Reproduce the free-chiral response

The previous lesson expressed a current two-point coefficient as a derivative of a partition function. We will now compute that derivative from the free matter integral. The target is $\partial_b^2\operatorname{Re}F|_{b=1}=\pi^2/8$. Reaching it requires more than differentiating an infinite product: we must specify the sources, retain the ultraviolet subtraction, and distinguish a physical change of geometry from a change of R-current.

Prepare with what squashing measures and the free chiral determinant. The former supplies the background and current normalization. The latter supplies mode cancellation and the finite normalization of the round determinant.

State the complete free problem

The theory is one massless, free three-dimensional $\mathcal N=2$ chiral multiplet, with no gauge integral or superpotential. Its superconformal scalar R-charge is $r=1/2$, held fixed throughout the physical calculation. The parent has four real Poincaré supercharges; the chosen Euclidean background preserves the four rigid directions described in the preceding lesson.

Let $\eta^a$ be the global round forms with $d\eta^1=2\eta^2\wedge\eta^3$ cyclically and positive $\eta^1\wedge\eta^2\wedge\eta^3$ orientation. At radius scale $\ell$, use

$$\begin{aligned} h&=\frac{b+b^{-1}}2,\qquad d=b-b^{-1},\qquad b>0,\\ g_b&=\ell^2\bigl[(\eta^1)^2+(\eta^2)^2+h^2(\eta^3)^2\bigr],\\ A&=V=dh\,\eta^3,\qquad C=\frac{i\ell d}{2}\eta^3,\qquad H=\frac{ih}{\ell}. \end{aligned} \tag{O09.1}$$

This is the positive-$H$ homogeneous family of P, on the ordinary unquotiented sphere with its unique spin structure and trivial R bundle. The sources are fixed, not integrated over. At the round point $H=+i/\ell$, and the paper’s Ricci scalar is $-6/\ell^2$. We impose no extra flavor mass. The negative-$H$ round calculation in the determinant lesson supplies the same massless real normalization; its massive phase dictionary must not be imported without translation.

Use the conformal supersymmetric subtraction prescription of the preceding lesson. Its allowed finite contact contributions on this family are imaginary, so the observable is $\operatorname{Re}F=-\log|Z|$. Below we write a real determinant representative; this does not remove the global meaning of its possible contact-term phase.

From unpaired modes to a convergent integral

Introduce a dimensionless spectral shift $\alpha$ temporarily. The net unpaired factors are

$$\omega_{mn}=mb+nb^{-1}+h,\qquad Z(b,\alpha)= \prod_{m,n\geq0}^{\rm reg} \frac{\omega_{mn}+\alpha}{\omega_{mn}-\alpha}, \qquad \alpha=h(1-r). \tag{O09.2}$$

The product is the localized mode-pairing result in Imamura–Yokoyama (86)–(87). We use that sourced spectral input and derive its regularization and response here. Their parameter $v$ equals $1/h$. Writing $w$ for their parameter $u$, $b=(1+iw)/v$, with $w=-id/(2h)$ on our real-$b$ branch. Thus their real geometric branch is analytically continued when needed; $w$ is unrelated to the $\log b$ used below. Interchanging the two integer labels implements $b\leftrightarrow b^{-1}$.

For fixed physical $r=1/2$, the shift is $\alpha=h/2$, not a constant. At $b=1$, $\omega_{mn}=m+n+1=k$. There are exactly $k$ pairs with this value, recovering the round product $\prod_{k\geq1}[(k+\alpha)/(k-\alpha)]^k$.

To obtain an integral, use the elementary identity $\log(A/B)=\int_0^\infty(e^{-Bt}-e^{-At})\,dt/t$ for positive $A,B$. Summing the two geometric series gives

$$\sum_{m,n\geq0} \bigl[e^{-(\omega_{mn}-\alpha)t} -e^{-(\omega_{mn}+\alpha)t}\bigr] =\frac{\sinh(\alpha t)} {2\sinh(bt/2)\sinh(t/2b)}.$$

The small-$t$ divergence is still present. With $t=2x$, its prescribed subtraction defines

$$\mathcal F(b,\alpha)= -\int_0^\infty\frac{dx}{2x} \left[ \frac{\sinh(2\alpha x)} {\sinh(bx)\sinh(b^{-1}x)} -\frac{2\alpha}{x} \right],\qquad F_{\rm phys}(b)=\mathcal F(b,h/2). \tag{O09.3}$$

Initially take real $|\alpha|<h$. Near zero the ratio is $2\alpha/x+O(x)$, so the subtracted integrand is finite. At infinity the ratio decays exponentially, while the subtraction contributes an integrable $x^{-2}$ tail. The same strip, with real parts, gives a complex continuation. No denominator mode vanishes there. The round Gaussian and its sufficiently small continuous deformations are nonsingular; the response requires only this neighborhood.

