Free-chiral squashing response

Can a calculation reproduce a current coefficient and tell you why an apparently convergent answer is wrong? This laboratory evaluates the free-chiral squashing response in three ways. One method integrates the differentiated determinant, another differentiates its unpaired modes, and a third evaluates the determinant at finite squashing before taking differences.

Read the free-chiral capstone first. It supplies the determinant and its subtraction prescription. Here the task is to implement those formulas, expose their numerical failure modes, and reproduce

$$F''(1)=\frac{\pi^2}{8},\qquad \tau_{RR}=\frac{2}{\pi^2}F''(1)=\frac14. \tag{LAB06.1}$$

Fix the physical path before taking a derivative

The theory is one massless three-dimensional $\mathcal N=2$ free chiral multiplet, with scalar R-charge $r=1/2$ held fixed. There is no gauge group, gauge-rank parameter, superpotential or flavor mass. The Lorentzian parent has four real Poincaré supercharges. We use the four rigid supersymmetry directions of the homogeneous sphere family; this does not assert that the curved background retains the full superconformal algebra.

For completeness, write its fixed sources using the global round forms $\eta^a$, with $d\eta^1=2\eta^2\wedge\eta^3$ cyclically and positive orientation. Define the dimensionless parameters $u=\log b$, $h=\cosh u$ and $d=b-b^{-1}$:

$$\begin{aligned} g_b&=\ell^2\bigl[(\eta^1)^2+(\eta^2)^2+h^2(\eta^3)^2\bigr],\\ A&=V=dh\,\eta^3,\qquad C=\frac{i\ell d}{2}\eta^3,\qquad H=\frac{ih}{\ell}. \end{aligned} \tag{LAB06.2}$$

Here $\ell>0$ is a fixed radius scale, $b>0$, and the ordinary sphere has its unique spin structure and trivial R bundle. These sources are not integrated over. In particular, the round auxiliary field is $H=+i/\ell$. The background construction and its normalization are in the source-response lesson.

We calculate the real representative $F=-\log|Z|$. In the inherited conformal supersymmetric prescription, the allowed contact-term changes on this family are imaginary and do not alter the response. Adding an arbitrary real subtraction would define a different problem. The dimensionless $F$ and $\tau_{RR}$ below do not depend on $\ell$.

The fixed-R determinant uses

$$\mathcal F(b,\alpha)=-\int_0^\infty\frac{dx}{2x} \left[\frac{\sinh(2\alpha x)}{\sinh(bx)\sinh(x/b)}-\frac{2\alpha}{x}\right], \qquad F_{\rm phys}(b)=\mathcal F(b,h/2). \tag{LAB06.3}$$

Its starting domain is real $|\alpha|<h$. The dimensionless integration variable $x$ is a rescaled proper time. The subtracted term removes the short-time divergence; it is retained at every squashing value.

The mode product in Imamura–Yokoyama (86)–(87) fixes $\alpha=h(1-r)$. In their parameters, $v=1/h$ and $w=-i(b-b^{-1})/(2h)$, where $w$ denotes their $u$, not our $\log b$. Then $b=(1+iw)/v$. The real-$b$ branch uses the continuation specified in the capstone; it is not a claim that different squashed metrics have identical sources.

By contrast, P (8.11) literally prints $\sinh x$ and $1/x$, so it holds $\alpha=1/2$ fixed. Call that function $G(b)$. At the round point, $\partial_\alpha\mathcal F(1,1/2)=0$ and $h'(1)=0$. The chain rule gives $F_{\rm phys}''(1)=G''(1)$. The program checks this common Hessian and also shows that the two finite-$b$ functions differ. It never silently replaces the printed numerator.

Evaluate the differentiated integral stably

Differentiating $G$ produces the convergent integral

$$G''(1)=\int_0^\infty J(x)\,dx,\qquad J(x)=\frac{\sinh x\cosh x-x}{\sinh^3x} =\frac23-\frac{x^2}{5}+O(x^4). \tag{LAB06.4}$$

At small $x$, the numerator subtracts two numbers of order $x$ to recover a quantity of order $x^3$. Ordinary floating-point evaluation therefore loses accuracy. Increasing the quadrature’s sampling density does not repair a wrongly evaluated integrand.

The code instead defines $s(x)=\log(\sinh x/x)$ and uses

$$J(x)=\frac{\operatorname{expm1}(s(2x))}{x^2}e^{-3s(x)}. \tag{LAB06.5}$$

The function expm1(t) evaluates $e^t-1$ without first rounding $e^t$ to one. For $|x|<0.01$, log_sinhc evaluates the convergent Taylor series of $s$ through degree 48. Larger positive arguments use exponential expressions. The tested working precisions are stated below; this finite series is not advertised as an arbitrary-precision algorithm at every requested precision.

