What squashing the sphere measures

Before this lesson: Superconformal currents and two-point functions; Squashed spheres and their full background data; R-mixing and F-maximization.

Changing a metric probes the stress tensor. A supersymmetric change of geometry also changes the sources that accompany the metric. Those additional changes can make a current correlator the most direct way to compute the response. We will follow one complete background and derive the coefficient, including the subtraction needed when two current insertions coincide.

Prepare with the current two-point normalization, squashed spheres and their full background data, and the physical R-symmetry selected by F-maximization. The first supplies the flat correlator; the second supplies the supersymmetric family. Here we compute what its partition-function curvature measures.

Fix the observable before differentiating

Work with a unitary three-dimensional $\mathcal N=2$ SCFT and its superconformal R-multiplet. Its Lorentzian superconformal algebra has four real Poincaré and four real conformal supercharges. The background below is the four-rigid-supercharge family of P, section 5.2; this specifies preserved curved generators, not a new count for the parent theory. No gauge group or rank is needed.

Fix the theory at its conformal point, set additional flavor masses to zero, and hold the physical R-current fixed as the geometry changes. Write

F(b)=-\log Z(b),\qquad \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)\frac1{x^2}. \tag{O08.1}

The second equation fixes the charge and coefficient convention. It first holds at separated points; a distributional extension is needed to integrate it. For a nonzero physical current, unitarity gives $\tau_{RR}>0$ . Take the continuous logarithm near the round background. Its real part is $-\log|Z|$ and does not depend on a logarithm branch.

The prescription is the conformal, supersymmetric source prescription of P: power divergences are removed consistently with its Ward identities, while allowed finite background Chern–Simons terms retain their prescribed global meaning. We do not add arbitrary finite terms with new mass scales. This restriction matters when we take a real part.

Follow the complete background

Set the round radius to one, so all coordinates below are dimensionless. Let $g_0$ be the round metric and $v_0$ the unit Hopf one-form, with the orientation chosen so that $dv_0=2*_0v_0$ . Introduce

h=\frac{b+b^{-1}}2,\qquad d=b-b^{-1},\qquad b>0, \qquad h^2-1=\frac{d^2}{4}.

The full background of P (5.11), (8.4)–(8.5) can be written

g_b=g_0+(h^2-1)v_0\otimes v_0,\qquad v_b=h v_0, \qquad A=V=dhv_0,\qquad C=\frac{id}{2}v_0,\qquad H=ih. \tag{O08.2}

Here $v_b$ has unit norm in $g_b$ . The field $A$ is the background R-connection; $V,H,C$ are its rigid-supergravity companions, with $V=-i*_b dC$ . They are fixed sources, not dynamical gauge fields. Indeed $*_b dv_0=2h v_0$ verifies that dual relation. The metric and $A,V$ are real, while $C,H$ are imaginary. Restoring radius $\ell$ multiplies the metric by $\ell^2$ , replaces $H$ by $ih/\ell$ and $C$ by $i\ell d v_0/2$ , and leaves the displayed connection one-forms $A,V$ unchanged. Their orthonormal components have mass dimension one.

This family has $H=+i$ at $b=1$ . Earlier round-sphere examples used the other Killing-spinor branch, $H=-i/\ell$ . We use the source’s positive branch throughout this calculation; retaining the old value of $H$ while varying (O08.2) would not specify this family. Its round metric still has the paper’s scalar curvature $R=-6$ .

For $b=1+\epsilon$ ,

h=1+\frac{\epsilon^2}{2}+O(\epsilon^3),\quad d=2\epsilon-\epsilon^2+O(\epsilon^3),\quad g_b=g_0+\epsilon^2v_0\otimes v_0+O(\epsilon^3).

Thus the metric and $H$ have no first derivative. The actual first-order source is

A-\frac32V=-\epsilon v_0+O(\epsilon^2),\qquad C=i\epsilon v_0+O(\epsilon^2),\qquad S'(1)=\int_{S^3}\!\sqrt{g_0}\,v_{0\mu} \bigl(ij^{(Z)\mu}-j^{(R)\mu}\bigr). \tag{O08.3}

The source for the R-current is $A-3V/2$ , not the matter covariant derivative’s connection $A-V/2$ . The operator $j^{(Z)}$ belongs to the R-multiplet. In the SCFT with its superconformal improvement it is redundant: its separated-point correlators vanish. Its contact terms must still be considered. These are precisely the source identifications in P (8.6)–(8.8).

