Recover the linearized flat-space description

Before this lesson: The curved supersymmetry algebra; The R-multiplet through its physical currents; Stress tensors and improvement terms.

Suppose a curved-space action has been proposed. Before using it at finite curvature, we can ask a simpler question: does its first variation reproduce the flat-space currents that couple to the background? Supersymmetry makes this test more restrictive. The stress tensor must occur together with particular R-current and central-current partners.

The preparation is the curved algebra, the R-multiplet and stress-tensor improvements. We will recover one explicit deformation in components, including an operator whose supersymmetry variation generates it. This is the kind of consistency test performed in section 7 of the three-manifold paper. It is a first-order statement.

Fix the expansion and the supercharge

Work in three-dimensional Euclidean space with coordinates $(\tau,x,y)=(x^1,x^2,x^3)$ and orientation $\epsilon^{123}=1$ . The parent theory is N=2, with four real Lorentzian Poincaré supercharges and a conserved R-symmetry. No dynamical gauge group is assumed.

Introduce a real dimensionless expansion parameter $\varepsilon$ :

g_{\mu\nu}=\delta_{\mu\nu}+2\varepsilon h_{\mu\nu}+O(\varepsilon^2), \qquad S(\varepsilon)=S_0+\varepsilon S_1+O(\varepsilon^2).

Below, $A,V,C,H$ denote the coefficients of the first-order sources; their actual backgrounds are $\varepsilon A,\varepsilon V,\varepsilon C,\varepsilon H$ at this order. The bosonic current coupling is

S_1=\int d^3x\left[ -h^{\mu\nu}T_{\mu\nu} +j_R^\mu\left(A_\mu-\frac32V_\mu\right) +j_Z^\mu C_\mu+J_ZH\right]. \tag{R08.1}

This is P (7.1), in the Euclidean convention $\delta S=-\frac12\int T^{\mu\nu}\delta g_{\mu\nu}$ . All operators on the right are flat-space operators, evaluated at zero background.

Choose the commuting spinor direction $u_+=(1,0)^T$ and $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$ . Strip off the constant supersymmetry parameter normalization and call the corresponding odd generator $Q$ . It has mass dimension $1/2$ and R-charge $-1$ . Its same-charge algebra gives $Q^2=0$ on the operator algebra, with the usual equations-of-motion and contact-term qualifications. Multiplying by an external odd parameter would instead give an even variation.

An operator is Q-closed if $Q\mathcal O=0$ . It is Q-exact if $\mathcal O=Q\mathcal V$ for some operator $\mathcal V$ . Exact operators are closed when the nilpotency relation applies. These are algebraic definitions; neither definition by itself says that an expectation value vanishes.

Find the fermionic preimage in components

Let $\widetilde{\mathcal S}_{\mu,\alpha}$ be the opposite-R-charge supercurrent: $\mu$ is a vector index, while $\alpha=1,2$ is a lower spinor index. It has dimension $5/2$ and R-charge $+1$ . The needed transformation from P (7.3) is

Q\widetilde{\mathcal S}_\mu =u_+\left(2j_{Z\mu} +i\epsilon_{\mu\nu\rho}\partial^\nu j_R^\rho\right) +\gamma^\nu u_+ \left(2iT_{\mu\nu}+\partial_\nu j_{R\mu} -\epsilon_{\mu\nu\rho}\partial^\rho J_Z\right). \tag{R08.2}

Take the second spinor component. Since $\gamma_1u_+=u_+$ , $\gamma_2u_+=(0,-1)^T$ and $\gamma_3u_+=(0,-i)^T$ , the term proportional to $j_Z$ drops out. The two relevant rows are

\begin{aligned} Q\widetilde{\mathcal S}_{2,2} &=-2iT_{22}+2T_{23}-\partial_xj_{R2} -i\partial_yj_{R2}+i\partial_\tau J_Z,\\ Q\widetilde{\mathcal S}_{3,2} &=-2iT_{23}+2T_{33}-\partial_xj_{R3} -i\partial_yj_{R3}-\partial_\tau J_Z . \end{aligned} \tag{R08.3}

Define the odd operator $\mathcal W=\widetilde{\mathcal S}_{2,2} -i\widetilde{\mathcal S}_{3,2}$ . The $T_{23}$ terms cancel, leaving

\begin{aligned} \mathcal O_\perp ={}&T_{22}+T_{33} -\frac i2(\partial_xj_{R2}+\partial_yj_{R3})\\ &-\frac12(\partial_xj_{R3}-\partial_yj_{R2}) -\partial_\tau J_Z =\frac i2Q\mathcal W . \end{aligned} \tag{R08.4}

This exhibits a Q-exact bosonic operator rather than merely declaring one exists. All its terms have dimension three and R-charge zero. A transverse stress-tensor combination alone would miss the derivative partners.