There is a simple independent bound on the omitted large-$x$ tail. The numerator of $J$ is positive because its derivative is $2\sinh^2x$. Also $J<\cosh x/\sinh^2x$. Consequently

$$0<\int_L^\infty J(x)\,dx<\frac1{\sinh L}. \tag{LAB06.6}$$

For $L=24$ the upper bound is $7.551\times10^{-11}$; for $L=48$ it is $2.851\times10^{-21}$. These are analytic tail bounds. Numerical quadrature on the retained interval still has its own error, checked by increasing arithmetic precision. A repeat at higher precision is useful evidence, not an interval proof.

Differentiate actual modes as a second check

The unpaired frequencies are $\omega_{mn}=mb+n/b+h$, with $m,n\geq0$. At $u=0$, group the modes by $k=m+n+1$. For each mode, $\omega=k$, $\partial_u\omega=m-n$ and $\partial_u^2\omega=k$. Writing $f(\omega)=\log(\omega-1/2)-\log(\omega+1/2)$, the program forms the literal shell sum

$$S_k=\sum_{m=0}^{k-1} \left[f'(k)k+f''(k)(2m-k+1)^2\right]. \tag{LAB06.7}$$

This tests the mode multiplicity and both derivatives. The large-$k$ limit is $1/3$. The capstone’s continued spectral prescription includes the finite contribution $(1/3)\zeta(0)=-1/6$, so the finite approximation is

$$R_K=-\frac16+\sum_{k=1}^K(S_k-\tfrac13). \tag{LAB06.8}$$

The known partial-fraction identity converts this into an odd-square sum. Define $c_K=2/[3(2K+1)]-1/[2(2K+1)^2]$. The exact infinite answer lies strictly between

$$R_K+c_K+\frac1{2(2K+1)} \quad\hbox{and}\quad R_K+c_K+\frac1{2(2K-1)}. \tag{LAB06.9}$$

At $K=128$, the program obtains $1.23369296036<H_0<1.23370821939$, where $H_0$ denotes the Hessian, not the auxiliary field $H$. This slower but independently controlled calculation contains $\pi^2/8$. Increasing precision alone does not narrow a fixed-$K$ tail interval.

The separate --omit-finite-part mode removes only $-1/6$. Its intervals converge toward $\pi^2/8+1/6$ and exclude the correct answer. The program exits successfully after detecting this intended error; it has not validated the altered prescription.

Reproduce the output and its error scales

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

Save the editable calculation, pinned requirement and complete expected output in one folder. The tested environment is Python 3.13.7 with mpmath 1.3.0. From that folder in PowerShell:

python -m venv .venv
.venv/Scripts/python.exe -m pip install -r requirements.txt
.venv/Scripts/python.exe squashing_response.py
.venv/Scripts/python.exe squashing_response.py --omit-finite-part

On macOS or Linux, create the environment with python3 -m venv .venv and use .venv/bin/python afterward. Installation needs registry access; running the calculation needs no network or repository files. The optional --dps 70 increases working precision. The program accepts 40–80 working digits and repeats selected evaluations with 20 additional digits.

Quantity	Code name
Dimensionless log-squashing $u$	u
Stable differentiated integrand	response_kernel
Actual subtracted finite-squashing integral	free_energy
Differentiated individual modes	literal_shell
Subtracted sum and analytic tail bounds	shell_interval

The default result includes

{
  "integral_Hessian": "1.233700550136169827354",
  "tau_RR": "0.25",
  "working_decimal_digits": 50,
  "stability_decimal_digits": 70,
  "all_high_precision_comparisons_pass": true
}

Five integral, symmetry and precision comparisons pass a scaled tolerance of $10^{-35}$. This does not make every number in the output accurate to 35 digits. To see why, first evaluate the physical integral, then form

$$D_\delta=\frac{2[F_{\rm phys}(e^\delta)-F_{\rm phys}(1)]}{\delta^2}, \qquad \mathcal R_\delta=\frac{4D_{\delta/2}-D_\delta}{3}. \tag{LAB06.10}$$

Exchanging $m,n$ makes this determinant even in $u$, so the one-sided-looking expression equals the centered difference. The subtraction in (LAB06.3) is integrated explicitly: its tail from one to infinity contributes $+\alpha$. It is not dropped to make the integral easier.

$\delta$	Error in $D_\delta$	Error in $\mathcal R_\delta$
$0.1$	$3.4272\times10^{-3}$	$1.9432\times10^{-6}$
$0.05$	$8.5534\times10^{-4}$	$1.2157\times10^{-7}$
$0.025$	$2.1374\times10^{-4}$	$7.6001\times10^{-9}$

These are finite-step errors: the successive Richardson errors fall by approximately sixteen. A 50-digit arithmetic setting does not remove the $O(\delta^4)$ bias. At $u=0.1$, the physical and printed fixed-shift free energies differ by $3.5838361\times10^{-5}$, despite their equal round Hessians. That difference is a source-path distinction, not numerical noise.

