Paper guide: Appendices C and D, four-dimensional origin

Why does a three-dimensional R-charged field see both a shifted connection and the combination $z-rH$? Appendices C and D of Supersymmetric Field Theories on Three-Manifolds, version 3 answer this by reducing a four-dimensional construction. The calculation is particularly useful for finding missing signs: a single four-dimensional derivative must account for both terms.

Read the four-dimensional new minimal background and twisted dimensional reduction first. Those lessons derive the background equation and complete reduction. This guide supplies a reading order, separates the paper’s overlapping notation, and changes one geometric assumption to test the argument’s limits.

The parent theory has four-dimensional $\mathcal N=1$ supersymmetry, or four real Poincaré generators. Its three-dimensional description has $\mathcal N=2$, also four real generators. Euclidean spinors of opposite chirality are independent complex fields; their complex solution count does not add physical supercharges. The construction does not specify a dynamical gauge group or rank. A chosen background can preserve fewer generators.

Read the background before the component table

Start with (C.1), not the long transformation table. A rigid background consists of a metric, an R connection and a conserved auxiliary vector. Its sources are fixed, and its gravitino variation must vanish. They are not required to solve gravitational equations of motion. A conserved continuous R-symmetry admitting the appropriate current multiplet is an input. Festuccia–Seiberg, section 2 and section 6

Then read Appendix D in three passes: geometry in D.1, spinor equations in D.2, and multiplets in D.3. The following map covers all numbered equations. Site equation labels identify the relevant calculation inside the linked teaching home.

Paper landmark	Question it answers	Canonical calculation
(C.1)	Which fixed sources preserve a spinor?	New minimal equations, (B02.4)–(B02.5).
(C.2)–(C.3)	What transforms, and which algebra closes?	Twisted reduction, (B03.12)–(B03.15); read the component dictionary before applying the table.
(D.1)–(D.2)	Which direction is removed, and how are dotted indices identified?	Twisted reduction, (B03.4) and its frame discussion.
(D.3)–(D.7)	How do the metric, frame and horizontal components reduce?	Twisted reduction, (B03.1)–(B03.2), (B03.5).
(D.8)–(D.11)	Where do central charge, spin rotations and curvature terms come from?	Twisted reduction, (B03.3), (B03.6), and the curvature check following (B03.14).
(D.12)–(D.16)	Which source identification avoids an extra spinor projection?	Reduction source dictionary, (B03.7)–(B03.9).
(D.17)–(D.18)	Which three-dimensional fields and derivatives enter the transformations?	Twisted reduction, (B03.12)–(B03.14).
(D.19)	How does translation along the fiber become a central transformation?	Twisted reduction, (B03.15).

The source’s (D.14) is a sufficient choice that makes the vertical equation an identity in the spinor. It is not a classification of every background preserving one selected spinor. Also, the reduction retains spinors independent of the fiber coordinate in the stated frame. A fiber-dependent four-dimensional solution is not automatically retained.

The transformations in (C.3) and (D.19) are odd operations labeled by commuting spinors, as in the teaching lessons. Their bracket is an anticommutator. Multiplying by external odd parameters gives ordinary even variations with a correspondingly translated commutator; one must not change the bracket while keeping its parameter signs unchanged.

Separate the symbols before reducing them

Write four-dimensional background sources as $\mathcal A,\mathcal V$, and the three-dimensional ones as $A,V,H,C$. Write the vector component of a general matter multiplet as $\mathcal B_M$; it is not the background R connection. Denote the multiplet’s bottom scalar by $X$.

Printed symbol and context	Meaning here
$A_M,V_M$ in (C.1)	Background $\mathcal A_M,\mathcal V_M$.
$C,A_M,D$ in the multiplet of (C.2)	Matter components $X,\mathcal B_M,\mathcal D$.
$c_\mu,v_\mu,a_\mu$ in D.1–D.2	Backgrounds $C_\mu,V_\mu,A_\mu$.
$a_\mu,\sigma,D^{(3)}$ in D.3	Matter components $b_\mu=\mathcal B_\mu-C_\mu\mathcal B_y$, $\sigma=\mathcal B_y$, $D=\mathcal D-H\sigma$.
$D_\mu^{(3)}$ acting on a component of R-charge $q$	$\nabla_\mu-iq(A_\mu-V_\mu/2)-izC_\mu$.

