Turn the matrix integral into a Fermi gas

The sphere integral contains interacting eigenvalues and an oscillatory Chern–Simons phase. Yet the same answer can be organized as the partition function of independent fermions in one auxiliary dimension. The interactions do not disappear from the three-dimensional theory: their information becomes a particular one-particle operator.

Use the two large-rank limits and eigenvalue measures as preparation. We will perform the exact transformation at finite rank and obtain two numerical checkpoints. No large-$N$ approximation or phase-space estimate is used here.

Keep the same undeformed round-sphere observable of compact $U(N)_k\times U(N)_{-k}$, positive integer $k$, rank $N$ per factor, and no extra quotient. The generic full supersymmetry is $\mathcal N=6$, enhanced at $k=1,2$; the localization calculation uses the established $\mathcal N=2$ sphere subalgebra. All integration variables below are dimensionless. The original Gaussian Abel prescription and $F=-\log Z$ remain fixed.

Replace the eigenvalue interaction by a determinant

Set $f(v)=1/[2\cosh(\pi v)]$. The normalized matrix integral contains the square of

$$\mathcal D(s,t)= \frac{\prod_{i<j}2\sinh\pi(s_i-s_j)\,2\sinh\pi(t_i-t_j)} {\prod_{i,j}2\cosh\pi(s_i-t_j)} =\det_{i,j}f(s_i-t_j). \tag{AB18.1}$$

To derive this Cauchy identity, put $a_i=e^{2\pi s_i}$ and $b_j=e^{2\pi t_j}$. Then $f(s_i-t_j)=\sqrt{a_i b_j}/(a_i+b_j)$. After multiplying the determinant of $1/(a_i+b_j)$ by its common denominator, its numerator is alternating separately in the $a$‘s and $b$‘s. It must contain their two Vandermonde products. Their degrees exhaust the numerator degree, leaving a constant, fixed to one by a pole residue and induction from $N=1$. Converting those differences back to hyperbolic sines gives (AB18.1), including its sign. The identity is FG (3.1).

Expand both determinants as permutation sums. In each term relabel the $t$ variables so that the first permutation becomes the identity. The Gaussian phase and regulator are symmetric in that list. There are $N!$ equivalent copies, canceling one, and only one, of the original Weyl factorials. Thus

$$Z(N,k)=\frac1{N!}\sum_{\pi\in S_N}(-1)^\pi \int\prod_i ds_i\,dt_i\, e^{i\pi k\sum_i(s_i^2-t_i^2)} \prod_i f(s_i-t_i)f(s_i-t_{\pi(i)}), \tag{AB18.2}$$

where the integral still means its regulated limit. The sign $(-1)^\pi$ is positive for an even permutation and negative for an odd one. It will become the fermionic exchange sign.

Do the Fourier and Gaussian integrations

The required transform has no omitted $2\pi$:

$$f(v)=\int_{\mathbb R}dp\,f(p)e^{2\pi ipv}. \tag{AB18.3}$$

For example, substitute $u=e^{2\pi p}$ on the right. It becomes $\frac1{2\pi}\int_0^\infty u^{-1/2+iv}/(1+u)\,du$. The beta integral $\int_0^\infty u^{a-1}/(1+u)\,du=\pi/\sin\pi a$, continued within $0<\operatorname{Re}a<1$, gives precisely $f(v)$.

Apply this transform to the two factors in each term, with Fourier variables $p_i,q_i$. The linear arguments of $s_i,t_i$ are respectively $p_i+q_i$ and $-(p_i+q_{\pi^{-1}(i)})$. The paired Fresnel integrations are

$$\begin{aligned} \int ds\,e^{i\pi ks^2+2\pi ias} &=k^{-1/2}e^{i\pi/4}e^{-i\pi a^2/k},\\ \int dt\,e^{-i\pi kt^2-2\pi ibt} &=k^{-1/2}e^{-i\pi/4}e^{i\pi b^2/k}. \end{aligned} \tag{AB18.4}$$

Their constant phases cancel. Summing the squares cancels all $p_i^2$ and $q_i^2$ terms, leaving $\exp[(2\pi i/k)\sum_i p_i(q_{\pi^{-1}(i)}-q_i)]$. Integrate each $p_i$ with (AB18.3) and replace the summed permutation by its inverse. The result is

$$Z(N,k)=\frac1{N!k^N}\sum_{\pi\in S_N}(-1)^\pi \int\prod_i dq_i\, \prod_i f(q_i)f\!\left(\frac{q_i-q_{\pi(i)}}k\right). \tag{AB18.5}$$

