Chern-Simons and Fayet-Iliopoulos terms

A three-dimensional gauge field admits a Chern–Simons coupling as well as a Maxwell kinetic term. Supersymmetry fixes the scalar, auxiliary, and fermion terms that accompany it. An FI coupling instead shifts the equation imposed by the vector auxiliary. We will derive both, then ask which background terms are needed on a curved space.

The specified Abelian gauge theory is sufficient preparation for the flat entry and Exercises 1–2. The later curved section uses gauge and matter actions from multiplets. These are separate entrances: the first calculation uses constant spinor labels and no supergravity sources.

Begin in flat space

Work with three-dimensional $\mathcal N=2$ supersymmetry, with four real Lorentzian supercharges before Euclidean continuation. The vector multiplet is $(a_\mu,\sigma,\lambda,\widetilde\lambda,D)$. Its gauge group is compact $U(1)$, normalized so a unit-charge field has derivative $\partial_\mu-ia_\mu$, gauge parameters have period $2\pi$, and $\int f/(2\pi)\in\mathbb Z$. The two Euclidean gauginos are independent. Their R-charges are $+1,-1$; the three bosonic entries have R-charge zero.

Use $\epsilon^{123}=+1$, $f=da$, and $b^\mu=\epsilon^{\mu\nu\rho}f_{\nu\rho}/2$. Thus $b=\operatorname{curl}a$. The engineering dimensions are $[a]=[\sigma]=1$, $[\lambda]=3/2$, and $[D]=2$.

For clarity, all transformation data needed here are displayed locally. The matrices obey $\gamma_\mu\gamma_\nu=\delta_{\mu\nu}+i\epsilon_{\mu\nu\rho}\gamma_\rho$. Write $uv=u^{\mathsf T}Cv$, with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ and $\gamma_\mu^{\mathsf T}C=-C\gamma_\mu$. The labels $\zeta,\widetilde\zeta$ commute; the operations they label are odd and obey the graded product rule. In particular, $\delta(\lambda\widetilde\lambda)=(\delta\lambda)\widetilde\lambda-\lambda(\delta\widetilde\lambda)$. The flat rules are

$$\begin{aligned} \delta a_\mu&=-i(\zeta\gamma_\mu\widetilde\lambda+ \widetilde\zeta\gamma_\mu\lambda),& \delta\sigma&=-\zeta\widetilde\lambda+\widetilde\zeta\lambda,\\ \delta\lambda&=i\zeta D-i\gamma^\mu\zeta\,b_\mu -i\gamma^\mu\zeta\,\partial_\mu\sigma,& \delta\widetilde\lambda&=-i\widetilde\zeta D-i\gamma^\mu\widetilde\zeta\,b_\mu +i\gamma^\mu\widetilde\zeta\,\partial_\mu\sigma,\\ \delta D&=\partial_\mu(\zeta\gamma^\mu\widetilde\lambda -\widetilde\zeta\gamma^\mu\lambda). \end{aligned} \tag{A06.1}$$

They are the zero-background specialization of P v3, (6.20).

Derive the supersymmetric partners

Start with $i\,a\cdot b$, and abbreviate $U_\mu=\zeta\gamma_\mu\widetilde\lambda$, $T_\mu=\widetilde\zeta\gamma_\mu\lambda$. The vector identity $\operatorname{div}(a\times\delta a)=\delta a\cdot b-a\cdot\operatorname{curl}\delta a$ gives

$$\delta(i\,a\cdot b) =2(U+T)\cdot b-\partial_\mu[a\times(U+T)]^\mu.$$

The first term is not a boundary term. Try the neutral dimension-three partners $\sigma D$ and $i\lambda\widetilde\lambda$. Their variations, using (A06.1), combine as

$$\begin{aligned} \delta_\zeta(-2\sigma D+2i\lambda\widetilde\lambda) &=-2U\cdot b-2\partial_\mu(\sigma U^\mu),\\ \delta_{\widetilde\zeta}(-2\sigma D+2i\lambda\widetilde\lambda) &=-2T\cdot b+2\partial_\mu(\sigma T^\mu). \end{aligned}$$

