Gauge and matter actions from multiplets

A gauge coupling introduces more than a covariant derivative. Supersymmetry also couples the charged scalar to the vector scalar and auxiliary, and pairs the matter fermion with the gaugino. We will obtain these terms from one product, then construct the gauge kinetic term by the same method.

Use the invariant D-term prescription and the curved vector multiplet. The theory is three-dimensional $\mathcal N=2$, with four real Lorentzian supercharges before Euclidean continuation. The fixed sources $(g,A,V,H,C)$ admit the Killing spinors under consideration. The Abelian vector multiplet $(a,\sigma,\lambda,\widetilde\lambda,D)$ is dynamical and remains off shell. We do not impose its gaugino equations or auxiliary equation while constructing the action.

Take a compact $U(1)$ gauge group with a unit-charge probe convention, so the matter charge $q$ is an integer. The chiral multiplet $\Phi=(\phi,\psi,F)$ has this gauge charge, central charge $z$, and bottom R-charge $r$. Its independent antichiral partner has opposite gauge and central charges. Component R-charges are

Component	$\phi$	$\psi$	$F$	$\widetilde\phi$	$\widetilde\psi$	$\widetilde F$
R-charge	$r$	$r-1$	$r-2$	$-r$	$1-r$	$2-r$

The engineering dimensions are $1/2,1,3/2$ for a matter scalar, fermion, and auxiliary. For the vector they are $[a]=[\sigma]=1$, $[\lambda]=[\widetilde\lambda]=3/2$, and $[D]=2$. Also $[z]=[A]=[V]=[H]=1$ and $[R]=2$. R-charge is not engineering dimension; $r$ is not assumed to be a superconformal value.

Expand the gauge-invariant product

Write $\mathcal V$ for the vector multiplet, to distinguish it from the rigid vector source $V_\mu$. Before component gauge fixing, the gauge-invariant product is

$$\mathcal K=\widetilde\Phi\,e^{-2q\mathcal V}\Phi.$$

The sign follows from $\delta\mathcal V=\Omega+\widetilde\Omega$ and $\delta\Phi=2q\Omega\Phi$, $\delta\widetilde\Phi=2q\widetilde\Omega\widetilde\Phi$. Their three variations cancel. In the residual component gauge this is the usual $a\mapsto a+d\Lambda$, $\phi\mapsto e^{iq\Lambda}\phi$. The exponential construction is the route stated in P v3, footnote 10 following (6.25).

The general multiplet product from P, (6.16) is an algebraic operation on components, with covariant derivatives where required. Denote its bottom, vector, scalar, and highest entries by $C_{\mathcal S},a_{\mathcal S\mu},\sigma_{\mathcal S},D_{\mathcal S}$. They need not be independent elementary fields.

In Wess–Zumino gauge, $\mathcal V$ has only $a,\sigma,\lambda,\widetilde\lambda,D$ nonzero. Its square has only one nonzero entry:

$$D_{\mathcal V^2}=-a_\mu a^\mu-\sigma^2.$$

For example, these are the $-a_1\cdot a_2-\sigma_1\sigma_2$ terms in the product rule; all terms involving bottom components or the lower spinors vanish. Multiplication by another $\mathcal V$ then gives zero. Thus $E=e^{-2q\mathcal V}=1-2q\mathcal V+2q^2\mathcal V^2$ has

$$\begin{aligned} C_E&=1,& a_{E\mu}&=-2qa_\mu,& \sigma_E&=-2q\sigma,\\ \lambda_E&=-2q\lambda,& \widetilde\lambda_E&=-2q\widetilde\lambda,& D_E&=-2qD-2q^2(a_\mu a^\mu+\sigma^2). \end{aligned}$$

The lower spinors $\chi_E,\widetilde\chi_E$ and the two $M$ components vanish. The apparently nongauge-invariant $a^2$ term is essential: it will complete the charged scalar derivative.