These steps can be justified for the specific Abel prescription, rather than by an unregulated interchange. Retain $e^{-\epsilon\sum(s_i^2+t_i^2)}$, $\epsilon>0$, through the Fourier transforms. A plus Gaussian is then $\sqrt{\pi/(\epsilon-i\pi k)}\exp[-\pi^2a^2/(\epsilon-i\pi k)]$, with the conjugate denominator for the minus Gaussian. For real Fourier arguments their exponentials have modulus at most one, and each paired prefactor has modulus at most $1/k$. The remaining integrand is bounded by $k^{-N}\prod_i f(p_i)f(q_i)$, an integrable function since $\int f=1/2$. Dominated convergence therefore gives (AB18.4)–(AB18.5) at each fixed finite $N,k$. This also establishes the existence of this undeformed finite-rank Abel limit. It does not establish every other regulator or deformed contour prescription.

Identify the one-particle operator

Set $x_i=2\pi q_i$. Split each factor $[2\cosh(x_i/2)]^{-1}$ into two square roots. A permutation uses each variable once at each end, so this changes no product. Equation (AB18.5) becomes

$$\begin{aligned} Z(N,k)&=\frac1{N!}\int_{\mathbb R^N}\prod_i dx_i\, \det[\rho(x_i,x_j)],\\ \rho(x,y)&=\frac1{2\pi k} \frac1{\sqrt{2\cosh(x/2)}} \frac1{2\cosh((x-y)/(2k))} \frac1{\sqrt{2\cosh(y/2)}}. \end{aligned} \tag{AB18.6}$$

This is FG (3.2)–(3.5). The Hilbert space is $L^2(\mathbb R,dx)$ and $(\widehat\rho\psi)(x)=\int dy\,\rho(x,y)\psi(y)$. The prefactor belongs to this measure. The new $x$ is a Fourier-transformed integration variable, not an eigenvalue of the original gauge scalar or a physical spatial coordinate of ABJM.

To see the operator structure, take $[\widehat x,\widehat p]=i\hbar$, $\hbar=2\pi k$, and $\langle x|p\rangle=e^{ixp/\hbar}/\sqrt{2\pi\hbar}$. Define $U(x)=\log[2\cosh(x/2)]$ and $T(p)=\log[2\cosh(p/2)]$. The Fourier integral (AB18.3) gives

$$\widehat\rho=e^{-U(\widehat x)/2}e^{-T(\widehat p)}e^{-U(\widehat x)/2} \equiv e^{-\widehat H}. \tag{AB18.7}$$

This is an exact factorization, FG (3.27)–(3.34). Since position and momentum do not commute, $\widehat H$ is not exactly $U(\widehat x)+T(\widehat p)$. No small-$k$ approximation has entered.

Positivity follows by writing $\widehat\rho=Q^\dagger Q$ with $Q=e^{-T(\widehat p)/2}e^{-U(\widehat x)/2}$. Its squared Hilbert–Schmidt norm, meaning the integral of the squared mixed position/momentum kernel, is

$$\|Q\|_{\rm HS}^2 =\frac1{2\pi\hbar}\int dx\,dp\,e^{-U(x)-T(p)} =\frac{\pi^2}{2\pi\hbar}=\frac1{4k}. \tag{AB18.8}$$

Both elementary integrals equal $\pi$. Finiteness proves that $\widehat\rho$ is trace class: its nonnegative eigenvalues $\lambda_j$ have finite sum. Both factors of $Q$ are injective, since their multiplication functions are strictly positive in their respective representations; hence there is no zero eigenvector. Also $\|\widehat\rho\|\leq1/4$, since both $e^{-U}$ and $e^{-T}$ are bounded by $1/2$. Thus it has discrete summable positive eigenvalues, tending to zero; write $\lambda_j=e^{-E_j}$ to define auxiliary energies $E_j$. A “density operator” here is a Boltzmann operator, not a probability matrix normalized to trace one.

