Large N and what a 't Hooft limit holds fixed

Increasing the size of a gauge group does not specify a limit of a physical observable. The coupling might decrease, remain fixed, or grow. A Chern–Simons level might stay at an integer or grow with the rank. Even the same matrix expression can acquire a different limit if its normalization changes.

We will specify these choices and then perform an exact finite-dimensional calculation that distinguishes a limiting answer from its corrections. Ordinary QFT, gauge theory, and Gaussian integration suffice; no supersymmetry or Lie-theory lesson is a prerequisite.

A limit is a sequence of theories and measurements

For $SU(N)$ or $U(N)$, $N$ is the dimension of the fundamental color space. Their mathematical ranks are $N-1$ and $N$, respectively. “Large rank” conventionally means taking this color parameter to infinity. It does not change the number of spacetime dimensions or the amount of supersymmetry.

Write a measured quantity as $\mathcal O_N(g,mR,\ldots)$, where $R$ is a specified length and $m$ a mass. A limit requires a sequence of allowed parameters, together with a prescription for comparing observables. Keeping $m$ fixed while changing $R$ does not keep $mR$ fixed. Changing the representation of a Wilson loop or letting the number of fields in an operator grow with $N$ likewise changes the problem.

We use the ordinary fundamental trace and its explicitly normalized counterpart:

$$\operatorname{Tr}_N I_N=N,\qquad \operatorname{tr}_N M=\frac1N\operatorname{Tr}_N M.$$

Replacing one by the other requires a compensating factor of $N$. The optional matrix-valued matter lesson explains their color-index meaning in more detail.

First calculation: keep the Yang–Mills coupling combination fixed

Consider four-dimensional $SU(N)$ $\mathcal N=4$ supersymmetric Yang–Mills theory, with sixteen real Poincaré supercharges and vanishing theta angle. Its supersymmetric interactions are part of the theory’s definition; only the gauge kinetic term is needed for this scaling calculation. Use flat Euclidean space, Hermitian generators $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$, and a connection with the coupling absorbed:

$$D_\mu=\partial_\mu-iA_\mu,\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-i[A_\mu,A_\nu].$$

Both $A_\mu$ and a derivative have mass dimension one. Define the dimensionless coupling $g_{\rm YM}$ by

$$S_{\rm gauge} =\frac1{2g_{\rm YM}^2}\int d^4x\, \operatorname{Tr}_N(F_{\mu\nu}F_{\mu\nu}). \tag{T15.1}$$

For comparison with canonical components, set $A_\mu=g_{\rm YM}A_{\mu,\rm can}$. Then $F_{\mu\nu}=g_{\rm YM}(\partial_\mu A_{\nu,\rm can}-\partial_\nu A_{\mu,\rm can}-ig_{\rm YM}[A_{\mu,\rm can},A_{\nu,\rm can}])$. Using the generator norm, (T15.1) has canonical component coefficient $1/4$. Its cubic and quartic gauge vertices carry $g_{\rm YM}$ and $g_{\rm YM}^2$.

The ‘t Hooft coupling in this theory is $\lambda_{\rm YM}=g_{\rm YM}^2N$. Holding it fixed gives the exact rewriting

$$g_{\rm YM}^2=\frac{\lambda_{\rm YM}}N,\qquad S_{\rm gauge} =\frac{N}{2\lambda_{\rm YM}}\int d^4x\,\operatorname{Tr}_N F^2 =\frac{N^2}{2\lambda_{\rm YM}}\int d^4x\,\operatorname{tr}_N F^2. \tag{T15.2}$$

The last two expressions are the same action. They do not imply two competing large-$N$ scalings.

Although $g_{\rm YM}\to0$, an internal color sum can compensate a power of the coupling: $g_{\rm YM}^2N=\lambda_{\rm YM}$ remains finite. Minahan’s discussion after (5.12) identifies this combination in planar perturbation theory. “Planar” refers to the leading color-contraction organization; its detailed diagrammatic derivation is a further method. A fixed value such as $\lambda_{\rm YM}=100$ is not a weak-coupling limit merely because $N$ is large.

Holding $g_{\rm YM}^2=g_0^2>0$ instead gives $\lambda_{\rm YM}=Ng_0^2\to\infty$. That is a different path through theories. For an observable expanded at fixed $\lambda_{\rm YM}$, taking the coupling large is an additional approximation whose order and error must be specified.