At $b=1$, (O09.3) agrees with the symmetric zeta prescription already derived for the round determinant: $\mathcal F(1,\alpha)=-\mathcal L(\alpha)$ and $\partial_\alpha\mathcal F(1,\alpha)=\pi\alpha\cot(\pi\alpha)$. In particular, $F_{\rm phys}(1)=\tfrac12\log2$. Dropping the subtraction, or changing its finite normalization as $b$ varies, would change the problem.

Match the paper’s printed integral carefully

P (8.11) prints a numerator $\sinh x$ and subtraction $1/x$. Read literally, it is

$$G(b)=\mathcal F(b,\tfrac12) =-\int_0^\infty\frac{dx}{2x} \left[\frac{\sinh x}{\sinh(bx)\sinh(b^{-1}x)}-\frac1x\right]. \tag{O09.4}$$

The fixed-R product instead has $\sinh(hx)$ and $h/x$. These are different functions away from the round point. For the second derivative in question, however, the distinction has an exact resolution. Since $h'(1)=0$, $h''(1)=1$ and $\partial_\alpha\mathcal F(1,\tfrac12)=0$, the chain rule gives

$$F_{\rm phys}''(1) =G''(1)+\frac12\, \partial_\alpha\mathcal F(1,\tfrac12) =G''(1). \tag{O09.5}$$

Thus the printed integral correctly reproduces the required round Hessian. We use it for that purpose, without identifying the full functions. Holding $\alpha=1/2$ corresponds to the varying assignment $r(b)=1-1/(2h)$ in the product; its effect first disappears at this order because the round free energy is stationary at the physical R charge.

Differentiate and evaluate

Set $u=\log b$ and $D(u,x)=\sinh(e^ux)\sinh(e^{-u}x)$. At $u=0$,

$$\partial_uD=0,\qquad \partial_u^2D=2x\sinh x\cosh x-2x^2.$$

The vanishing first derivative also gives $\partial_u^2G|_0=\partial_b^2G|_1$. Differentiating the subtracted expression yields

$$G''(1)=\int_0^\infty J(x)\,dx,\qquad J(x)=\frac{\sinh x\cosh x-x}{\sinh^3x} =\frac{\cosh x}{\sinh^2x}-\frac{x}{\sinh^3x}. \tag{O09.6}$$

This is an ordinary convergent integral. Its endpoints are $J(x)=2/3-x^2/5+O(x^4)$ and $J(x)=2e^{-x}+O(xe^{-3x})$. The numerator is positive: its derivative is $2\sinh^2x$ and it vanishes at zero. Differentiation under the integral is justified in a compact neighborhood of $u=0$ by the subtracted small-$x$ expansion and a uniform exponentially decaying bound at infinity.

Do not integrate the two terms in the last expression separately without a common lower cutoff; each diverges there. At cutoff $\varepsilon>0$, integration by parts gives

$$\int_\varepsilon^\infty J(x)\,dx =\frac12\left[ \frac1{\sinh\varepsilon} -\frac{\varepsilon\cosh\varepsilon}{\sinh^2\varepsilon} \right] +\frac12\int_\varepsilon^\infty\frac{x}{\sinh x}\,dx. \tag{O09.7}$$

The boundary bracket tends to zero. Now expand the remaining positive integrand: $1/\sinh x=2\sum_{n\geq0}e^{-(2n+1)x}$. Positivity permits termwise integration, and $\int_0^\infty xe^{-cx}dx=c^{-2}$ gives

$$\boxed{ F_{\rm phys}''(1) =\sum_{n=0}^\infty\frac1{(2n+1)^2} =\left(1-\frac14\right)\frac{\pi^2}{6} =\frac{\pi^2}{8}. } \tag{O09.8}$$

All radius factors have canceled. This is a free Gaussian calculation, exact in the prescribed response; it is neither a large-rank limit nor a truncation to the constant mode.

