Paper guide: Section 8, physical observables

Section 8 of Supersymmetric Field Theories on Three-Manifolds, version 3 turns a curved-space partition function into a flat-space current measurement. The main reading question is: which derivative measures which current? A correct determinant evaluated along the wrong source path can answer a different question.

Begin with R-mixing and F-maximization and source derivatives and contact terms. They explain the two choices made at the start of the paper’s argument: select the superconformal R-current, and fix the source prescription. The later response capstone uses three additional completed calculations.

First pass: identify what is held fixed

The setting is a unitary three-dimensional $\mathcal N=2$ SCFT in its conformal vacuum. The Lorentzian algebra has four real Poincaré and four real conformal supercharges. The Euclidean sphere family preserves four rigid supersymmetry directions; it does not change the parent theory’s supersymmetry. There is no universal gauge group or rank in the response theorem. The free-chiral example is gauge neutral.

Read the opening of section 8.1 before its formulas. The R-multiplet is the superconformal one. In a theory with mixing flavor currents, determining it is a separate task; an arbitrary trial charge is not interchangeable with the physical charge. Keep all additional physical masses zero and hold this current fixed while squashing.

Next distinguish $F=-\log Z$ from $\operatorname{Re}F=-\log|Z|$. A continuous logarithm is sufficient near the round point. The conformal supersymmetric source prescription fixes power subtractions and retains permitted background contact terms. Its reality properties on this particular family make the real response useful. This is not permission to discard all phases, or to add arbitrary real local terms.

These choices explain why F-maximization and the squashing measurement have opposite local curvatures. The former varies the R-current; the latter varies a complete background while holding that current fixed.

Response capstone: follow the equation map

For this section prepare with superconformal current correlators, what squashing measures, and the free-chiral response. Those are the derivation homes. The following map identifies the role of every numbered equation in the paper’s section 8.

Paper equations	Role in the argument	Where to reconstruct the step
(8.1)	Define the complex free energy.	Source derivatives, (O06.1), fixes the response sign for $-\log Z$.
(8.2)	Relate its real round-point Hessian to $\tau_{RR}$.	Squashing response, (O08.9).
(8.3)	Define the R-current coefficient and its stress-tensor partner.	Current correlators, (O07.2), (O07.7)–(O07.9); the last equations define any conversion to $C_T$.
(8.4)–(8.5)	Specify the metric and auxiliary sources, in a positive-$H$ branch.	Squashing response, (O08.2), (O08.6).
(8.6)	Identify the first-order current insertion.	Squashing response, (O08.3); use $A-3V/2$, not the matter derivative's $A-V/2$.
(8.7)–(8.8)	Convert the Hessian into an integrated correlation with local terms.	Source derivatives, (O06.1), and squashing response, (O08.4)–(O08.5).
(8.9)–(8.10)	Weyl-map the correlator and introduce chordal distance.	Squashing response, (O08.7)–(O08.9), includes the regulated integral omitted in the paper.
(8.11)	Give the printed free-chiral integral.	Free-chiral response, (O09.2)–(O09.5), distinguishes its fixed shift from fixed R-charge.
(8.12)	Evaluate the free round Hessian.	Free-chiral response, (O09.6)–(O09.11), gives the integral, spectral subtraction and $\tau_{RR}=1/4$.
(8.13)–(8.14)	Check a separate leading large-$N$ supergravity regime.	Exit calculation 2 below checks the implication and its limits; these are not general free-field formulas.

The notation $\tau_{rr}$ in the paper is $\tau_{RR}$ here. Both use

$$\langle j_\mu^{(R)}(x)j_\nu^{(R)}(0)\rangle =\frac{\tau_{RR}}{16\pi^2} (\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)|x|^{-2}, \qquad \tau_{RR}=\frac2{\pi^2} \left.\partial_b^2\operatorname{Re}F\right|_{b=1}. \tag{PC07.1}$$

The first expression initially concerns separated points. The second requires their local completion. A coefficient called $C_T$ has no numerical meaning until its tensor normalization is also stated.

