The free chiral theory on the round sphere

A supersymmetric background supplies spinors that can generate transformations. It does not make every action on that background supersymmetric. We now construct an action whose kinetic, auxiliary, and curvature terms cancel against one another under the transformations of the curved chiral multiplet.

Our background is the one solved in the round-sphere spinor lesson. The calculation is three-dimensional and Euclidean. The parent flat theory has $\mathcal N=2$, or four real Lorentzian Poincaré supercharges; we use its complete algebra after complexification. There is one gauge-neutral chiral multiplet, its independent anti-chiral partner, zero central charge, and no superpotential. No gauge rank or gauge-multiplet field enters. The background preserves two complex solution directions in each spinor family.

The result will be a complete classical free action. “Free” means quadratic in the matter fields; its curvature couplings are still essential.

Fix the background and the fields

Let $\ell>0$ be the sphere radius. Keep all background fields fixed:

$$A_\mu=V_\mu=C_\mu=0,\qquad H=-\frac{i}{\ell}, \qquad R=-\frac6{\ell^2}.$$

Here $R$ uses the target paper’s curvature trace. The scalar $H$ is a background auxiliary field, not the matter auxiliary field $F_\Phi$. We never impose a gravitational equation of motion on $H$.

In the global oriented orthonormal frame of the sphere lesson,

$$\nabla_a\chi=E_a\chi+\frac{i}{2\ell}\gamma_a\chi,\qquad \gamma_a\gamma_b=\delta_{ab}\mathbf1+i\epsilon_{abc}\gamma_c, \qquad \epsilon_{123}=1.$$

The commuting spinor labels $\zeta,\widetilde\zeta$ have arbitrary constant components in that frame. Consequently,

$$\nabla_\mu\zeta=k\gamma_\mu\zeta,\qquad \nabla_\mu\widetilde\zeta=k\gamma_\mu\widetilde\zeta, \qquad k=\frac{i}{2\ell}.$$

They are covariantly nonconstant. Setting their covariant derivatives to zero would remove terms needed below.

The independent matter fields are $(\phi,\psi,F_\Phi)$ and $(\widetilde\phi,\widetilde\psi,\widetilde F_\Phi)$. Their R-charges are respectively $(r,r-1,r-2)$ and $(-r,1-r,2-r)$. Their engineering mass dimensions are $(1/2,1,3/2)$ in each triplet. The scalar R-charge $r$ is a fixed real parameter here; it need not equal its free conformal value.

We use lower-column spinors and the contraction

$$uv=u^T C v,\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix},\qquad \gamma_a^T C=-C\gamma_a.$$

The matter spinors are Grassmann odd. The labels are commuting, so the operations $\delta_\zeta,\delta_{\widetilde\zeta}$ are odd and obey

$$\delta(XY)=(\delta X)Y+(-1)^{|X|}X\,\delta Y.$$

In particular, $\delta(\widetilde\psi\psi)=(\delta\widetilde\psi)\psi-\widetilde\psi\,\delta\psi$. The exchange rules used below are $\widetilde\psi\zeta=-\zeta\widetilde\psi$ and $\widetilde\psi\gamma_\mu\zeta=\zeta\gamma_\mu\widetilde\psi$. Multiplication on the left by an external odd parameter turns these operations into physical even variations.

Obtain the action and the transformations

On this background the matter derivative P, (6.6) and the kinetic derivative (6.31) both reduce to the Levi-Civita derivative, including its spin connection on spinors. The gauge-neutral, zero-central-charge specialization of (6.32) is

$$\begin{aligned} \mathcal L={}& \nabla_\mu\widetilde\phi\,\nabla^\mu\phi -i\widetilde\psi\gamma^\mu\nabla_\mu\psi -\widetilde F_\Phi F_\Phi\\ &+\left[-\frac r4 R+r\left(r-\frac12\right)H^2\right]\widetilde\phi\phi -\left(r-\frac12\right)H\,i\widetilde\psi\psi . \end{aligned}$$

Substitute the actual background, retaining $i^2=-1$:

$$-\frac r4 R+r\left(r-\frac12\right)H^2 =\frac{3r}{2\ell^2}-\frac{r(r-\frac12)}{\ell^2} =\frac{r(2-r)}{\ell^2}.$$

Similarly $-(r-\tfrac12)Hi=-(r-\tfrac12)/\ell$. Define three coefficients to keep the cancellation readable:

$$\alpha=\frac1\ell,\qquad b=r\alpha,\qquad m=\left(r-\frac12\right)\alpha,\qquad U=r(2-r)\alpha^2.$$

