The Airy function and the corrections it omits

Before this lesson: Derive the N to the three-halves free energy.

The phase-space calculation gave the leading sphere free energy, $F\sim\pi\sqrt{2k}\,N^{3/2}/3$ . Can we package its power corrections into one function, and how would we tell whether that function is the complete answer?

Keep the same compact three-dimensional $U(N)_k\times U(N)_{-k}$ theory, positive integer $k$ , rank $N$ per factor, and no further quotient. The round-sphere normalization and phase prescription are fixed. The generic theory has twelve real Poincaré supercharges, enhanced to sixteen at $k=1,2$ ; the localization calculation uses a chosen supercharge in its manifest $\mathcal N=2$ description. Here $N\to\infty$ means fixed $k$ , and $F=-\log Z$ . The auxiliary Fermi gas is ungraded: its canonical partition function is the sphere observable, not a supersymmetric index.

Three coefficients with different origins

Let $\mu$ be the dimensionless chemical potential. The perturbative branch of the grand potential is

J_{\rm pert}(\mu)=\frac C3\mu^3+B\mu+A, \qquad C=\frac{2}{\pi^2k},\qquad B=\frac{k}{24}+\frac{1}{3k}. \tag{AB20.1}

The coefficient $C$ came from the phase-space area. The linear and constant terms require more information. In the quantum spectral calculation, the smoothed high-energy count has the form $n(E)=CE^2+n_0+\cdots$ , with $n_0=-1/(3k)+k/24$ . This is a statement about an averaged staircase, not an identity for every level. The quantum input and its perturbative WKB scope are given in FG (5.64)–(5.67).

To see how the count enters, integrate the occupation formula by parts. Near the Fermi edge, set $t=E-\mu$ :

J'(\mu)=\int dE\,n(E)\frac{e^{E-\mu}}{(1+e^{E-\mu})^2} \sim C\mu^2+n_0+\frac{\pi^2C}{3}. \tag{AB20.2}

The edge weight is even, integrates to one, and has second moment $\pi^2/3$ . For example, integration by parts reduces that moment to $4\int_0^\infty t/(e^t+1)\,dt=\pi^2/3$ . This supplies the extra $2/(3k)$ in $B=n_0+\pi^2C/3$ . Extending the edge integral to the whole real line is a large- $\mu$ step; the omitted low-energy and exponential terms are not being set to zero in the exact spectrum.

The integration constant $A=A(k)$ cannot be recovered from $J'$ or the leading diamond. For an absolute numerical prediction we adopt the constant-map prescription displayed in HMO (3.4):

\begin{aligned} A(k)={}&-\frac{\zeta(3)k^2}{8\pi^2}+\frac16\log\frac{4\pi}{k} +2\zeta'(-1)\\ &-\frac13\int_0^\infty\frac{dx}{e^{kx}-1} \left(\frac3{x^3}-\frac1x-\frac3{x\sinh^2x}\right). \end{aligned} \tag{AB20.3}

Here $\zeta$ is the Riemann zeta function and the prime differentiates its argument. HMO identifies this formula as a conjectured constant-map input and tests it against exact data. Our inversion below derives its consequences; it does not prove that input. In particular, we do not fit $A$ to the data used to test the answer.

The three terms in parentheses must be combined: their apparent singularities cancel, leaving $-x/5+O(x^3)$ . The full integrand therefore tends to $-1/(5k)$ at the origin. Two useful evaluations are

A(1)=\frac{\log2}{4}-\frac{\zeta(3)}{8\pi^2} \simeq0.158062566611, \qquad A(2)=-\frac{\zeta(3)}{2\pi^2} \simeq-0.060896914117. \tag{AB20.4}

A periodic function and a nonperiodic branch

The exact Fredholm determinant is $\Xi(e^\mu)=\det(1+e^\mu\widehat\rho)=\sum_{N\geq0}Z_Ne^{N\mu}$ . It is periodic under $\mu\mapsto\mu+2\pi i$ . The exponential of the cubic in (AB20.1) is not. Ignoring this distinction would turn a finite coefficient contour into an unexplained infinite one.

