Twisted reduction to three dimensions

A circle reduction does more than remove a coordinate. If the circle is fibered over space, momentum around it behaves as charge under a three-dimensional connection. Supersymmetry also ties that connection to the fixed auxiliary sources. We will derive both effects and see why the scalar combination $z-rH$ appears in curved transformation rules.

Use the four-dimensional new minimal equation and the three-dimensional background dictionary. The parent matter theory has four-dimensional $\mathcal N=1$ supersymmetry, four real Lorentzian Poincaré generators, all manifest. Its ordinary reduction has three-dimensional $\mathcal N=2$, also four real generators. Euclidean Killing-spinor solutions are counted over $\mathbb C$; neither changing dimension nor choosing independent Euclidean columns doubles the physical algebra. No dynamical gauge group or gauge rank is specified here.

This is the classical source and transformation-law construction of Appendix D. It is not an assertion that integrating out every nonzero circle mode leaves an unchanged quantum action.

The metric determines which components are horizontal

Let $x^\mu$ be three-dimensional coordinates and $y$ a length coordinate along the fiber. Start with

$$ds_4^2=g_{\mu\nu}dx^\mu dx^\nu+\vartheta^2, \qquad \vartheta=dy+c_\mu dx^\mu . \tag{B03.1}$$

All metric coefficients are independent of $y$. The fiber has fixed unit coefficient in this expression; if it is a circle, its radius is specified by $y\sim y+2\pi R$. We have not included a varying radius scalar.

The vectors dual to the base one-forms are horizontal lifts: $E_a=e_a{}^\mu(\partial_\mu-c_\mu\partial_y)$, while $E_f=\partial_y$. Consequently a one-form decomposes as

$$W=W_\mu dx^\mu+W_y\,dy =(W_\mu-W_yc_\mu)dx^\mu+W_y\vartheta . \tag{B03.2}$$

Its horizontal component is $W_\mu-W_yc_\mu$, not its coordinate component $W_\mu$. For a vector, the corresponding vertical coefficient is $U^y+c_\mu U^\mu$. These two rules are compatible with contraction and raising indices. In particular $\det G=\det g$.

Under the change of fiber coordinate $y'=y-\lambda(x)$, $c'=c+d\lambda$ keeps $\vartheta$ fixed. A mode

$$\Phi(x,y)=e^{izy}\phi(x)$$

then has $\phi'=e^{iz\lambda}\phi$. Applying a horizontal derivative gives

$$(\partial_\mu-c_\mu\partial_y)\Phi =e^{izy}(\partial_\mu-izc_\mu)\phi . \tag{B03.3}$$

Thus $c$ is the graviphoton connection, and fiber momentum $z$ becomes its charge. Here $[z]=1$ and $[c_\mu]=0$ when $x^\mu$ has length units. For an ordinary periodic scalar on a circle, $z=n/R$ with integer $n$. A formal local ansatz permits a continuous parameter; a genuine compactification additionally requires the chosen global boundary conditions.

Rotate the spinors with the fibration

To keep the matrix conventions explicit, use the four-dimensional frame ordering of Appendix D: the fiber is direction $2$, and the ordered base directions are $(1,3,4)$. This is an even permutation of the orientation used in the preceding lesson. Its positive Euclidean blocks give

$$\gamma_1=\tau_3,\qquad \gamma_2=-\tau_1,\qquad \gamma_3=-\tau_2,\qquad \gamma_a\gamma_b=\delta_{ab}+i\varepsilon_{abc}\gamma_c .$$

These are exactly the three-dimensional matrices of the paper. Identify the opposite-chirality four-dimensional column by $\widetilde\zeta_4=S\widetilde\zeta$, where $S=i\tau_2$, and leave the undotted column unchanged. This fixed index identification gives

$$\begin{aligned} \Sigma_{ab}&=\frac i2\varepsilon_{abc}\gamma_c, &\Sigma_{af}&=\frac i2\gamma_a,\\ S^{-1}\widetilde\Sigma_{ab}S&=\frac i2\varepsilon_{abc}\gamma_c, &S^{-1}\widetilde\Sigma_{af}S&=-\frac i2\gamma_a . \end{aligned} \tag{B03.4}$$

It also implements the dotted-to-undotted contraction convention in (D.2). This is a component dictionary; it does not rephase a pair of adjoint Lorentzian supercharges. Tilded Euclidean fields remain independent.

