Put ABJM on a supersymmetric three-sphere

The flat ABJM action has a particular relation between its Chern–Simons levels and matter interactions. Moving it to a sphere should preserve that relation. Merely changing ordinary derivatives to covariant derivatives would miss the scalar curvature coupling.

Use the full symmetry and field organization, the free sphere action, the gauge–matter component construction, and the curved Chern–Simons completion. We will assemble their interacting counterpart without choosing a localization locus.

The theory remains three-dimensional compact $U(N)_k\times U(N)_{-k}$, equal finite $N\geq1$, integer $k>0$, ordinary fundamental traces, and canonical matter normalization. There is no additional common-center quotient, mass deformation, FI coupling, or defining Yang–Mills term. Its generic flat-space supersymmetry has twelve real Poincaré supercharges; the chosen $\mathcal N=2$ presentation displays four. The exceptional levels one and two have the known further enhancement. These counts are distinct from the particular rigid generators constructed below.

Fix every background field

Let the round sphere have radius $\ell>0$. In the paper’s curvature convention choose

$$g=g_{S^3_\ell},\qquad A_\mu=V_\mu=C_\mu=0,\qquad H=-\frac{i}{\ell},\qquad R=-\frac6{\ell^2}. \tag{AB11.1}$$

All background gravitini vanish. The R line bundle is trivial, and the sphere has its unique spin structure. Capital $A_\mu$ is the rigid R connection, unrelated to the matter matrices $A_i$ or the two dynamical connections $a,\widehat a$. No flavor vector multiplet is turned on. The two-form potential underlying $H$ is understood in patches: since $H=i*dB$, its field strength is $dB=-\operatorname{vol}/\ell$. It is not a single global two-form with nonzero integral of its exterior derivative.

The two independent commuting labels solve

$$\nabla_\mu\zeta=\frac{i}{2\ell}\gamma_\mu\zeta,\qquad \nabla_\mu\widetilde\zeta=\frac{i}{2\ell}\gamma_\mu\widetilde\zeta. \tag{AB11.2}$$

The gamma matrices obey $\gamma_a\gamma_b=\delta_{ab}+i\epsilon_{abc}\gamma_c$, $\epsilon_{123}=1$. In the global sphere frame of the prerequisite, either label has arbitrary constant complex components; its covariant derivative is still (AB11.2). There are two complex solution directions in each R-charge family. This is the complexified four-generator rigid $\mathcal N=2$ subalgebra, conventionally described as $SU(2|1)$ together with the other sphere $SU(2)$ isometry. Its odd generators are sphere combinations of flat Poincaré and conformal generators, not four new flat translations.

This construction guarantees that subalgebra. The conformal ABJM theory can have additional sphere symmetries, but solving these two equations alone does not count all of them or prove that a deformed metric preserves twelve supercharges. Euclidean labels and tilded fields remain independent; no pointwise conjugacy condition is imposed.

Concretely, for a gauge-invariant scalar $\mathcal O$ of R-charge $r$ and zero central charge, put $s=\zeta\widetilde\zeta$ and $K^\mu=\zeta\gamma^\mu\widetilde\zeta$. The rigid algebra, P (6.1), becomes

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\mathcal O =-2iK^\mu\partial_\mu\mathcal O +\frac{2r}{\ell}s\,\mathcal O.$$

The same-family anticommutators vanish. The mixed operation is an isometry plus an R transformation; gauge-charged components additionally require the compensating gauge transformation. This is the algebra being used to construct the action.

Keep the two color actions visible

The four chiral multiplets have scalars $A_i:V_R\to V_L$ and $B_i:V_L\to V_R$, $i=1,2$. Every scalar has R-charge $1/2$, its fermion has $-1/2$, and its auxiliary has $-3/2$. Antichiral partners have the opposite charges. Their engineering dimensions are $1/2,1,3/2$. Both vector multiplets have fields $(a,\sigma,\lambda,\widetilde\lambda,D)$, with hats for the right node.