For example, in the first line the $2D\zeta\widetilde\lambda$ from the auxiliary product cancels the $-2D\zeta\widetilde\lambda$ from the gaugino product; the remaining scalar-gradient terms complete $\partial_\mu(\sigma U^\mu)$. This fixes the relative coefficients:

$$\boxed{\mathcal L_{\rm CS} =\frac{k}{4\pi} \left(i\epsilon^{\mu\nu\rho}a_\mu\partial_\nu a_\rho -2\sigma D+2i\lambda\widetilde\lambda\right),\qquad \mathcal L_{\rm FI}=\xi D.} \tag{A06.2}$$

Here $k$ is dimensionless and $[\xi]=1$. The FI variation is already a divergence. For the combined density,

$$\delta(\mathcal L_{\rm CS}+\mathcal L_{\rm FI}) =\partial_\mu\left[ -\frac{k}{4\pi}\{a\times(U+T)+2\sigma(U-T)\}^\mu +\xi(U-T)^\mu\right]. \tag{A06.3}$$

No field equation was used. On a closed space, or with vanishing boundary flux, the action is invariant. With a physical boundary this flux must be treated explicitly. The result agrees with P, (6.27)–(6.28).

What makes the level an integer

The local expression $a\wedge da$ is insufficient on a nontrivial bundle, where no single global potential exists. A spin Chern–Simons functional uses the connection and spin structure. When extended over a bounding spin four-manifold $X$, its connection part is represented by

$$S_{{\rm CS},E}=\frac{ik}{4\pi}\int_X f\wedge f.$$

Two extensions differ by a closed spin four-manifold $Y$. Their difference is $i\pi k\int_Y c_1^2$, where $c_1=[f/(2\pi)]$ is the integral flux class. The self-intersection $\int_Yc_1^2$ is even on a spin four-manifold, so an integer $k$ changes the action by $2\pi i$ times an integer. The weight $e^{-S_E}$ is therefore well defined. The global functional, including its bundle information, is the definition; applying a single-patch $a\,da$ formula to a nontrivial bundle can give a spurious even-level restriction. Without spin structure, an ordinary purely bosonic $U(1)$ term requires even diagonal level. CT v2, (1.5)–(1.7)

There is a separate quantum issue in a theory with charged fermions. In the regulator convention used for SQED,

$$k+\frac12\sum_fq_f^2\in\mathbb Z,\qquad k_{\rm eff}=k+\frac12\sum_fq_f^2\operatorname{sign}(M_f). \tag{A06.4}$$

Here $M_f$ is the signed physical Dirac mass in the inherited SQED regulator convention. For gauge charge $q$ and central mass parameter $z$, it is $M_f=q\sigma-z$. An optional translation for readers of the curved matter action: its shorthand $M=z-q\sigma$ is the negative of this physical mass in flat space. The second formula applies when those fermions are massive and integrated out. The first says that a possible half-integer local offset must be combined with its determinant prescription; it is not a stand-alone half-integer spin Chern–Simons functional. The $+1,-1$ pair permits $k=0$; a single unit-charge fermion does not. These quantum statements are sourced inputs, not consequences of the classical variation. AHISS v1, (2.10)–(2.12)

Worked example: a common topological mass

Remove charged matter for this example and retain a Maxwell term with coupling $e$, where $[e^2]=1$. Its auxiliary term is $-D^2/(2e^2)$. Adding (A06.2) gives

$$\mathcal L_{\rm aux} =-\frac{D^2}{2e^2} +D\left(\xi-\frac{k\sigma}{2\pi}\right),\qquad D_*=e^2\left(\xi-\frac{k\sigma}{2\pi}\right).$$

The shifted imaginary Euclidean contour $D=D_*+id$ gives a positive Gaussian in real $d$. Eliminating the auxiliary leaves $U=e^2(\xi-k\sigma/(2\pi))^2/2$. For $k\ne0$, the supersymmetric homogeneous configuration has $\sigma_*=2\pi\xi/k$ and $D=0$.