Put $f=dc=\tfrac12f_{ab}e^a\wedge e^b$ and

$$v^a=-\frac i2\varepsilon^{abc}f_{bc}, \qquad f_{ab}=i\varepsilon_{abc}v^c .$$

For the ordinary Cartan connection, $de^A+\Omega^A{}_B\wedge e^B=0$ gives

$$\Omega^{(4)}_{ab}=\Omega^{(3)}_{ab}-\frac12 f_{ab}\vartheta, \qquad \Omega^{(4)}_{fa}=\frac12 f_{ab}e^b . \tag{B03.5}$$

For example, the second formula cancels $d\vartheta=f$ in Cartan’s equation; the first cancels its induced term in the equation for $e^a$. The paper’s input-first connection has the opposite all-lowered sign, as explained in the prerequisite.

Insert (B03.4)–(B03.5) into $\nabla=\partial+\Omega_{AB}\Sigma^{AB}/2$, summing both ordered frame pairs. For a fiber-independent label,

$$\begin{aligned} \nabla_f^{(4)}\zeta&=\frac14v^a\gamma_a\zeta,& \nabla_{h,a}^{(4)}\zeta &=\nabla_a^{(3)}\zeta+\frac14\varepsilon_{abc}v^b\gamma_c\zeta,\\ \nabla_f^{(4)}\widetilde\zeta&=\frac14v^a\gamma_a\widetilde\zeta,& \nabla_{h,a}^{(4)}\widetilde\zeta &=\nabla_a^{(3)}\widetilde\zeta-\frac14\varepsilon_{abc}v^b\gamma_c\widetilde\zeta . \end{aligned} \tag{B03.6}$$

The subscript $h$ means the horizontal lift. A mode $e^{izy}$ adds $iz$ to the vertical derivative and $-izc_\mu$ to the horizontal one. The two signs in the horizontal spin rotations are consequential.

Solve the vertical equation before reducing the horizontal one

Denote the four-dimensional background one-forms by $\mathcal A,\mathcal V$ and the final three-dimensional sources by $A,V,H,C$. The first four-dimensional equation is

$$(\nabla_M^{(4)}-i\mathcal A_M)\zeta =-i\mathcal V_M\zeta+i\mathcal V^N\Sigma_{MN}\zeta .$$

The background and supersymmetry labels have zero fiber charge. Using (B03.6), its vertical component becomes

$$\frac14v\cdot\gamma\,\zeta =i(\mathcal A_f-\mathcal V_f)\zeta +\frac12\mathcal V_h\cdot\gamma\,\zeta . \tag{B03.7}$$

The opposite chirality reverses the sign of the scalar difference. To make both vertical equations identities on all spinor components, choose $\mathcal A_f=\mathcal V_f=H$ and $\mathcal V_h=v/2$. This is the ansatz used in the reduction. It is sufficient to avoid an extra vertical projection; it is not a necessary condition for one specially selected four-dimensional spinor.

Define the remaining source by $\mathcal A_h=A-v/2$. The complete dictionary is

$$\boxed{ C=c,\qquad V=v=-i*_3dC,\qquad \mathcal A=A-\frac12V+H\vartheta,\qquad \mathcal V=\frac12V+H\vartheta .} \tag{B03.8}$$

The Hodge star uses the base metric and orientation. In particular, $A_\mu=\mathcal A_\mu-Hc_\mu+V_\mu/2$ in coordinate components. The four-dimensional divergence constraint reduces to $\nabla_\mu^{(3)}V^\mu=0$, which follows from $d^2C=0$ locally. The four-dimensional two-form source $B^{(4)}$ is distinct from the metric connection $C$. In the preceding lesson’s orientation its field strength can be written $dB^{(4)}=-H\,\mathrm{vol}_3-\tfrac i2\,dC\wedge\vartheta$. Taking its four-dimensional dual with (B02.1) gives exactly $\mathcal V$ in (B03.8). Local existence does not settle global potential patches or any chosen compact period lattice.