There is also a useful check on what “closed” means here. Applying the remaining transformations in (7.3) gives

Q\mathcal O_\perp=\frac12\partial^\mu\mathcal S_{\mu,2}=0. \tag{R08.5}

The equality to zero uses supercurrent conservation. The currents are composite operators satisfying their Ward identities; this is not an identity on every off-shell matter configuration. In correlation functions, coincident insertions and charged operators can supply contact terms. Those qualifications must survive when we integrate the deformation.

Derive a small supersymmetric background

Let $f(\tau,x,y)$ be a smooth real dimensionless function, and choose $h_{22}=h_{33}=f$ , with all other $h_{\mu\nu}$ zero. The metric is therefore

ds^2=d\tau^2+(1+2\varepsilon f)(dx^2+dy^2)+O(\varepsilon^2).

Start from $S_1=-\int f\mathcal O_\perp$ . Assume periodic fields, adequate decay, or compact support so that the integrations by parts have no boundary contribution. Equation (R08.4) yields

\boxed{\displaystyle S_1=Q\Psi,\qquad \Psi=-\frac i2\int d^3x\,f\mathcal W .} \tag{R08.6}

The source profile is fixed under $Q$ . Moving the derivatives in $-f\mathcal O_\perp$ onto $f$ and comparing with (R08.1) determines

\begin{gathered} C=V=0,\qquad H=-\partial_\tau f,\qquad A_\tau=0,\\ A_x=\frac12(\partial_yf-i\partial_xf),\qquad A_y=-\frac12(\partial_xf+i\partial_yf). \end{gathered} \tag{R08.7}

For example, the term $f\,\partial_\tau J_Z$ becomes $-(\partial_\tau f)J_Z$ , fixing the sign of $H$ . The curl term fixes the real part of the transverse R-connection, and the divergence term fixes its imaginary part. A real metric can require complex Euclidean sources. We have not performed a complex gauge transformation.

The source constraints hold: $V=-i*dC=0$ and $\nabla\cdot V=0$ . A local two-form coefficient $B=i f\,dx\wedge dy$ gives $H=i*dB=-\partial_\tau f$ at this order. A derivative of $f$ has dimension one, as do $A,H$ .

Independently check the Killing equation. In the orthonormal coframe $e^1=d\tau$ , $e^2=(1+\varepsilon f)dx$ , $e^3=(1+\varepsilon f)dy$ , the spin-connection matrices to first order are

\begin{aligned} \Gamma_\tau&=0,\\ \Gamma_x&=\frac{i\varepsilon}{2} (\partial_y f\,\gamma_1-\partial_\tau f\,\gamma_3),\\ \Gamma_y&=\frac{i\varepsilon}{2} (-\partial_x f\,\gamma_1+\partial_\tau f\,\gamma_2). \end{aligned}

Substitute $\zeta=(1+\varepsilon f/2)u_+$ into $(\partial_\mu+\Gamma_\mu-i\varepsilon A_\mu)\zeta =-\varepsilon H\gamma_\mu u_+/2+O(\varepsilon^2)$ . The $\tau$ equation reads $\varepsilon(\partial_\tau f)u_+/2$ on both sides. In the $x$ equation, the terms along $u_+$ cancel:

\frac{\partial_xf}{2} +\frac i2\partial_yf-iA_x=0.

The remaining term is $-i(\partial_\tau f)\gamma_3u_+/2 =(\partial_\tau f)\gamma_2u_+/2$ , as required; the $y$ equation works similarly. Thus the current derivation and the geometric equation agree. We have constructed one preserved charge; we have not established a second independent solution.

Check the curvature interaction on a chiral scalar

Take the free gauge-neutral massless chiral, with scalar R-charge $r$ and $\rho=\widetilde\phi\phi$ . The Euclidean tilded field is independent until an integration contour is chosen. Its improvement from P (7.13) is

\Delta T_{\mu\nu} =\frac r2(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)\rho.