With canonical scalar $\sigma_c=\sigma/e$, its mass squared is $(ke^2/(2\pi))^2$. Independently, variation of the gauge action gives

$$\operatorname{curl}b+i\frac{ke^2}{2\pi}b=0,\qquad \operatorname{div}b=0.$$

Apply $\operatorname{curl}-ike^2/(2\pi)$: the result is $[-\partial^2+(ke^2/(2\pi))^2]b=0$. After Lorentzian continuation this describes the one massive photon polarization, with mass $|k|e^2/(2\pi)$. Finally, putting $\lambda=e\lambda_c$ and $\chi_c=-i\widetilde\lambda_c^{\mathsf T}C$ gives the gaugino operator $\gamma\cdot\partial-ke^2/(2\pi)$. Its mass magnitude agrees with the photon and scalar. This tests the factor of two in the auxiliary and gaugino terms.

A pure Chern–Simons vector sector has no Maxwell kinetic term: $\sigma,D,\lambda,\widetilde\lambda$ are then auxiliary, and the gauge field has no local propagating polarization. It can still have global topological states. Do not count it as a propagating Yang–Mills multiplet. KWY v4, section 2, (2.1) uses this pure-CS field content; its symbol “Tr” already includes $k/(4\pi)$, whereas our coefficient is explicit.

A background source gives an FI coupling

Take a second compact vector multiplet $(B,\nu,\lambda_B,\widetilde\lambda_B,D_B)$. Polarizing (A06.2) gives the level-$p$ mixed coupling

$$\mathcal L_{\rm mixed} =\frac{p}{2\pi}\left[ iB\cdot b-\nu D-\sigma D_B +i(\lambda\widetilde\lambda_B+\lambda_B\widetilde\lambda) \right],\qquad p\in\mathbb Z, \tag{A06.5}$$

up to a divergence. In particular, $B\wedge da$ and $a\wedge dB$ have equal integrals on a closed manifold; both cross terms are needed to obtain $1/(2\pi)$. The global mixed functional has extension $ip(2\pi)^{-1}\int_X f_B\wedge f$, whose extension ambiguity is $2\pi ip$ times an integer.

If the second multiplet is dynamical, this couples two gauge sectors. If it is a fixed flavor source, only the first multiplet is integrated. In flat space its constant configuration $B=D_B=\lambda_B=\widetilde\lambda_B=0$, $\nu\in\mathbb R$, preserves all four supercharges. It induces $\xi_{\rm induced}=-p\nu/(2\pi)$. For the unit topological source, $p=1$ and $\xi=-\zeta=-\nu/(2\pi)$, exactly the FI sign used in the deformation calculation. That comparison is optional preparation.

If both multiplets are backgrounds, the same polynomial specifies contact terms rather than dynamical equations. Adding an allowed integer background Chern–Simons counterterm changes contact responses and sometimes partition-function phases; it does not add a propagating gauge field. A chosen quantum comparison must keep those counterterms and determinant phases fixed.

Complete the couplings on a rigid background

This section uses the general-multiplet product and neutral D-term prescription from the gauge-and-matter action lesson. The fixed rigid sources are $(g_{\mu\nu},A_\mu,V_\mu,H,C_\mu)$; capital $A_\mu$ is the R-symmetry connection, distinct from the dynamical $a_\mu$. Keep the same full theory, and retain the supercharges whose labels solve the background Killing equations. Background auxiliaries are not eliminated.

For a neutral general multiplet $\mathcal S$, that lesson gave

$$\mathcal L_D[\mathcal S]=-\frac12 (D_{\mathcal S}-a_{\mathcal S}\cdot V-\sigma_{\mathcal S}H).$$

Apply it first to $-2\xi\mathcal V$, where $\mathcal V$ denotes the vector multiplet:

$$\boxed{\mathcal L_{\rm FI}=\xi(D-a\cdot V-\sigma H).} \tag{A06.6}$$

This is also a direct variation check. In the curved rules,

$$\delta D=\nabla_\mu(U-T)^\mu-iV\cdot(U+T)+H\delta\sigma.$$

The last two terms cancel the variations of $-a\cdot V-\sigma H$, leaving $\xi\nabla_\mu(U-T)^\mu$. Minimal replacement of $\partial$ by $\nabla$ in $\xi D$ would miss these cancellations.