Following the periodization idea in HMO (3.8), define a perturbative periodic expression on $\operatorname{Re}\mu=a>0$ :

\Xi_{\rm pert}^{\rm per}(\mu) =\sum_{j\in\mathbb Z}\exp J_{\rm pert}(\mu+2\pi i j). \tag{AB20.5}

For $\mu=a+iy$ , the real part of the exponent is $Ca^3/3-Cay^2+Ba+A$ . The sum converges by Gaussian decay. Extracting an integer- $N$ coefficient over one period and translating each term gives

\begin{aligned} Z_N^{\rm pert} &=\int_{a-i\pi}^{a+i\pi}\frac{d\mu}{2\pi i} e^{-N\mu}\Xi_{\rm pert}^{\rm per}(\mu)\\ &=\int_{a-i\infty}^{a+i\infty}\frac{d\mu}{2\pi i} \exp\left[\frac C3\mu^3-(N-B)\mu+A\right]. \end{aligned} \tag{AB20.6}

The translated intervals tile the infinite line, and $e^{2\pi i jN}=1$ . This is an exact manipulation of the defined perturbative expression. It is not an assertion that (AB20.5) equals the exact determinant. Nor need its continuation to small or negative $N$ have the properties of a particle partition function. In particular, imposing $Z_0^{\rm pert}=1$ would not be a valid way to determine $A$ .

The Airy integral and its normalization

Define the Airy function by the convergent vertical contour

\operatorname{Ai}(z)=\int_{a_t-i\infty}^{a_t+i\infty} \frac{dt}{2\pi i}\,e^{t^3/3-zt},\qquad a_t>0. \tag{AB20.7}

This contour can be deformed into the usual two decaying rays. Differentiation twice under the integral inserts $t^2$ , while integrating $\partial_t e^{t^3/3-zt}$ gives zero. Thus $\operatorname{Ai}''(z)=z\operatorname{Ai}(z)$ . The contour selects the solution that decays for large positive $z$ .

Set $t=C^{1/3}\mu$ in (AB20.6). The measure, as well as the exponent, changes:

\boxed{Z_N^{\rm pert}=e^{A(k)}C^{-1/3} \operatorname{Ai}\!\left[C^{-1/3}(N-B)\right].} \tag{AB20.8}

Neither $e^A$ nor $C^{-1/3}$ is optional. This agrees with FG (5.37). It resums the algebraic large- $N$ corrections belonging to the cubic branch. Calling it an Airy function does not turn it into the complete finite- $N$ answer.

For a direct saddle check, put $n=N-B>0$ , $\mu_* =\sqrt{n/C}$ and $\mu=\mu_*+iv/\sqrt{C\mu_*}$ . The exponent becomes exactly

J_{\rm pert}-N\mu =A-\frac{2n^{3/2}}{3\sqrt C}-v^2 -\frac{i v^3}{3\sqrt C\,\mu_*^{3/2}}. \tag{AB20.9}

The Gaussian term supplies the fluctuation prefactor; the odd cubic term integrates to zero at first order. Consequently,

\begin{aligned} Z_N^{\rm pert}&\sim \frac{e^{A-2n^{3/2}/(3\sqrt C)}}{2\sqrt\pi\,C^{1/4}n^{1/4}} \left[1-\frac{5\sqrt C}{48n^{3/2}}+O(n^{-3})\right],\\ F_N^{\rm pert}&= \frac{2n^{3/2}}{3\sqrt C}+\frac14\log n+\frac14\log C +\log(2\sqrt\pi)-A+O(n^{-3/2}). \end{aligned} \tag{AB20.10}

The first coefficient is again $\pi\sqrt{2k}/3$ . The shift $B$ , Gaussian logarithm and constant $A$ now have distinct visible roles. The expansion of the Airy function is asymptotic; the contour formula specifies its resummation.

A finite-rank comparison that can fail

Using (AB20.3), compare (AB20.8) with the independent exact formulas in HMO Appendix B. Write $\delta=Z_N/Z_N^{\rm pert}-1$ :

$k$	$N$	Exact $Z_N$	$\delta$
$1$	$1$	$1/4$	$5.23042678\times10^{-6}$
$1$	$2$	$1/(16\pi)$	$4.97171529\times10^{-7}$
$1$	$4$	$(10-\pi^2)/(1024\pi^2)$	$1.04575758\times10^{-8}$
$2$	$1$	$1/8$	$2.51945178\times10^{-3}$
$2$	$2$	$1/(32\pi^2)$	$4.31440996\times10^{-4}$
$2$	$4$	$(24-32\pi^2+3\pi^4)/(49152\pi^4)$	$2.57017888\times10^{-5}$

The $k=2$ entries are on PDF page 41. These discrepancies are much larger than numerical evaluation error. An additional calculation discretizes the positive kernel $\rho(x,y)$ , computes its eigenvalues $\lambda_j$ , and obtains $Z_N$ as the coefficient of $z^N$ in $\prod_j(1+z\lambda_j)$ . Increasing the quadrature from 768 to 1024 nodes and comparing position windows $[-80,80]$ and $[-96,96]$ changes these six entries by less than $1.5\times10^{-13}$ relatively. The final values agree with the exact formulas within $10^{-13}$ . The analytic window-tail bounds are below $10^{-18}$ for these entries; quadrature and floating-point errors are separate from those bounds.