The paper also uses a different one-index sigma package from the positive Euclidean blocks in the four-dimensional lesson: $\sigma_P=-is$, $\widetilde\sigma_P=-i\widetilde s$. Consequently its two-index matrices equal minus the site’s $\Sigma,\widetilde\Sigma$. Its input-first frame connection equals minus the ordinary Cartan connection, so the two signs cancel in the spin derivative. Use the dotted-column conversion in the reduction lesson as well; replacing a dotted index by an undotted one without that conversion loses signs in (D.17).

The reduction direction is frame direction $2$ in the four-dimensional ordering $(x^1,y,x^2,x^3)$. Keep this ordering when using epsilon tensors. Curvature has the paper’s sign, negative on a round sphere. Euclidean bilinears such as $V_\mu V^\mu$ contain no complex conjugation.

Finally, the four-dimensional two-form potential is distinct from the metric connection $C$. The four-dimensional lesson uses $\mathcal V^M=\varepsilon^{MNPQ}\partial_N B_{PQ}/2$; Festuccia–Seiberg print coefficient $1/4$. Equal $\mathcal V$ therefore requires $B_{\rm FS}=2B$ in these conventions. Neither this normalization nor the local reduction assigns a compact two-form charge lattice.

Reconstruct the missing horizontal step

Take the paper’s unit-fiber metric and write its source dictionary as one-forms:

$$\begin{gathered} ds_4^2=g_{\mu\nu}dx^\mu dx^\nu+\theta^2,\qquad \theta=dy+C,\\ \mathcal A=A-\frac12V+H\theta,\qquad \mathcal V=\frac12V+H\theta,\qquad V=-i*_3dC. \end{gathered} \tag{PC09.1}$$

The horizontal vector is $E_\mu=\partial_\mu-C_\mu\partial_y$. Thus $\mathcal A(E_\mu)=A_\mu-V_\mu/2$, whereas the coordinate component is $\mathcal A_\mu=A_\mu-V_\mu/2+HC_\mu$. This distinction is compressed into (D.16) by the preceding vector-reduction rules.

For a scalar component of R-charge $q$, let $\Phi(x,y)=e^{izy}\phi(x)$ and $\mathscr D_M=\partial_M-iq\mathcal A_M$. Direct subtraction gives

$$\begin{aligned} e^{-izy}\mathscr D_y\Phi&=i(z-qH)\phi,\\ e^{-izy}(\mathscr D_\mu-C_\mu\mathscr D_y)\Phi &=\left[\partial_\mu-iq(A_\mu-V_\mu/2+HC_\mu) -iC_\mu(z-qH)\right]\phi\\ &=\left[\partial_\mu-iq(A_\mu-V_\mu/2)-izC_\mu\right]\phi . \end{aligned} \tag{PC09.2}$$

The $HC_\mu$ terms cancel. Using the coordinate component as a horizontal one would retain a spurious term. For the bottom field, $q=r$, and the vertical line supplies precisely the combination $z-rH$ in (D.17) and (D.19). For spinors, (D.9) adds the frame-rotation terms; the scalar calculation is not a substitute for them.

Here coordinates have length units, $[A_\mu]=[V_\mu]=[H]=[z]=1$, $[C_\mu]=0$, and $q$ is dimensionless. This also distinguishes $C$ from an ordinarily normalized compact gauge connection: if $y$ has period $2\pi R_f$, a periodic scalar has $z=n/R_f$ and couples with integer $n$ to $C/R_f$.

The connection in (PC09.2) governs matter transformations. The linearized current coupling uses $j^{(R)\mu}(A_\mu-3V_\mu/2)$ in (2.10). These are different statements about the same sources. One cannot replace the matter connection by the current-source combination.