For Chern–Simons, multiply $\mathcal V$ by its linear field-strength multiplet $\Sigma$. The general-multiplet entries required from $\Sigma$ are

$$C_\Sigma=\sigma,\quad a_\Sigma=ib-\sigma V,\quad \sigma_\Sigma=-(D+H\sigma),\quad \chi_\Sigma=i\widetilde\lambda,\quad \widetilde\chi_\Sigma=-i\lambda.$$

Since the bottom and lower spinors of $\mathcal V$ vanish in component gauge, the product rule reduces to

$$\begin{aligned} a_{\mathcal V\Sigma}&=\sigma a,& \sigma_{\mathcal V\Sigma}&=\sigma^2,\\ D_{\mathcal V\Sigma} &=\sigma D-a\cdot(ib-\sigma V)+\sigma(D+H\sigma) -i\lambda\widetilde\lambda-i\widetilde\lambda\lambda\\ &=2\sigma D-i a\cdot b+\sigma a\cdot V+H\sigma^2 -2i\lambda\widetilde\lambda. \end{aligned}$$

The two odd scalar contractions are equal. Subtracting the product’s $a\cdot V$ and $\sigma H$ entries cancels both explicit background terms. Consequently $\mathcal L_D[k\mathcal V\Sigma/(2\pi)]$ is exactly (A06.2), with the curved volume measure. The curved gaugino and auxiliary variations likewise give (A06.3) with $\partial_\mu$ replaced by $\nabla_\mu$. No extra $H\sigma^2$ term belongs in this convention. This derives P, (6.27)–(6.28); compare the unchanged sphere CS density in KWY, (2.1)–(2.3), with its different auxiliary and action conventions.

The mixed gauge/flavor term (A06.5) also retains its form, but a frozen source must now satisfy its curved gaugino equations. For the four-supercharge backgrounds considered here, P (6.26) gives $B=-\nu C+B^{(0)}$, constant $\nu$, and $D_B=-\nu H$, with $B^{(0)}$ flat. Freezing $B=D_B=0$ on a general curved background would be a different, usually nonsupersymmetric choice.

A mixed coupling to the R source

An R source is part of the rigid multiplet, so its completion is not obtained by treating $A_\mu$ alone as an ordinary flavor connection. For a neutral linear multiplet with entries $(J,j_\mu,K)$, the invariant prescription is $j\cdot(A-V/2)+HK-(R-2V^2+2H^2)J/4$. Insert $-(k_{gR}/(2\pi))\Sigma$, whose three entries are proportional to $(\sigma,-ib,D+H\sigma)$. After integration by parts this gives

$$\mathcal L_{gR}=\frac{k_{gR}}{2\pi} \left[ i\epsilon^{\mu\nu\rho}a_\mu\partial_\nu \left(A_\rho-\frac12V_\rho\right) -DH+\frac{\sigma}{4}(R-2V^2-2H^2) \right]. \tag{A06.7}$$

In particular, $-H(D+H\sigma)+\sigma(R-2V^2+2H^2)/4$ produces the displayed $-2H^2$ coefficient. This is P, (6.15), (6.21), and (6.29). It couples a dynamical field to fixed sources and is distinct from pure-background contact terms. Its global quantization depends on the specified R-bundle and charge lattice; a freely varied trial R-charge is not itself a choice of an integer-normalized compact connection.

Local supersymmetry and global gauge consistency remain separate requirements. For example, $\nabla_\mu V^\mu=0$ makes the FI term’s infinitesimal gauge variation a divergence. For its global definition we may take $C$ globally defined, using $V=-i*dC$. More general flux sectors require an explicit global mixed-coupling definition and compatible periods; the local divergence identity alone does not establish large-gauge invariance.

Exercises

1. Freeze one gauge multiplet

Keep a dynamical Maxwell–CS multiplet with $k\ne0$, FI coefficient $\xi_0$, and a mixed coupling of integer level $p$ to a constant flat supersymmetric background scalar $\nu$. Find the effective FI coefficient, homogeneous supersymmetric scalar value, and fluctuation mass. Does freezing the source remove the gaugino CS term of the dynamical multiplet?