Thus the test checks the absolute normalization and resolves the omitted contributions. It does not establish their full functional form, or prove the prescription for $A(k)$ at every level.

What an Airy resummation leaves out

The sourced nonperturbative sectors include terms of the schematic form

\delta J(\mu) \supset (a_m\mu^2+b_m\mu+c_m)e^{-2m\mu} +d_l e^{-4l\mu/k},\qquad m,l\geq1. \tag{AB20.11}

They are called membrane and worldsheet sectors, respectively; these names refer to their dual geometric interpretation. The relevant exponential scales and qualifications are discussed in HMO (1.1)–(1.5). At the leading saddle, the first possible actions become

e^{-2\mu_*}\sim e^{-\pi\sqrt{2kN}},\qquad e^{-4\mu_*/k}\sim e^{-2\pi\sqrt{2N/k}}. \tag{AB20.12}

At fixed $k$ both are smaller than every inverse power of $N$ . No finite power series, and no instruction to sum only that power series, determines their coefficients. Some coefficients vanish at special levels. When actions coincide, separately singular sector coefficients must be combined before taking the level limit; the exact positive-kernel partition function is finite. The detailed cancellation is extra information, as in HMO §5.

One transformation can already be derived without that extra information. If a branch receives a small term $\varepsilon d e^{-s\mu}$ with $s>0$ , expand its exponential once. Then

\delta Z_N=\varepsilon d\,Z_{N+s}^{\rm pert}+O(\varepsilon^2). \tag{AB20.13}

The shifted subscript means evaluation of the analytic Airy expression, not a theory of fractional rank. Polynomial factors of $\mu$ become derivatives with respect to $N$ . This explains why Airy functions can also express individual nonperturbative corrections without making the uncorrected Airy formula exact.

The distinction of limits remains essential. If $k$ grows with $N$ at fixed $\lambda=N/k$ , the worldsheet scale becomes $e^{-2\pi\sqrt{2\lambda}}$ , not a correction that vanishes with rank. The fixed- $k$ hierarchy cannot simply be reused as a uniform error estimate in the other limit.

Practice

1. Changing the density-operator normalization

Someone replaces $\widehat\rho$ by $e^\eta\widehat\rho$ , with real $\eta$ , calling this a harmless normalization. Derive the change of $Z_N$ . Then invert a general cubic branch $C\mu^3/3+D\mu^2/2+B\mu+A$ , and use it to check the same result. What changes in the sphere free energy?

Hint 1

Each term in the $N$ th exterior-power trace contains exactly $N$ eigenvalues. In the grand partition function the rescaling shifts $\mu$ .

Hint 2

For the cubic with a quadratic term, translate $\nu=\mu+D/(2C)$ . Track its constant term, not just its Airy argument.

Solution

The exterior coefficient gives $Z_N\mapsto e^{\eta N}Z_N$ , and $\Xi(e^\mu)\mapsto\Xi(e^{\mu+\eta})$ . Completing the cubic yields

Z_N^{\rm cub}= \exp\left[A+\frac{D(N-B)}{2C}+\frac{D^3}{12C^2}\right]C^{-1/3} \operatorname{Ai}\!\left[C^{-1/3}\left(N-B+\frac{D^2}{4C}\right)\right].

The vertical contour can be translated to a positive real part in $\nu$ before using the Airy definition. Expanding the shifted original branch gives

D=2C\eta,\qquad B_{\rm new}=B+C\eta^2, \qquad A_{\rm new}=A+B\eta+\frac C3\eta^3.

Substitution cancels $\eta$ from the Airy argument and leaves $A+\eta N$ in the prefactor. Thus $F_N\mapsto F_N-\eta N$ . The spectral normalization is physical input for this fixed sphere prescription. Setting $\operatorname{Tr}\rho$ to one would change its partition functions, not merely their presentation. The exact exterior-coefficient argument checks the contour manipulation independently.

2. Which correction produces which power?

Derive the coefficient $-5/48$ in the first line of (AB20.10) directly from (AB20.9). Expand $F_N^{\rm pert}$ at fixed $k$ through its $N$ -independent term. Does keeping only $C$ determine the $\sqrt N$ or constant term? Could a term $e^{-c\sqrt N}$ be inferred from this expansion?

