p,x)=\int_0^x e^{-u}u^{p-1}du$, the pointwise inequalities give the exact bounds

$$\frac{\gamma(\tfrac32,2z)-\gamma(\tfrac52,2z)/(2z)} {\Gamma(\tfrac32)} \leq A(z)\leq \frac{\gamma(\tfrac32,2z)-\gamma(\tfrac52,2z)/(4z)} {\Gamma(\tfrac32)}.$$

They hold for every $z>0$; the lower integral is nonnegative because its integrand is. For large $z$ they approach $1-3/(4z)$ and $1-3/(8z)$ up to exponentially small tails. These are bounds, not equal asymptotic coefficients.

The sharper expansion has $a=-3/8$ and $b=-15/128$. Thus $b-a^2/2=-3/16$, and

$$\log W_0=z-\frac32\log z+\frac12\log\frac2\pi -\frac{3}{8z}-\frac{3}{16z^2}+O(z^{-3}).$$

The logarithm divided by $z$ tends to one, whereas $W_0/e^z\sim\sqrt{2/\pi}\,z^{-3/2}\to0$. A relative statement about logarithms does not become the same relative statement after exponentiation.

2. Design an accuracy comparison

At $N=100$ and $\lambda=400$, estimate separately the error from using only the leading endpoint formula and the first relative rank corrections for $U(N)$ and $SU(N)$. Is the leading endpoint formula plausibly a one-percent approximation to the planar answer? What can improve that error without changing $N$? State what must still be checked before claiming a one-percent approximation to the finite-rank observable.

Hint 1

Here $z=20$. The first omitted endpoint term is $-3/(8z)$. The finite-rank comparison uses $I_2(z)/I_1(z)$, and the central term is $\lambda/(8N^2)$.

Hint 2

After including the first endpoint correction, the next relative term has size approximately $15/(128z^2)$. Error estimates with different denominators should not be silently added as exact equalities.

Solution

The leading endpoint formula misses an approximately $1.875\%$ correction; its actual error relative to $W_0$ is about $1.94270\%$. It fails the proposed one-percent criterion. Including the first correction gives an actual planar error of about $0.031275\%$, consistent with the next-term estimate $15/(128\cdot400)\approx0.029297\%$. No change in $N$ was needed.

The first relative rank corrections are

$$\delta_U=\frac{400^{3/2}}{96\cdot100^2} \frac{I_2(20)}{I_1(20)} \approx0.00771656,\qquad \delta_{SU}=\delta_U-\frac{400}{8\cdot100^2} \approx0.00271656.$$

They refer to the finite-rank answer divided by the planar answer, minus one. In particular the central subtraction is $0.005$, large enough to matter at the requested precision. The exact $U(N)$ ratio is approximately $1.00774397$, corroborating the estimate here.

For a finite-rank accuracy claim, combine the chosen approximations with their correct ratios and check the remaining genus terms, for example against the exact finite polynomial. Opposite-signed first corrections can cancel at a selected input; that cancellation alone does not bound omitted terms. Raising $N$ controls rank corrections, while retaining another endpoint term controls the separate strong-coupling truncation.

3. Increase the winding while holding the theory fixed

Replace the insertion by $N^{-1}\operatorname{Tr}e^{pM}$ for a positive integer winding $p$. Derive its planar answer and its leading large-coupling rank diagnostic. Can a rank that approximates the singly wound circle well be inadequate for a multiply wound one? Evaluate this question for $N=100,\lambda=400$ and $p=1,2,4$. This is one wound trace, not a product of $p$ separate traces.

Hint 1

Either replace $z$ by $pz$ in the density integral or rescale the entire Gaussian matrix by $p$. The effective argument is $p^2\lambda$.

Hint 2

The relative genus-one diagnostic grows as the three-halves power of that argument. The central factor changes as its first power.

Solution

The density remains that of the original theory; only the insertion changes. Substitution in (N16.3) gives

$$W_{0,p}(\lambda)=\frac{2I_1(p\sqrt\lambda)}{p\sqrt\lambda}.$$

Equivalently, the exact finite-rank answer is $W_{U(N)}(p^2\lambda)$, with central correction $e^{-p^2\lambda/(8N^2)}$ for $SU(N)$. This agrees with Drukker–Gross (4.1)–(4.2).

For $p\sqrt\lambda\gg1$, the leading relative $U(N)$ rank term is $p^3\lambda^{3/2}/(96N^2)$ times $I_2(p\sqrt\lambda)/I_1(p\sqrt\lambda)$. Its parametric control requires $p\ll N^{2/3}/\sqrt\lambda$, up to the displayed numerical coefficient. The central exponent is instead $p^2\lambda/(8N^2)$. A result proved with fixed winding cannot be assumed uniform in a growing winding.

For the specified inputs, evaluating the finite polynomials gives

$$\frac{W_{U(N),p}}{W_{0,p}}-1 \approx \begin{cases} 0.00774397,&p=1,\\ 0.06619345,&p=2,\\ 0.68269941,&p=4. \end{cases}$$

The fundamental result is within one percent of its planar value, while winding four differs by about $68\%$. The exact result, rather than a truncated genus series, was used for these comparisons. Increasing winding actually improves the endpoint expansion parameter $1/(p\sqrt\lambda)$ while worsening the rank expansion. This is a concrete example of two approximation errors moving in opposite directions.

The planar circle is therefore an exact function of the fixed ‘t Hooft coupling, with a separately derived strong-coupling expansion. Before using an exponential growth law, specify whether the target is the expectation value, its logarithm, or only the logarithm’s leading coefficient—and which parameters remain fixed.

Compare exact, planar, and gravitational predictions builds a rank-extrapolation test from this circle and contrasts it with a fixed-level ABJM observable.