A charged probe of the fibration

Use the local supersymmetric example already solved in the reduction lesson: flat base coordinates, real nonzero $\beta$ of dimension one, and

$$C=\beta x^1dx^2,\qquad V=-i\beta\,dx^3,\qquad H=\beta,\qquad A=-\frac{3i\beta}{2}dx^3 . \tag{PC09.3}$$

It has one fiber-invariant solution in each spinor sector. For a scalar of charge $r$ and fiber momentum $z$, the four-dimensional source is $\mathcal A=-i\beta\,dx^3+\beta\,dy+\beta^2x^1dx^2$. Therefore $\mathscr D_2=\partial_2-ir\beta^2x^1$ and $\mathscr D_y=i(z-r\beta)$ on the mode. Subtracting $\beta x^1\mathscr D_y$ gives $D_2^{(3)}=\partial_2-iz\beta x^1$. Likewise $D_3^{(3)}=\partial_3-r\beta$. The R contribution cancels from the first expression, but remains in the second. The latter is a real shift produced by an imaginary Euclidean source, not a unitary momentum shift. No positive action or integration contour follows from this algebra alone.

Test the fixed-fiber assumption

The formal source construction does not by itself establish a quantum compactification or a real global uplift. Appendix D ends D.2 by allowing complexification of the three-dimensional graviphoton after reduction; a real four-dimensional metric initially gives real $C$. Circle periods, spin structure, source holonomy and patching must still be specified globally. The retained fields and labels must have compatible boundary conditions.

There is a separate local limitation worth calculating. Replace the metric, solely as a geometric test, by

$$ds_4^2=g_{\mu\nu}dx^\mu dx^\nu+\rho(x)^2\theta^2,\qquad \rho>0. \tag{PC09.4}$$

The dimensionless function $\rho$ changes the proper fiber radius. The triangular coframe now gives $\sqrt G=\rho\sqrt g$. For a $y$-independent vector $\mathcal U=U^\mu E_\mu+W\partial_y$, the coordinate divergence is

$$\nabla_M^{(4)}\mathcal U^M =\frac1{\rho\sqrt g}\partial_\mu(\rho\sqrt g\,U^\mu) =\nabla_\mu^{(3)}U^\mu+U^\mu\partial_\mu\log\rho . \tag{PC09.5}$$

Consequently, keeping $U^\mu=V^\mu/2$ with $\nabla_\mu V^\mu=0$ no longer guarantees conservation. On the flat base of (PC09.3), choose $\rho=e^{\kappa x^3}$, with real $\kappa$ of dimension one. The proposed conserved source would instead have divergence $-i\beta\kappa/2$. It already fails a defining condition of Appendix C.

This diagnoses a failed reuse of (D.14), not a no-go theorem for varying-radius supersymmetric reductions. The spin connection changes too. A consistent extension would have to revisit both sources and spinor equations. The last exit calculation checks the changed measure independently using a scalar probe.

Exit calculations

1. Which R-charge belongs in the derivative?

In (C.2), the bottom field has charge $r$ and $\zeta,\widetilde\zeta$ have charges $+1,-1$. Infer the charges of $\chi,\widetilde\chi,M,\widetilde M$. All components have fiber dependence $e^{izy}$. Find the vertical scalar derivatives of $X,M,\widetilde M$, and specialize their differences to $H=-i/\ell$.

Hint 1

The variation preserves total R-charge. First use the two terms in $\delta X$, then $\delta\chi=\zeta M+\cdots$.

Hint 2

The supersymmetry labels have zero fiber momentum. Central charge remains $z$ even when R-charge changes.

Solution

Charge conservation gives $R(\chi)=r-1$, $R(\widetilde\chi)=r+1$, $R(M)=r-2$, $R(\widetilde M)=r+2$. Equation (PC09.2) gives vertical coefficients $i(z-rH)$, $i[z-(r-2)H]$, and $i[z-(r+2)H]$. Relative to the bottom coefficient, the last two differ by $+2iH$ and $-2iH$, hence $+2/\ell$ and $-2/\ell$ on the stated sphere background.

The same symbol $D_\mu^{(3)}$ in the transformation table acts with each operand’s charge. Using the bottom charge on every component breaks R covariance. These derivative shifts are not independent mass-shell assertions.