Here $m$ names the algebraic fermion coefficient, not an independently introduced real-mass deformation. The action is

$$\boxed{ S=\int_{S^3}\!d^3x\,\sqrt g\, \left( \nabla_\mu\widetilde\phi\,\nabla^\mu\phi -i\widetilde\psi\gamma^\mu\nabla_\mu\psi -\widetilde F_\Phi F_\Phi +U\widetilde\phi\phi-m\widetilde\psi\psi \right). }$$

Every term has mass dimension three and R-charge zero. In particular $[b]=[m]=1$, $[U]=2$, and $\sqrt g\,d^3x$ has dimension $-3$ in physical coordinates.

The specialization of P, (6.8) and (6.10) gives

$$\begin{aligned} \delta\phi&=\sqrt2\,\zeta\psi,\\ \delta\psi&=\sqrt2\,\zeta F_\Phi +\sqrt2\,b\,\widetilde\zeta\phi -\sqrt2 i\gamma^\mu\widetilde\zeta\,\nabla_\mu\phi,\\ \delta F_\Phi&=-\sqrt2\,m\,\widetilde\zeta\psi -\sqrt2 i\widetilde\zeta\gamma^\mu\nabla_\mu\psi,\\[2pt] \delta\widetilde\phi&=-\sqrt2\,\widetilde\zeta\widetilde\psi,\\ \delta\widetilde\psi&=\sqrt2\,\widetilde\zeta\widetilde F_\Phi -\sqrt2\,b\,\zeta\widetilde\phi +\sqrt2 i\gamma^\mu\zeta\,\nabla_\mu\widetilde\phi,\\ \delta\widetilde F_\Phi&=-\sqrt2\,m\,\zeta\widetilde\psi -\sqrt2 i\zeta\gamma^\mu\nabla_\mu\widetilde\psi . \end{aligned}$$

An easy mistake is to replace the auxiliary coefficient $r-\tfrac12$ by $r-2$. In the paper, the derivative in $\delta F_\Phi$ acts on the entire product $\widetilde\zeta\gamma^\mu\psi$. The transpose identity gives

$$(\nabla_\mu\widetilde\zeta)\gamma^\mu\psi =-3k\,\widetilde\zeta\psi.$$

Its contribution to $\delta F_\Phi$ is $-\sqrt2 i(-3k)\widetilde\zeta\psi =-3\sqrt2\alpha\,\widetilde\zeta\psi/2$. Combining it with $-\sqrt2(r-2)\alpha\,\widetilde\zeta\psi$ gives $-\sqrt2 m\,\widetilde\zeta\psi$. The same calculation applies to $\delta\widetilde F_\Phi$.

Vary every term: the first family

Set $\widetilde\zeta=0$. Write the five action terms as $\mathcal L_s,\mathcal L_\psi,\mathcal L_F,\mathcal L_U,\mathcal L_m$ in the order displayed above, and abbreviate $p_\mu=\nabla_\mu\phi$, $\widetilde p_\mu=\nabla_\mu\widetilde\phi$. These are gradients, not momenta.

All computations may be done in a normal orthonormal frame at a point because the resulting expressions are tensorial. Use $\nabla_\mu C=\nabla_\mu\gamma^\nu=0$ and the Killing equation, but no field equation. Direct differentiation gives

$$\begin{aligned} \frac{\delta_\zeta\mathcal L_s}{\sqrt2} ={}&\widetilde p^\mu\zeta\nabla_\mu\psi -k\widetilde p^\mu\zeta\gamma_\mu\psi,\\ \frac{\delta_\zeta\mathcal L_\psi}{\sqrt2} ={}&-\widetilde p_\nu\zeta\gamma^\nu\gamma^\mu\nabla_\mu\psi +ib\widetilde\phi\,\zeta\gamma^\mu\nabla_\mu\psi\\ &+i(\nabla_\mu F_\Phi)\zeta\gamma^\mu\widetilde\psi +\frac{3\alpha}{2}F_\Phi\zeta\widetilde\psi,\\ \frac{\delta_\zeta\mathcal L_F}{\sqrt2} ={}&mF_\Phi\zeta\widetilde\psi +iF_\Phi\zeta\gamma^\mu\nabla_\mu\widetilde\psi,\\ \frac{\delta_\zeta\mathcal L_U}{\sqrt2} ={}&U\widetilde\phi\,\zeta\psi,\\ \frac{\delta_\zeta\mathcal L_m}{\sqrt2} ={}&mb\widetilde\phi\,\zeta\psi +im\widetilde p_\mu\zeta\gamma^\mu\psi -mF_\Phi\zeta\widetilde\psi . \end{aligned}$$

For example, the plus sign multiplying $i\widetilde\psi\gamma^\mu\nabla_\mu(\delta\psi)$ inside $\delta\mathcal L_\psi$ follows from the odd product rule. Replacing it by a minus sign ruins the cancellation.