The horizontal equation before removing the extra spin rotation is

$$\nabla_{h,a}^{(4)}\zeta =i(A_a-V_a)\zeta-\frac12H\gamma_a\zeta -\frac14\varepsilon_{abc}V^b\gamma_c\zeta .$$

Subtract the $+\tfrac14$ term in (B03.6). For the other chirality, subtract its $-\tfrac14$ term. The result is

$$\begin{aligned} (\nabla_a^{(3)}-iA_a)\zeta &=-\frac12H\gamma_a\zeta-iV_a\zeta -\frac12\varepsilon_{abc}V^b\gamma_c\zeta,\\ (\nabla_a^{(3)}+iA_a)\widetilde\zeta &=-\frac12H\gamma_a\widetilde\zeta+iV_a\widetilde\zeta +\frac12\varepsilon_{abc}V^b\gamma_c\widetilde\zeta . \end{aligned} \tag{B03.9}$$

This restores the step between the paper’s (D.13) and (D.15). The coefficient one-half comes partly from the four-dimensional source term and partly from the fiber’s spin connection.

For a positive real four-dimensional metric, $c$ is real, so $v=-i*_3dc$ is imaginary. General Euclidean three-dimensional backgrounds allow complex $C$. Accordingly Appendix D first reduces the equations and then complexifies their sources. A complex $C$ need not be the connection of a real Riemannian circle fibration. Nor does a locally defined $C$ automatically provide global circle transition functions with an integral charge lattice.

A nontrivial fibration with explicit surviving labels

Take a flat base with coordinates $(x^1,x^2,x^3)$ and a real nonzero constant $b$ of mass dimension one:

$$c=b x^1dx^2,\qquad V=-ib\,dx^3,\qquad H=b,\qquad A=-\frac{3ib}{2}\,dx^3 . \tag{B03.10}$$

The four-dimensional metric is real and positive, $ds_4^2=\sum_a(dx^a)^2+(dy+b x^1dx^2)^2$. The sources (B03.8) are complex, as Euclidean rigid supersymmetry permits. For this example one smooth two-form potential is $B^{(4)}=-b x^1dx^2\wedge dx^3-\tfrac{ib}{2}x^1dx^2\wedge dy$; it remains a globally defined form if the fiber is a circle.

Choose constant columns with $\gamma_3u_\pm=\pm u_\pm$: $u_\pm=(1,\mp i)^T/\sqrt2$. For the positive-charge equation, the three right-hand-side matrices acting on $\zeta$ are

$$M_1=-\frac b2(\gamma_1+i\gamma_2),\qquad M_2=-iM_1,\qquad M_3=\frac b2(1-\gamma_3).$$

They all annihilate $u_+$. The opposite-charge matrices annihilate $u_-$. Thus $\zeta=u_+$ and $\widetilde\zeta=u_-$ solve the complete equations, including their vertical components. Since $[M_1,M_3]=bM_1$ has one-dimensional kernel, the curvature condition permits at most the $u_+$ line in the first sector; the other sector similarly has one line. The solution dimensions in this fiber-independent sector are exactly $(1,1)$ on this connected background. If $y$ is a circle, choose its periodic spin structure so these constant components are global. No sphere geometry has been used.

An independent geometric check probes the same connection with an R-neutral scalar. Substituting $\Phi=e^{izy}\phi$ into the ordinary four-dimensional Laplacian gives

$$-\Delta_4\longrightarrow -\partial_1^2-(\partial_2-izb x^1)^2-\partial_3^2+z^2 . \tag{B03.11}$$

For $\phi=e^{ikx^2+ipx^3}f(x^1)$ and $zb\ne0$, this is a harmonic oscillator centered at $x^1=k/(zb)$. Its eigenvalues are $(2n+1)|zb|+p^2+z^2$, $n=0,1,\ldots$. This checks both the horizontal charge and the vertical mass term. It is a scalar geometry probe, not the full supersymmetric matter kinetic operator or a localization determinant.

How a general multiplet reorganizes

Let a four-dimensional general multiplet have bottom scalar $X$ of R-charge $r$, spinors $\chi,\widetilde\chi$, scalars $M,\widetilde M$, vector component $\mathcal B_M$, spinors $\lambda,\widetilde\lambda$, and top auxiliary $\mathcal D$. The vector component is not the background R connection. Give every component the same fiber dependence $e^{izy}$; the supersymmetry labels remain independent of $y$.