For any bosonic node pair $(X_L,X_R)$, define its action on matter by

$$X_{\mathcal R}A_i=X_LA_i-A_iX_R,\qquad X_{\mathcal R}B_i=X_RB_i-B_iX_L. \tag{AB11.3}$$

Thus $\mathcal D_\mu A_i=\nabla_\mu A_i-ia_\mu A_i+iA_i\widehat a_\mu$, with the nodes exchanged for $B_i$. The spin connection is included on fermions. Define $\Sigma=\sigma_{\mathcal R}$ and $\mathsf D=D_{\mathcal R}$ as operators on the direct sum $\mathcal R$ of these four representations. They are not rigid sources.

For the action, collect the scalar entries into $\phi^I$; $I$ includes species and both color labels. Its dual is $\widetilde\phi_I$, and repeated entries are summed. Choose Hermitian generators $T_\alpha$ of the product group in this representation, and write $\Sigma^I{}_J=\sigma^\alpha(T_\alpha)^I{}_J$ and $\mathsf D^I{}_J=D^\alpha(T_\alpha)^I{}_J$. The components $\sigma^\alpha,D^\alpha$ include both nodes with their actual representation signs. This is an exact way to keep matrix order, not a replacement by four Abelian charges.

Spinors use $uv=u^{\mathsf T}Cv$, with $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. All dynamical spinors anticommute. For example, $\lambda^\alpha\psi^J$ in the next equation is this spinor contraction; its color generator is written separately. The operations labeled by commuting $\zeta,\widetilde\zeta$ obey the graded product rule.

Derive the curvature completion

For a scalar of R-charge $r$ and zero central charge, the gauge–matter formula contains, as an operator on $\mathcal R$,

$$\mathsf D+\Sigma H+\Sigma^2 +2(r-1)H\Sigma-\frac r4R +r\left(r-\frac12\right)H^2. \tag{AB11.4}$$

Here the paper’s combination $z-q\sigma$ has become $-\Sigma$. At $r=1/2$, the first and second terms linear in $H\Sigma$ cancel: $\Sigma H+2(r-1)H\Sigma=0$. The explicit $H^2$ coefficient vanishes. The result is

$$\mathsf D+\Sigma^2-\frac18R =\mathsf D+\Sigma^2+\frac3{4\ell^2}. \tag{AB11.5}$$

The fermion scalar-bilinear coefficient similarly becomes $-i\Sigma$, with no separate $1/\ell$ mass. Its kinetic operator still has a spin connection. This cancellation is the $r=1/2,z=0$ specialization of P, (6.31)–(6.32). In particular, one must not retain $\Sigma H$ while dropping its partner.

The complete matter density is therefore

$$\begin{aligned} \mathcal L_{\rm mat}={}& (\mathcal D_\mu\widetilde\phi)_I(\mathcal D^\mu\phi)^I -i\widetilde\psi_I\gamma^\mu(\mathcal D_\mu\psi)^I -\widetilde F_I F^I\\ &+\widetilde\phi_I \left(\mathsf D+\Sigma^2+\frac3{4\ell^2}\mathbf1\right)^I{}_J\phi^J -i\widetilde\psi_I\Sigma^I{}_J\psi^J\\ &+\sqrt2 i\,(T_\alpha)^I{}_J \left[\widetilde\phi_I(\lambda^\alpha\psi^J) +(\widetilde\lambda^\alpha\widetilde\psi_I)\phi^J\right]. \end{aligned} \tag{AB11.6}$$

For example, $\Sigma^2 A=\sigma^2A-2\sigma A\widehat\sigma+A\widehat\sigma^2$. The middle term is required. The second Yukawa term uses the same displayed generator indices as the first; replacing the dual representation by a second copy of the fundamental would reverse a node coupling.

