Responses on deformed spheres

A fourth derivative can be numerically stable and still differentiate the wrong mass. This laboratory reproduces a published sphere response, then changes the geometry while retaining the same flavor generator. A separate probability-density calculation checks the derivatives without differentiating the partition function numerically.

Prepare with ABJM masses and squashing and current two-point normalizations. The localization derivation belongs to those lessons. Here the aim is to turn their precisely normalized answer into a reproducible calculation and a useful failure test.

Fix the source before calculating

Use compact $U(1)_1\times U(1)_{-1}$ ABJM, retaining both gauge circles and adding no quotient of their common center. The rank and positive level are exactly $N=k=1$. The undeformed flat conformal theory has sixteen real Poincaré supercharges and sixteen conformal partners. Its manifest three-dimensional $\mathcal N=2$ formulation has four real Poincaré generators. Our generic toric background supplies one Euclidean rigid-label direction of each R charge; this is not a count of all symmetries at a special conformal point.

Let $L>0$ be a length and $b>0$ the dimensionless squashing parameter. Set $\ell_1=Lb$, $\ell_2=L/b$, and use $0\leq\chi\leq\pi/2$, with both angles of period $2\pi$. The complete background inherited from the prerequisite is

$$\begin{aligned} ds^2&=f^2d\chi^2+\ell_1^2\cos^2\chi\,d\varphi^2 +\ell_2^2\sin^2\chi\,d\vartheta^2,\\ f^2&=\ell_1^2\sin^2\chi+\ell_2^2\cos^2\chi,\qquad H=\frac{i}{f},\qquad V=C=0,\\ A^{(R)}&=\frac12\left(1-\frac{\ell_1}{f}\right)d\varphi +\frac12\left(1-\frac{\ell_2}{f}\right)d\vartheta. \end{aligned} \tag{LAB12.1}$$

Use the orientation of the ordered diagonal coframe, the unique spin structure on the unquotiented sphere and the trivial R line bundle. The two-form source has patch potentials with $dB=d\mathrm{vol}/f$; it need not have one global potential. The round limit has $H=+i/L$. This positive branch gives the first-node Cartan phase $e^{-i\pi s^2}$ and the opposite second-node phase, as derived in the prerequisite. It does not change the inherited coefficient $+ik/(4\pi)$ of the Euclidean gauge Chern–Simons action.

In the ordered chiral list $(A_1,A_2,B_2,B_1)$ take scalar R-charge $1/2$ and the flavor generator

$$T=\frac12\operatorname{diag}(1,1,-1,-1),\qquad \operatorname{Tr}_4T^2=1,\qquad x=L\sigma_F. \tag{LAB12.2}$$

Here $x$ is the dimensionless source and $\sigma_F$ its dimensional value. Freeze the flavor multiplet at $a_F=0$, $\sigma_F=x/L$, $D_F=-H\sigma_F$, and zero gauginos. Each chiral therefore feels mass $+x/(2L)$ or $-x/(2L)$. The auxiliary source changes with the mass. It is not eliminated by a field equation.

This is the first generator in Binder–Chester–Pufu (3.7). In integer-charge units the source is $M=x/2$. The prerequisite’s displayed charge table chooses a different Cartan direction; the interchange symmetry in Kubo–Nosaka–Pang (2.5) gives the same one-mass partition function. The conversion of charge units does not identify the two charge tables field by field.

Keep the inherited double-sine prescription, the Abel-defined Cartan integral, and no added integer background Chern–Simons counterterms or new subtraction scale. Define $F_b(x)=-\log Z_b(x)$ on the branch continuous from real zero mass. The rank-one result is

$$Z_b(x)=D_b(x/2)^2. \tag{LAB12.3}$$

The function $D_b$ pairs two chiral determinants. This is KNP (4.11), with its ABJM masses $(m_1,m_2,m_3)=(x,0,0)$ and dual variables $(\zeta,m)=(x/2,-x/2)$. No large-rank approximation is used. We keep real $x$; we do not follow that paper’s special complex adjoint-mass tuning $m_b=i(b^2-3)/(4b)$.