In our transverse warp, its contribution is

-\int f(\Delta T_{22}+\Delta T_{33}) =-r\int \rho\left[\partial_\tau^2f +\frac12(\partial_x^2+\partial_y^2)f\right]. \tag{R08.8}

Two integrations by parts explain the sign. The paper uses negative scalar curvature on a sphere. For $g=\delta+2\varepsilon h$ ,

R_P=\varepsilon R_P^{(1)}+O(\varepsilon^2),\qquad R_P^{(1)}=4\partial_\tau^2 f +2(\partial_x^2+\partial_y^2)f.

Equation (R08.8) is exactly $-\frac r4\int R_P^{(1)}\rho$ , the first variation of the curvature coupling in P (6.32).

One can check more than this single coefficient. The bosonic action at $V=C=z=0$ contains

\sqrt g\left[ g^{\mu\nu}D_\mu\widetilde\phi D_\nu\phi -\widetilde F_\Phi F_\Phi -\frac r4R_P\rho+r\left(r-\frac12\right)H^2\rho \right],

where $D_\mu\phi=(\partial_\mu-irA_\mu)\phi$ and the tilded scalar has charge $-r$ . Its first variation is $2f(\partial_\tau\widetilde\phi\,\partial_\tau\phi-\widetilde F_\Phi F_\Phi) +A_\mu j_{R,\mathrm{scalar}}^\mu-rR_P^{(1)}\rho/4$ , with $j_{R,\mathrm{scalar}}^\mu=ir(\widetilde\phi\partial^\mu\phi-\partial^\mu\widetilde\phi\,\phi)$ . The first term is the minimally coupled transverse stress response; the second is the R-source response; the third is the improvement. The $H^2$ term starts at second order. This bosonic coefficient check complements the full current-multiplet argument; the bosonic density by itself is not claimed to be Q-exact.

What first-order exactness does and does not establish

Under an anomaly-free $Q$ Ward identity, an invariant integration cycle and regulator, and suitable boundary and insertion conditions, $\langle Q\Psi\rangle_0=0$ . Then $F=-\log Z$ has $F'(0)=\langle S_1\rangle_0=0$ . Convergence and a well-defined partition function are additional assumptions, not consequences of (R08.6).

To see the missing information, write

S(\varepsilon)=S_0+\varepsilon S_1+\varepsilon^2S_2+\cdots, \qquad Q(\varepsilon)=Q_0+\varepsilon Q_1+\cdots .

The metric, volume factor, source interactions and preserved spinor all change. Neither $S_2$ nor the corrections to the transformation law were determined by identifying $S_1=Q_0\Psi$ . In a fixed measure and source prescription,

F''(0)=2\langle S_2\rangle_0-\langle S_1S_1\rangle_{0,\mathrm c}. \tag{R08.9}

The subscript means the connected correlator. Coincident insertions require the corresponding contact terms and source counterterms. A formal manipulation of separated-point Q-exact operators cannot discard those terms automatically.

This is the explicit limitation stated after P (7.7). We have verified a tangent deformation and its linearized background, not proved independence throughout a finite family. The later localization argument explains the stronger hypotheses under which an entire deformation can be controlled.

Exercises

1. Test a periodic warp with a nonzero curvature response

Let $f=\cos(k\tau)$ , with $k>0$ , on a periodic interval of length $2\pi/k$ and unit transverse coordinate area. Determine $A,H,\zeta$ and $R_P^{(1)}$ . For a test scalar profile $\rho=\rho_0[1+a\cos(k\tau)]$ , $\rho_0>0$ , $|a|<1$ , evaluate the integrated curvature contribution to $S_1$ by both sides of (R08.8). Must this individual contribution vanish because the complete deformation is Q-exact?

Hint 1

There are no transverse derivatives of $f$ . Keep track of whether $H$ denotes its first-order coefficient or the full background.

Hint 2

The period integral of $\cos(k\tau)$ vanishes, while that of its square is $\pi/k$ . On the stress side differentiate $\rho$ , not $f$ , before integrating.

Solution

The first-order coefficients and normalized spinor are

A=0,\qquad H=k\sin(k\tau),\qquad \zeta=\left[1+\frac{\varepsilon}{2}\cos(k\tau)\right]u_+ +O(\varepsilon^2),\qquad R_P^{(1)}=-4k^2\cos(k\tau).

The physical scalar source is $\varepsilon H+O(\varepsilon^2)$ . The curvature term gives

-\frac r4\int_0^{2\pi/k}R_P^{(1)}\rho\,d\tau =r k^2\rho_0a\,\frac{\pi}{k} =\pi r a k\rho_0 .