The derivation in P §6.3 is displayed for Abelian fields and explicitly permits its non-Abelian extension. Here the extension is implemented by the representation matrices and the ordinary invariant node traces. The canonical gauge–matter lesson supplies the neutral D-term construction; on a closed sphere its integration-by-parts divergence integrates to zero.

Add the two Chern–Simons terms and the superpotential

Write the connection three-form as $\omega_3(a)=\operatorname{Tr}(a\,da-\tfrac{2i}{3}a^3)$. With the Euclidean weight $e^{-S_E}$, the vector action is

$$\begin{aligned} S_{\rm vec}={}&\frac{ik}{4\pi} \int_{S^3}\bigl[\omega_3(a)-\omega_3(\widehat a)\bigr]\\ &+\frac{k}{4\pi}\int_{S^3}\!d^3x\sqrt g\, \left[ \operatorname{Tr}_L(-2\sigma D+2i\lambda\widetilde\lambda) -\operatorname{Tr}_R(-2\widehat\sigma\widehat D +2i\widehat\lambda\widetilde{\widehat\lambda}) \right]. \end{aligned} \tag{AB11.7}$$

In the second line the trace includes the ordered color product and spinor contraction. The level multiplies the ordinary trace explicitly; it is not a normalized trace divided by $N$. The compact functional is the spin Chern–Simons functional at the declared integer levels, with its extension or patch definition when required. A local three-form alone is not a general definition on a nontrivial bundle.

There is no additional $H\sigma^2$ term. The curved multiplet construction cancels that apparent term, leaving P (6.28). Adding it would change the auxiliary constraints and the theory. These vector fields have no Yang–Mills kinetic term; the gauginos, vector scalars and $D$ fields are algebraic.

Keep the previously fixed holomorphic interaction

$$W=\frac{4\pi}{k}\operatorname{Tr}_L (A_1B_1A_2B_2-A_1B_2A_2B_1). \tag{AB11.8}$$

Define $W_I=\partial W/\partial\phi^I$ and $W_{IJ}=\partial_I\partial_JW$ as entry derivatives. Let $\widetilde W(\widetilde\phi)$ be the independent antichiral polynomial which equals $W^*$ on the physical scalar slice; conjugating a matrix word reverses its order. Its derivatives carry upper entry indices. The full superpotential density is

$$\mathcal L_W= W_I F^I-\frac12W_{IJ}\psi^I\psi^J +\widetilde W^{\,I}\widetilde F_I +\frac12\widetilde W^{\,IJ}\widetilde\psi_I\widetilde\psi_J. \tag{AB11.9}$$

The opposite signs of the two Yukawa terms are the established chiral and antichiral component convention. A tilde is not an instruction to complex-conjugate an odd product without reversing its order. Equations (AB11.6)–(AB11.9) specify every interaction:

$$\boxed{S_E=S_{\rm vec}+\int_{S^3}d^3x\sqrt g\, (\mathcal L_{\rm mat}+\mathcal L_W).} \tag{AB11.10}$$

No superpotential curvature term has been omitted. Each quartic word has R-charge two and central charge zero. Its highest chiral component $F_W$ has variation $\delta F_W=-\sqrt2 i\nabla_\mu(\widetilde\zeta\gamma^\mu\psi_W)$, where $\psi_W=W_I\psi^I$. The corresponding antichiral top component is also a divergence. Thus P (6.14) makes (AB11.9) invariant on the closed sphere, without using any matter equation of motion. In the unsimplified rule the extra algebraic term is proportional to $z_W-(r_W-2)H=0$. This explains why a guessed term proportional to $HW$ would be wrong.

All terms have dimension three. The dimensionless coefficient $4\pi/k$ has not acquired a radius factor. Gauge neutrality, R-neutrality and the neutral-multiplet construction check different parts of the action; none alone would fix all coefficients.

Read the source comparison with its phase convention

KWY (2.14) has the same unit-radius conformal coefficient $+3/4$ and displays all the gauge–matter partners. Its auxiliary and spinor continuation is different: its printed scalar term is $-\phi^\dagger\sigma^2\phi$, so those symbols cannot simply be substituted into (AB11.6).

