Paper guide: Section 7, the linearized check

Section 7 tests the curved construction by returning to familiar flat-space currents. A correct first-order coupling must reproduce the stress tensor, R current and central-current partners of the original theory. Supersymmetry relates those operators, so checking one curvature coefficient alone is insufficient.

Begin with the linearized flat-space description, which derives a transverse metric deformation and its fermionic preimage. Keep P v3, Section 7, printed pages 30–33 / PDF pages 31–34, open beside this guide. We will reconstruct a different direction: a change of auxiliary sources with the metric held flat. Its FI coupling gives a concrete sign and normalization test.

The theory is three-dimensional $\mathcal N=2$, with four real Lorentzian Poincaré supercharges before Euclidean continuation. We select one positive-charge complex Euclidean spinor direction. The current argument needs a conserved R-multiplet but no particular dynamical gauge group. The gauge example below specifies a compact $U(1)$ of rank one.

Read the comparison in two stages

First read (7.1)–(7.7), which determine source combinations from the abstract current multiplet. Then read the two models in (7.8)–(7.13), which test the previously constructed actions. The model formulas are applications of the current method, not a second derivation of all curved transformation rules.

P v3 passage	What to establish	Canonical calculation
(7.1)–(7.2), printed 30/PDF 31	Fix the metric variation, the three source couplings and the selected spinor.	Expansion and supercharge, R08.1
(7.3)–(7.5), printed 30–31/PDF 31–32	Distinguish closed operators from explicit supersymmetry variations.	A fermionic preimage, R08.2–R08.5; the additional preimage below
(7.4), footnote 11	Translate complex components without losing factors of two.	Adapted coordinates and the explicit dictionary below
(7.6)–(7.7), printed 31/PDF 32	Integrate derivatives onto the sources and check the field-strength constraint.	The small metric background, R08.6–R08.7; the flat-metric source direction below
Paragraph after (7.7), printed 32/PDF 33	Identify the unproved nonlinear step.	The limits of first-order exactness, R08.9
(7.8)–(7.10), printed 32/PDF 33	Compare the vector currents with the YM and FI actions; understand the absence of an explicit CS contribution to this multiplet.	Gauge kinetic terms and curved CS/FI terms
(7.11)–(7.13), printed 33/PDF 34	Keep $z=-m$, the chosen R improvement and the nonminimal stress tensor together.	The chiral curvature check, R08.8, and stress-tensor improvements

The absence of one stress component from the Q-closed combinations in (7.5) restricts the compatible real metric perturbation. It does not assert that the stress tensor of every supersymmetric theory vanishes. Likewise, “the CS term does not contribute to the R-multiplet” in this example does not make the CS coupling physically irrelevant: it still occurs in the action and field equations.

Restore a different Q-exact direction

Use real coordinates $(\tau,x,y)$, orientation $\epsilon^{\tau xy}=1$, and

$$z=x-iy,\qquad \partial_z=\frac12(\partial_x+i\partial_y), \qquad j_z=\frac12(j_x+ij_y). \tag{PC06.1}$$

The last expression is a covector component. It is not the contravariant component $j^z=j^x-ij^y$. Choose the site matrices $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$ and $u_+=(1,0)^T$. As in R08, $Q$ is the odd generator with that direction and $\widetilde{\mathcal S}_{\mu,\alpha}$ is the opposite-charge supercurrent.

Take the indicated three components of (7.3), and define

$$\mathcal W_Z=\widetilde{\mathcal S}_{x,1} +i\widetilde{\mathcal S}_{y,1} +\widetilde{\mathcal S}_{\tau,2}.$$

The first two terms give

$$\begin{aligned} Q(\widetilde{\mathcal S}_{x,1}+i\widetilde{\mathcal S}_{y,1}) ={}&2(j_{Zx}+ij_{Zy})+2iT_{\tau x}-2T_{\tau y}\\ &+(\partial_x+i\partial_y)j_{R\tau} +\partial_yJ_Z-i\partial_xJ_Z. \end{aligned}$$

The $\tau,2$ component cancels both stress terms and the R-current derivative and supplies another $\partial_yJ_Z-i\partial_xJ_Z$. Thus

$$\boxed{\mathcal O_Z=j_{Zz}-i\partial_zJ_Z =\frac14 Q\mathcal W_Z.} \tag{PC06.2}$$

This restores a preimage for the last operator in (7.5). It uses the stated component transformation, rather than guessing exactness from a charge count. The two sides have dimension three and R-charge zero.