Reconstruct the source insertion before integrating

Use unit radius, a round metric $g_0$, and a unit Hopf form $v_0$ with $dv_0=2*_0v_0$. With $h=(b+b^{-1})/2$ and $d=b-b^{-1}$, the full source family is

$$g_b=g_0+(h^2-1)v_0^2,\quad A=V=dhv_0,\quad C=\frac{id}{2}v_0,\quad H=ih. \tag{PC07.2}$$

Here $v_0^2$ means $v_0\otimes v_0$. The paper’s unit form at finite $b$ is $v_b=hv_0$. Its round auxiliary value is $H=+i$, while its round Ricci scalar is $-6$. These are fixed sources on the ordinary sphere with its unique spin structure, not fields to integrate out.

For $b=1+\epsilon$, $h=1+O(\epsilon^2)$ and $d=2\epsilon+O(\epsilon^2)$. Thus the first metric and $H$ variations vanish, but $A-3V/2=-\epsilon v_0+O(\epsilon^2)$ and $C=i\epsilon v_0+O(\epsilon^2)$. They give

$$S'(1)=\int\sqrt{g_0}\,v_{0\mu} \bigl(ij^{(Z)\mu}-j^{(R)\mu}\bigr),\qquad F''=\langle S''\rangle-\langle S'S'\rangle_c. \tag{PC07.3}$$

This reconstructs the sign and source combination in (8.6)–(8.8). The superconformally improved $j^{(Z)}$ is redundant at separated points; its contact terms need not vanish. Likewise, a vanishing noncontact one-point function does not justify dropping the local part of $S''$. The current lesson supplies the tensor, and the squashing lesson supplies the Ward-preserving coincident subtraction. Reading (8.8) as an ordinary unregulated double integral would miss that step.

Compare the two free-chiral source paths

Now use one massless free chiral with physical scalar charge $r=1/2$. Write $u=\log b$, so $h=\cosh u$. Denote the regulated determinant by

$$\begin{aligned} \mathcal F(u,\alpha) &=-\int_0^\infty\frac{dx}{2x} \left[ \frac{\sinh(2\alpha x)} {\sinh(e^ux)\sinh(e^{-u}x)} -\frac{2\alpha}{x} \right],\\ F_{\rm phys}(u)&=\mathcal F(u,\tfrac12\cosh u), \qquad G(u)=\mathcal F(u,\tfrac12). \end{aligned} \tag{PC07.4}$$

The variable $x$ here is dimensionless proper time, not a spacetime coordinate. Initially take real $|\alpha|<\cosh u$, so there is no zero denominator mode. The ultraviolet subtraction is part of the definition.

The literal numerator in P (8.11) is $\sinh x$, so it defines $G$. The actual fixed-R mode product has frequencies $me^u+ne^{-u}+h$ and shift $\alpha=h(1-r)$, hence $\alpha=h/2$. This follows from IY (86)–(87), with the continuation detailed in the free-chiral lesson: $v_{\rm IY}=1/h$, $w_{\rm IY}=-i\sinh u/h$, and $(1+iw_{\rm IY})/v_{\rm IY}=b$. Here $w_{\rm IY}$ names IY’s $u$; it is not our $\log b$. Real positive $b$ uses this continued branch.

At the common round point, $\mathcal F_\alpha(0,1/2)=0$ and $\mathcal F_{\alpha\alpha}(0,1/2)=-\pi^2/2$. Because $\alpha'(0)=0$, both paths have the same Hessian

$$F_{\rm phys}''(0)=G''(0)=\frac{\pi^2}{8}, \qquad \tau_{RR}=\frac14. \tag{PC07.5}$$

The derivative with respect to $u$ equals the derivative with respect to $b$ at this stationary point because $db/du=1$ there. Equality in (PC07.5) does not assert $F_{\rm phys}=G$ away from it.