Reproduce the round answer and change one input

At $b=1$, $D_1(w)=1/(2\cosh\pi w)$. Consequently

$$\begin{aligned} Z_1(0)&=\frac14,\\ F_1(x)&=\log4+\frac{\pi^2x^2}{4}-\frac{\pi^4x^4}{96}+O(x^6),\\ F_{xx}(1;0)&=\frac{\pi^2}{2},\qquad \boxed{F_{xxxx}(1;0)=-\frac{\pi^4}{4}.} \end{aligned} \tag{LAB12.4}$$

The last line reproduces BCP (3.43). It is the derivative of the complete source-dependent action. Quadratic scalar-mass terms contribute, so it is not just an integrated four-point function of the linear insertion. The correlator calculation derives that distinction. In BCP’s normalization, (3.9) gives $c_T=32\operatorname{Re}F_{xx}/\pi^2=16$ at this round point, where one free real scalar has $c_T=1$.

For the controlled change $b^2=3$, the finite product KNP (A.8) gives

$$D_{\sqrt3}(w)=\frac1{1+2\cosh(2\pi w/\sqrt3)},\qquad F_{\sqrt3}(x)=2\log[1+2\cosh(\pi x/\sqrt3)]. \tag{LAB12.5}$$

To obtain the denominator, multiply the two factors $2\cosh(u+i\pi/6)$ and $2\cosh(u-i\pi/6)$. Their product is $4\cosh^2u-1=1+2\cosh2u$. Expanding the logarithm at $x=0$ gives

Fixed geometry	Zero-mass partition function	Second mass derivative	Fourth mass derivative
$b^2=1$	$1/4$	$\pi^2/2$	$-\pi^4/4$
$b^2=3$	$1/9$	$4\pi^2/9$	$-4\pi^4/27$

Every derivative in this table is with respect to $x$ at fixed $b,L$. Comparing the rows changes the background; it does not compute $\partial_b^2F$. Nor does inserting the second row into the round-sphere formula define a new flat-space $c_T$.

Use an integral that can be tested

The subtraction at zero mass turns the regulated determinant representation into a convergent real integral. Define

$$Q_b=b+b^{-1},\qquad K_b(t)=\frac{\sinh(Q_bt/2)}{\sinh(bt)\sinh(t/b)}.$$

Subtract KNP (A.10) at $w=0$, combine the two halves of its contour, and put $w=x/2$. The result and its differentiated forms are

$$\begin{aligned} \Delta F_b(x)&\equiv F_b(x)-F_b(0) =4\int_0^\infty\frac{K_b(t)}t\sin^2\frac{xt}{2}\,dt,\\ F_{xx}(b;0)&=2\int_0^\infty tK_b(t)\,dt,\\ F_{xxxx}(b;0)&=-2\int_0^\infty t^3K_b(t)\,dt. \end{aligned} \tag{LAB12.6}$$

Near zero, $K_b(t)=Q_b/(2t)+O(t)$. Thus the three integrands approach $Q_bx^2/2$, $Q_b$, and zero, respectively. At infinity they decay exponentially for fixed positive $b$. These bounds also justify the mass derivatives on a bounded real interval. The unsubtracted real integral would diverge; it must not be used to guess the absolute normalization. We get $Z_b(0)$ from the exact finite product instead.

The program supports $b^2=1,3,5,7,9$. For $b^2=2n-1$, that product is

$$D_b(w)=\prod_{j=1}^{n} \left[2\cosh\left(\frac{\pi w}{b}+i\theta_j\right)\right]^{-1}, \qquad \theta_j=\frac{\pi}{b^2}\left(\frac{n+1}{2}-j\right). \tag{LAB12.7}$$

Conjugate factors pair, giving a positive determinant for real $w$. Its analytic derivatives test the integral at each supported geometry. The implementation evaluates $K_b$ using exponentials and expm1, rather than overflowing hyperbolic functions or subtracting nearly equal numbers. It uses $\sin^2(xt/2)$ rather than $1-\cos xt$ near zero.

Mathematical quantity	Program name
$b^2$, dimensionless geometry	b_squared
$x=L\sigma_F$, dimensionless test mass	mass
$D_b(w)$, absolutely normalized paired determinant	determinant(w, b_squared)
$F_{xx},F_{xxxx}$ at zero mass	Fxx, Fxxxx
Upper integration limit	cutoff

Run and interpret the downloads

Download these four files into one folder:

Download the complete lab (ZIP) with the program, pinned dependencies, reference output and run instructions.