Why this is a gas of independent fermions

Expand the determinant in (AB18.6). It is the kernel of the antisymmetrizer acting on $N$ copies of the one-particle operator. In an orthonormal eigenbasis, a repeated one-particle level gives zero because an antisymmetric state cannot occupy it twice. Distinct occupied levels give

$$Z(N,k)=\sum_{j_1<\cdots<j_N} \lambda_{j_1}\cdots\lambda_{j_N} =\sum_{j_1<\cdots<j_N}e^{-(E_{j_1}+\cdots+E_{j_N})}. \tag{AB18.9}$$

“Independent” means that the total auxiliary energy is a sum of one-particle energies. The exclusion rule remains. These fermions are not being identified with ABJM’s elementary fermion fields, and this ungraded thermal-style trace is not a supersymmetric index. The auxiliary inverse temperature is fixed to one by the definition of $\widehat H$.

Let $t_m=\operatorname{Tr}\widehat\rho^{\,m}$. A cycle of length $m$ in a permutation contributes $t_m$; its sign is $(-1)^{m-1}$. In particular,

$$Z(1,k)=t_1,\qquad Z(2,k)=\frac12(t_1^2-t_2). \tag{AB18.10}$$

The subtraction removes the forbidden equal-level pair. It is not an interaction correction. Positivity is manifest in (AB18.9), and $Z(N,k)\leq t_1^N/N!$ follows by allowing repeated levels in the positive sum.

Allowing every occupation number $n_j=0,1$ and weighting total rank by $z$ gives

$$\Xi(z,k)=\sum_{N\geq0}Z(N,k)z^N =\prod_j(1+z\lambda_j)=\det(1+z\widehat\rho), \quad \log\Xi=\sum_{m\geq1}\frac{(-1)^{m-1}}m z^m t_m. \tag{AB18.11}$$

The infinite product converges for finite $z$ because $\sum\lambda_j<\infty$; the logarithm’s displayed Taylor series requires $|z|\|\rho\|<1$. Positive $z$ beyond that disk is handled by the product, not by insisting the Taylor series converge. This supplies the generating function used in the previous lesson. It also explains the word grand-canonical: rank can vary in the sum, whereas the canonical $Z(N,k)$ fixes it. See FG (3.9)–(3.24).

Worked check: the first two particle numbers

The kernel diagonal gives

$$t_1=\int dx\,\rho(x,x) =\frac1{4\pi k}\int\frac{dx}{2\cosh(x/2)} =\frac1{4k}. \tag{AB18.12}$$

This independently reproduces the direct Fourier-constraint calculation of the rank-one sphere integral. The determinant’s factors of two and the operator measure are both tested.

For $t_2$, set $d=x-y$, $c=(x+y)/2$, whose absolute Jacobian is one. The elementary convolution is

$$\int_{\mathbb R}\frac{dc}{2[\cosh c+\cosh(d/2)]} =\frac{d}{2\sinh(d/2)}. \tag{AB18.13}$$

One can obtain it by substituting $v=\tanh(c/2)$; its value at $d=0$ is the continuous limit one. Therefore

$$t_2(k)=\frac1{16\pi^2k^2}\int_0^\infty \frac{d\,dd}{\sinh(d/2)\cosh^2(d/(2k))}. \tag{AB18.14}$$

At $k=1$, put $d=2u$ and use $1/(\sinh u\cosh^2u)=1/\sinh u-\sinh u/\cosh^2u$. The first integral is $\int_0^\infty u/\sinh u\,du=2\sum_{r\geq0}(2r+1)^{-2}=\pi^2/4$; the second is $\int_0^\infty\operatorname{sech}u\,du=\pi/2$ after integration by parts. Consequently

$$t_2(1)=\frac1{16}-\frac1{8\pi},\qquad \boxed{Z(2,1)=\frac1{16\pi}=0.0198943678865\ldots .} \tag{AB18.15}$$

Treating the particles as independent distinguishable objects and then merely dividing by $2!$ would instead give $t_1^2/2=1/32$. The missing exchange term is measurable even though the auxiliary Hamiltonian has no pair potential.

There is a useful error bound for general positive $k$. If (AB18.14) is truncated at $d=D>0$, its omitted positive tail is at most

$$\frac{e^{-aD}}{2\pi^2k^2(1-e^{-D})} \left(\frac D a+\frac1{a^2}\right), \qquad a=\frac12+\frac1k. \tag{AB18.16}$$

Use $\operatorname{csch}(d/2)\leq2e^{-d/2}/(1-e^{-D})$ and $\operatorname{sech}^2(d/(2k))\leq4e^{-d/k}$, then integrate $d e^{-ad}$. This is a cutoff bound, separate from quadrature or floating-point errors. The direct kernel is also suitable for a positive weighted quadrature matrix; comparing its trace and squared trace with (AB18.12)–(AB18.14) tests that discretization without assuming its eigenvalues are exact.