The terms $mF_\Phi\zeta\widetilde\psi$ cancel explicitly. The remaining auxiliary terms are precisely

$$\nabla_\mu\left(\sqrt2 i F_\Phi\zeta\gamma^\mu\widetilde\psi\right),$$

because $i(\nabla_\mu\zeta)\gamma^\mu\widetilde\psi =3\alpha\zeta\widetilde\psi/2$. The auxiliary field has therefore participated in the proof; we have not set it to zero.

To combine the first derivative terms, use $\gamma^\nu\gamma^\mu=g^{\nu\mu}+i\epsilon^{\nu\mu\rho}\gamma_\rho$. The metric part cancels the first term of $\delta\mathcal L_s$. Introduce

$$B_{\zeta,0}^{\mu} =-\sqrt2 i\epsilon^{\nu\mu\rho}\widetilde p_\nu\zeta\gamma_\rho\psi +\sqrt2 iF_\Phi\zeta\gamma^\mu\widetilde\psi .$$

When differentiating its first term, the scalar Hessian drops out: $\epsilon^{\nu\mu\rho}\nabla_\mu\nabla_\nu\widetilde\phi=0$. Differentiating the label and using the gamma product leaves

$$\begin{aligned} \delta_\zeta\mathcal L-\nabla_\mu B_{\zeta,0}^{\mu} ={}&\sqrt2 i b \left(\widetilde p_\mu\zeta\gamma^\mu\psi +\widetilde\phi\,\zeta\gamma^\mu\nabla_\mu\psi\right)\\ &+\sqrt2(U+mb)\widetilde\phi\,\zeta\psi. \end{aligned}$$

Here the coefficient of $\widetilde p_\mu\zeta\gamma^\mu\psi$ has become $b$ because

$$b=m+\frac{\alpha}{2}.$$

One final divergence completes the cancellation:

$$\begin{aligned} \nabla_\mu\left(\sqrt2 i b\widetilde\phi\,\zeta\gamma^\mu\psi\right) ={}&\sqrt2 i b \left(\widetilde p_\mu\zeta\gamma^\mu\psi +\widetilde\phi\,\zeta\gamma^\mu\nabla_\mu\psi\right)\\ &+\frac{3\sqrt2 b\alpha}{2}\widetilde\phi\,\zeta\psi. \end{aligned}$$

The needed scalar identity is

$$U+mb =\left[r(2-r)+r\left(r-\frac12\right)\right]\alpha^2 =\frac{3b\alpha}{2}.$$

We have proved the full local result

$$\boxed{ \delta_\zeta\mathcal L=\nabla_\mu B_\zeta^\mu,\qquad B_\zeta^\mu=B_{\zeta,0}^\mu +\sqrt2 i b\widetilde\phi\,\zeta\gamma^\mu\psi . }$$

This also explains how one could determine the curvature completion: with the transformations fixed, the derivative terms require $m=b-\alpha/2$, and the scalar terms require $U=3b\alpha/2-mb$.

Check the independent second family

The second family must be checked independently in Euclidean signature. Set $\zeta=0$, and define $\nabla^2\phi=\nabla_\mu\nabla^\mu\phi$. The five variations are

$$\begin{aligned} \frac{\delta_{\widetilde\zeta}\mathcal L_s}{\sqrt2} ={}&k p_\mu\widetilde\zeta\gamma^\mu\widetilde\psi -p_\mu\widetilde\zeta\nabla^\mu\widetilde\psi,\\ \frac{\delta_{\widetilde\zeta}\mathcal L_\psi}{\sqrt2} ={}&-i\widetilde F_\Phi\widetilde\zeta\gamma^\mu\nabla_\mu\psi +\frac{3b\alpha}{2}\phi\,\widetilde\zeta\widetilde\psi\\ &+i\left(b-\frac{\alpha}{2}\right)p_\mu \widetilde\zeta\gamma^\mu\widetilde\psi -(\nabla^2\phi)\widetilde\zeta\widetilde\psi,\\ \frac{\delta_{\widetilde\zeta}\mathcal L_F}{\sqrt2} ={}&m\widetilde F_\Phi\widetilde\zeta\psi +i\widetilde F_\Phi\widetilde\zeta\gamma^\mu\nabla_\mu\psi,\\ \frac{\delta_{\widetilde\zeta}\mathcal L_U}{\sqrt2} ={}&-U\phi\,\widetilde\zeta\widetilde\psi,\\ \frac{\delta_{\widetilde\zeta}\mathcal L_m}{\sqrt2} ={}&-m\widetilde F_\Phi\widetilde\zeta\psi -mb\phi\,\widetilde\zeta\widetilde\psi -im p_\mu\widetilde\zeta\gamma^\mu\widetilde\psi . \end{aligned}$$

