Two different large-N limits

An ABJM free energy can grow as $N^{3/2}$ or as $N^2$ without a contradiction. The two statements compare different sequences of theories. More subtly, substituting one sequence into the leading term of another can discard a contribution that remains important at every rank.

We will make that failure quantitative using an actual ABJM correction. The preparation is the exact rank-one partition function, what a large-rank limit holds fixed, and saddle points. We will use one explicitly sourced asymptotic input before deriving its consequences. Deriving that input from the matrix integral is the purpose of the subsequent Fermi-gas and density-of-states lessons.

Fix the theory before choosing the limit

Keep the compact three-dimensional theory $U(N)_k\times U(N)_{-k}$ with positive integer $k$, equal ranks, and no additional quotient of its centers. Each factor has rank $N$; the product has rank $2N$. In all formulas below $N$ means the rank of one factor. Replacing it by the total rank would change numerical coefficients.

The generic full theory has twelve real Poincaré supercharges; levels one and two have the established enhancement to sixteen. The sphere calculation uses its manifest $\mathcal N=2$ subalgebra and the complete round background already specified in the prerequisite. We keep zero real masses, zero FI deformation, and the inherited absolute measure and phase prescription. The real sphere free energy is

$$F(N,k)=-\log Z(N,k). \tag{AB17.1}$$

This is a sphere observable, not the thermal free energy on flat space. The radius has disappeared after using dimensionless variables. Neither changing rank nor changing level is a change of radius.

Define $\lambda=N/k$. The two standard limits are:

Limit	Sequence	Coupling behavior
Fixed level	$N\to\infty$, $k=k_0\in\mathbb Z_{>0}$	$\lambda\to\infty$
‘t Hooft, or planar	$N,k\to\infty$, $N/k\to\lambda_0>0$	$\lambda_0$ remains a parameter

For example, $(N,k)=(n,2)$ and $(N,k)=(2n,n)$ have very different limits. Integer levels are part of the family, not a regulator to discard. At fixed rational $\lambda_0=p/q$, the sequence $(pn,qn)$ realizes the ratio exactly. As explained in AB, section 2.2, perturbative weakness requires $\lambda\ll1$. A large integer $k$ by itself does not ensure it.

“Planar” organizes the leading color contractions. Schematically $F/N^2$ tends to a function $f_0(\lambda)$; nonplanar terms are lower order in rank at fixed coupling. The function $f_0$ need not be approximated by its weak-coupling series. Strong coupling, $\lambda\gg1$, is a separate limit of that function. Possible logarithmic normalization terms below order $N^2$ do not change this distinction.

An asymptotic input and the saddle it determines

It is useful to collect the partition functions for different ranks into a generating function. Introduce an auxiliary positive weight $z=e^\mu$ and set $Z(0,k)=1$:

$$\Xi(z,k)=\sum_{n=0}^{\infty} Z(n,k)z^n, \qquad J(\mu,k)=\log\Xi(e^\mu,k). \tag{AB17.2}$$

The dimensionless variable $\mu$ is conjugate to rank. It is not a real mass or chemical potential added to the original three-dimensional action. In a statistical reformulation it will become a chemical potential for auxiliary fermions. Here it is enough to understand coefficient extraction: integrating $z^{n-N-1}$ once around the origin gives zero except when $n=N$.

Thus, on a circle of radius $e^{\mu_*}$ where the generating function is analytic,

$$Z(N,k)=\frac1{2\pi i}\oint\frac{\Xi(z,k)}{z^{N+1}}\,dz =\frac1{2\pi}\int_{-\pi}^{\pi} \exp[J(\mu_*+it,k)-N(\mu_*+it)]\,dt. \tag{AB17.3}$$

The exponential, rather than a globally single-valued logarithm, is what the contour requires.

The needed physical input is the large-positive-$\mu$ polynomial part

$$J_{\rm pol}(\mu,k)=\frac{C(k)}3\mu^3+B(k)\mu+A(k), \qquad C(k)=\frac{2}{\pi^2k},\quad B(k)=\frac{k}{24}+\frac1{3k}. \tag{AB17.4}$$

Here $A(k)$ is independent of $\mu$. Marino–Putrov obtain the cubic coefficient in (4.25)–(4.27) and its polynomial corrections in (5.65) from the spectrum of the matrix integral. We use that result in this lesson. We have not derived it from the action coefficient, nor replaced the full generating function by a polynomial. Exponentially small terms and the correct continuation around the contour matter beyond the algebraic saddle expansion. The source explicitly discusses the finite contour in (5.84)–(5.86).

