The first exact partition-function checks

We now have a complete integral and can ask for an exact number. The first useful case is one color at each node. It retains the Chern–Simons coupling, both gauge-scalar integrations and all four matter multiplets, while removing the non-Abelian root products.

Start from the normalized sphere matrix integral. We keep three-dimensional compact $U(1)_k\times U(1)_{-k}$, integer $k>0$, the round background $H=-i/\ell$, and no real masses, FI terms or additional quotient. The generic full theory has twelve real Poincaré supercharges, with sixteen at levels one and two; the calculation uses the established $\mathcal N=2$ sphere subalgebra. The total gauge rank is two.

The exact result will be

$$\boxed{Z(1,k)=\frac1{4k},\qquad F_{S^3}(1,k)=-\log Z(1,k)=\log(4k).} \tag{AB13.1}$$

It is an answer at finite rank and fixed level. No large-rank, weak-coupling or saddle approximation is used.

Begin with the actual two-variable integral

Let $s=\ell\sigma$ and $t=\ell\widehat\sigma$ be the two real, dimensionless gauge scalars. At rank one the two Weyl groups are trivial and there are no roots. The inherited expression is

$$Z(1,k)= \int_{\mathbb R^2}ds\,dt\, \frac{e^{\,i\pi k(s^2-t^2)}}{[2\cosh\pi(s-t)]^2}, \tag{AB13.2}$$

with the limiting prescription to be supplied. There is no $1/2!$ and no extra integration over a compact angle. The factors of two in the denominator belong to the absolute determinant normalization.

For comparison, Marino–Putrov (2.1) has precisely this phase after $\mu=2\pi s$, $\nu=2\pi t$. Its measures are $d\mu/(2\pi)$ and $d\nu/(2\pi)$. The original KWY (3.68), (4.4) uses the opposite first-node phase; exchanging the nodes gives the present convention. We keep the action convention throughout instead of silently changing the sign of one level.

Define a relative variable and a common-center variable:

$$x=s-t,\qquad y=\frac{s+t}{2},\qquad s=y+\frac x2,\quad t=y-\frac x2. \tag{AB13.3}$$

Their absolute Jacobian is one, and $s^2-t^2=2xy$. Write $f(x)=1/[4\cosh^2(\pi x)]$. Formally, the integrand becomes $f(x)e^{2\pi ikxy}$.

The problem is already visible: $\int dx\,dy\,|f(x)e^{2\pi ikxy}|=\infty$, because the absolute value is independent of $y$. Integrating $y$ first as an ordinary improper integral is therefore unjustified. Its useful interpretation is a Fourier distribution, which we will derive as a limit of ordinary convergent integrals.

Integrate first, then remove the regulator

Multiply (AB13.2) by $e^{-\epsilon(s^2+t^2)}$, with $\epsilon>0$, and call the result $Z_\epsilon(1,k)$. This is the positive Gaussian Abel regulator specified in the prerequisite. At fixed $\epsilon$ the integral is absolutely convergent, so changes of variables and either order of integration are legitimate.

Since $s^2+t^2=2y^2+x^2/2$, the common-center integral is

$$\int_{\mathbb R}dy\, e^{-2\epsilon y^2+2\pi ikxy} =\sqrt{\frac{\pi}{2\epsilon}}\, e^{-\pi^2k^2x^2/(2\epsilon)}. \tag{AB13.4}$$

This follows by the Fourier transform of a real Gaussian. One can derive that transform by differentiating with respect to its frequency and integrating by parts, fixing its value at zero frequency by the ordinary Gaussian integral. No unregulated interchange has occurred.

Set $C_\epsilon=\pi^2k^2+\epsilon^2$. The remaining answer is

$$Z_\epsilon(1,k)= \sqrt{\frac{\pi}{2\epsilon}} \int_{\mathbb R}dx\, e^{-C_\epsilon x^2/(2\epsilon)}f(x). \tag{AB13.5}$$

To make the limit transparent, introduce the normalized Gaussian density

$$p_\epsilon(x)= \sqrt{\frac{C_\epsilon}{2\pi\epsilon}}\, e^{-C_\epsilon x^2/(2\epsilon)},\qquad \int_{\mathbb R}p_\epsilon(x)\,dx=1. \tag{AB13.6}$$

Its variance is $\epsilon/C_\epsilon$. Therefore

$$Z_\epsilon(1,k)= \frac{\pi}{\sqrt{C_\epsilon}} \int_{\mathbb R}p_\epsilon(x)f(x)\,dx. \tag{AB13.7}$$

For fixed nonzero $k$, the variance tends to zero and the prefactor tends to $1/|k|$. More explicitly, put $x=\sqrt{\epsilon/C_\epsilon}\,u$. The integral in (AB13.7) is $\int du\,e^{-u^2/2}f(\sqrt{\epsilon/C_\epsilon}\,u)/\sqrt{2\pi}$. The function $f$ is continuous at zero and bounded by $1/4$. The integrable Gaussian bound allows the dominated-convergence theorem, giving

$$\lim_{\epsilon\downarrow0}Z_\epsilon(1,k) =\frac{f(0)}{|k|}=\frac1{4|k|}. \tag{AB13.8}$$

For the specified $k>0$ this proves (AB13.1).