In the current convention $\langle j_\mu^{(R)}(x)j_\nu^{(R)}(0)\rangle =\tau_{RR}(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu) x^{-2}/(16\pi^2)$, the preceding lesson therefore gives

$$\boxed{\tau_{RR} =\frac2{\pi^2}\operatorname{Re}F_{\rm phys}''(1) =\frac14.} \tag{O09.9}$$

No convention for a coefficient called $C_T$ has been introduced. A conversion would require its tensor normalization.

Check through spectral shells and finite squashing

There is a second derivation which tests the subtraction. For the fixed-shift function, group the modes by $k=m+n+1$. At $u=0$, $\omega=k$, $\omega'=m-n$ and $\omega''=k$, while $\sum_{m+n=k-1}(m-n)^2=k(k^2-1)/3$. Direct differentiation of the finite shell’s negative logarithm gives

$$\begin{aligned} S_k&=\frac{k^2}{k^2-\frac14} -\frac{2k^2(k^2-1)}{3(k^2-\frac14)^2}\\ &=\frac13+\frac23\left(\frac1{2k-1}-\frac1{2k+1}\right) +\frac12\left[\frac1{(2k-1)^2}+\frac1{(2k+1)^2}\right]. \end{aligned} \tag{O09.10}$$

Its constant large-$k$ term cannot be summed ordinarily. Introduce $\mathscr D(s,u)=\sum_{m,n}[(\omega_{mn}-\alpha)^{-s}-(\omega_{mn}+\alpha)^{-s}]$ first, with $\operatorname{Re}s>2$, and define $\mathcal F=-\partial_s\mathscr D|_{s=0}$ by continuation. Differentiating twice in $u$ in that convergent domain and then continuing gives the shell sum below. The two shifted branches cancel the possible pole before setting $s=0$; its constant term contributes $\tfrac13\zeta_{\rm R}(0)=-1/6$; the remaining series converges. Thus

$$G''(1)=-\frac16+\sum_{k=1}^\infty(S_k-\tfrac13). \tag{O09.11}$$

The first fractions telescope to $2/3$. The squared fractions sum to $\pi^2/8-1/2$. Their sum with $-1/6$ is again $\pi^2/8$. Deleting only the linear divergence would miss the finite $-1/6$.

As a numerical test of the actual fixed-R determinant, evaluate (O09.3) at $b=e^{\pm\delta}$ before taking any derivative:

$$D_\delta= \frac{F_{\rm phys}(e^\delta)-2F_{\rm phys}(1) +F_{\rm phys}(e^{-\delta})}{\delta^2}, \qquad R_\delta=\frac{4D_{\delta/2}-D_\delta}{3}.$$

The mode product is even in $u$ after exchanging $m,n$, so $D_\delta=F''(1)+O(\delta^2)$ and $R_\delta=F''(1)+O(\delta^4)$. Direct quadrature gives:

$\delta$	$D_\delta$	$\lvert R_\delta-\pi^2/8\rvert$
$0.1$	$1.237127720332114$	$1.9432\times10^{-6}$
$0.05$	$1.234555885298328$	$1.2157\times10^{-7}$
$0.025$	$1.233914292748376$	$7.6001\times10^{-9}$

The successive errors decrease by about sixteen, as the fourth-order estimate predicts. The target is $\pi^2/8=1.233700550136169827\ldots$. The calculation used 75-digit arithmetic to keep cancellation below these finite-step errors; the table does not claim 75-digit finite-difference accuracy. The spectral sum supplies a check with a different regulator and elementary tail bounds.

Exercises

1. Control a finite mode sum

Let $R_K=-1/6+\sum_{k=1}^K(S_k-1/3)$ and $A_K=\sum_{n=0}^{K-1}(2n+1)^{-2}$. Express $R_K$ in terms of $A_K$. Use a monotone-integral bound to enclose the infinite answer from a finite $K$. Finally explain the error made by subtracting $K/3$ from the bare shell sum and omitting the finite term.

Hint 1

The simple fractions telescope. The two squared-fraction sums share all but their first and last entries.

Hint 2

For decreasing $f(n)=(2n+1)^{-2}$, $\int_K^\infty f(x)dx<\sum_{n=K}^\infty f(n)<\int_{K-1}^\infty f(x)dx$.