First retain only $C\mu^3/3$. Stationarity of the exponent gives $C\mu_*^2=N$, hence the positive saddle and its exponent are

$$\mu_* =\sqrt{N/C}=\pi\sqrt{kN/2},\qquad J(\mu_*)-N\mu_*=-\frac{2N^{3/2}}{3\sqrt C}. \tag{AB17.5}$$

Because our free energy is minus the logarithm, this yields

$$\boxed{F(N,k)\sim\frac{\pi\sqrt{2k}}3N^{3/2}, \qquad N\to\infty,\quad k>0\ \text{fixed}.} \tag{AB17.6}$$

FG calls $\log Z$ its free energy, so its displayed (4.27) has the opposite sign. A positive cubic $J$ does not make $\log Z$ positive: coefficient extraction subtracts $N\mu_*$.

There is also a local stability check. Put $\mu=\mu_*+it$ in the cubic exponent. Exactly,

$$J_{\rm pol}(\mu_*+it)-N(\mu_*+it) =-\frac{2N^{3/2}}{3\sqrt C} -C\mu_*t^2-\frac{iC}3t^3 \tag{AB17.7}$$

when $A=B=0$. The quadratic term decreases along the vertical contour; it would increase along the real direction. The Gaussian variance of $t$ is $1/(2C\mu_*)$, so its width relative to $\mu_*$ is of order $(C\mu_*^3)^{-1/2}$. At fixed $k$, $\mu_*\propto N^{1/2}$, the absolute width shrinks as $N^{-1/4}$, and the relative width shrinks as $N^{-3/4}$. This checks a genuine saddle, while the physical justification for discarding the rest of the exact contour remains part of the sourced asymptotics.

In the subsequent spectral representation the auxiliary Planck constant is $\hbar=2\pi k$, by FG (3.28)–(3.29). Fixed positive integer $k$ is not a small-$\hbar$ limit. The relevant high-energy comparison is instead $\hbar/\mu_*\sim2\sqrt{2k/N}$, which tends to zero at fixed level. At fixed $\lambda$ it becomes $2\sqrt{2/\lambda}$ and requires strong coupling to become small. This diagnostic explains why a large-$N$ limit and a semiclassical expansion of the auxiliary problem cannot be identified without specifying the path.

A correction changes its status when the level grows

Now retain the linear term. Define $n_*=N-B(k)$ and assume $n_*>0$. The polynomial saddle gives

$$\mu_* =\sqrt{n_*/C},\qquad F_{\rm saddle}=\frac{2}{3\sqrt C}n_*^{3/2}-A(k). \tag{AB17.8}$$

Neither $n_*$ nor $N-B$ is a new gauge rank. The rank stays the integer $N$. This is a shift in an asymptotic formula.

The relative change in its first term, compared with (AB17.6), is controlled by

$$\frac{(N-B)^{3/2}}{N^{3/2}} =\left(1-\frac BN\right)^{3/2} =1-\frac{3B}{2N}+O((B/N)^2). \tag{AB17.9}$$

At fixed $k$, $B/N\to0$. The corresponding absolute correction to $F$ is of order $N^{1/2}$, smaller than $N^{3/2}$. The Gaussian prefactor adds a logarithmic correction: within this saddle approximation, $Z\simeq e^{-F_{\rm saddle}}/[2\sqrt{\pi C\mu_*}]$. It does not remove the rank shift.

At fixed $\lambda$, however,

$$\frac{B(k)}N=\frac1{24\lambda}+\frac{\lambda}{3N^2}. \tag{AB17.10}$$

The first term survives at infinite rank. For the diagnostic ratio in (AB17.9), level four has $B=1/4$, while the sequence $N/k=2$ has a nonzero limiting shift:

Sequence	$B/N$	$(1-B/N)^{3/2}$
$N=100$, $k=4$	$0.0025$	$0.996252345$
$N=400$, $k=4$	$0.000625$	$0.999062647$
$N,k\to\infty$, $N/k=2$	$1/48$	$0.96891333$

These are values of a specified correction, not numerical evaluations of the complete ABJM free energy at those ranks. In particular the table is not a claim that a strong-coupling approximation is accurate at $\lambda=2$.

