Reading ABJM as a Fermi gas

Marcos Mariño and Pavel Putrov’s ABJM theory as a Fermi gas, arXiv:1110.4066v3, dated 14 March 2012, asks how an interacting matrix integral can become a tractable one-particle spectral problem. Its answer has two stages: an exact change of description, then an asymptotic analysis. Keeping those stages distinct is the main reading skill to practice here.

Begin with the two large-rank limits and the Fermi-gas transformation. The section on asymptotics additionally uses phase-space counting and the Airy resummation. You can complete the finite-rank reconstruction first.

We retain the undeformed round-sphere observable of compact three-dimensional $U(N)_k\times U(N)_{-k}$, with positive integer $k$, rank $N$ per factor and no further quotient. Its generic full supersymmetry is $\mathcal N=6$, with enhancement at $k=1,2$; the localization calculation uses the established $\mathcal N=2$ description. The auxiliary fermions below are neither the elementary fermions of ABJM nor states in a supersymmetric index. The absolute measure and Gaussian Abel prescription are those of the canonical matrix-integral derivation.

A reading order with questions to answer

Page numbers in this table are the paper’s printed numbers; the PDF counter is one larger. Equation numbers refer only to v3.

Passage	Reading question	Canonical calculation
Introduction (1.3)–(1.4), p. 3; Section 2, especially (2.1), p. 5	Which parameter is fixed, and which normalized integral is being transformed?	Specify the limit
Section 3.1, (3.1)–(3.5), pp. 9–10	Where do the exchange sign, the single remaining factorial and the kernel measure come from?	Cauchy identity and Fourier integrations
(3.6)–(3.16), pp. 10–11	Why is a cycle trace different from a fixed-particle partition function?	Occupation statistics
(3.17)–(3.34), pp. 12–13	What operator has the required eigenvalues, and why is its Hamiltonian not exactly a sum of two functions?	The one-particle operator
Sections 4.1–4.2, (4.1)–(4.27), pp. 17–20	How does a spectral count lead to a filled-state free energy?	Count and fill levels
Section 5.2, (5.25)–(5.37), pp. 25–27	Which three coefficients enter the cubic grand potential, and what information is lost on discarding its remainder?	Separate the three inputs and invert the cubic
Section 5.3, (5.38), p. 27, and (5.64)–(5.68), p. 31; footnote 3, p. 32	What is established in the WKB expansion, and which proposal for the constant term is withdrawn?	Quantum counting input and absolute normalization
Section 5.4, especially (5.74)–(5.80), p. 33	What lies beyond a power expansion in the auxiliary Planck constant?	The omitted exponential sectors

On a first pass, use the result of the quantum calculation and then return to Section 5.1’s Wigner expansion to study its mechanism. A Wigner transform represents an operator by a phase-space function; products of operators become a derivative-corrected product, not ordinary multiplication. Appendix A checks the next quantum order. Sections 3.2 and 6 extend the method to other matter contents and quivers; they are further reading, not assumptions about the particular ABJM operator used here.

Translate the notation before expanding

Paper notation or convention	Meaning used here
$Z_\ell$ in (3.7)	$t_\ell=\operatorname{Tr}\rho^\ell$, a one-particle spectral trace; it is generally not $Z(\ell,k)$
$Z(N)$	$Z(N,k)$ at the stated fixed level, with $Z(0,k)=1$
Statistical sign $\eta$ in (3.8)	$\eta=-1$ for this fermion gas
$q,p$ in the auxiliary problem	Dimensionless position and momentum; our position is $x$, and $[x,p]=i\hbar$, $\hbar=2\pi k$
$J=\log\Xi$	Same sign here; positive fugacity is $z=e^\mu$
Source free energy in (3.16)	$\log Z$ to leading saddle order; our $F=-\log Z$ has the opposite sign
$n(E)$ and its smooth expansion	Exact level counting is a staircase; the polynomial used in the WKB argument describes its smoothed asymptotic information

The density operator acts on $L^2(\mathbb R,dx)$ with kernel

$$\rho(x,y)=\frac{u(x)u(y)}{4\pi k} \operatorname{sech}\!\left(\frac{x-y}{2k}\right), \qquad u(x)=[2\cosh(x/2)]^{-1/2}. \tag{PC15.1}$$

The factor $1/(4\pi k)$ includes the factor two in the middle hyperbolic cosine of source (3.3). Its trace is $t_1=1/(4k)$, not one. In operator form it is $e^{-U(x)/2}e^{-T(p)}e^{-U(x)/2}$, with $U(v)=T(v)=\log[2\cosh(v/2)]$. Source (3.34) explicitly warns that $-\log\rho\ne U(x)+T(p)$ exactly. Likewise, the phase-space cell is $2\pi\hbar=4\pi^2k$.

These statements resolve two literal normalization slips in the asymptotic formulas: the first measure in printed (4.13) should agree with its own $\mathrm{Vol}/(2\pi\hbar)$, and the leading count in (5.64) is $2E^2/(\pi^2k)$, as required by (4.25) and (5.65). Differentiating (4.8) also requires a minus before the polylogarithm in (4.9). These are checks against the neighboring equations, not changes of physical convention.