This also explains the compact distributional calculation:

$$\int_{\mathbb R}dy\,e^{2\pi ikxy} =\delta(kx)=\frac1{|k|}\delta(x). \tag{AB13.9}$$

An equality of distributions means equality after integration against suitable test functions. The Gaussian argument establishes its use here with the actual matter factor. The common gauge scalar imposes the relative-scalar constraint; it does not contribute a removable infinite volume.

What the number tests

At $k=1$, $Z=1/4$. The product of four zero-mass R-charge-$1/2$ chiral determinants is also $(2^{-1/2})^4=1/4$. This is a useful normalization check, not by itself a proof that two complete quantum theories or their global operator spectra coincide.

The result is radius independent because the original measure used the dimensionless variables $\ell\sigma,\ell\widehat\sigma$. Keeping a dimensional measure while dropping its radius factors would spoil that statement.

The phase prescription and normalization remain part of the answer. We use $F=-\log Z$; the free-energy convention discussed after Marino–Putrov (3.11) has the opposite sign. Our result is positive in this prescription, but the original integrand is not a positive probability weight.

The same paper gives a spectral reformulation in (3.3)–(3.7). Deriving that reformulation provides a later independent comparison, rather than an input into the direct calculation here. Nothing in the rank-one proof establishes convergence of every higher-rank contour or licenses extrapolating a large-rank asymptotic formula down to $N=1$.

Control the regulator error

A numerical value at a small positive $\epsilon$ is not yet the limit. The inequality $0\leq\tanh^2z\leq z^2$ gives, using the variance in (AB13.6),

$$0\leq \frac{\pi}{4\sqrt{C_\epsilon}}-Z_\epsilon(1,k) \leq \frac{\pi^3\epsilon}{4C_\epsilon^{3/2}}. \tag{AB13.10}$$

Combining this with $1-(1+u)^{-1/2}\leq u/2$ for $u\geq0$ yields the convenient relative bound

$$0\leq1-4kZ_\epsilon(1,k) \leq \frac{\epsilon}{k^2}+ \frac{\epsilon^2}{2\pi^2k^2}, \qquad k>0. \tag{AB13.11}$$

This bounds the bias from retaining the regulator. It is separate from numerical quadrature error and floating-point error.

For example, at level one:

$\epsilon$	$Z_\epsilon(1,1)$	Relative regulator bias	Bound from (AB13.11)
$0.1$	$0.2288487331$	$0.08460507$	$0.10050661$
$0.03$	$0.2429049864$	$0.02838005$	$0.03004559$
$0.01$	$0.2475474031$	$0.00981039$	$0.01000507$

The values converge to $0.25$, but $\epsilon=0.01$ still has about a one-percent bias. Reporting many quadrature digits would not remove it.

For efficient evaluation use the rescaled Gaussian variable $u$ above, whose width stays fixed. A fixed grid in $x$ can miss the peak as its width shrinks like $\sqrt{\epsilon}/(\pi k)$.

The small-regulator expansion follows from $\operatorname{sech}^2z=1-z^2+\tfrac23z^4+O(z^6)$ and the Gaussian moments $\mathbb E[u^2]=1$, $\mathbb E[u^4]=3$:

$$Z_\epsilon(1,k)=\frac1{4k} \left[ 1-\frac{\epsilon}{k^2} +\left(\frac2{k^4}-\frac1{2\pi^2k^2}\right)\epsilon^2 +O(\epsilon^3) \right], \quad \epsilon\to0,\quad k>0\ \text{fixed}. \tag{AB13.12}$$

The remainder estimate is justified because the sixth derivative of $\operatorname{sech}^2z$ is bounded on the real line and the sixth Gaussian moment is finite. Expanding the prefactor as well as the integrand is essential. For instance, $2Z_{\epsilon/2}-Z_\epsilon$ cancels the linear bias, leaving a quadratic one. This is a numerical extrapolation in a removable regulator, not a perturbation expansion in a physical coupling.

Exercises

1. Recover the measure in another set of variables

Start instead with

$$\int_{\mathbb R^2}\frac{d\mu\,d\nu}{(2\pi)^2}\, \frac{e^{\,ik(\mu^2-\nu^2)/(4\pi)}} {[2\cosh((\mu-\nu)/2)]^2}.$$

Derive the same answer with $X=\mu-\nu$, $Y=(\mu+\nu)/2$ and the regulator inherited from $\epsilon(s^2+t^2)$. State what changes for negative integer $k$, and why the argument fails at $k=0$.

Hint 1

The Jacobian of this linear change is again one. Keep the factor $(2\pi)^{-2}$ outside the two integrations.

Hint 2

The Fourier identity is $\int dY\,e^{i aXY}=2\pi\delta(aX)$. The regulator must be removed only after the convergent integral is formed.