The constant $A(k)$ cannot simply be called “order one” in both limits either. It is constant in $N$ only when $k$ stays fixed. Nor can a small-$k$ expansion for it be substituted at $k=N/\lambda\to\infty$.

What actually agrees with the planar strong-coupling answer

Drukker–Marino–Putrov (5.7), (5.34)–(5.37) obtain the planar answer independently. On the equal-rank slice define $\widehat\lambda=\lambda-1/24$. Translating their $g_s=2\pi i/k$ and their free-energy sign, the displayed algebraic terms are

$$\left.F(N,k)\right|_{\rm planar,strong} =k^2\left[ \frac{\pi\sqrt2}{3}\widehat\lambda^{3/2} +\frac{\zeta(3)}{8\pi^2} +\text{terms exponentially small at large }\widehat\lambda \right]. \tag{AB17.11}$$

Here $\zeta(3)=\sum_{j\geq1}j^{-3}$. The notation first selects the leading planar answer, then expands it at strong coupling; lower orders in rank are not written. The exponential terms have scale $e^{-2\pi\sqrt{2\widehat\lambda}}$ with their own prefactors. Taking $N$ larger at a fixed moderate $\lambda$ does not suppress these relative to other planar contributions.

To check the normalization, the source’s leading planar function is $4\pi^3\sqrt2\,\widehat\lambda^{3/2}/3+\zeta(3)/2$. Its contribution to $\log Z$ is multiplied by $g_s^{-2}=-k^2/(4\pi^2)$; changing to $F=-\log Z$ gives (AB17.11). The constant term is present in the inspected v4, whose footnote 7 records its omission in an earlier version.

Consequently the leading strong-coupling behavior of the planar function is

$$f_0(\lambda)\sim\frac{\pi\sqrt2}{3\sqrt\lambda}, \qquad \lambda\to\infty. \tag{AB17.12}$$

This agrees with the algebraic substitution $k=N/\lambda$ into (AB17.6), but the agreement is supported by a separate planar calculation. The fixed-level result alone did not justify changing the limit. Even within (AB17.11), the shift gives a relative correction $-1/(16\lambda)$, and the constant gives one of order $\lambda^{-3/2}$, irrespective of rank.

The labels “M-theory limit” and “string limit” often accompany the two sequences. They name dual descriptions; no gravity calculation was needed here. A weakly curved, weakly coupled type-IIA supergravity approximation imposes the extra inequalities $1\ll\lambda$ and $\lambda^{5/2}/N^2\ll1$, as in AB (4.7). A planar limit at arbitrary fixed $\lambda$ does not automatically satisfy the first.

Exercises

1. Test an intermediate sequence

Take the allowed integer sequence $N=n^4$, $k=n^2$ with $n\to\infty$. Find $\lambda$, $B/N$, the power of $n$ in the candidate leading free energy, and the relative Gaussian width $\delta t/\mu_*$. Is this either standard limit? Do these tests prove a uniform asymptotic formula along the sequence?

Hint 1

Keep both terms in $B=k/24+1/(3k)$. Use $C\propto1/k$ and $\mu_*\sim\sqrt{kN}$.

Hint 2

The width is $\delta t=(2C\mu_*)^{-1/2}$. Passing local tests is different from controlling every omitted sector.

Solution

Here $\lambda=n^2\to\infty$, while $k=n^2\to\infty$: neither $k$ nor $\lambda$ is fixed. The correction is

$$\frac BN=\frac1{24n^2}+\frac1{3n^6}.$$

The candidate leading term is $\pi\sqrt2\,n^7/3$; its shift changes it by a relative leading amount $-1/(16n^2)$. Also $C\propto n^{-2}$ and $\mu_*\propto n^3$, so $C\mu_*^3\propto n^7$ and $\delta t/\mu_*\propto n^{-7/2}$.