First form $\mathcal K_0=\widetilde\Phi\Phi$ with the unmodified product rule. At this intermediate step use

$$\mathring D_\mu X =\nabla_\mu X-ir_X\left(A_\mu-\frac12V_\mu\right)X-iz_XC_\mu X, \qquad m_0=z-rH,$$

where $r_X,z_X$ are the charges of the component acted upon. Define $\rho=\widetilde\phi\phi$, $S=\widetilde\psi\psi$, $j_\mu=\widetilde\psi\gamma_\mu\psi$. Spinor contractions retain $uv=u^TCv$ and the odd-field signs of the prerequisites. The chiral embeddings (6.7), (6.9) give

$$\begin{aligned} \chi_0&=-\sqrt2 i\,\widetilde\phi\psi, &\widetilde\chi_0&=\sqrt2 i\,\phi\widetilde\psi,\\ a_{0\mu}&=-i(\widetilde\phi\mathring D_\mu\phi -\phi\mathring D_\mu\widetilde\phi)+j_\mu, &\sigma_0&=2m_0\rho+iS. \end{aligned}$$

To see the highest component’s origin, set $Q_{\rm geom}=R-2V_\mu V^\mu-6H^2$. The two embedded highest components contribute $[rQ_{\rm geom}/2-2m_0H]\rho$. The two scalar-component/charge products contribute $-2m_0^2\rho$. The auxiliary product contributes $2\widetilde F F$. Together with the derivative and odd-spinor products, this gives

$$\begin{aligned} D_0={}&-2\mathring D_\mu\widetilde\phi\,\mathring D^\mu\phi +i\widetilde\psi\gamma^\mu\mathring D_\mu\psi -i(\mathring D_\mu\widetilde\psi)\gamma^\mu\psi +2\widetilde F F\\ &-2\left[m_0H+m_0^2-\frac r4Q_{\rm geom}\right]\rho -2im_0S+2V^\mu j_\mu. \end{aligned}$$

Now multiply $E\mathcal K_0$. Since $C_E=1$ and its lower spinors vanish, the relevant general product rule reduces to

$$\begin{aligned} D_{\mathcal K} ={}&D_0+\rho D_E-a_E^\mu a_{0\mu}-\sigma_E\sigma_0 -\lambda_E\chi_0+\widetilde\lambda_E\widetilde\chi_0,\\ a_{\mathcal K\mu}&=a_{0\mu}-2qa_\mu\rho,\qquad \sigma_{\mathcal K}=\sigma_0-2q\sigma\rho. \end{aligned}$$

This short formula exposes the interactions:

Product contribution to $D_{\mathcal K}$	What it supplies
$-2qD\rho$	The vector auxiliary coupling
$-2q^2a^2\rho+2qa^\mu a_{0\mu}$	Charged scalar and fermion derivatives
$-2q^2\sigma^2\rho+2q\sigma\sigma_0$	The shift $m_0\mapsto m_0-q\sigma$
$-\lambda_E\chi_0+\widetilde\lambda_E\widetilde\chi_0$	$-2\sqrt2 iq(\widetilde\phi\lambda\psi+\phi\widetilde\lambda\widetilde\psi)$

In particular, the last line fixes both Yukawa signs and their factor of $\sqrt2$. It is not an independent guess.

Put

$$D_\mu X=\mathring D_\mu X-iq_Xa_\mu X,\qquad m=z-q\sigma-rH.$$

The quadratic and linear $a$ terms recombine into these derivatives. The result is

$$\begin{aligned} a_{\mathcal K\mu} &=-i(\widetilde\phi D_\mu\phi-\phi D_\mu\widetilde\phi)+j_\mu, &\sigma_{\mathcal K}&=2m\rho+iS,\\ D_{\mathcal K} ={}&-2D_\mu\widetilde\phi D^\mu\phi +i\widetilde\psi\gamma^\mu D_\mu\psi -i(D_\mu\widetilde\psi)\gamma^\mu\psi+2\widetilde F F\\ &-2\left[qD+(z-rH)H+m^2-\frac r4Q_{\rm geom}\right]\rho -2imS+2V^\mu j_\mu\\ &-2\sqrt2 iq(\widetilde\phi\lambda\psi +\phi\widetilde\lambda\widetilde\psi). \end{aligned}$$

This reconstructs P, (6.25). The factor $(z-rH)H$ has not become $mH$ at this stage; the distinction matters in the next step.