To obtain the fermion-kinetic row, use the symmetric scalar Hessian and $\sum_a\gamma_a\gamma_b\gamma_a=-\gamma_b$. For instance, the derivative of the label in $\widetilde\psi\gamma^\mu\gamma^\nu\widetilde\zeta\,p_\nu$ produces $-k p_\nu\widetilde\psi\gamma^\nu\widetilde\zeta$. This accounts for its $-\alpha/2$ coefficient.

The auxiliary terms cancel between rows. The scalar terms cancel by $U+mb=3b\alpha/2$. The remaining terms are

$$\delta_{\widetilde\zeta}\mathcal L =\sqrt2\left[ k p_\mu\widetilde\zeta\gamma^\mu\widetilde\psi -p_\mu\widetilde\zeta\nabla^\mu\widetilde\psi -(\nabla^2\phi)\widetilde\zeta\widetilde\psi \right].$$

They are another exact divergence:

$$\boxed{ \delta_{\widetilde\zeta}\mathcal L=\nabla_\mu B_{\widetilde\zeta}^{\mu}, \qquad B_{\widetilde\zeta}^{\mu} =-\sqrt2(\widetilde\zeta\widetilde\psi)\nabla^\mu\phi . }$$

The two displayed $B$ expressions have different forms because we wrote the fermion kinetic term with its derivative acting on $\psi$. They are boundary expressions for this choice of Lagrangian, not a claim that the two symmetries are inequivalent.

For smooth global fields and the global spinor labels constructed above, each $B^\mu$ is a smooth global vector. The sphere has no boundary, and the metric is fixed under these rigid transformations. Thus

$$\delta S=\int_{S^3}\!d^3x\,\sqrt g\,\nabla_\mu B^\mu=0.$$

No scalar, fermion, or auxiliary matter equation has been used. This is off-shell classical invariance. A quantum change-of-variables argument would also require a compatible measure, regulator, and integration prescription.

Worked example: a nonzero local variation that integrates to zero

Use the dimensionless ambient coordinates $y^I$ of the sphere lesson, $\sum_{I=0}^3(y^I)^2=1$. In the global frame,

$$E_a y^0=-\alpha y^a,\qquad E_a y^b=\alpha\left(\delta_{ab}y^0-\epsilon_{abc}y^c\right), \qquad \operatorname{div}E_a=0 .$$

Choose the $\widetilde\zeta$ family and the smooth field profile

$$\widetilde\zeta=t\begin{pmatrix}1\\0\end{pmatrix},\qquad \phi=v y^0,\qquad \widetilde\psi=\vartheta\begin{pmatrix}0\\1\end{pmatrix}, \qquad \widetilde\phi=\psi=F_\Phi=\widetilde F_\Phi=0.$$

Here $t,v$ commute and $\vartheta$ is a single Grassmann-odd amplitude. Their dimensions are $-1/2,1/2,1$, respectively. Constant spinor components are globally defined in this frame. The profile is an algebraic test of the local identity; it need not lie on a bosonic conjugation contour.

The contraction is $\widetilde\zeta\widetilde\psi=-t\vartheta$. Therefore

$$B_{\widetilde\zeta}^{a}=-\sqrt2\,tv\vartheta\,\alpha y^a,\qquad \nabla_aB_{\widetilde\zeta}^{a} =-3\sqrt2\,tv\vartheta\,\alpha^2 y^0.$$

Check this directly from the action. Set $Y=y^a\gamma_a$ and $e_1=(1,0)^T$. The frame identities give

$$\begin{aligned} \gamma^a\nabla_a(y^0e_1) &=-\alpha Y e_1+\frac{3i\alpha}{2}y^0e_1,\\ \gamma^a\nabla_a(Y e_1) &=3\alpha y^0e_1-\frac{i\alpha}{2}Y e_1 . \end{aligned}$$