• Python program
• Pinned dependency
• Complete round-sphere output
• Complete changed-geometry output

The tested environment is Python 3.13.7 with mpmath 1.3.0. No notebook, repository imports or network access are used during the calculation. From that folder run:

python -m venv .venv
.venv\Scripts\python -m pip install -r requirements.txt
.venv\Scripts\python -I sphere_responses.py
.venv\Scripts\python -I sphere_responses.py --b-squared 3

On systems using a bin directory for virtual environments, replace .venv\Scripts\python with .venv/bin/python. The default run uses 50 decimal digits and a separate mass probe $x=0.2$; the reported derivatives remain at zero mass. Its key values are

Z_zero_mass    = 0.25
Z_probe       = 0.2268631269963737601119551
Fxx           = 4.934802200544679309417245
Fxxxx         = -24.35227275850060930911008
round_cT      = 16

The JSON distinguishes the integral and product evaluations. Six comparisons, including a repeat at 70 digits, must satisfy a scaled tolerance of $10^{-35}$. The program stops with an error if they fail. This tolerance applies to those comparisons, not to finite differentiation steps.

Run the deliberate error separately:

.venv\Scripts\python -I sphere_responses.py --wrong-mass-normalization

It differentiates $D_b(x)^2$ instead of $D_b(x/2)^2$. The second derivative is four times too large and the fourth is sixteen times too large. The output must say detected: true. This is a changed source, not a rounding error. To test arithmetic precision without changing the source, use --dps 40 or --dps 70.

Separate three numerical errors

First, truncating the integral has an analytic bound. For $t\geq T$, put $c=Q_b/2$ and $d_T=(1-e^{-2bT})(1-e^{-2T/b})$. The exponential form of the kernel gives $K_b(t)\leq2e^{-ct}/d_T$. Hence

$$\begin{aligned} |\mathrm{tail}(F_{xx})|&\leq\frac{4\Gamma(2,cT)}{d_Tc^2},& |\mathrm{tail}(F_{xxxx})|&\leq\frac{4\Gamma(4,cT)}{d_Tc^4},\\ |\mathrm{tail}(\Delta F)|&\leq\frac{8e^{-cT}}{d_TcT}. \end{aligned} \tag{LAB12.8}$$

Here $\Gamma(a,z)=\int_z^\infty u^{a-1}e^{-u}du$ is the upper incomplete gamma function. The default round cutoff is about $149.668$. Its fourth-response tail is bounded by $1.369\times10^{-58}$. This bound controls the omitted interval, not the quadrature on the retained interval.

Second, the precision repeat tests arithmetic and adaptive integration. It is not an interval-certified quadrature theorem. The independent product comparison makes an undetected integration error less likely.

Third, finite differences retain their own bias. The program prints steps $h=0.08,0.04,0.02$ even though it does not need them for its derivative result. At the last step the round fourth-difference estimate is about $-24.32027$, still about $0.03201$ away from the derivative. More working digits at the same $h$ do not remove that error.

Check with a normalized density

The exact Fourier identity KNP (A.7) supplies a different test. Since $D_b(s)>0$ here and $\int D_b(s)ds=D_b(0)$, define the probability density

$$p_b(s)=\frac{D_b(s)}{D_b(0)},\qquad \frac{Z_b(x)}{Z_b(0)}= \left[\int_{-\infty}^{\infty}p_b(s)e^{i\pi xs}\,ds\right]^2. \tag{LAB12.9}$$

The frequency is $\pi x$, not $2\pi x$, because the determinant argument is $x/2$. Expanding the characteristic function and then its logarithm yields

$$F_{xx}=2\pi^2\langle s^2\rangle, \qquad F_{xxxx}=-2\pi^4 \bigl(\langle s^4\rangle-3\langle s^2\rangle^2\bigr). \tag{LAB12.10}$$

Direct integration of this positive density checks both normalization and the connected fourth moment. It gives $(\langle s^2\rangle,\langle s^4\rangle)=(1/4,5/16)$ at $b^2=1$ and $(2/9,2/9)$ at $b^2=3$, reproducing the table. The independent calculation also checks $b^2=5,7,9$. This auxiliary probability measure is a representation of the normalized determinant, not a claim that the original oscillatory gauge integral is a positive measure.