Apply the curved D-term prescription

The composite is neutral under gauge, R, and central-charge symmetries. The preceding lesson therefore supplies

$$\mathcal L_D[\mathcal K] =-\frac12(D_{\mathcal K}-a_{\mathcal K\mu}V^\mu-\sigma_{\mathcal K}H).$$

For the scalar terms, $H\sigma_{\mathcal K}/2$ adds $Hm\rho$. Since $m_0=m+q\sigma$,

$$qD+H(m_0+m)=q(D+\sigma H)+2Hm.$$

This is how the shifted auxiliary $D+\sigma H$ reappears in matter interactions. For the fermions, integration by parts puts the derivative on $\psi$ and leaves the explicit divergence $\tfrac i2\nabla_\mu j^\mu$.

There are also terms linear in $V$. They can be organized by a modified derivative,

$$\mathscr D_\mu X=D_\mu X+i r_{0X}V_\mu X,\qquad (r_{0\phi},r_{0\psi},r_{0\widetilde\phi},r_{0\widetilde\psi}) =\left(\frac12,-\frac12,-\frac12,\frac12\right).$$

These fixed numbers are a convenient definition, not a claim that the interacting theory has those R-charges. To check the rearrangement, let $J_\mu=\widetilde\phi D_\mu\phi-\phi D_\mu\widetilde\phi$. Then

$$\begin{aligned} \mathscr D_\mu\widetilde\phi\,\mathscr D^\mu\phi &=D_\mu\widetilde\phi D^\mu\phi-\frac i2V^\mu J_\mu+\frac14V^2\rho,\\ -i\widetilde\psi\gamma^\mu\mathscr D_\mu\psi &=-i\widetilde\psi\gamma^\mu D_\mu\psi-\frac12V^\mu j_\mu. \end{aligned}$$

In particular, the first identity subtracts $V^2\rho/4$ from the remaining scalar coefficient. Define $M=z-q\sigma$ and obtain

$$\begin{aligned} \mathcal L_D[\mathcal K] &=\mathcal L_{\rm matter}+\frac i2\nabla_\mu j^\mu,\\ \mathcal L_{\rm matter} ={}&\mathscr D_\mu\widetilde\phi\,\mathscr D^\mu\phi -i\widetilde\psi\gamma^\mu\mathscr D_\mu\psi-\widetilde F F\\ &+\left[ q(D+\sigma H)+M^2-2(r-1)HM-\frac r4R +\frac12\left(r-\frac12\right)V^2 +r\left(r-\frac12\right)H^2 \right]\rho\\ &+i\left[M-\left(r-\frac12\right)H\right]\widetilde\psi\psi +\sqrt2 iq(\widetilde\phi\lambda\psi +\phi\widetilde\lambda\widetilde\psi). \end{aligned}$$

Every term has dimension three. The first Yukawa term, for example, has R-charge $-r+1+(r-1)=0$ and gauge charge $-q+0+q=0$. The full expression agrees with P, (6.31)–(6.32). Its classical off-shell invariance follows from the neutral-multiplet D-term construction, provided its boundary flux vanishes and the fields patch globally. On a closed manifold the displayed divergence does not change the action. On a manifold with boundary it must be retained or accompanied by suitable boundary data.