In particular $\delta\psi=\sqrt2\,tv(b y^0e_1+i\alpha Y e_1)$. After removing the common factor $\sqrt2\,tv\vartheta$, the five action variations are

Term	Variation divided by $\sqrt2\,tv\vartheta$
Scalar kinetic term	$0$
Fermion kinetic term	$-(3+\tfrac32r)\alpha^2y^0-im\alpha y^1$
Auxiliary term	$0$
Scalar curvature term	$U y^0$
Fermion algebraic term	$mb y^0+im\alpha y^1$

The $y^1$ terms cancel, while $U+mb=3r\alpha^2/2$ leaves exactly $-3\alpha^2y^0$. The local variation is nonzero. Its integral vanishes because $y^0$ has zero average on the sphere: the isometry $(y^0,y^1)\mapsto(-y^0,-y^1)$ preserves the measure and reverses the integrand.

Curvature, auxiliaries, and integration contours

At the free conformal R-charge $r=1/2$, P, (6.31) and the discussion after (6.32) identify the conformal specialization. Here it gives

$$m=0,\qquad U=\frac{3}{4\ell^2}=-\frac R8.$$

The sphere still has a scalar curvature coupling. The vanishing of the explicit fermion algebraic term does not remove its spin connection.

The proof treated tilded and untilded fields as independent Euclidean variables. To discuss a bosonic Gaussian, one possible convergent choice is

$$\widetilde\phi=\phi^*,\qquad \widetilde F_\Phi=-F_\Phi^*.$$

On this contour its bosonic quadratic form is

$$S_{\rm bos} =\int\!d^3x\,\sqrt g\, \left(|\nabla\phi|^2+U|\phi|^2+|F_\Phi|^2\right).$$

For $0<r<2$, $U>0$ and this form is positive. At $r=0$ or $2$, the constant scalar mode is a flat direction. Outside that interval, the constant scalar direction is unbounded below on this contour. These statements concern this explicit Gaussian prescription; the polynomial supersymmetry identity itself holds for every real $r$. Fermions remain independent Berezin integration variables. We have not asserted that a conjugation relation between the two commuting labels is required, or that this bosonic contour alone establishes all quantum Ward identities.

The classical algebraic matter equations set $F_\Phi=\widetilde F_\Phi=0$ in this free model. One may use those equations after proving invariance, but an auxiliary Gaussian still has fluctuations. Eliminating the matter auxiliary also changes whether the transformation algebra is realized off shell. None of these operations eliminate the fixed background $H$.

Take the flat limit while holding the theory fixed

Take $\ell\to\infty$ at fixed $r$, fixed physical field normalization, and fixed distances in a local patch. Then

$$b,m=O(\ell^{-1}),\qquad U=O(\ell^{-2}),\qquad \nabla_\mu\zeta,\nabla_\mu\widetilde\zeta=O(\ell^{-1}).$$

In a local frame approaching a Cartesian frame, the transformations become the flat chiral transformations and

$$\mathcal L\longrightarrow \partial_\mu\widetilde\phi\,\partial^\mu\phi -i\widetilde\psi\gamma^\mu\partial_\mu\psi -\widetilde F_\Phi F_\Phi.$$

The extra term $\sqrt2 i b\widetilde\phi\,\zeta\gamma^\mu\psi$ in $B_\zeta^\mu$ vanishes as well. This is a local large-radius limit of the action and symmetry. It does not assert convergence of an unregulated infinite-volume partition function.

Exercises

1. Change only the curvature sign

Suppose someone keeps $H=-i/\ell$ and the displayed transformations but substitutes $R=+6/\ell^2$ into the source action. Find the resulting scalar coefficient $U_{\rm wrong}$ and the uncancelled $\zeta$ variation. Evaluate its integral on $\widetilde\phi=v$, $\zeta=t(1,0)^T$, $\psi=\vartheta(0,1)^T$, with all other fields zero and constant components. Use $\operatorname{Vol}(S^3)=2\pi^2\ell^3$.

Hint 1

Compare the two Lagrangians before varying. Only the coefficient of $\widetilde\phi\phi$ changes.

Hint 2

The correct action variation integrates to zero. The difference is $\sqrt2(U_{\rm wrong}-U)\widetilde\phi\,\zeta\psi$, and $\zeta\psi=-t\vartheta$.