Restore the two-cycle contribution

At particle number two, source (3.8) has only two cycle types. Two one-cycles have weight $t_1^2/2!$. One two-cycle has weight $-t_2/2$: the denominator two counts its cyclic relabelings, and the minus sign is fermionic exchange. Hence

$$Z(2,k)=\frac12(t_1^2-t_2). \tag{PC15.2}$$

The subtraction is essential even for independent particles. It removes two fermions occupying the same one-particle state. It does not represent a residual pair potential.

Here is a way to apply that formula which also avoids subtracting close numerical answers. For $d=x-y$, the diagonal product minus the exchanged product of (PC15.1) is proportional to $1-\operatorname{sech}^2(d/(2k))=\tanh^2(d/(2k))$. Set $c=(x+y)/2$, with unit Jacobian. The canonical two-particle calculation gives the convolution $\int dc\,u(c+d/2)^2u(c-d/2)^2=d/[2\sinh(d/2)]$. Thus the required fixed-rank answer is the positive integral

$$\boxed{Z(2,k)=\frac1{32\pi^2k^2}\int_0^\infty \frac{d\,\tanh^2(d/(2k))}{\sinh(d/2)}\,dd.} \tag{PC15.3}$$

At $k=1$ it yields $1/(16\pi)$; at $k=2$ it yields $1/(32\pi^2)$. These are finite-rank checks of the exact kernel, not evaluations of an Airy approximation.

Change the limit: hold two particles fixed

Now let $k\to\infty$ through positive integers, keeping $N=2$. This is neither of the paper’s large-rank limits. The leading kernel is proportional to $u(x)u(y)$, which has rank one: it offers only one occupied orbital at this order. Its two-fermion determinant vanishes. Consequently the seemingly natural estimate $Z(2,k)\sim t_1^2/2\propto k^{-2}$ is wrong.

To find the first surviving term, use $\tanh v\sim v$ in (PC15.3). The needed moments follow by expanding $1/\sinh(d/2)=2\sum_{j\geq0}e^{-(j+1/2)d}$ and integrating each positive term:

$$\int_0^\infty\frac{d^3\,dd}{\sinh(d/2)}=2\pi^4, \qquad \int_0^\infty\frac{d^5\,dd}{\sinh(d/2)}=16\pi^6. \tag{PC15.4}$$

For $v\geq0$, $0\leq v-\tanh v=\int_0^v\tanh^2t\,dt\leq v^3/3$. Multiplying by $v+\tanh v\leq2v$ controls the replacement throughout the integral, including its far tail. We obtain

$$0\leq\frac{\pi^2}{64k^4}-Z(2,k) \leq\frac{\pi^4}{48k^6}, \qquad Z(2,k)\sim\frac{\pi^2}{64k^4}. \tag{PC15.5}$$

The bound is valid for all positive $k$, though its lower estimate is useful only once $k$ is sufficiently large. It establishes the asymptotic limit without assuming a uniform Taylor series for all $d$. In particular, $F(2,k)=4\log k-\log(\pi^2/64)+O(k^{-2})$. The coefficient four comes from exchange canceling the leading rank-one contribution.

Read the asymptotic answer with its remainder

For this section use the additional phase-volume and Airy lessons linked above. At fixed $k$ and large $N$, the leading count is $n(E)\sim CE^2$, $C=2/(\pi^2k)$. Filling levels gives $N\sim C\mu_*^2$ and

$$F\sim\frac{2}{3\sqrt C}N^{3/2} =\frac{\pi\sqrt{2k}}3N^{3/2}. \tag{PC15.6}$$

Source (5.38) and the argument through (5.64) organize the quantum corrections in the WKB expansion. This is not an assertion that the exact integer-valued staircase equals a quadratic polynomial plus a uniformly exponentially small remainder. The resulting perturbative cubic branch and its inversion are

$$\begin{aligned} J_{\rm pert}(\mu)&=C\mu^3/3+B\mu+A(k), \qquad B=\frac{k}{24}+\frac1{3k},\\ Z_N^{\rm pert}&=e^{A(k)}C^{-1/3} \operatorname{Ai}\!\left[C^{-1/3}(N-B)\right]. \end{aligned} \tag{PC15.7}$$

The exact Fredholm determinant $\Xi(e^\mu)=\det(1+e^\mu\rho)$ is periodic under $\mu\mapsto\mu+2\pi i$. The exponential of a cubic branch is not. The canonical contour explanation constructs a periodic sum of the branch’s translates, then unfolds an integer-rank coefficient contour. That construction specifies the perturbative expression; it is not an equality between the cubic and the full determinant. The source’s (3.12) is instead the small-fugacity Taylor series of the exact logarithm, with convergence restricted by $|z|\|\rho\|<1$.

Nor is $A(k)$ fixed by the leading phase volume or by demanding that the Airy continuation at $N=0$ equal one. Read (5.68) together with footnote 3: v3 withdraws the proposed coefficient pattern (5.70) underlying (5.71). The later, explicitly sourced constant-map input used for the lesson’s absolute comparisons is additional information; it is not proved by the cubic inversion or supplied by that withdrawn formula.