Solution

Summing (O09.10) gives

$$R_K=A_K-\frac{2}{3(2K+1)} +\frac{1}{2(2K+1)^2}.$$

The positive omitted odd-square tail $T_K$ satisfies

$$\frac1{2(2K+1)}<T_K<\frac1{2(2K-1)},\qquad K\geq1.$$

Therefore add $2/[3(2K+1)]-1/[2(2K+1)^2]$ to $R_K$ and then add these two bounds to obtain an interval containing $\pi^2/8$. At $K=640$ the interval is $1.23370024527809<\pi^2/8<1.23370085563003$. The uncorrected $R_K=1.23278980316521$ lies lower, consistent with its leading error $7/(12K)$.

Subtracting only $K/3$ instead gives the limit $\pi^2/8+1/6$. Converting that number to the current normalization would incorrectly produce $1/4+1/(3\pi^2)$. The discrepancy is a finite normalization error, not evidence against the current-response relation.

2. Change the source path

Instead of keeping the physical R charge fixed, take $b=e^u$ and $\alpha(u)=1/2+qu$, with real constant $q$. Compute the real second derivative at $u=0$. Why need it not remain positive? Compare this change with the physical path $\alpha=h/2$.

Hint 1

The function $\mathcal F(e^u,\alpha)$ is even in $u$ for fixed $\alpha$, so its mixed first derivatives with one $u$ derivative vanish at zero.

Hint 2

Differentiate $\partial_\alpha\mathcal F(1,\alpha)=\pi\alpha\cot(\pi\alpha)$ once more and evaluate at $\alpha=1/2$.

Solution

At the round point, $\mathcal F_{\alpha\alpha}=-\pi^2/2$ and $\mathcal F_{u\alpha}=0$. The chain rule therefore gives

$$\left.\frac{d^2}{du^2}\mathcal F(e^u,\tfrac12+qu)\right|_0 =\frac{\pi^2}{8}-\frac{\pi^2q^2}{2}.$$

It vanishes at $|q|=1/2$ and is negative beyond that value. This does not imply a negative physical $\tau_{RR}$. Since $r=1-\alpha/h$, the chosen path has $r'(0)=-q$: it varies the trial R assignment, hence the associated flavor-source direction, at first order. It is not the fixed-superconformal-current squashing measurement.

For the physical path $\alpha=h/2$, $\alpha'(0)=0$ and $\alpha''(0)=1/2$. The only extra chain-rule term is then $\alpha''\mathcal F_\alpha$, which vanishes at the physical round point. This reproduces (O09.5).

3. Distinguish an infrared pole from the ultraviolet subtraction

At fixed positive $b$, approach the edge of the strip by setting $\alpha=h-\varepsilon$ with $\varepsilon>0$. Identify the singular mode and determine the leading behavior of $Z$ and $\mathcal F$ as $\varepsilon\to0$. Can the subtraction in (O09.3) remove this singularity?

Hint 1

The smallest $\omega_{mn}$ occurs at $m=n=0$. All other denominator factors remain separated from zero.

Hint 2

The large-$x$ heat ratio behaves as $2e^{-2\varepsilon x}$. Its contribution changes when an arbitrarily long proper-time interval becomes relevant.

Solution

The $(0,0)$ denominator is $\omega_{00}-\alpha=\varepsilon$, while its numerator tends to $2h$. Removing this one factor leaves a regular nonzero determinant near the endpoint. Thus

$$Z(b,h-\varepsilon)\sim\frac{\text{nonzero constant}}{\varepsilon}, \qquad \mathcal F(b,h-\varepsilon)=\log\varepsilon+O(1).$$

The same conclusion follows from the long-time term $-\int^\infty e^{-2\varepsilon x}dx/x$ in the free energy: its singular part is $\log\varepsilon$. This is a bosonic zero-mode divergence, not a fermion zero making $Z$ vanish.

The subtraction $2\alpha/x$ fixes the short-time ultraviolet behavior and is regular as $\alpha\to h$. It cannot remove an infrared zero mode. An integral at the endpoint requires new treatment of that mode; it cannot be justified by continuing the nonsingular Gaussian calculation without qualification.

The benchmark now has two independent routes: a convergent differentiated determinant and a regulated spectral sum. Their agreement fixes the numerical coefficient only together with the declared R-current, complete source family and subtraction prescription.

The Section 8 paper guide returns these calculations to the original argument and compares two families that agree at second order but differ at fourth order.

Implement the determinant and distinguish numerical errors from source-path changes in the free-chiral squashing laboratory.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Supersymmetric Field Theories on Three-Manifolds
• Paper guide: Section 8, physical observables