Worked extension: find the first difference

We can test that distinction at the next even order, without redoing the determinant derivation. For fixed $\alpha$, $\mathcal F$ is even in $u$. Let $A=\mathcal F_{uu\alpha}(0,1/2)$ and $B=\mathcal F_{\alpha\alpha}(0,1/2)$. The shift is $\delta\alpha=\cosh u/2-1/2=u^2/4+O(u^4)$. Taylor expansion, including the mixed derivative, gives

$$F_{\rm phys}(u)-G(u) =\left(\frac A8+\frac B{32}\right)u^4+O(u^6). \tag{PC07.6}$$

To calculate the new derivative, differentiate the subtracted kernel:

$$A=2\int_0^\infty \frac{x\cosh x\,(\sinh x\cosh x-x)}{\sinh^4x}\,dx.$$

This integral is convergent. Its integrand tends to $4/3$ at zero and decays exponentially at infinity. For an explicit integration check define

$$B_0(x)=\frac13\left[ \frac{x\cosh x}{\sinh^2x} +\frac1{\sinh x}-\frac{2x^2}{\sinh^3x}\right].$$

Direct differentiation makes the preceding integrand $5x/(3\sinh x)-B_0'(x)$. Both endpoint values of $B_0$ vanish; its individually divergent small-$x$ terms must be kept together. Using $\int_0^\infty x/\sinh x\,dx=\pi^2/4$, we obtain $A=5\pi^2/12$. With $B=-\pi^2/2$ this yields the new comparison

$$\boxed{F_{\rm phys}(u)-G(u) =\frac{7\pi^2}{192}u^4+O(u^6).} \tag{PC07.7}$$

This finite-order result is derived here; it is not an equation asserted by either source. It quantifies why a successful round Hessian cannot identify the full determinant family.

Directly integrating the difference of the two kernels gives $(F_{\rm phys}-G)/u^4=0.356186387111,\ 0.358900661893,\ 0.359596018166$ at $u=0.16,\ 0.08,\ 0.04$, approaching $7\pi^2/192=0.359829327123\ldots$. Removing the leading $O(u^2)$ error with two successive steps leaves errors $2.39\times10^{-5},\ 1.52\times10^{-6},\ 9.57\times10^{-8}$. Their roughly sixteenfold decrease tests the next error order. The quadrature used 75-digit arithmetic, checked at 95 digits; these finite-step errors, rather than that arithmetic precision, describe the displayed approximation.

Three exit calculations

1. Recover the coefficient in a different coordinate

Suppose a calculation supplies the same fixed-R determinant in the coordinate $b(t)=\exp(2t+t^2)$, and its complex free energy also contains an allowed fixed-prescription contribution $i\chi(t)$ with real $\chi$. Find the formula that recovers $\tau_{RR}$ from its real $t$ Hessian. Does $b''(0)=6$ add a term? Give a controlled finite-step estimate for the free chiral.

Hint 1

Differentiate the composite function twice. A coordinate acceleration multiplies the first derivative of the original function.

Hint 2

The speed is $b'(0)=2$. In a symmetric difference, odd powers of $t$ cancel even though $b(t)$ is not even.

Solution

Let $\widehat F(t)=F(b(t))+i\chi(t)$. Then

$$\operatorname{Re}\widehat F''(0) =4\operatorname{Re}F''(1)+6\operatorname{Re}F'(1) =4\operatorname{Re}F''(1).$$

The last equality uses stationarity, not an assumption that the coordinate acceleration vanishes. The imaginary contribution drops out for real $t$, without being declared absent from the complex partition function. Therefore

$$\tau_{RR}=\frac{1}{2\pi^2}\operatorname{Re}\widehat F''(0) =\frac{1}{2\pi^2}\frac{\pi^2}{2}=\frac14.$$

For a finite step $\delta$, calculate

$$T_\delta= \frac{\operatorname{Re}[ \widehat F(\delta)+\widehat F(-\delta)-2\widehat F(0)]} {2\pi^2\delta^2},\qquad \widetilde T_\delta=\frac{4T_{\delta/2}-T_\delta}{3}.$$