Finally, source (5.67) retains $Z_{\rm np}(N)$. Its Section 5.3 describes membrane terms with powers multiplying $e^{-2m\mu}$, computed in the WKB expansion; Section 5.4 discusses effects beyond that expansion with scale $e^{-4l\mu/k}$. At $\mu_*\sim\pi\sqrt{kN/2}$ these have possible scales $e^{-\pi m\sqrt{2kN}}$ and $e^{-2\pi l\sqrt{2N/k}}$. Coefficients, their possible vanishing, mixed sectors and cancellations at special levels require further input. An Airy resummation of the power corrections does not determine them. It also does not certify a fixed-$N$, growing-$k$ approximation such as our two-particle calculation.

Exit calculations

1. Keep the kernel and change the statistics

As a diagnostic, replace the auxiliary antisymmetrizer by a symmetrizer. Derive its two-boson answer and large-$k$ leading term. Why would getting that result from the ABJM integral expose an error?

Hint 1

The identity and transposition now both enter with positive signs.

Hint 2

Use $t_2=t_1^2-2Z(2,k)$ and the controlled fermionic limit.

Solution

The symmetrizer gives $Z_B(2,k)=(t_1^2+t_2)/2=t_1^2-Z(2,k)$. Therefore

$$Z_B(2,k)=\frac1{16k^2}-\frac{\pi^2}{64k^4}+O(k^{-6}).$$

Two bosons can occupy the leading single orbital. Thus their answer survives at $k^{-2}$, whereas the fermionic answer first survives at $k^{-4}$. Merely dividing the distinguishable-particle answer by $2!$ gives the third, different result $1/(32k^2)$. Neither replacement reproduces the determinant obtained by the Cauchy transformation. This comparison changes an auxiliary mathematical rule, not the physical spin of the matter fields in ABJM.

2. Delete one level before taking the logarithm

Suppose an approximate spectral computation omits one eigenvalue $a>0$, retaining the others. Express its $\Xi_{\rm kept}$, $t_1,t_2$, and $Z(2,k)$ in terms of the full values and $a$. If both trace errors are small, is their relative size by itself a good bound on the relative error in $Z(2,k)$?

Hint 1

Remove the single factor $1+az$ from the exact product, not a constant from every energy.

Hint 2

Every missing pair consists of the deleted level and one retained level.

Solution

The exact finite-deletion identities are

$$\Xi_{\rm kept}=\frac{\Xi}{1+az},\qquad t_{1,\rm kept}=t_1-a,\qquad t_{2,\rm kept}=t_2-a^2, \qquad Z_{2,\rm kept}=Z_2-a(t_1-a).$$

Substitution into (PC15.2) gives the last equality; alternatively sum the omitted distinct pairs directly. The change in the logarithm is $-\log(1+az)$, not a constant grand-potential shift. At large $k$, $t_1^2$ and $t_2$ are individually of order $k^{-2}$ but their difference is of order $k^{-4}$. Uncorrelated relative errors in those traces can therefore be amplified by order $k^2$ in the subtraction. A useful estimate must control the absolute omitted pair weight relative to $Z_2$, or use a positive representation such as (PC15.3). Small trace errors alone do not resolve an exponentially small difference from an Airy approximation.

3. Decide which limit an exponential survives

Take the first possible worldsheet scale $e^{-2\pi\sqrt{2N/k}}$. Compare fixed $k$ with $k=N/\lambda$ at fixed positive rational $\lambda$, along an integer sequence. Also evaluate $B/N$ along the second sequence. Does increasing $N$ alone make the same Airy error estimate small in both cases? What happens if instead $N=2$ and $k$ grows?

Hint 1

An exponential in $\sqrt N$ need not stay an exponential in $\sqrt N$ when its coefficient varies with $N$.

Hint 2

Substitute into both terms of $B=k/24+1/(3k)$ before estimating their ratio to $N$.

Solution

At fixed $k$, the displayed scale tends to zero faster than any inverse power of $N$. At fixed $\lambda=N/k$, it is $e^{-2\pi\sqrt{2\lambda}}$, independent of $N$, while

$$\frac BN=\frac1{24\lambda}+\frac{\lambda}{3N^2}.$$

Large $\lambda$ can suppress this worldsheet scale; large $N$ alone cannot. Its coefficient still has to be checked, so an action estimate is not a complete error bound. Also $A(k)$ varies on this sequence, even though it is independent of $N$ when $k$ is fixed.

With $N=2$ and growing $k$, $B$ eventually exceeds $N$, and the positive large-$N-B$ Airy saddle is unavailable. The exact answer is instead controlled by (PC15.5), with $F=4\log k-\log(\pi^2/64)+O(k^{-2})$. Substituting this sequence into a fixed-$k$ asymptotic formula has changed its assumptions. The distinction is about control of an approximation, not a contradiction between the two exact descriptions of the sphere integral.

Return to (3.8) and (5.67) after these calculations. The first organizes exact exchange statistics at every finite rank. The second keeps an explicit nonperturbative remainder. Reading both with their assumptions intact is what lets the paper connect a matrix integral to a useful large-rank expansion.