KWY’s “Tr” in (2.1) includes $k/(4\pi)$, while ours is ordinary. Its covariant derivative is written with $+iA_{\rm KWY}$, compared with our $-ia$; matching a charged derivative requires $A_{\rm KWY}=-a$. In addition, its classical oscillatory exponent (3.7) is $-i\pi k_{\rm KWY}\sigma_0^2$. The inherited P convention gives the opposite sign at matched real dimensionless scalar and level. Comparing these phases requires $k_{\rm KWY}=-k$, or exchanging the two opposite-level nodes. This is a comparison of the displayed phase and normalization, not a universal off-shell field dictionary. We retain (AB11.7) and its weight throughout.

Eliminate auxiliaries without changing their meaning

On the physical bosonic slice put $\widetilde\phi=\phi^\dagger$ and take $\sigma,\widehat\sigma$ Hermitian. The curvature term in (AB11.6) contains neither $D$ nor $\sigma$. Thus their equations are exactly those of the sextic-potential calculation:

$$\begin{gathered} t=\frac{2\pi}{k},\quad \mu_L=\sum_i(A_iA_i^\dagger-B_i^\dagger B_i),\quad \mu_R=\sum_i(B_iB_i^\dagger-A_i^\dagger A_i),\\ \sigma=t\mu_L,\qquad \widehat\sigma=-t\mu_R. \end{gathered} \tag{AB11.11}$$

They do not impose $\mu_L=\mu_R=0$. The remaining vector-scalar equations determine $D,\widehat D$, rather than impose extra constraints on matter. Likewise $\widetilde F_I=W_I$, $F^I=\widetilde W^{\,I}$. Consequently

$$\begin{aligned} V_{S^3}={}&\frac3{4\ell^2} \sum_i(\|A_i\|^2+\|B_i\|^2)+V_6,\\ V_6={}&t^2\sum_i\left( \|\mu_LA_i+A_i\mu_R\|^2+ \|\mu_RB_i+B_i\mu_L\|^2\right) +\sum_I|W_I|^2. \end{aligned} \tag{AB11.12}$$

The norm is the ordinary sum of squared matrix entries. The ordered cubic gradients inside $W_I$ and their coefficient $4\pi/k$ are unchanged. Eliminating algebraic gauginos similarly gives the familiar quartic matter interactions with two scalars and two fermions; the off-shell expression (AB11.6) already specifies them without suppressing their signs.

An algebraic stationary point is not an integration contour. For the chiral auxiliary pair one may use the shifted direction $F^I=\widetilde W^{\,I}+f^I$, $\widetilde F_I=W_I-(f^I)^*$ on the physical scalar slice. Its action becomes $|f|^2+\sum_I|W_I|^2$, which is a decaying Gaussian in $f$. For the purely linear vector auxiliary, an imaginary Hermitian direction represents a Fourier constraint, not a positive real Gaussian. The gauge Chern–Simons factor is oscillatory. A regulated path integral must retain these prescriptions, its determinant phases and any specified background counterterms; a real-valued minimum of the off-shell density does not define it. We add no FI, gauge–R, or pure-background counterterm in the displayed classical action.

Worked check: what becomes of a flat direction?

For diagonal constant scalar matrices the flat sextic potential vanishes, but the sphere contribution is

$$S_{\rm scalar} =\frac{3\pi^2\ell}{2} \sum_i(\|A_i\|^2+\|B_i\|^2), \qquad \operatorname{Vol}(S^3_\ell)=2\pi^2\ell^3. \tag{AB11.13}$$

Thus a nonzero constant point of the flat moduli space is not a stationary scalar configuration on the sphere. More generally, keep any smooth gauge connection fixed and rescale every scalar by a real parameter $u$. After bosonic auxiliary elimination the real matter action has the form $S_{\rm matter}(u)=u^2S_2+u^6S_6$, where $S_2$ includes covariant gradients and the positive curvature mass, and $S_6\geq0$. At a stationary physical bosonic configuration, $0=\partial_u S|_{u=1}=2S_2+6S_6$. Therefore every scalar vanishes.