Solution

The phase becomes $ikXY/(2\pi)$ and the regulator is $\epsilon(2Y^2+X^2/2)/(4\pi^2)$. The regulated Gaussian calculation therefore has the distributional limit

$$\int dY\,e^{ikXY/(2\pi)} =\frac{4\pi^2}{|k|}\delta(X).$$

Its $4\pi^2$ cancels the original measure denominator. Evaluating the matter factor at zero gives $1/(4|k|)$. A negative level complex-conjugates the original phase; it does not make the partition function negative.

At $k=0$ there is no Fourier constraint. The regulated common-center Gaussian contributes a factor proportional to $\epsilon^{-1/2}$, while the relative matter integral tends to $\int dX/[4\cosh^2(X/2)]=1$. Thus the full answer diverges rather than tending to the nonzero-level formula. Level zero is outside the fixed ABJM model under discussion, and taking $k\to0$ is not a uniform extension of this proof.

2. Change the damping function

In the $x,y$ variables, replace the Gaussian regulator by $e^{-a|y|}$ with $a>0$, leaving the integrable matter factor unchanged. Show that each integral is absolutely convergent, evaluate the $y$ integral, and prove that $a\downarrow0$ gives the same answer. Does this prove independence from every possible prescription?

Hint 1

Split the Fourier integral at $y=0$ and integrate the two decaying exponentials. The resulting function of $x$ is positive.

Hint 2

Set $x=a u/(2\pi|k|)$. The remaining normalized weight is proportional to $1/(1+u^2)$.

Solution

The absolute double integral is $(2/a)\int f(x)dx=1/(\pi a)$, finite for each $a>0$. The Fourier transform gives

$$Z_a^{\rm L}(1,k)= \int_{\mathbb R}dx\,f(x) \frac{2a}{a^2+4\pi^2k^2x^2} =\frac1{\pi|k|} \int_{\mathbb R} \frac{f(a u/(2\pi|k|))}{1+u^2}\,du .$$

The bound $f\leq1/4$ supplies an integrable dominant function independent of $a$. Thus $Z_a^{\rm L}\to f(0)/|k|=1/(4|k|)$. The Gaussian and Laplace regulators give two explicit approximate identities with the same mass and support in the limit.

This proves agreement of these two prescriptions for this integrand, not a theorem about all cutoffs, all complex contour rotations or all ranks. The finite-regulator values and rates of convergence need not agree. The regulator is an integration device and has not been identified as a new supersymmetric term in the field-theory action.

3. Move the Fourier constraint

Insert the bounded factor $e^{2\pi i(\eta y+\xi x)}$, with real dimensionless $\eta,\xi$, into the Gaussian-regulated rank-one integral. Treat this as a mathematical test of the distribution; no physical mass or FI normalization is being assumed. Find its limit and explain how it tests the sign of the inherited classical phase.

Hint 1

The $y$ frequency is now $kx+\eta$, so its Gaussian Fourier transform peaks at a nonzero value of $x$.

Hint 2

The extra $x$ phase must be evaluated at the peak together with the matter factor. Do not drop it merely because the uninserted answer is real.

Solution

At positive regulator, integrate $y$ first:

$$Z_\epsilon(\eta,\xi)= \sqrt{\frac{\pi}{2\epsilon}} \int dx\,f(x)e^{-\epsilon x^2/2+2\pi i\xi x} e^{-\pi^2(kx+\eta)^2/(2\epsilon)}.$$

Shift to $x=-\eta/k+\sqrt{\epsilon}\,u/(\pi|k|)$. The fixed Gaussian in $u$ dominates the bounded remaining factors. Its normalization gives

$$\lim_{\epsilon\downarrow0}Z_\epsilon(\eta,\xi) =\frac{e^{-2\pi i\xi\eta/k}} {4|k|\cosh^2(\pi\eta/k)}.$$

At $\eta=\xi=0$ this reduces to the partition function. Reversing the sign of the original Chern–Simons phase moves the support to $x=+\eta/k$ and conjugates the displayed source phase while leaving its magnitude unchanged. This diagnostic can detect a phase-sign error that the real uninserted answer alone cannot reveal. It is not a derivation of a Wilson-loop insertion or a physical background-source dictionary.

The first exact check is now complete: the direct rank-one integral has a defined contour prescription, a proved limit, and controlled regulator errors. Higher-rank evaluation and the independent spectral comparison require their own constructions.

Real masses and squashed spheres changes the calculation by specifying a complete geometry and flavor multiplet. Track the auxiliary branch, recover this round answer, and calculate a finite-rank mass response.

Bosonic supersymmetric Wilson loops inserts an exponentially growing character into this same measure. The new contour argument explains why the partition-function limit alone does not settle the loop expectation.

Two different large-N limits next specifies which sequence of theories an asymptotic answer describes. Its sourced saddle input is a different kind of result from the exact finite-rank calculation above.

Turn the matrix integral into a Fermi gas independently reproduces this normalization as an operator trace and calculates the first two-particle answers with the required exchange subtraction.

What localization shares across three and four dimensions compares the two completed sphere calculations. Identify their different charge squares, surviving measures and observables, then calculate a physical coupling response that the deformation identity does not set to zero.