Exercises

1. Change units or change the source

At $b=1$, let the dimensional source be $\mu$ and $x=L\mu$. Find the second and fourth $\mu$ derivatives at zero. Then add an allowed integer-level background counterterm in the compact integer-charge cover, $\Delta F=i\pi\kappa M^2$, where $M=x/2$ and $\kappa\in\mathbb Z$. Which of the reported real responses changes? Compare with the deliberate wrong-normalization mode.

Hint 1

Each $\mu$ derivative contributes a factor of $L$. Distinguish rescaling the variable from adding a polynomial to the function.

Hint 2

The counterterm is $i\pi\kappa x^2/4$. It has no fourth derivative, and its second derivative is imaginary.

Solution

At fixed $L$, $F_{\mu\mu}=L^2\pi^2/2$ and $F_{\mu\mu\mu\mu}=-L^4\pi^4/4$. These powers restore the dimensions of derivatives with respect to a mass. The specified counterterm shifts $F_{xx}$ by $i\pi\kappa/2$ and leaves $F_{xxxx}$ unchanged. Thus neither real response changes in this contact-term comparison. It does change a phase-sensitive second source response. The integer condition refers to the stated compact cover, not an arbitrary redefinition of the faithful flavor group.

By contrast, the wrong-normalization mode replaces the entire function by $F(2x)$, multiplying its real second and fourth derivatives by four and sixteen. No imaginary quadratic counterterm repairs that mismatch. The lab fixes $\kappa=0$ throughout its outputs.

2. Predict the derivative bias

For an even analytic $F$, write $\Delta F(h)=c_2h^2/2!+c_4h^4/4!+c_6h^6/6!+\cdots$. Derive the leading errors in

$$D_2(h)=\frac{2\Delta F(h)}{h^2},\qquad D_4(h)=\frac{2\Delta F(2h)-8\Delta F(h)}{h^4}.$$

Evaluate the round-theory prediction for the fourth-difference error at $h=0.02$. If each of the five raw free energies in the symmetric fourth-difference stencil has absolute numerical error at most $\epsilon$, bound its contribution to the derivative error.

Hint 1

The coefficients of $h^2$ in the fourth-difference numerator cancel. Keep the $h^6$ term.

Hint 2

The round expansion has $c_6=\pi^6/2$. The five stencil coefficients are $1,-4,6,-4,1$.

Solution

Substitution gives $D_2=c_2+c_4h^2/12+O(h^4)$ and $D_4=c_4+c_6h^2/6+O(h^4)$. Therefore the round fourth-difference bias starts at $\pi^6h^2/12$, about $0.0320463$ at the chosen step; the actual $0.0320060$ includes the next terms. Halving $h$ reduces the leading bias by four. This is a convergence prediction, not an exact finite-step equality.

The triangle inequality bounds the numerical-error contribution by $16\epsilon/h^4$. Decreasing the step eventually magnifies fixed input errors, so step and working precision must be varied separately. The released derivative integrals avoid this subtraction problem; the stencil remains a diagnostic.

3. Recover a connected response from moments

At $b^2=3$, use the density moments $\langle s^2\rangle=2/9$ and $\langle s^4\rangle=2/9$ to recover both mass responses. What wrong fourth response results from using the raw fourth moment? Explain why the correct negative answer is compatible with a positive current two-point coefficient.

Hint 1

Expand the characteristic function to fourth order. Its logarithm subtracts the square of the quadratic term.

Hint 2

The fourth cumulant is $2/9-3(2/9)^2=2/27$, not $2/9$.

Solution

Equation (LAB12.10) gives $F_{xx}=4\pi^2/9$ and $F_{xxxx}=-4\pi^4/27$. Replacing the cumulant with the raw moment instead gives $-4\pi^4/9$, three times the correct answer. The logarithm of the partition function selects connected combinations. Positivity fixes the sign of the appropriate two-point norm; it does not require every fourth derivative of $F$ to be positive. The positive density has a positive variance and a positive fourth cumulant here, which produce a positive second and negative fourth response because their Fourier and logarithm signs differ.

These tests establish a specified rank-one response and its controlled background variation. Higher rank, another level, a different mass path or a separated-point four-point function requires additional physics. The deformed-sphere correlator study explains how to pose that next calculation without discarding the nonlinear source terms.