Solution

The inconsistent substitution gives

$$U_{\rm wrong} =-\frac{3r}{2\ell^2} -\frac{r(r-\frac12)}{\ell^2} =-\frac{r(r+1)}{\ell^2},\qquad U_{\rm wrong}-U=-\frac{3r}{\ell^2}.$$

Thus the extra local variation is $-3\sqrt2 r\widetilde\phi\,\zeta\psi/\ell^2$. On the specified global profile it is the constant $3\sqrt2 rvt\vartheta/\ell^2$. Its integral is

$$\delta_\zeta S_{\rm wrong} =6\sqrt2\pi^2r\ell\,vt\vartheta .$$

For $r\ne0$ this is a nonzero Grassmann polynomial. The accidental insensitivity of this particular test at $r=0$ does not repair a convention change for arbitrary $r$. A consistent translation to an opposite curvature convention must translate the curvature coefficient as well.

2. An auxiliary equation is not a vanishing Gaussian fluctuation

Restrict the bosonic contour above to constant fields $\phi=c$, $F_\Phi=d$. With $V_3=2\pi^2\ell^3$ and Lebesgue measure $d^2c\,d^2d$, calculate the finite-dimensional Gaussian integral and $\langle|c|^2\rangle$, $\langle\widetilde F_\Phi F_\Phi\rangle$. For which real $r$ does it converge? Evaluate the scalar moment at $r=1/2$.

Hint 1

The constant-field action is $S_0=V_3(U|c|^2+|d|^2)$. Use polar coordinates on each complex plane.

Hint 2

For $a>0$, $\int_{\mathbb C}d^2z\,e^{-a|z|^2}=\pi/a$. Differentiate with respect to $a$ to obtain the moment. On this contour $\widetilde F_\Phi F_\Phi=-|d|^2$.

Solution

The radial integral is $2\pi\int_0^\infty \rho e^{-a\rho^2}d\rho=\pi/a$. Therefore

$$Z_0=\frac{\pi^2}{V_3^2 U},\qquad \langle|c|^2\rangle=\frac1{V_3U},\qquad \langle\widetilde F_\Phi F_\Phi\rangle=-\frac1{V_3}.$$

Convergence requires $U>0$, or $0<r<2$. At either endpoint the $c$ integral has infinite volume; outside, its integrand grows along large $|c|$. At $r=1/2$,

$$\langle|c|^2\rangle=\frac{2}{3\pi^2\ell}.$$

The nonzero auxiliary moment is consistent with the algebraic saddle equation $d=0$. A saddle location and a fluctuating integration variable are different statements. $Z_0$ is only the specified bosonic constant-field truncation, not the sphere partition function of the complete theory.

3. Change the assumption behind the flat limit

Instead of holding $r$ fixed, choose a family $r(\ell)=\tfrac12+M\ell$, where $M$ is a fixed nonzero real quantity of mass dimension one. Calculate $b,m,U$ and their large-radius limits. Does this family approach the massless flat action proved above? Does the stated bosonic contour remain convergent?

Hint 1

Substitute $r(\ell)$ before taking the limit. A coefficient proportional to $\ell^{-1}$ need not vanish if its numerator grows with $\ell$.

Hint 2

Compute $(\tfrac12+M\ell)(\tfrac32-M\ell)$ for the scalar coefficient. Test convergence with the constant scalar direction, without invoking the fermion determinant.

Solution

The coefficients are

$$b=M+\frac1{2\ell},\qquad m=M,\qquad U=-M^2+\frac M\ell+\frac3{4\ell^2}.$$

Thus $b\to M$, $m\to M$, and $U\to-M^2$. Neither the algebraic transformation terms nor the quadratic action approach the fixed-$r$ massless result. The family changes the R-charge assignment while changing the geometry, so it changes a premise of that limit.

For sufficiently large $\ell$, $r(\ell)$ lies outside $(0,2)$, and the bosonic constant scalar Gaussian on the stated contour diverges. A surviving coefficient called $m$ does not by itself establish a healthy real-mass deformation; that construction requires its own background and reality prescription.

Check your understanding

Starting from the displayed transformations alone, explain why the derivative terms force $m=b-\alpha/2$ and the remaining scalar terms force $U=3b\alpha/2-mb$. Then identify exactly where the compactness of $S^3$ enters. It is used to discard the integral of a global divergence, not to replace the local cancellation.

You now have a complete example of the rigid-background construction: specified global spinors, an explicit off-shell matter theory, both local variation identities, and a controlled flat limit. Later action constructions can organize these same cancellations using multiplet products; this example provides a direct component check against them.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 6, curved multiplets and actions