Turn a derivative into an integrated correlator

Differentiating the path integral with its complete action gives

F'=\langle S'\rangle,\qquad F''=\langle S''\rangle-\langle S'S'\rangle_c. \tag{O08.4}

The second term has a minus sign because the weight is $e^{-S}$ . The first term includes the second-order source variation and terms from differentiating the action twice, often called seagull terms. Omitting it before specifying the regulator can change the answer.

Conformal mapping from the flat vacuum makes the noncontact one-point functions vanish on the round sphere. Accordingly $\operatorname{Re}F'(1)=0$ , and the noncontact part of $\langle S''\rangle$ vanishes in this prescription. Removing redundant separated insertions leaves

\operatorname{Re}F''(1)=-\mathcal I_{\rm ren},\qquad \mathcal I=\int\!d^3x\sqrt{g_0(x)}\int\!d^3y\sqrt{g_0(y)}\, v_{0\mu}(x)v_{0\nu}(y) \langle j^{(R)\mu}(x)j^{(R)\nu}(y)\rangle. \tag{O08.5}

The subscript means that the coincident singularity is treated with the Ward-preserving prescription, not that an ordinary positive integral has been assigned a finite value by declaration. We now compute it.

Evaluate the round-sphere integral

Use stereographic coordinates $x=(x^1,x^2,x^3)$ , $\rho=|x|$ and $\Omega=2/(1+\rho^2)$ , so $g_0=\Omega^2\delta$ . The required Hopf form is

v_0=\Omega^2\left[ -x^2dx^1+x^1dx^2+\frac{1-\rho^2}{2}dx^3 +x^3(x^1dx^1+x^2dx^2+x^3dx^3)\right]. \tag{O08.6}

It obeys $g_0^{-1}(v_0,v_0)=1$ , and at the origin $v_0=2dx^3$ . The subgroup of round isometries preserving this Hopf form acts transitively. We may therefore fix $y=0$ and multiply by $\operatorname{Vol}(S^3)=2\pi^2$ .

For contravariant coordinate components the current’s Weyl factor is $\Omega^{-3}$ . Thus the flat correlator in (O08.1) gives

\langle j^\mu(x)j^\nu(0)\rangle_{S^3} =\frac{\tau_{RR}}{4\pi^2\Omega(x)^3\Omega(0)^3} \frac{\delta^{\mu\nu}\rho^2-2x^\mu x^\nu}{\rho^6}. \tag{O08.7}

This is P (8.9), with chordal distance $s(x,0)=\sqrt{\Omega(x)\Omega(0)}\,\rho$ . Coordinate products here use $\delta$ ; physical tensor norms use $g_0$ .

To do the contraction explicitly, note that $v_{0\mu}x^\mu=\Omega x^3$ and $v_{03}=\Omega^2[(1-\rho^2)/2+(x^3)^2]$ . Its angular average uses $\langle(x^3)^2\rangle=\rho^2/3$ . Excluding $\rho<\varepsilon$ then reduces the full integral to

\mathcal I_\varepsilon =\frac{\pi\tau_{RR}}3\int_\varepsilon^\infty \frac{1-3\rho^2}{\rho^2(1+\rho^2)^2}\,d\rho =\frac{\pi\tau_{RR}}3 \left[\frac1\varepsilon+\frac{2\varepsilon}{1+\varepsilon^2} +3\arctan\varepsilon-\frac{3\pi}{2}\right]. \tag{O08.8}

For example, the integrand decomposes as $\rho^{-2}-(1+\rho^2)^{-1}-4(1+\rho^2)^{-2}$ , which checks the antiderivative and its sign. The divergence is local: it comes entirely from coincident insertions.