2. Translate the auxiliary and its curvature coefficient

Use the unit-fiber dictionary to reduce $\mathcal R-6\mathcal V_M\mathcal V^M$ in (C.2), where $\mathcal R$ is the paper’s four-dimensional Ricci scalar. Evaluate it on (PC09.3). Also find $\delta D$ in terms of $\delta\mathcal D,\delta\sigma$ at fixed $H$. Does setting $D=0$ follow from the field definition?

Hint 1

Equation (D.11) gives $\mathcal R=R_3-V^2/2$. The horizontal and vertical parts of $\mathcal V$ in (PC09.1) are orthogonal.

Hint 2

Keep $V^2$ bilinear. In the worked example it equals $-\beta^2$, not $+\beta^2$.

Solution

Since $\mathcal V^2=V^2/4+H^2$,

$$\mathcal R-6\mathcal V^2 =R_3-\frac12V^2-6\left(\frac14V^2+H^2\right) =R_3-2V^2-6H^2.$$

For (PC09.3), $\mathcal R=\beta^2/2$ and $\mathcal V^2=3\beta^2/4$, so both forms give $-4\beta^2$. This checks the coefficient in (D.17), including its continuation convention.

At fixed $H$, $\delta D=\delta\mathcal D-H\delta\sigma$. This second term is essential when reducing the transformation of the auxiliary. Defining $D=\mathcal D-H\sigma$ is an invertible field redefinition. Setting $D=0$ is an additional field restriction, equivalent to $\mathcal D=H\sigma$; whether it solves an equation of motion requires an action. No such equation was used in the reduction.

3. Measure a varying-radius scalar probe

In (PC09.4), take a flat base with one nonconstant coordinate $x$, set $C=0$, and let $\rho=e^{\kappa x}$. A neutral scalar has fiber dependence $e^{iny/R_f}$ and momentum $k_f=n/R_f$. Per unit transverse volume and per coordinate fiber length, its norm and kinetic form are

$$N=\int_{\mathbb R}\rho|\phi|^2dx,\qquad E=\int_{\mathbb R}\rho\left(|\phi'|^2+k_f^2\rho^{-2}|\phi|^2\right)dx.$$

Set $\phi=\rho^{-1/2}\psi$. Derive the operator acting on $\psi$ and evaluate $E/N$ for $\psi=(a/\pi)^{1/4}e^{-a(x-x_0)^2/2+ipx}$, with $a>0$ and real $p,x_0,\kappa$.

Hint 1

Expand $|\psi'-\kappa\psi/2|^2$. Its cross term is a derivative of $|\psi|^2$.

Hint 2

The normalized Gaussian has variance $1/(2a)$ and $\langle e^{-2\kappa x}\rangle=e^{-2\kappa x_0+\kappa^2/a}$.

Solution

The norm becomes $\int|\psi|^2dx$. Decay removes the boundary term $-\kappa|\psi|^2/2$ from the energy. Thus the unitarily transformed operator and this normalized probe’s energy are

$$\widehat L=-\partial_x^2+\frac{\kappa^2}{4} +k_f^2e^{-2\kappa x},\qquad \frac EN=\frac a2+p^2+\frac{\kappa^2}{4} +k_f^2e^{-2\kappa x_0+\kappa^2/a}.$$

The original weighted operator is $-\rho^{-1}\partial_x(\rho\partial_x)+k_f^2\rho^{-2}$. Its first derivative is exactly what the rescaling removes. At $\kappa=0$ the result reduces to the ordinary Gaussian gradient energy plus $k_f^2$. Direct integration of the original weighted form gives the same answer for nonzero $\kappa$: in a fixed length unit, $(a,x_0,p,\kappa,n,R_f)=(1.3,0.4,0.7,0.6,2,1.7)$ gives $E/N=2.35970996983615$ in inverse-square length units.

This is a Rayleigh expectation, not an asserted eigenvalue or a supersymmetric determinant. It demonstrates independently why retaining the unit-fiber measure would change the reduced theory.

Return to (D.17) with the charges, horizontal components and auxiliary shift written beside it. Then (D.19) should have a concrete interpretation: the retained four-dimensional translation splits into a three-dimensional covariant motion and a fiber-charge transformation. The canonical reduction lesson derives that algebra; global compactification and quantum matching require additional data beyond these local equations.