Hint 1

Only $-\nu D$ survives from the mixed density when the other background entries vanish.

Hint 2

Collect the coefficient of $D$ before eliminating it. A constant shift of the minimum does not change the second derivative of the potential.

Solution

The coefficient is $\xi_{\rm eff}=\xi_0-p\nu/(2\pi)$. Thus

$$D_*=e^2\left(\xi_{\rm eff}-\frac{k\sigma}{2\pi}\right), \qquad \sigma_*=\frac{2\pi\xi_0-p\nu}{k}, \qquad m_{\rm vec}=\frac{|k|e^2}{2\pi}.$$

At the supersymmetric homogeneous configuration, $D=0$ and the scalar is constant. The mixed gaugino terms vanish because the background gauginos vanish, but the dynamical term $ik\lambda\widetilde\lambda/(2\pi)$ remains. Removing it would destroy the common mass and the variation cancellation. All expressions have the correct dimensions: $[\sigma_*]=[\xi_0]=[\nu]=1$.

2. Follow the induced level across a threshold

Return to one-flavor SQED, with its paired $k=0$ regulator, no superpotential, a real axial mass $\mu\ne0$, and constant $\sigma$. The two physical fermion masses are $M_+=\mu+\sigma$, $M_-=\mu-\sigma$. Find the effective gauge level away from the two massless walls. Why is this a statement about an effective action, not a discontinuity in the microscopic definition?

Hint 1

Both squared charges are one. Opposite electric charges do not by themselves make equal-sign real masses cancel.

Hint 2

Compare $|\sigma|>|\mu|$ with $|\sigma|<|\mu|$. At a wall, do not integrate out the massless fermion.

Solution

Equation (A06.4) gives

$$k_{\rm eff}=\frac12[ \operatorname{sign}(\mu+\sigma)+\operatorname{sign}(\mu-\sigma)] = \begin{cases} 0,&|\sigma|>|\mu|,\\ \operatorname{sign}\mu,&|\sigma|<|\mu|. \end{cases}$$

These are integer levels, consistent with the unit-charge paired theory. The result applies at energies small compared with both nonzero masses. At $|\sigma|=|\mu|$ this approximation fails: one charged fermion must remain in the low-energy description. The microscopic regulator and couplings have not jumped; two different massive effective descriptions meet at a gapless threshold. This checks only the gauge level, not every flavor or R contact coefficient.

3. Evaluate the curved terms on a sphere

Use the curved section. On a round sphere of radius $\ell$, take $A=V=C=0$, $H=-i/\ell$, and $R=-6/\ell^2$. A four-supercharge invariant Abelian configuration has $a=0$, constant $\sigma=s$, vanishing gauginos, and $D=-Hs=is/\ell$. The volume is $2\pi^2\ell^3$. Evaluate $S_{\rm CS}$ and $S_{\rm FI}$. What would be lost by keeping only $\xi D$?

Hint 1

The connection term vanishes, but $\sigma D$ does not. In the FI term both $D$ and $-\sigma H$ contribute equally.

Hint 2

Integrate the constant densities and introduce the dimensionless scalar $t=\ell s$. Use the stated Euclidean weight $e^{-S_E}$.

Solution

The densities are $-iks^2/(2\pi\ell)$ and $2i\xi s/\ell$. Hence

$$S_{\rm CS}=-i\pi k\,t^2,\qquad S_{\rm FI}=4i\pi^2\xi\ell\,t,\qquad e^{-S_{\rm CS}-S_{\rm FI}} =e^{\,i\pi kt^2-4i\pi^2\xi\ell t}.$$

Minimal covariantization of $\xi D$ gives half the FI result and fails the curved variation check. Both exponents are dimensionless. These are the values of specified classical actions on a supersymmetric configuration, not a derivation of localization or of its integration contour. In a locally flat large-radius limit the density reduces to (A06.2); one must not hold the extensive sphere volume fixed while taking that limit.

The same local polynomial can govern a dynamical topological sector or a background response. Its use requires the compact charge normalization, quantum regulator, boundary conditions, and background profile to be specified alongside its supersymmetric completion.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 6, curved multiplets and actions
• Paper guide: Section 7, the linearized check