A hard exclusion does not preserve the current Ward identity by itself. Its restoring contact term removes $\pi\tau_{RR}/(3\varepsilon)$ . A finite real source-mass term cannot then be chosen freely: it would violate background gauge invariance, and a finite Maxwell term would introduce a new scale. The supersymmetric treatment of this kind of power divergence and its seagull is explained in FM (3.8) and footnote 9, the prescription cited by P’s footnote 13. We obtain

\mathcal I_{\rm ren}= \lim_{\varepsilon\to0}\left(\mathcal I_\varepsilon- \frac{\pi\tau_{RR}}{3\varepsilon}\right) =-\frac{\pi^2}{2}\tau_{RR},\qquad \boxed{\displaystyle \tau_{RR}=\frac{2}{\pi^2} \left.\frac{d^2\operatorname{Re}F}{db^2}\right|_{b=1}}. \tag{O08.9}

The sign change after subtraction is consistent with unitarity. It is the renormalized distribution, not the unregulated cutoff integrand, which enters the response.

As a separate check, conformally map the whole integral to flat space and use the transverse distribution derived in the current lesson. With $\widehat v(p)=\int e^{ip\cdot x}v(x)d^3x$ , $n=p/|p|$ and $P_T=1-nn^{\mathsf T}$ , differentiating the transform $\widehat{(1+\rho^2)^{-2}}=\pi^2e^{-|p|}$ gives

P_T\widehat v_0=4\pi^2e^{-|p|} \left[e_3-n_3n+i(e_3\mathbin{\times}n)\right],\qquad \int\frac{d^3p}{(2\pi)^3}|p|\,|P_T\widehat v_0|^2=4\pi^2.

The last equality uses $\int d\Omega_n(1-n_3^2)=8\pi/3$ and $\int_0^\infty p^3e^{-2p}dp=3/8$ . The conserved Fourier kernel is $-\tau_{RR}|p|P_T/8$ , so it independently gives $\mathcal I_{\rm ren}=-\pi^2\tau_{RR}/2$ without a hard exclusion. The integral is convergent at both momentum endpoints.

Explain the real part and the limits of the statement

Local terms in the background fields encode contact terms in the R-multiplet. P’s analysis after (8.10) establishes that their allowed finite contributions on this complete family are imaginary. The real second derivative therefore isolates the separated coefficient. This is an input about supersymmetry, the source prescription and reality conditions, not a claim that every real response on every background is independent of local terms.

The result has no residual radius: $F$ , $b$ and $\tau_{RR}$ are dimensionless, and restoring $\ell$ cancels between volumes, current factors and normalized sources. It is exact at the round point, with no large-rank or weak-coupling approximation. Localization is a possible way to compute $F$ ; the response derivation itself used current correlators.

F-maximization selects the physical R-current before this measurement, subject to its usual accidental-symmetry qualifications. It does not say that $b$ should maximize $F$ . Here $\operatorname{Re}F''(1)>0$ : the round point is a local minimum along this particular background direction. Changing a trial R-current, turning on an extra flavor source, or varying only the metric changes the problem.

The remaining practical question is how to obtain the same coefficient by differentiating an actual localized determinant. The free-chiral calculation provides that next test; the result above explains which observable such a differentiation must reproduce.

Exercises

1. A parameter can hide the linear source

Set $u=\log b$ , so $h=\cosh u$ . Express the leading real response in $u$ and in $h-1$ . Is $d_h^2\operatorname{Re}F$ at $h=1$ the coefficient in (O08.9)? Explain why the metric alone does not distinguish the two signs of $u$ .

Hint 1

For a regular reparameterization, use the chain rule and $\operatorname{Re}F'(1)=0$ . Here $db/du=1$ at $u=0$ .

Hint 2

The metric depends on $h^2-1$ , whereas $A,V,C$ also depend on $d=2\sinh u$ . The map from $u$ to $h$ has zero first derivative at the round point.

Solution

Writing $F_0=\operatorname{Re}F(1)$ , (O08.9) gives

\operatorname{Re}F(e^u)=F_0+\frac{\pi^2\tau_{RR}}4u^2+O(u^3), \qquad h-1=\frac{u^2}{2}+O(u^4).

Hence along either local branch $\operatorname{Re}F=F_0+(\pi^2\tau_{RR}/2)(h-1)+O((h-1)^{3/2})$ . The one-sided first derivative with respect to $h$ has the stated coefficient. The second derivative with respect to $h$ is not fixed by the round-point theorem and need not even be finite without further information about higher terms. We have not assumed an exact evenness of the full answer in $u$ .