The new three-dimensional components are

$$b_\mu=\mathcal B_\mu-c_\mu\mathcal B_y,\qquad \sigma=\mathcal B_y,\qquad D=\mathcal D-H\sigma . \tag{B03.12}$$

Other components retain their names, with the dotted-column identification above. Their R-charges are $r-1,r+1$ for $\chi,\widetilde\chi$; $r-2,r+2$ for $M,\widetilde M$; $r+1,r-1$ for $\lambda,\widetilde\lambda$; and $r$ for $b,\sigma,D$. They all carry central charge $z$.

For a component of R-charge $q$, the four-dimensional R-covariant derivative reduces to

$$\begin{aligned} \mathscr D_h^{(4)} &=\nabla^{(3)}-iq\left(A-\frac12V\right)-izC,\\ \mathscr D_f^{(4)}&=i(z-qH). \end{aligned} \tag{B03.13}$$

For spinors add the horizontal and vertical spin rotations in (B03.6); for a scalar these extra terms are absent. This directly explains the matter connection $A-V/2$ and the effective scalar $z-rH$. The separate R-current source $A-3V/2$ discussed in the prerequisite does not replace the matter connection.

For example, the undotted rule in (C.2) contains $\sigma_P^M\widetilde\zeta_4(i\mathcal B_M+\mathscr D_MX)$. The matrices satisfy $\sigma_{P,h}S=-\gamma$ and $\sigma_{P,f}S=i1$. Its horizontal and vertical parts therefore give

$$\begin{aligned} \delta\chi&=\zeta M -\gamma^\mu\widetilde\zeta(\mathscr D_\mu X+ib_\mu) -\widetilde\zeta\,[\sigma+(z-rH)X],\\ \mathscr D_\mu&=\nabla_\mu^{(3)} -ir(A_\mu-V_\mu/2)-izC_\mu . \end{aligned} \tag{B03.14}$$

The same index identification yields $\delta X=i\zeta\chi+i\widetilde\zeta\widetilde\chi$ and

$$\delta\widetilde\chi =\widetilde\zeta\widetilde M -\gamma^\mu\zeta(\mathscr D_\mu X-ib_\mu) -\zeta\,[\sigma-(z-rH)X].$$

Here bilinears use the antisymmetric contraction of the three-dimensional spinors. The sign change in the tilded term of $\delta X$ includes the dotted contraction convention, not just a renamed component.

The auxiliary shift in (B03.12) is a definition needed to match the three-dimensional general-multiplet convention; it is not an elimination of an auxiliary by its equation of motion. The four-dimensional gaugino term $i\zeta\mathcal D$ becomes $i\zeta(D+H\sigma)$. Another useful check is the curvature combination in the top-component rule. In the paper’s Ricci convention,

$$R_4=R_3-\frac12V^2,\qquad \mathcal V^2=H^2+\frac14V^2,\qquad R_4-6\mathcal V^2=R_3-2V^2-6H^2 .$$

No complex conjugation occurs in these squares. These are precisely the combinations in the complete reduced rules (D.17)–(D.18).

Finally let $s=\zeta\widetilde\zeta$ and $K^\mu=\zeta\gamma^\mu\widetilde\zeta$. The same matrix map gives the four-dimensional bilinear vector as $K_4=-K^h+is\,\partial_y$. Applying its R-covariant Lie derivative to a scalar mode reduces the algebra (C.3) to

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi =-2i\left[ K^\mu\mathscr D_\mu\phi+s(z-rH)\phi \right]. \tag{B03.15}$$

Thus the central transformation is the vertical translation in the chosen mode. Tensor and spinor fields additionally carry the appropriate Lie derivative on their indices; the scalar calculation already fixes the charge term and its sign.

Practice

1. A flat connection can still change the modes

Set all R and auxiliary sources to zero and take a flat base with $x^1\sim x^1+L$, $y\sim y+2\pi R$, and $c=\beta\,dx^1$. For an ordinary periodic scalar, find the eigenvalue of $-\Delta_4$ on $\exp(2\pi i m x^1/L+i n y/R)$, with no dependence on the other coordinates. When can a globally defined fiber coordinate change remove $c$? Explain the corresponding relabeling of modes.

Hint 1

The horizontal momentum is $2\pi m/L-\beta n/R$, even though $dc=0$.