Hint 1

The first nonzero correction to the Gaussian integral is the square of the imaginary cubic term. Use the normalized moment $\langle v^6\rangle=15/8$ .

Hint 2

Expand $(N-B)^{3/2}$ and $\log(N-B)$ separately. The determinant prefactor contributes a logarithm even though $J_{\rm pert}$ is polynomial.

Solution

Writing the cubic coefficient as $\beta=1/(3\sqrt C\mu_*^{3/2})$ , the normalized Gaussian integral is $\langle e^{-i\beta v^3}\rangle=1-\beta^2\langle v^6\rangle/2+O(\beta^4)$ . Its correction is $-5/(48C\mu_*^3)=-5\sqrt C/(48n^{3/2})$ , fixing both sign and power. Taking the negative logarithm changes this correction’s sign in $F$ .

At fixed $k$ , hence fixed $A,B,C$ ,

F_N^{\rm pert}= \frac{2}{3\sqrt C}N^{3/2}-\frac B{\sqrt C}N^{1/2} +\frac14\log N+\left[\frac14\log C+\log(2\sqrt\pi)-A\right] +O(N^{-1/2}).

Thus $C$ determines the leading coefficient, while $B$ changes the first subleading power and $A$ changes the constant. For example, the next algebraic term is $B^2N^{-1/2}/(4\sqrt C)$ , from expanding $n^{3/2}$ . Adding $e^{-c\sqrt N}$ for $c>0$ leaves every coefficient of the inverse-power asymptotic expansion unchanged. Such a term requires additional information. Also, $A(k)$ is constant in this expansion only because $k$ is held fixed.

3. A supplied exponential correction

Suppose the supplied correction is $\delta J=\varepsilon(p_2\mu^2+p_1\mu+p_0)e^{-s\mu}$ . Derive its first-order canonical contribution. Apply the result to the leading combined $k=2$ term proposed and tested in HMO (1.5), namely $(4\mu^2+2\mu+1)e^{-2\mu}/\pi^2$ . Explain why this tests a sourced correction without deriving the full nonperturbative answer.

Hint 1

Differentiate the coefficient kernel $e^{-N\mu}$ with respect to its real analytic parameter $N$ before setting $N$ to an integer.

Hint 2

After the shift by $s$ , use $\partial_N^2 Z_N^{\rm pert}=(N-B)Z_N^{\rm pert}/C$ , which follows from the Airy differential equation.

Solution

Multiplication by $\mu$ is $-\partial_N$ , so the contribution is

\delta Z_N=\varepsilon\left[p_2\partial_N^2-p_1\partial_N+p_0\right] Z_{N+s}^{\rm pert}+O(\varepsilon^2).

For $k=2$ put $z_2=C^{-1/3}(N+2-B)$ . The displayed sector contributes

\Delta Z_N= \frac{e^A C^{-1/3}}{\pi^2} \left[\left(\frac{4(N+2-B)}C+1\right)\operatorname{Ai}(z_2) -2C^{-1/3}\operatorname{Ai}'(z_2)\right].

For $N=1$ , the original relative discrepancy is $2.51945178\times10^{-3}$ ; after subtracting this contribution, $(Z_N-Z_N^{\rm pert}-\Delta Z_N)/Z_N^{\rm pert}\simeq7.43715\times10^{-8}$ . At $N=4$ the corresponding residual is $2.60503\times10^{-11}$ . These values use the exact finite-rank formulas, not a fit of the polynomial coefficients.

The small- $\varepsilon$ statement is exact to first order. For the physical asymptotic expansion, omitted terms include both the square of the supplied exponential and new terms at higher action. The Airy ratio behaves as $Z_{N+s}^{\rm pert}/Z_N^{\rm pert}\sim e^{-s\sqrt{(N-B)/C}}$ times algebraic factors, explaining the exponential scale. Recovering one such sector is a strong check of its normalization; it does not determine the remaining sectors or establish their full cancellation structure.

Compare exact, planar, and gravitational predictions turns these distinctions into numerical diagnostics. Use neighboring ranks to test the shift while canceling an unknown rank-independent normalization.

The Fermi-gas source companion maps this approximation to the original equations and tests why a different large-level limit needs a different estimate.

Next, resolve the first nonperturbative correction at level four. Compare exact finite-rank data with independent spectral calculations and determine which residuals the numerical resolution can actually test.