Let $F(\tau,x,y)$ be a prescribed smooth dimensionless complex function, unrelated to a chiral auxiliary. On a periodic space or with suitable support, integrate by parts in

$$S_1=\int d^3x\,F\mathcal O_Z =Q\left(\frac14\int d^3x\,F\mathcal W_Z\right). \tag{PC06.3}$$

It gives the source coefficients

$$\begin{gathered} C=\frac F2(dx+i\,dy),\qquad \kappa=\partial_zF,\qquad w=-\partial_\tau F,\\ V=\kappa\,d\tau+\frac w2(dx+i\,dy),\qquad H=i\kappa,\qquad A=\frac32V. \tag{PC06.4} \end{gathered}$$

Here the actual sources are $\varepsilon$ times these coefficients and the metric is unchanged. In particular, $A-3V/2=0$: there is no R-current term in this direction. Yet the matter connection $A-V/2=V$ need not vanish.

Direct differentiation gives $V=-i*dC$ and $\partial_\tau\kappa+\partial_z w=0$. With $U=w\partial_z$ this is precisely the flat version of the source freedom after (7.7), with its divergence condition retained. Substitution into the Killing equation annihilates the constant $u_+$: the $U$ terms cancel by $\Phi U=iU$, and the $\kappa$ terms cancel with $H=i\kappa$. No second independent solution is assumed.

This choice of $C$ displays a potential, not just its curl. A flat addition to $C$ is additional data on a space with closed noncontractible curves; the local equation for $V$ cannot recover it.

Worked check: recover the FI response twice

Take the dynamical Abelian vector multiplet $(a,\sigma,\lambda,\widetilde\lambda,D)$, with FI parameter $\xi$ of dimension one. Let $f_a=da$ and $b_a=*f_a$; this $b_a$ is not the background vector $V$. The FI parts of (7.10) are

$$j_Z^\mu=i\xi b_a^\mu,\qquad J_Z=-\xi\sigma. \tag{PC06.5}$$

Consequently the current coupling predicts

$$S_{1,\mathrm{FI}} =\xi\int d^3x\,(i C\cdot b_a-\sigma H) =-\xi\int d^3x\,(a\cdot V+\sigma H). \tag{PC06.6}$$

The second equality follows from $\int C\cdot\operatorname{curl}a =\int a\cdot\operatorname{curl}C$ and $\operatorname{curl}C=iV$, with the boundary contribution absent. This matches the linear part of $\mathcal L_{\rm FI}=\xi(D-a\cdot V-\sigma H)$ in (6.27). The sign comes from two factors of $i$, not from changing the orientation.

For a nonzero integrated test, use a three-torus of coordinate lengths $2\pi/p,2\pi/q,L_y$, with $p,q>0$, and choose

$$F=\cos(p\tau)\cos(qx),\quad a=a_0\cos(p\tau)\sin(qx)\,d\tau,\quad \sigma=\sigma_0\cos(p\tau)\sin(qx).$$

All functions are periodic. These dynamical fields are test profiles; they are not asserted to solve matter field equations or BPS conditions. Put $\mathcal V_3=(2\pi/p)(2\pi/q)L_y$. Since $\kappa=-(q/2)\cos(p\tau)\sin(qx)$, the final form in (PC06.6) gives

$$S_{1,\mathrm{FI}} =\frac{\xi q\mathcal V_3}{8}(a_0+i\sigma_0). \tag{PC06.7}$$

Alternatively $b_{a,y}=-q a_0\cos(p\tau)\cos(qx)$. The $C\cdot b_a$ term therefore averages $\cos^2(qx)$, while the $H\sigma$ term averages $\sin^2(qx)$. Both averages are $1/2$, reproducing (PC06.7) with the same relative imaginary coefficient. The response is dimensionless, as an action must be.

The YM part of the same source check works similarly: $j_Z^\mu=e^{-2}\partial_\nu(\sigma f_a^{\mu\nu})+\cdots$ gives $i\sigma\epsilon^{\mu\nu\rho}V_\mu f_{a\nu\rho}/(2e^2)$ after integration by parts, while $J_ZH$ contains $-\sigma DH/e^2$. These are the first-order terms in (6.30). The terms quadratic in $V,H$ and the changes of transformation law require additional information from the full curved construction.

Read the conclusion at its actual order

Equation (PC06.3) is an operator-level first-order identity with the same current-conservation and contact qualifications explained in R08. Turning it into a vanishing expectation value also requires an invariant measure, regulator, integration cycle and boundary conditions. The statement is about the complete deformation, not its FI contribution on an arbitrary test field.

The paragraph after (7.7) explicitly leaves the nonlinear analysis for further work. A first-order test cannot fix terms beginning at $\varepsilon^2$, nor does it establish a Ward identity for every finite background. In this particular flat-metric example the Killing equation happens to remain linear in the sources for the fixed constant spinor. That fact alone does not provide the missing quantum or action-level hypotheses.