Hint 2

The proposed coordinate shift $y'=y+\beta x^1$ must change by an integer multiple of the fiber circumference when $x^1$ completes its circle.

Solution

The eigenvalue is

$$\left(\frac{2\pi m}{L}-\frac{\beta n}{R}\right)^2+\frac{n^2}{R^2}.$$

A single-valued fiber redefinition removes $c$ if $\beta L=2\pi Rw$ for an integer $w$. Reexpressing the same function in $y'=y+\beta x^1$ gives $m'=m-nw$ and the usual untwisted eigenvalue $(2\pi m'/L)^2+n^2/R^2$. For generic $\beta$ the local connection is flat but its global holonomy cannot be removed this way. The integer fiber charge, the flat connection and the radius all affect a genuine mode spectrum.

2. A vertical kernel is weaker than the reduction ansatz

Retain a given fibration but change $\mathcal A_f-\mathcal V_f=\delta$ and $\mathcal V_h=v/2+u$, with complex scalar $\delta$ and vector $u$. Derive the two vertical constraints. For $\delta\ne0$ and $u=2i\delta\,e^3$, determine their kernels. Have you thereby proved that this changed background preserves supersymmetry?

Hint 1

Cancel the common $v\cdot\gamma/4$ term on both sides of the vertical equation.

Hint 2

The remaining matrices are linear combinations of $1$ and $\gamma_3$. The horizontal differential equations still have to be solved.

Solution

The constraints are $(i\delta+u\cdot\gamma/2)\zeta=0$ and $(-i\delta+u\cdot\gamma/2)\widetilde\zeta=0$. Both determinants equal $-\delta^2-u^2/4$. With the assigned complex $u$, these become $i\delta(1+\gamma_3)\zeta=0$ and $i\delta(-1+\gamma_3)\widetilde\zeta=0$. Their respective kernels have $\gamma_3=-1$ and $+1$.

The vertical equations now permit only lines, instead of being identities on both components. This is a necessary local test, not a solution of the horizontal equations or their global holonomy; the changed sources must also satisfy their divergence constraint. In particular it does not establish surviving supercharges. Setting $\delta=u=0$ was a deliberate way to avoid imposing such an extra projection, not a classification of all possible reductions.

3. Track the auxiliary on a sphere

Reduce the preceding lesson’s $S^3_\ell\times S^1$ background: $c=0$, $\mathcal A=\mathcal V=-i\,dy/\ell$. Consider an Abelian vector multiplet with constant vertical component $\mathcal B_y=\sigma$, vanishing horizontal gauge field and gauginos, and zero central charge. Determine the three-dimensional sources and the auxiliary needed to make both gaugino variations vanish. Why is simply setting the three-dimensional auxiliary to zero incorrect?

Hint 1

The R-neutral vector multiplet has $i\zeta\mathcal D=i\zeta(D+H\sigma)$ in its gaugino variation.

Hint 2

All field-strength and derivative terms vanish for this configuration. Use both signs of the tilded and untilded variations.

Solution

The sources are $A=V=C=0$, $H=-i/\ell$. The four-dimensional gauge field has zero field strength, so the gaugino equations give $\mathcal D=0$. In three-dimensional variables this is

$$D=-H\sigma=\frac{i\sigma}{\ell}.$$

Indeed $\delta\lambda=i\zeta(D+H\sigma)$ and $\delta\widetilde\lambda=-i\widetilde\zeta(D+H\sigma)$ both vanish. Setting $D=0$ would fail for nonzero $\sigma$. These equations select a supersymmetric fixed configuration; they are not the matter theory’s auxiliary equation of motion. The imaginary Euclidean value is consistent with the source convention and does not prescribe a path-integral contour for an unspecified theory.

Ordinary untwisted reduction is recovered when the fibration and relevant source shifts vanish. The more general construction preserves a specified fiber-invariant set of supersymmetry labels, assigns consistent charges to the matter modes, and reorganizes their transformations. Interactions must respect those charge assignments. A finite-radius quantum compactification also has a tower of modes, threshold effects and global data; none is computed merely by this local dictionary.

The Appendices C/D paper guide maps every reduction equation, tracks each component charge, and tests what changes when the fiber radius varies.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Supersymmetric Field Theories on Three-Manifolds
• Paper guide: Appendices C and D, four-dimensional origin