The metric and $H$ are the same at $u$ and $-u$ , but the three one-form sources reverse sign. Thus $h$ labels the metric but is not a regular signed parameter for the complete background near $u=0$ . For a regular parameter $t$ with $b(0)=1$ , the measurable Hessian instead transforms as $\partial_t^2\operatorname{Re}F|_0=[b'(0)]^2\pi^2\tau_{RR}/2$ .

2. Check the subtraction with physical distance

Replace the coordinate exclusion in (O08.8) by a geodesic ball of radius $\delta$ on the unit sphere. Show that $\varepsilon=\tan(\delta/2)$ , find the divergent term, and verify that the finite answer is unchanged. Estimate the leading error after subtraction at small $\delta$ .

Hint 1

Along a radial line the proper length element is $2\,d\rho/(1+\rho^2)$ .

Hint 2

Use $2\varepsilon/(1+\varepsilon^2)=\sin\delta$ and $\arctan\varepsilon=\delta/2$ . Expand $\cot(\delta/2)$ before extracting a constant.

Solution

The radial integral gives $\delta=2\arctan\varepsilon$ , so

\mathcal I_\delta=\frac{\pi\tau_{RR}}3 \left[\cot\frac\delta2+\sin\delta+\frac{3\delta}{2} -\frac{3\pi}{2}\right].

The required divergence is $2\pi\tau_{RR}/(3\delta)$ . Since $\cot(\delta/2)=2/\delta-\delta/6+O(\delta^3)$ ,

\mathcal I_\delta-\frac{2\pi\tau_{RR}}{3\delta} =-\frac{\pi^2\tau_{RR}}2+\frac{7\pi\tau_{RR}}9\delta+O(\delta^3).

The finite coefficient agrees. Halving a sufficiently small cutoff halves the leading residual; the unsubtracted integral does not converge. Restoring a radius $\ell$ replaces $\delta$ by physical distance divided by $\ell$ , so the finite response remains dimensionless. This checks a controlled change of cutoff, not permission to add arbitrary finite counterterms.

3. Diagnose a different subtraction prescription

A calculation adds $\Delta F=c(4M)^{-1}\int\sqrt g\, f_{\mu\nu}f^{\mu\nu}$ , where $a=A-3V/2$ , $f=da$ , $c$ is real and dimensionless, and $M$ is a fixed positive mass. Find its effect on the round second derivative at radius $\ell$ . Can taking a real part remove it? Is this a counterexample to (O08.9)?

Hint 1

At first order $a=-\epsilon v_0$ , while $dv_0=2*_0v_0$ on the unit sphere. Metric corrections start at order $\epsilon^2$ and multiply an already quadratic field strength.

Hint 2

The norm of a two-form is $|f|^2=f_{\mu\nu}f^{\mu\nu}/2$ . Restoring the radius gives two inverse metrics and one volume factor.

Solution

The unit-radius form has $|dv_0|^2=4$ , hence $f_{\mu\nu}f^{\mu\nu}=8\epsilon^2+O(\epsilon^3)$ . The volume is $2\pi^2$ , and the radius dependence is $\ell^{3-4}=\ell^{-1}$ . Therefore

\Delta F=\frac{4\pi^2c}{M\ell}\epsilon^2+O(\epsilon^3),\qquad \Delta\!\left[\partial_b^2\operatorname{Re}F\right]_{b=1} =\frac{8\pi^2c}{M\ell}.

This change is real, so taking a real part does not remove it. The new mass scale and the chosen Maxwell subtraction lie outside the conformal supersymmetric prescription in the theorem; the displayed term alone is not asserted to be its supersymmetric completion. It is a diagnostic of a changed source functional, not a contradiction. Its radius dependence also exposes the difference from the dimensionless universal coefficient. Finite imaginary contact contributions within the prescribed family require a different analysis and do not change (O08.9).

Reproduce the free-chiral response now calculates this curvature directly from the matter spectrum. Follow the fixed R-charge through the changing eigenvalues, then compare a convergent integral with a carefully subtracted spectral sum.