Analyticity gives $T_\delta=1/4+O(\delta^2)$ and $\widetilde T_\delta=1/4+O(\delta^4)$. Evaluating the full fixed-R integral before taking differences gives

$\delta$	$T_\delta$	$\widetilde T_\delta$
$0.08$	$0.252201449803$	$0.249993088781$
$0.04$	$0.250545179036$	$0.249999571322$
$0.02$	$0.250135973251$	$0.249999973259$

The extrapolated errors decrease by about sixteen. Using $2\widehat F''/\pi^2$ without the squared coordinate speed would instead give $1$, an error of four. This check changes the coordinate, not the physical source family or the R-current.

2. Decide which example permits a large-rank shortcut

In the leading large-$N$ regime specified after P (8.12), take the source’s two inputs $F(b)=h^2F(1)$ and $F(1)=\pi^2\tau_{RR}/4$. Check their agreement with (PC07.1). Then try the first input on a free chiral, whose $F(1)=\log2/2$. Diagnose the resulting discrepancy.

Hint 1

At $b=1$, $h'=0$ and $h''=1$, so differentiate $h^2$ before substituting.

Hint 2

A free chiral has no gauge rank that can be sent to infinity. The paragraph describes a supergravity regime, not an identity for every three-dimensional SCFT.

Solution

The first formula gives $F''(1)=2F(1)$. The second then gives $F''(1)=\pi^2\tau_{RR}/2$, exactly the stated response. This is the consistency check in (8.13)–(8.14), conditional on the large-$N$ theory with an $\mathrm{AdS}_4\times X_7$ supergravity dual described there. It is not a derivation of either input from the response theorem.

For the free chiral the shortcut would instead yield

$$\tau_{\rm shortcut} =\frac{4F(1)}{\pi^2} =\frac{2\log2}{\pi^2} \simeq0.140460986,$$

which disagrees with $1/4$. The exact free answer has $F''(1)=\pi^2/8$, not $\log2$. There is no contradiction: the large-rank simplification was applied outside its stated regime. A successful check in one example cannot promote its finite-$b$ formula to a universal law.

3. How much does agreement through fourth order determine?

Use the same regulated two-variable function, but choose $\alpha(u)=1/2+cu^2$ with real constant $c$ near $u=0$. Find the coefficient of $u^4$ in $\mathcal F(u,\alpha(u))-G(u)$. Besides $c=0$, is there a path that agrees with $G$ through fourth order? Which $c$ agrees with the physical fixed-R path through order $u^2$?

Hint 1

Retain both $\tfrac12\mathcal F_{uu\alpha}u^2\delta\alpha$ and $\tfrac12\mathcal F_{\alpha\alpha}(\delta\alpha)^2$.

Hint 2

The trial charge along this path is $r(u)=1-\alpha(u)/\cosh u$. Compare its quadratic term with a constant charge.

Solution

The common round first derivative in $\alpha$ vanishes, so

$$\mathcal F(u,\tfrac12+cu^2)-G(u) =\pi^2\left(\frac{5c}{24}-\frac{c^2}{4}\right)u^4+O(u^6).$$

This coefficient vanishes at $c=0$ and $c=5/6$. The latter is a different source path whose two contributions cancel at this order; agreement through fourth order does not identify the whole function.

In contrast, $r(u)=1/2+(1/4-c)u^2+O(u^4)$, so only $c=1/4$ keeps the physical R-charge fixed through order $u^2$. Inserting it recovers $7\pi^2/192$. Exact fixed R requires the full $\alpha(u)=\cosh u/2$, not merely its quadratic truncation. For any finite $c$, a sufficiently small neighborhood remains inside $|\alpha|<\cosh u$; no zero-mode singularity was crossed in these comparisons. These changes of the curvature/source prescription do not change the free field’s flat-space scaling dimension.

A completed reading of Section 8 now has a numerical exit test and a scope test: recover $1/4$ with the source-normalized Hessian, and explain why neither that number nor a shared low-order expansion identifies an arbitrary finite-squashing partition function.