The integer level changes the three-dimensional question

Now consider equal-rank ABJM theory with compact gauge group $U(N)_k\times U(N)_{-k}$, positive integer $k$, and no extra global quotient. This three-dimensional theory has bifundamental matter and Chern–Simons gauge terms, with no defining Maxwell kinetic term. Its generic full supersymmetry has twelve real Poincaré supercharges. At $k=1,2$ the count increases to sixteen; the ABJM definition explains the enhancement and low-rank qualifications. These facts identify the family; no supersymmetry transformation is needed below.

With Hermitian connections $a,\widehat a$ and the ordinary trace, its Euclidean connection action contains

$$S_{{\rm CS},E} =\frac{ik}{4\pi}\int d^3x\,\epsilon^{\mu\nu\rho} \operatorname{Tr}_N \left(a_\mu\partial_\nu a_\rho-\frac{2i}{3}a_\mu a_\nu a_\rho -\widehat a_\mu\partial_\nu\widehat a_\rho +\frac{2i}{3}\widehat a_\mu\widehat a_\nu\widehat a_\rho\right). \tag{T15.3}$$

Here $\epsilon^{123}=+1$. The connection has dimension one, so the integrand has dimension three and $k$ is dimensionless. The compact theory requires the integer level and its global Chern–Simons definition; continuously varying $k$ at finite $N$ would leave this declared family. The opposite levels and fundamental-trace convention come from AB, sections 2.1–2.2.

Its useful coupling ratio is $\lambda_{\rm AB}=N/k$, not $g_{\rm YM}^2N$. The two symbols called “‘t Hooft coupling” belong to different theories. AB’s large-$N$ paragraph describes perturbative weakness when $k\gg N$, rather than when $k$ is merely a large integer.

Family of limits	Held fixed	What changes
Four-dimensional Yang–Mills ‘t Hooft limit	$\lambda_{\rm YM}>0$	$N\to\infty$, $g_{\rm YM}^2=\lambda_{\rm YM}/N$
ABJM ‘t Hooft limit	$\lambda_{\rm AB}>0$	$N,k\to\infty$ with $N/k\to\lambda_{\rm AB}$
ABJM fixed-level limit	Integer $k=k_0>0$	$N\to\infty$, $\lambda_{\rm AB}=N/k_0\to\infty$

FG, (1.3)–(1.4) distinguishes the last two limits explicitly. For rational $\lambda_{\rm AB}=p/q>0$, choose $N=pn$, $k=qn$ with positive integers $n$. For an irrational target, integer ratios can approach it; they cannot equal it at every finite rank.

There is also a direct action-coefficient check. Replacing the ordinary trace in (T15.3) by $\operatorname{tr}_N$ changes its prefactor to $ikN/(4\pi)$. This coefficient grows as $N$ at fixed $k$, and as $N^2$ at fixed $N/k$. The configurations and their fluctuations have not yet been integrated. Consequently this coefficient comparison alone does not determine the free-energy power: typical fields and the measure also depend on the limit.

The mass dimensions offer another safeguard. In four dimensions $g_{\rm YM}^2$ is dimensionless. A three-dimensional Yang–Mills coupling instead has $[g_3^2]=1$, so its dimensionless color-enhanced parameter at energy $E$ would be $g_3^2N/E$. Neither quantity can replace ABJM’s dimensionless integer level.

Worked finite-rank check: a normalized trace fluctuates less

To see what a finite-rank correction means without assuming a field-theory solution, define a Gaussian model on all Hermitian $N$-by-$N$ matrices:

$$Z_N(s)=\int dV(M)\, e^{-\frac{N}{2s}\operatorname{Tr}_N M^2}, \qquad O_N=\frac1N\operatorname{Tr}_N M^2,\qquad s>0. \tag{T15.4}$$

All entries and $s$ are dimensionless. The measure is induced by $ds_M^2=\operatorname{Tr}_N(dM\,dM)$. This is a defined finite-dimensional model, not an asserted localization formula for either field theory.