Give the vector a kinetic term

The field-strength multiplet $\Sigma$ from the vector lesson has linear-multiplet entries

$$(J,j,\widetilde j,j_\mu,K) =(\sigma,i\widetilde\lambda,-i\lambda,-ib_\mu,D+H\sigma), \qquad b^\mu=\frac12\epsilon^{\mu\nu\rho}f_{\nu\rho}.$$

The symbols $j,\widetilde j$ in this tuple are spinor entries; they are distinct from the matter bilinear $j_\mu$ above. Only this paragraph uses them. Embed $\Sigma$ in a general multiplet using P, (6.11). Writing $K_D=D+H\sigma$, its bosonic entries become

$$C_\Sigma=\sigma,\quad a_{\Sigma\mu}=ib_\mu-\sigma V_\mu,\quad \sigma_\Sigma=-K_D,\quad D_\Sigma=iV\cdot b-HK_D-\nabla^2\sigma-V^2\sigma.$$

Apply the D-term prescription to $-\Sigma^2/e^2$, where $[e^2]=1$. The bosonic highest component of the square is

$$D_{\Sigma^2}\big|_{\rm bos} =2\sigma D_\Sigma-a_\Sigma^2-K_D^2-(\partial\sigma)^2.$$

Subtracting $a_{\Sigma^2}\cdot V+\sigma_{\Sigma^2}H$, with $a_{\Sigma^2}=2\sigma a_\Sigma$ and $\sigma_{\Sigma^2}=-2\sigma K_D$ in this bosonic calculation, gives

$$b^2+2i\sigma V\cdot b-\sigma^2V^2-K_D^2 -2\sigma\nabla^2\sigma-(\partial\sigma)^2.$$

Integrating the Laplacian by parts turns the last two terms into $(\partial\sigma)^2$, up to $-2\nabla_\mu(\sigma\partial^\mu\sigma)$. Also $b^2=f_{\mu\nu}f^{\mu\nu}/2$.

For the odd part, the embedding has $\chi_\Sigma=i\widetilde\lambda$, $\widetilde\chi_\Sigma=-i\lambda$, and

$$\begin{aligned} \lambda_\Sigma &=\gamma^\mu\left(D_\mu+\frac i2V_\mu\right)\lambda-\frac H2\lambda,\\ \widetilde\lambda_\Sigma &=\gamma^\mu\left(D_\mu-\frac i2V_\mu\right)\widetilde\lambda-\frac H2\widetilde\lambda. \end{aligned}$$

The product also contributes $-\widetilde\lambda\gamma_\mu\lambda$ to $a_{\Sigma^2\mu}$ and $i\widetilde\lambda\lambda$ to $\sigma_{\Sigma^2}$. Substituting these entries into the odd terms of (6.16) yields

$$\begin{aligned} \bigl(D_{\Sigma^2}-a_{\Sigma^2}\cdot V-\sigma_{\Sigma^2}H\bigr)_{\rm ferm} ={}&-i\widetilde\lambda\gamma^\mu D_\mu\lambda +i(D_\mu\widetilde\lambda)\gamma^\mu\lambda\\ &+V_\mu\widetilde\lambda\gamma^\mu\lambda +iH\widetilde\lambda\lambda. \end{aligned}$$

Combining the two derivative terms by parts now gives the complete density

$$\begin{aligned} \mathcal L_{\rm YM} =\frac1{e^2}\bigg[ &\frac14f_{\mu\nu}f^{\mu\nu} +\frac12\partial_\mu\sigma\partial^\mu\sigma -i\widetilde\lambda\gamma^\mu \left(D_\mu+\frac i2V_\mu\right)\lambda\\ &+i\sigma V\cdot b-\frac12V^2\sigma^2 -\frac12(D+H\sigma)^2 +\frac i2H\widetilde\lambda\lambda \bigg]. \end{aligned}$$

Here $D_\mu\lambda=\nabla_\mu\lambda-i(A_\mu-V_\mu/2)\lambda$; the Abelian gaugino is neutral under $a_\mu$. The exact product density differs from this form by

$$\mathcal L_D[-\Sigma^2/e^2]-\mathcal L_{\rm YM} =-\frac1{e^2}\nabla_\mu(\sigma\partial^\mu\sigma) +\frac{i}{2e^2}\nabla_\mu(\widetilde\lambda\gamma^\mu\lambda).$$

The full fermion completion is P, (6.30); it comes from the same product, with the scalar-derivative and fermion-derivative boundary terms shown explicitly. Thus $\int\sqrt g\,(\mathcal L_{\rm YM}+\mathcal L_{\rm matter})$ is the desired classical Abelian gauge-matter action under the stated boundary conditions. Neither its negative auxiliary squares nor its complex coefficients select a Euclidean integration contour.