Exit calculations

1. The curl misses a flat central connection

Keep $g$ flat and $A=V=H=0$, but add $C=c_\tau d\tau$ with real constant $c_\tau$ on a circle of length $L$. Suppose the compact central symmetry has a minimal positive charge $z_0$, so an allowed gauge transformation acts as $e^{iz_0\Lambda}$. For a periodic scalar of charge $z_0$, calculate the eigenvalues of the positive circle operator $-(\partial_\tau-iz_0c_\tau)^2$. When can $C$ be removed globally? What does this test say about using only $V$ to specify a background?

Hint 1

Use the periodic modes $e^{2\pi in\tau/L}$ before attempting a gauge transformation.

Hint 2

The local choice $\Lambda=-c_\tau\tau$ must also have a single-valued charge-$z_0$ exponential.

Solution

The eigenvalues are $(2\pi n/L-z_0c_\tau)^2$, $n\in\mathbb Z$. The connection is removable when $z_0c_\tau L\in2\pi\mathbb Z$; then an integer relabeling of $n$ restores the zero-connection spectrum. Otherwise the holonomy $e^{iz_0c_\tau L}$ shifts the spectrum despite $V=-i*dC=0$. For example a holonomy $-1$ gives a lowest eigenvalue $\pi^2/L^2$, whereas trivial holonomy permits a zero mode.

The central charges and their global period were specified for this exercise; a general formal complex source need not have this compact interpretation. This scalar spectral test does not compute a full supersymmetric partition function. It demonstrates the information lost by replacing a connection with its local curl. The supercharge itself has zero central charge, so its local Killing equation misses this datum.

2. Check the massive chiral current without changing the mass sign

In (7.11)–(7.13), set the dynamical gauge charge to zero and retain real mass $m$ with the source convention $z=-m$. Write the coefficient of $H\widetilde\phi\phi$ and of $iH\widetilde\psi\psi$ predicted by $J_ZH$. Compare them with (6.32). What remains at $r=1/2$?

Hint 1

Combine the two scalar terms in $J_Z=-m\rho+(r-1/2)(2m\rho-i\widetilde\psi\psi)$, where $\rho=\widetilde\phi\phi$.

Hint 2

The scalar action contains $-2(r-1)Hz\rho$ and the fermion action contains $[z-(r-1/2)H]i\widetilde\psi\psi$.

Solution

The coefficients are $2(r-1)m$ for $H\rho$ and $-(r-1/2)$ for $iH\widetilde\psi\psi$. Substituting $z=-m$ into the full action reproduces both. At $r=1/2$, the H-dependent fermion coefficient vanishes, but the scalar response is still $-mH\rho$. The real mass breaks the massless conformal limit; choosing the massless free-field R-charge does not remove the mass deformation.

The same improvement parameter multiplies the curl term $-i(r-1/2)\epsilon^{\mu\nu\rho}\partial_\nu j_\rho$ in $j_Z$ and the nonminimal stress tensor. It is inconsistent to retain one such term while changing the R assignment in the others. The field $m$ here is the stated real-mass parameter, whereas $z$ is its negative in this source convention.

3. A term invisible to every first-order check

Add $\Delta S=\eta\int d^3x\sqrt g\,H^2\widetilde\phi\phi$ to a proposed curved chiral action, with a dimensionless constant $\eta$. Does the flat action or its first-order current comparison determine $\eta$? In the periodic background used above, evaluate the coefficient of $\varepsilon^2$ on a constant scalar profile $\widetilde\phi\phi=\rho_0$. Explain why this is not a family of equally valid supersymmetric actions.

Hint 1

The actual source is $\varepsilon H$ and $H=i\kappa$ here. Average $\kappa^2$ over the torus before multiplying by the volume.

Hint 2

Under the selected charge, $Q\phi=\sqrt2\,u_+\psi$ and $Q\widetilde\phi=0$. Fixed bosonic sources do not vary.

Solution

The added term starts at order $\varepsilon^2$, so all values of $\eta$ pass the flat and first-order comparison. Its test-profile contribution is

$$\Delta S= -\varepsilon^2\frac{\eta\rho_0 q^2\mathcal V_3}{16} +O(\varepsilon^3).$$

The minus sign is $H^2=-\kappa^2$, not a Hermitian norm. This is a coefficient diagnostic on a chosen field profile, not a quantum response. For a general chiral field, the extra variation is $\sqrt2\eta\int\sqrt g\,H^2\widetilde\phi\,u_+\psi$. It is not canceled if the rest of the action and transformations are held fixed. The full supersymmetry check therefore fixes information that the linearized test cannot see. We have not produced a counterexample to a finite-deformation theorem whose hypotheses hold.

You should now be able to identify the flat operator coupled to a source change, exhibit its fermionic preimage when available, and test a component coefficient independently. Section 7 supplies this consistency test; the physical response calculation in Section 8 needs its own source path, contact prescription and normalization.

Continue with the Section 8 guide to distinguish the first source variation from the finite sphere family and its response.