Exercises

1. Fill three levels, or try to fill only two

Derive $Z(3)$ from permutation cycles in terms of $t_1,t_2,t_3$. Check the formula for a toy operator with exactly two nonzero eigenvalues $a,b>0$. Why must its three-fermion answer vanish even though $t_1^3$ is positive?

Hint 1

In $S_3$ there is one identity, three transpositions and two three-cycles. A three-cycle is even.

Hint 2

Insert $t_m=a^m+b^m$, or take the coefficient of $z^3$ in $(1+za)(1+zb)$.

Solution

The three cycle types give $t_1^3$, $-3t_1t_2$ and $+2t_3$. Dividing by $3!$ gives

$$Z(3)=\frac{t_1^3-3t_1t_2+2t_3}{6}.$$

For two levels, expand $(a+b)^3-3(a+b)(a^2+b^2)+2(a^3+b^3)$. The $a^3,b^3,a^2b,ab^2$ terms all cancel. There is no choice of three distinct levels. The positive $t_1^3$ counts repetitions and ordered choices, so it is not the fermionic canonical partition function. The actual ABJM operator has infinitely many positive levels; the two-level example checks the statistics, not a finite-rank truncation of the gauge theory.

2. Change the level to two

Evaluate (AB18.14) at $k=2$ and obtain $Z(2,2)$. Use the identity $1/(\sinh u\cosh^3u)=\coth u-\tanh u-\tanh u\operatorname{sech}^2u$. Keep the one-particle trace at the new level.

Hint 1

Set $d=4u$. Then $t_2(2)=(8\pi^2)^{-1}\int_0^\infty u/(\sinh u\cosh^3u)\,du$.

Hint 2

Use $\coth u-\tanh u=2/\sinh(2u)$ and integrate the last term by parts with $(\operatorname{sech}^2u)'=-2\tanh u\operatorname{sech}^2u$.

Solution

The first combined integral is $\int_0^\infty 2u/\sinh(2u)\,du=\pi^2/8$. The last is $\frac12\int_0^\infty\operatorname{sech}^2u\,du=1/2$; the boundary term vanishes at both ends. Hence

$$t_2(2)=\frac1{64}-\frac1{16\pi^2},\qquad t_1(2)=\frac18,\qquad Z(2,2)=\frac1{32\pi^2}.$$

Both $t_2$ and $Z(2,2)$ are positive, and $t_2<t_1^2$ as required by the positive distinct-pair sum. Keeping the old $t_1=1/4$ would mix two different theories.

3. Should the density operator have unit trace?

Suppose a numerical implementation replaces $\rho$ by $\rho/t_1$ to make its trace one. Determine the new canonical partition functions, free energy and auxiliary energies. Can one retain the original absolute sphere partition function after this replacement?

Hint 1

Every term in the $N$-particle determinant contains exactly $N$ kernel factors.

Hint 2

If $\lambda_j=e^{-E_j}$, divide the eigenvalue by $t_1$ before taking minus its logarithm.

Solution

The new eigenvalues are $\lambda_j/t_1$ and energies $E_j+\log t_1$. Therefore

$$Z_{\rm unit}(N,k)=t_1^{-N}Z(N,k),\qquad F_{\rm unit}=F+N\log t_1=F-N\log(4k).$$

The corresponding generating function is $\Xi_{\rm unit}(z)=\Xi(z/t_1)$. One may use this rescaling internally if the factor $t_1^N$ is restored. Without it the rank-one answer becomes one instead of $1/(4k)$, and every absolute partition function has changed. This is not an allowed removal of a gauge-center volume or a new choice of the physical Chern–Simons level. The Boltzmann operator in this calculation was never required to be a normalized quantum-state density matrix.

We now have an exact finite-rank statistical formulation with checked normalization and exchange signs. Extracting its high-energy spectrum and density of states is a further approximation; that next step will supply the leading large-rank coefficient used in the preceding lesson.

Derive the N to the three-halves free energy now uses this exact operator to count high-energy levels and obtain the leading fixed-level free energy. Classical state counting, its quantum input and finite-rank numerical evidence are kept distinct.

Reading ABJM as a Fermi gas maps the exact operator to the paper and derives a controlled two-particle limit at increasing level.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading ABJM as a Fermi gas