Check the sphere and flat limits

On the supplied round sphere take $A=V=C=0$, $H=-i/\ell$, and $R=-6/\ell^2$ in P’s curvature convention. The matter scalar coefficient reduces to

$$U_{S^3} =q\left(D-\frac{i\sigma}{\ell}\right) +M^2+\frac{2i(r-1)M}{\ell} +\frac{r(2-r)}{\ell^2},$$

and the fermion scalar-bilinear coefficient becomes $iM-(r-\tfrac12)/\ell$. The vector auxiliary square is $-(D-i\sigma/\ell)^2/(2e^2)$ and its gaugino mass coefficient is $1/(2e^2\ell)$. These are expressions for arbitrary dynamical vector fields; we have not set $D=i\sigma/\ell$.

Turning off the vector and taking $z=0$ recovers the free chiral coefficients derived by direct variation in the sphere action. For $r=1/2$ the scalar coefficient is $3/(4\ell^2)$, the conformal value in this convention. Taking $\ell\to\infty$ instead, with the fields fixed, leaves

$$\begin{aligned} \mathcal L_{\rm matter}^{\rm flat} ={}&D_\mu\widetilde\phi D^\mu\phi-i\widetilde\psi\gamma^\mu D_\mu\psi -\widetilde F F+[qD+(z-q\sigma)^2]\rho\\ &+i(z-q\sigma)\widetilde\psi\psi +\sqrt2 iq(\widetilde\phi\lambda\psi +\phi\widetilde\lambda\widetilde\psi). \end{aligned}$$

In this flat limit $D_\mu=\partial_\mu-iq_Xa_\mu$ on matter. The gauge, scalar, Yukawa, and auxiliary couplings all survive with the coefficients fixed by the original product.

Exercises

1. Find the missing gauge coupling

Set the rigid sources to zero and write $J_\mu^{(0)}=\widetilde\phi\,\partial_\mu\phi-\phi\,\partial_\mu\widetilde\phi$. Expand $D_\mu\widetilde\phi D^\mu\phi$ in powers of $a$. If the quadratic term from $E$ were omitted, could the remaining scalar kinetic terms be gauge invariant? Also verify the gauge, central, and R-charge neutrality of both Yukawa monomials.

Hint 1

The derivative acting on $\widetilde\phi$ has the opposite gauge sign. Include the product of the two connection terms.

Hint 2

At one point choose a gauge parameter with zero value and arbitrary derivative. The fields there are unchanged, but their ordinary derivatives acquire connection-like terms.

Solution

The expansion is

$$D_\mu\widetilde\phi D^\mu\phi =\partial_\mu\widetilde\phi\,\partial^\mu\phi +iq\,a^\mu J_\mu^{(0)}+q^2a^2\rho.$$

For $\phi'=e^{iq\Lambda}\phi$ and $a'=a+d\Lambda$, $D'\phi'=e^{iq\Lambda}D\phi$, with the opposite phase for the tilded field. Hence the complete expression is invariant. If its last term is removed, the candidate density changes by $-q^2[2a\cdot d\Lambda+(d\Lambda)^2]\rho$. This is nonzero in general. The seemingly nongauge-invariant quadratic component of the exponential was needed for a gauge-invariant product.

The monomial $\widetilde\phi\lambda\psi$ has charges $(-q)+0+q=0$, $(-z)+0+z=0$, and $(-r)+1+(r-1)=0$. For $\phi\widetilde\lambda\widetilde\psi$ they are $q+0-q=0$, $z+0-z=0$, and $r-1+(1-r)=0$. Both have dimension $1/2+3/2+1=3$.