Both the rank shift and relative local width become small. These are necessary diagnostics for using this saddle, not a proof that all omitted coefficients, other contour regions and exponential sectors are uniformly negligible as $k$ grows. The fixed-level theorem has different hypotheses; an extension needs additional control or an appropriate planar strong-coupling analysis.

2. A constant is not constant along every path

Suppose a term in the free energy is $\Delta F=a k^2$ with nonzero constant $a$. Compare it with $F_0=\pi\sqrt{2k}N^{3/2}/3$ at fixed $k$ and at fixed $\lambda$. Apply the result to the constant term in (AB17.11). Does it change the coefficient of $N^2$? Does it change the leading large-$\lambda$ power?

Hint 1

Form $\Delta F/F_0$ before taking either limit.

Hint 2

At fixed $\lambda$, $k^2=N^2/\lambda^2$. “Leading in rank” need not mean “leading in coupling.”

Solution

The relative contribution is $3a(k/N)^{3/2}/(\pi\sqrt2)$. It vanishes as $N^{-3/2}$ at fixed level. At fixed $\lambda$ it is $3a/(\pi\sqrt2\lambda^{3/2})$, independent of rank.

For $a=\zeta(3)/(8\pi^2)$, (AB17.11) therefore contains a term $N^2\zeta(3)/(8\pi^2\lambda^2)$. It contributes to the planar function $f_0(\lambda)$, but is subleading relative to its $\lambda^{-1/2}$ behavior at large coupling. This is a real example of a term that cannot be dropped just because it was independent of $N$ on another parameter path. It does not assert that the complete function $A(k)$ equals a single quadratic term.

3. Check the cubic fluctuation rather than its stationary point alone

Consider only the defined cubic contour integral obtained by extending (AB17.7) to all real $t$, with $C,\mu_*>0$ and $N=C\mu_*^2$. Divide it by its leading Gaussian value and call the ratio $R$. Show, using a standard normal real variable $U$, that

$$R=\mathbb E[\cos(\alpha U^3)],\qquad \alpha=\frac1{3\,2^{3/2}\sqrt{\xi}},\qquad \xi=C\mu_*^3.$$

Bound $1-R$ and the error after its first correction. Explain which physical errors those bounds do not estimate.

Hint 1

Rescale $t=U/\sqrt{2C\mu_*}$. Odd sine terms integrate to zero. The Gaussian moments are $\mathbb E U^6=15$ and $\mathbb E U^{12}=10395$.

Hint 2

For every real $v$, $0\leq1-\cos v\leq v^2/2$ and $|\cos v-1+v^2/2|\leq v^4/24$.

Solution

The Gaussian value is $e^{-2N\mu_*/3}/[2\sqrt{\pi C\mu_*}]$. After rescaling, its normalized weight is $e^{-U^2/2}/\sqrt{2\pi}$ and its residual phase is $-i\alpha U^3$. Taking the even part proves the representation. Since $\alpha^2=1/(72\xi)$,

$$0\leq1-R\leq\frac5{48\xi},\qquad \left|R-\left(1-\frac5{48\xi}\right)\right| \leq\frac{385}{4608\xi^2}.$$

For example, at $\xi=10$ the leading Gaussian has at most $1/96\simeq0.01042$ absolute error in the ratio, and the first corrected ratio has error at most $385/460800\simeq0.000836$. A lower bound on $R$ converts these to relative bounds if needed.

These are global bounds for the explicitly defined cubic integral. They do not bound replacing the physical grand potential by its polynomial part, extending its finite contour, or taking $k$ to grow with $N$. A numerical quadrature has another independent error: restricting the normalized Gaussian integration to $|U|\leq L$ loses at most $\operatorname{erfc}(L/\sqrt2)$ in absolute value. More digits cannot remove an omitted physical contribution.

Turn the matrix integral into a Fermi gas now constructs the statistical representation from the matrix integral itself. Its useful question is now precise: what spectral property supplies $C(k)$, and which corrections remain small for the chosen sequence of ranks and levels?

The ABJM holographic dictionary expresses both sequences in geometric parameters. Large radius, a large M-theory circle and weak string coupling are separate tests.

The Fermi-gas paper companion follows these limits into the original source and tests a different sequence with particle number held fixed.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading ABJM as a Fermi gas