This is a classical statement on the specified physical slice of a compact sphere. It does not classify complex saddles, solve the vector supersymmetry equations, or derive the localization integration cycle. The flat limit is local: at fixed physical fields and momenta the curvature density disappears as $\ell\to\infty$. The extensive sphere volume is not held fixed in that limit.

There is a second check of the coefficient which uses geometry instead of auxiliary algebra. In stereographic coordinates $x\in\mathbb R^3$, $g=\Omega^2\delta$ with $\Omega=2\ell^2/(\ell^2+|x|^2)$. For a scalar write $\phi_{S^3}=\Omega^{-1/2}f$. Direct expansion gives

$$\sqrt g\left( |\mathcal D\phi_{S^3}|^2+\frac3{4\ell^2}|\phi_{S^3}|^2 \right) =|\mathcal D f|^2 -\frac12\partial_\mu\left(|f|^2\partial_\mu\log\Omega\right). \tag{AB11.14}$$

The non-divergence remainder vanishes because $\tfrac12\partial^2\log\Omega+ \tfrac14(\partial\log\Omega)^2+ 3\Omega^2/(4\ell^2)=0$. For a smooth sphere field the boundary flux at the missing pole vanishes. The spinor rescaling is $\psi_{S^3}=\Omega^{-1}\psi_{\rm flat}$, with $\not{\mathcal D}_{S^3}=\Omega^{-2}\not{\mathcal D}_{\rm flat}\Omega$ in the corresponding conformal frame. A sextic scalar term and the induced two-scalar/two-fermion terms each scale as $\Omega^{-3}$, canceling the volume factor. The connection Chern–Simons form is metric independent. These checks recover the conformal sphere completion; they do not evaluate its quantum partition function.

Next, derive the sphere matrix integral. Use the supersymmetric deformation to find its constant scalar locus, then combine the classical phase, roots, matter weights and both Weyl factors.

Exercises

1. Change the R assignment carefully

Give all four chiral scalars charge $r=1/2+\delta$ while holding the quartic coefficient fixed and neutral. Find their sphere scalar and fermion coefficients from (AB11.4). Is the superpotential still an invariant F-term? Contrast this with assigning $r_A=1/2+\delta$, $r_B=1/2-\delta$.

Hint 1

Before substituting $H$, combine the two terms linear in $\Sigma H$. Then count the R-charge of the four scalar factors in $W$.

Hint 2

The gauge-invariant chiral multiplet whose top component is integrated must have bottom R-charge two. A changed source coupling need not preserve the full extended symmetry.

Solution

For any one chiral the coefficients are

$$U=\mathsf D+\Sigma^2-\frac{i(2r-1)}{\ell}\Sigma +\frac{r(2-r)}{\ell^2},\qquad M_\psi=-i\Sigma-\frac{r-1/2}{\ell}.$$

The common shift gives $U=\mathsf D+\Sigma^2-2i\delta\Sigma/\ell+ (3/4+\delta-\delta^2)/\ell^2$ and $M_\psi=-i\Sigma-\delta/\ell$. But $R(W)=2+4\delta$, so a fixed neutral coefficient fails the required F-term charge unless $\delta=0$. The gauge–matter D-terms alone do not repair this.

With opposite shifts, each word contains two A and two B fields, giving $R(W)=2$. Each species uses the same displayed formula with its own $r$ and its own two-sided $\Sigma$. This is an allowed manifest-$\mathcal N=2$ R/flavor choice with modified source couplings, not the undeformed superconformal assignment used in this lesson. It does not by itself establish the same full extended sphere symmetry. In either case all explicit curvature-dependent coefficients disappear in the fixed-field local flat limit.