2. Eliminate an auxiliary after constructing the action

In flat space take $z=0$, no superpotential, and a scalar reality slice $\widetilde\phi=\phi^*$. Keep the vector auxiliary $D$ complex while finding its algebraic stationary point. Using both kinetic densities, complete its square and determine the resulting scalar potential. Does this stationary-point calculation prove that the original Euclidean integral over real $D$ converges?

Hint 1

The relevant density is $-D^2/(2e^2)+qD\rho$, with $\rho=|\phi|^2$ and real positive $e^2$.

Hint 2

An algebraic stationary point and a convergent integration direction are different questions. Also retain the $(q\sigma)^2\rho$ term when identifying the scalar potential.

Solution

Completing the square gives

$$-\frac{D^2}{2e^2}+qD\rho =-\frac{(D-e^2q\rho)^2}{2e^2} +\frac{e^2q^2}{2}\rho^2.$$

Thus $D=e^2q\rho$. With no superpotential, $F=\widetilde F=0$ is the matter auxiliary stationary point. For real $\sigma$, the remaining scalar potential is

$$U=q^2\sigma^2|\phi|^2+\frac{e^2q^2}{2}|\phi|^4.$$

It is nonnegative on this scalar slice. But the original real-$D$ direction has a negative quadratic action, so $e^{-S_E}$ grows there. The stationary equation alone has not defined an integral. For the isolated auxiliary Gaussian, a shifted direction $D=e^2q\rho+id$, $d$ real, makes the quadratic term $+d^2/(2e^2)$. Choosing and justifying a contour for the full theory is additional data, distinct from the off-shell action construction and this algebraic elimination.

3. Check the conformal free field on the general background

Turn off the dynamical vector multiplet, set $q=z=0$, and choose $r=1/2$. Show that the kinetic density is independent of $H$ and depends on $A,V$ only through $A_\mu-\tfrac32V_\mu$. Give its scalar curvature coefficient and check it on the round sphere.

Hint 1

Do not inspect only the explicit $V^2$ term. Expand $\mathscr D$ using the actual component R-charges $r_\phi=1/2$ and $r_\psi=-1/2$.

Hint 2

The constants $r_{0X}$ in the modified derivative equal the component R-charges for this particular free-field choice.

Solution

All explicit $H$ terms vanish: $M=0$, $r-\tfrac12=0$, and the vector auxiliary term is absent. The explicit $V^2$ coefficient also vanishes. The two derivatives are

$$\mathscr D_\mu\phi =\left[\nabla_\mu-\frac i2\left(A_\mu-\frac32V_\mu\right)\right]\phi, \qquad \mathscr D_\mu\psi =\left[\nabla_\mu+\frac i2\left(A_\mu-\frac32V_\mu\right)\right]\psi.$$

The tilded derivatives have opposite connections. Therefore the surviving background dependence is precisely this R-connection, the metric, and the scalar curvature term $-R\rho/8$. On the sphere, $R=-6/\ell^2$ makes the latter $+3\rho/(4\ell^2)$, agreeing with the direct check.

This test would fail if the $V^2\rho/4$ generated by the derivative rearrangement had been forgotten. It checks the general-background coefficient, rather than only a background with $V=0$.

The action is now organized by neutral multiplets, with the origin of each interaction visible. Freezing an invariant flavor multiplet turns the same couplings into real masses; Chern–Simons and Fayet–Iliopoulos densities use other invariant products. Those constructions also require their own global and, at the quantum level, regularization conditions.

The Section 6 paper guide maps these action formulas to the original component construction. Its R/flavor-source example reconstructs the curvature-dependent auxiliary shift directly from the matter coefficient.

The A-twist dossier uses the curvature and auxiliary terms in this action to check a magnetic kinetic spectrum and its unpaired modes.

Boundaries, interfaces, and gluing keeps the boundary term in the action variation and uses it to identify compatible supersymmetric boundary conditions.

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