Change one input

1. Sample closer to the endpoint

Estimate how the relative error in the naive numerator of (LAB06.4) grows as $x$ decreases, if each elementary operation has relative rounding scale $\epsilon_{\rm mach}$. Why can evaluating it at $x=10^{-9}$ with ordinary double precision be worse than evaluating it at $10^{-5}$? What should a small-$x$ test of the stable routine approach?

Hint 1

Before subtraction both terms are of order $x$; their difference starts at order $x^3$.

Hint 2

Compare an absolute numerator error of order $\epsilon_{\rm mach}x$ with the desired numerator. Use the displayed Taylor series for the limit.

Solution

The numerator’s relative rounding error can scale as $\epsilon_{\rm mach}/x^2$. With $\epsilon_{\rm mach}$ of order $10^{-16}$, this estimate is about $10^{-6}$ at $10^{-5}$ but becomes order $10^2$ at $10^{-9}$. Exact cancellations or correlated library errors can change a particular observed value; there is no reliable small relative-error guarantee there.

The stable result must approach $2/3$, with its first deviation $-x^2/5$. A more aggressive quadrature that samples the naive expression closer to zero can amplify the failure. The repair is the stable expression or its controlled series, followed by a precision check, not an assertion that smaller coordinates are automatically better.

2. Balance step size against evaluation error

Suppose each of the three free energies used in $\mathcal R_\delta$ has an absolute error at most $\eta$. Keep their shared center value explicit. Derive an error bound for $\mathcal R_\delta$ from these evaluations. If its finite-step bias is bounded by $C\delta^4$, what step minimizes the resulting bound? Does comparing two working precisions by itself prove the assumed bound $\eta$?

Hint 1

Expand the Richardson combination as a linear combination of $F(e^{\delta/2})$, $F(e^\delta)$ and $F(1)$ before taking absolute values.

Hint 2

The two contributions scale as $\delta^4$ and $\delta^{-2}$. Differentiate their sum to find its minimum.

Solution

With $F$ denoting the physical path,

$$\mathcal R_\delta= \frac{32F(e^{\delta/2})-2F(e^\delta)-30F(1)}{3\delta^2}.$$

The triangle inequality bounds its evaluation error by $64\eta/(3\delta^2)$. Combining this with the assumed bias gives

$$E\leq C\delta^4+\frac{64\eta}{3\delta^2}, \qquad \delta_{\rm opt}^6=\frac{32\eta}{3C}.$$

For the illustrative inputs $C=0.01945$ and $\eta=10^{-30}$, the optimal scale is about $2.86\times10^{-5}$. The observed leading bias in the table motivates that value of $C$; the table alone does not prove a uniform bias bound at every step. Likewise, agreement at two precisions is evidence of stability, not a rigorous bound on either quadrature’s error. A certified combined interval would need certified bounds for both terms. The exercise explains why reducing the step indefinitely at fixed numerical accuracy eventually becomes counterproductive.

3. Differentiate a different squashing parameter

The original mode calculation also uses $v=1/h=\operatorname{sech}u$. Near the round point, write $F_{\rm phys}(e^u)=F_0+H_0u^2/2+O(u^4)$. Find the one-sided derivative $dF/dv$ at $v=1$. Why is it not another second derivative equal to $H_0$?

Hint 1

Here $v=1-u^2/2+O(u^4)$, so $v'(0)=0$. It is not an invertible linear coordinate near $u=0$.

Hint 2

Eliminate $u^2$ between the two expansions. The determinant is even in $u$, so either nearby branch gives the same answer.

Solution

For $v$ approaching one from below, $u^2=2(1-v)+O((1-v)^2)$ and $F=F_0+H_0(1-v)+O((1-v)^2)$. Therefore

$$\left.\frac{dF}{dv}\right|_{1^-}=-H_0=-\frac{\pi^2}{8}.$$

The singular coordinate change has turned the leading quadratic variation in $u$ into a linear variation in $1-v$. It is not legitimate to copy the $b$-Hessian formula with $b$ replaced by $v$. By contrast, $b=e^u$ has nonzero derivative at zero, and the vanishing first derivative of $F$ makes its $b$ and $u$ Hessians agree at the round point. A comparison of responses requires the parameter path as well as the numerical normalization.

This experiment supports the free, massless Gaussian result in the stated source prescription. It does not determine a contact-term phase, a massive determinant at an infrared pole, or the response of an interacting theory. The useful output is the coefficient together with a reproducible explanation of its subtraction, source path and numerical errors.