On the stress side, $\Delta T_{22}+\Delta T_{33}=r\partial_\tau^2\rho =-ra k^2\rho_0\cos(k\tau)$ , giving the same answer after multiplication by $-f$ . The result is generally nonzero. Q-exactness concerns the complete current deformation and, under additional Ward assumptions, its expectation value. It does not set each term to zero on an arbitrary field profile. This profile is a diagnostic of the metric coupling, not an asserted on-shell configuration or a quantum expectation value.

2. Restore the boundary term

Repeat the integration by parts leading to (R08.7) on a region with boundary. Find a vector $\mathcal B^\mu$ such that the source density in (R08.1) equals $-f\mathcal O_\perp-\partial_\mu\mathcal B^\mu$ . As a sign test, take a slab $0\le x\le L$ , unit transverse area, $f=f(x)$ , $j_{R3}=j_*$ constant, and all other current components in this integration-by-parts test zero. Compare the source integral and boundary flux.

Hint 1

Use the product rule separately for the divergence, curl and $\partial_\tau J_Z$ terms.

Hint 2

The slab has opposite outward normals at $x=0$ and $x=L$ . The relevant source is $A_y=-f'(x)/2$ .

Solution

The required boundary vector is

\mathcal B^\tau=fJ_Z,\qquad \mathcal B^x=\frac f2(ij_{R2}+j_{R3}),\qquad \mathcal B^y=\frac f2(ij_{R3}-j_{R2}).

Consequently $S_1=Q\Psi-\int_{\partial M}n_\mu\mathcal B^\mu$ . In the slab test the derivative-current operator $\mathcal O_\perp$ vanishes, whereas

\int_0^L A_yj_*\,dx=-\frac{j_*}{2}[f(L)-f(0)] =-\int_{\partial M}n_\mu\mathcal B^\mu .

This is a nonzero check of the sign. Supersymmetric boundary conditions, or compensating boundary interactions, must be specified before using the integrated Q-exact argument on a physical boundary. The test only checks the integration-by-parts identity; it does not construct a boundary supermultiplet.

3. Separate stationarity from finite independence

Derive (R08.9) by differentiating a convergent, fixed-measure integral. Then evaluate the elementary diagnostic

Z_{\rm toy}(\varepsilon) =\int_{-\infty}^{\infty}dx\, \exp\left[-\frac{x^2}{2}-\varepsilon^2x^2\right].

Find its first and second free-energy derivatives at zero. Explain exactly what this diagnostic does, and does not, say about the supersymmetric background calculation.

Hint 1

Use $F'=\langle S'\rangle$ and differentiate the normalized expectation value. Both the numerator and denominator vary.

Hint 2

Here $S_1=0$ and $S_2=x^2$ . The Gaussian variance at $\varepsilon=0$ is one.

Solution

Differentiating gives $F''=\langle S''\rangle-\langle(S')^2\rangle+\langle S'\rangle^2$ . At zero, $S'=S_1$ and $S''=2S_2$ , proving (R08.9). The Gaussian is

Z_{\rm toy}(\varepsilon) =\frac{\sqrt{2\pi}}{\sqrt{1+2\varepsilon^2}},\qquad F_{\rm toy}(\varepsilon)-F_{\rm toy}(0) =\frac12\log(1+2\varepsilon^2).

Thus $F'_{\rm toy}(0)=0$ but $F''_{\rm toy}(0)=2$ , also equal to $2\langle x^2\rangle_0$ . An exactly vanishing first variation does not determine the second. This ordinary integral is not a counterexample to a localization theorem and carries no claimed supercharge. It isolates the logical gap: the R-multiplet calculation determined $S_1$ , while control of a finite family requires its higher terms, transformations, measure, boundaries and contact prescription.

The exit test is concrete: for a proposed small background change, identify the flat currents it couples to, exhibit the fermionic preimage when one exists, and check the geometric equation independently. Each step is useful without promoting the resulting first-order identity to a finite-background theorem.

The Section 7 paper guide supplies the full source map and reconstructs another current preimage. Compare its FI and Yang-Mills checks with this metric deformation, then test what first-order agreement leaves undetermined.

Which geometric data does the partition function see? constructs an exact curved torus family, checks its tangent against this calculation and distinguishes a changing scalar spectrum from a protected response.