Choose an orthonormal Hermitian matrix basis $E_A$ with $\operatorname{Tr}_N(E_AE_B)=\delta_{AB}$. There are $d=N^2$ real coefficients in $M=\sum_Ax_AE_A$, and $\operatorname{Tr}M^2=\sum_Ax_A^2$, $dV=\prod_A dx_A$. Thus

$$Z_N(s)=\left(\frac{2\pi s}{N}\right)^{N^2/2}.$$

Every $x_A$ has variance $s/N$. More usefully, integrate with a source for the whole observable:

$$\begin{aligned} G_N(t)&=\left\langle e^{-tO_N}\right\rangle\\ &=\prod_{A=1}^{N^2} \frac{\int_{\mathbb R}dx_A\, e^{-[N/(2s)+t/N]x_A^2}} {\int_{\mathbb R}dx_A\,e^{-Nx_A^2/(2s)}}\\ &=\left(1+\frac{2st}{N^2}\right)^{-N^2/2}, \qquad t>-\frac{N^2}{2s}. \end{aligned} \tag{T15.5}$$

For this sign of the source, $-\partial_t\log G_N(0)$ is the mean and $\partial_t^2\log G_N(0)$ the variance. Therefore

$$\boxed{ \langle O_N\rangle=s,\qquad \operatorname{Var}(O_N)=\frac{2s^2}{N^2},\qquad \langle O_N^2\rangle=s^2\left(1+\frac2{N^2}\right). } \tag{T15.6}$$

The variance is the connected part $\langle O_N^2\rangle-\langle O_N\rangle^2$. It vanishes at fixed $s$, while the relative standard deviation is $\sqrt2/N$. For $s=3/2$:

$N$	Exact mean	Exact variance
$2$	$3/2$	$9/8$
$4$	$3/2$	$9/32$
$8$	$3/2$	$9/128$

This example demonstrates factorization for this particular observable. It does not prove factorization for every operator in an interacting gauge theory. No eigenvalue Jacobian or diagrammatic topology was needed; the matrix-integral lesson explains the alternative eigenvalue coordinates.

At fixed $s,t$, expanding the exact logarithm gives

$$\log G_N(t) =-st+\frac{s^2t^2}{N^2} -\frac{4s^3t^3}{3N^4}+O(N^{-6}). \tag{T15.7}$$

The exact formula and the truncated expansion are different statements. For $t\geq0$ and $2st/N^2\leq1$, the alternating logarithm series bounds the error after the $N^{-2}$ term by $4s^3t^3/(3N^4)$. This is an actual error criterion, not merely the instruction to choose a “large” rank.

If instead $s=Nh$ with fixed $h>0$, the same exact formulas give $\langle O_N\rangle=Nh$ and $\operatorname{Var}(O_N)=2h^2$. Relative fluctuations still shrink, but the observable does not approach a finite mean. Its normalization and the parameter path both matter.

A sourced asymptotic statement has a domain

An exact finite-$N$ formula includes all rank dependence within its stated model. A large-$N$ expansion approximates a specified sequence. A strong-coupling expansion concerns a coupling parameter and need not include finite-rank corrections. An approximation may require both conditions, separately.

For one physical benchmark, the undeformed supersymmetric round-sphere partition function of the stated ABJM theory has

$$F_{S^3}(N,k)\equiv-\log Z_{S^3}(N,k) \sim \frac{\pi\sqrt{2k}}3\,N^{3/2}, \qquad N\to\infty,\quad k>0\ \text{fixed}. \tag{T15.8}$$

This is the sourced leading result, not a derivation from (T15.3). FG, (4.27) and the following qualification give the fixed-level regime. Their free energy has the opposite sign, as also seen from their inverse transform and saddle exponent in (3.14)–(3.16).

Algebraically substituting $k=N/\lambda_{\rm AB}$ into the displayed leading expression produces $\pi\sqrt2\,N^2/(3\sqrt{\lambda_{\rm AB}})$. That substitution alone establishes no approximation at fixed $\lambda_{\rm AB}$: the omitted remainder was controlled at fixed $k$. A uniform estimate would have to remain valid over the growing range of levels being used. Exercise 3 shows how the inference can fail even for elementary positive functions.

The same care applies to changing an observable with rank. Minahan explicitly warns after (5.7) that the color expansion changes when the number of fields in a traced operator grows sufficiently large (PDF p. 12). A fixed-operator calculation is not automatically uniform in operator size.

Exercises

1. Remove one matrix direction

Repeat the Gaussian calculation on traceless Hermitian matrices $X$, with the same induced metric and the same coefficient $N/(2s)$. Find the exact mean and variance of $\operatorname{Tr}X^2/N$. Explain why the leading limit can agree with (T15.6) even though the finite-rank models differ.