2. Turn a stationary equation into an integration prescription

For one chiral auxiliary pair let $L_F=-\widetilde F F+wF+\widetilde w\widetilde F$, with $\widetilde w=w^*$ on the physical scalar slice. Verify the shifted contour described above and compute its normalized Gaussian integral $\int d^2f\,e^{-L_F}/\pi$. For one real component of the vector constraint $J=\mu-\sigma/t$, take $D=id$ and regulate the Fourier integral by $e^{-\epsilon d^2/2}$, $\epsilon>0$. Evaluate $\int dd\,e^{-\epsilon d^2/2-idJ}/(2\pi)$. What limiting operation is justified?

Hint 1

Substitute $F=\widetilde w+f$ and $\widetilde F=w-f^*$ before expanding. For the vector integral complete the ordinary Gaussian square or Fourier transform it.

Hint 2

The vector result is a normalized narrow Gaussian in $J$. Its limit is a distribution tested against a smooth integrable function, not a convergent integral at each point.

Solution

The terms linear in $f,f^*$ cancel and $L_F=|f|^2+|w|^2$. With the stipulated contour measure, the normalized integral is $e^{-|w|^2}$, reproducing the positive on-shell F potential. Direct integration over $\widetilde F=F^*$ instead would make the quadratic action negative.

The regulated vector integral is

$$\frac1{\sqrt{2\pi\epsilon}} \exp\left(-\frac{J^2}{2\epsilon}\right).$$

It tends to $\delta(J)$ in the distributional sense. This imposes $\sigma=t\mu$, with the opposite-node sign treated separately. It is not a minimization of a $D^2$ potential: the defining theory has no such term. These finite-dimensional checks justify the auxiliary steps under the stated prescriptions, not arbitrary interchanges of infinite-dimensional oscillatory integrals.

3. Test a noncommuting constant profile and its radius

Take $N=2$, real $v,b$, and $A_1=v1_2$, $A_2=B_2=0$, $B_1=bE_{12}$, where $(E_{12})_{ij}=\delta_{i1}\delta_{j2}$. Set the gauge connections to zero and eliminate the bosonic auxiliaries. Compute the matter action on the sphere. Then distinguish changing $\ell$ at fixed $v,b$ from changing it at fixed dimensionless $\bar v=\sqrt\ell\,v$, $\bar b=\sqrt\ell\,b$.

Hint 1

All F gradients vanish on this profile, but $\mu_L=v^2 1_2-b^2E_{22}$ and $\mu_R=b^2E_{11}-v^2 1_2$. Evaluate both kinds of vector cubic in (AB11.12).

Hint 2

The A cubic is $vb^2(E_{11}-E_{22})$. The B cubic cancels. The scalar norm is $2v^2+b^2$, and the sphere volume carries three powers of $\ell$.

Solution

The A cubic has norm squared $2v^2b^4$, while the B cubic and every F gradient vanish. Therefore

$$S_{\rm matter} =4\pi^2t^2\ell^3v^2b^4 +\frac{3\pi^2\ell}{2}(2v^2+b^2) =4\pi^2t^2\bar v^2\bar b^4 +\frac{3\pi^2}{2}(2\bar v^2+\bar b^2).$$

The two expressions describe the same action in different variables. At fixed physical field values the two terms scale as $\ell^3$ and $\ell$, respectively. At fixed dimensionless values the classical answer is independent of $\ell$. Dropping the first term just because the holomorphic F equations vanish would give the wrong interacting action. Dropping the second would retain a false constant flat direction. Neither radius comparison establishes an absolute quantum partition-function normalization or a determinant phase prescription.

The complete action is now available with its rigid background and integration qualifications. Deriving a finite matrix integral requires the next, separate steps: a chosen preserved operation, its gauge-fixed deformation, its locus, and the fluctuation measure.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading three-dimensional localization with matter