Hint 1

There are $N^2-1$ orthonormal real directions. Alternatively write $M=x_0I_N/\sqrt N+X$.

Hint 2

The identity direction is orthogonal to every traceless matrix. Its Gaussian variable has variance $s/N$.

Solution

Replace the exponent $N^2/2$ in (T15.5) by $(N^2-1)/2$, leaving $2st/N^2$ unchanged. Differentiating gives

$$\langle O_N\rangle_{\rm traceless} =s\left(1-\frac1{N^2}\right),\qquad \operatorname{Var}_{\rm traceless}(O_N) =\frac{2s^2}{N^2}\left(1-\frac1{N^2}\right).$$

Independently, orthogonality gives $O_N(M)=O_N(X)+x_0^2/N$, with independent variables. The removed mean is $s/N^2$ and the removed variance is $2s^2/N^4$, exactly reproducing both differences. At $N=2$ and $s=3/2$, the traceless mean is $9/8$ and variance $27/32$.

Both models approach mean $s$ at fixed $s$, but their finite-rank answers are unequal. This is a comparison of Hermitian vector-space integrals, not a statement that the global gauge theories $SU(N)$ and $U(N)$ are identical.

2. Choose allowed levels before taking the limit

Give an integer sequence with $\lambda_{\rm AB}=3/2$ exactly. Give one approaching $\lambda_{\rm AB}=\sqrt2$, with an error bound. Finally classify $(N,k)=(n^3,n^2)$: is it either standard limit, and how do $\lambda_{\rm AB}$ and the normalized-trace coefficient $kN/(4\pi)$ scale?

Hint 1

For the rational target, multiply its numerator and denominator by the same positive integer. For the irrational target use the defining inequality for the floor function.

Hint 2

The condition $k\to\infty$ alone does not fix $N/k$. Keep the two growth rates.

Solution

Choose $(N,k)=(3n,2n)$ for the first target. For the second take $k=n$, $N=\lfloor\sqrt2\,n\rfloor$. Then

$$0\leq \sqrt2-\frac Nk<\frac1n.$$

Every member has integer positive level, while its ratio approaches the desired irrational value. Exact equality at finite integers is impossible.

For $(n^3,n^2)$, the ratio is $\lambda_{\rm AB}=n$ and the coefficient is $n^5/(4\pi)$. Both rank and level grow, but neither $k$ nor $N/k$ stays fixed. It is a third scaling path, entering large effective coupling even though $1/k\to0$. The power $n^5$ belongs to this action coefficient; it does not compute the free energy.

3. Test an unjustified substitution

Consider the explicitly defined positive function

$$\mathcal A(N,k) =C\sqrt k\,N^{3/2}\left(1+\frac{k}{N}\right), \qquad C>0.$$

Show that it has the same leading fixed-$k$ scaling as its first factor. Along an allowed sequence with $N/k=\lambda_0>0$ fixed, find its ratio to that factor. Does large $N$ remove the relative error? What additional limit would remove it?

Hint 1

Use the relative remainder $k/N$, rather than the absolute difference of two growing quantities.

Hint 2

Substitute the entire exact function along the new sequence, not only its first factor.

Solution

At any fixed $k$, the ratio is $1+k/N\to1$, so $\mathcal A\sim C\sqrt k\,N^{3/2}$. Along a fixed rational $\lambda_0$ sequence,

$$\mathcal A =\frac{C}{\sqrt{\lambda_0}}N^2 \left(1+\frac1{\lambda_0}\right).$$

The fractional discrepancy from the substituted first factor is $1/\lambda_0$, independent of $N$. For $\lambda_0=3/2$, it is $2/3$, however large the ranks become. Sending $\lambda_0\to\infty$ afterwards removes it. An integer sequence approaching an irrational $\lambda_0$ has the same limiting ratio.

This constructed function is not an ABJM prediction. It proves that knowledge of a fixed-level leading term alone is insufficient to justify a fixed-ratio approximation. Additional control of the physical remainder would be needed.

A useful large-rank statement now includes the theory, its allowed parameter sequence, the normalized observable, and the approximation’s domain. Those choices must remain visible when color diagrams or eigenvalue methods supply the next calculation.