Build the action from three adjoint chiral multiplets

The field-content lesson organized four-dimensional $\mathcal N=4$ Yang–Mills as one $\mathcal N=1$ vector and three adjoint chirals. Which interactions make that collection one theory with sixteen real supercharges? We will construct the action and eliminate its auxiliaries, keeping every normalization visible. The result will treat the six real scalars equally, although the calculation initially groups them into three complex fields.

Use four-dimensional superspace and interactions for the F-term expansion and gauge-matter coupling. The calculation stays in flat Lorentzian signature $(+---)$ with gauge group $SU(N)$, $N\geq2$, real coupling $g>0$, zero theta angle, and no mass terms. Four real supercharges are manifest in the superfield notation; the specified theory has sixteen. No large-$N$ limit is taken.

Build the cubic interaction

Write the three traceless chiral matrices as $\Phi_i=\Phi_i^at^a$, $i=1,2,3$. A superscript $a$ is a gauge-algebra index, and

$$\operatorname{Tr}_N(t^at^b)=\frac12\delta^{ab}, \qquad [t^a,t^b]=if^{abc}t^c.$$

The generators are Hermitian and $f^{abc}$ is real and totally antisymmetric. The trace is the ordinary fundamental trace. For each chiral, $\Phi_i=\phi_i+\sqrt2\,\theta\psi_i+\theta^2F_i$ in chiral coordinates. Thus $2\operatorname{Tr}_N(\phi_i^\dagger\phi_i)=\sum_a|\phi_i^a|^2$ is the canonical component norm.

All three fields transform by conjugation. A trace of their ordered product is gauge invariant because adjacent group matrices cancel, leaving a conjugation of the entire word. This is the construction in Minahan, (4.1)–(4.4). The particular cubic we need is

$$W=2\sqrt2\,g\,\operatorname{Tr}_N \bigl(\Phi_1[\Phi_2,\Phi_3]\bigr) =i\sqrt2\,g\,f^{abc}\Phi_1^a\Phi_2^b\Phi_3^c. \tag{N02.1}$$

The equality follows from $\operatorname{Tr}_N(t^a[t^b,t^c])=if^{bca}/2$. Cyclically permuting the three chiral labels leaves this expression unchanged. The coefficient is essential: replacing $g$ here by an unrelated coupling produces a different, generically only $\mathcal N=1$, theory.

One way to fix it is the massless adjoint-hypermultiplet construction already used in the preceding lesson. In Tachikawa, (2.1.16), the three matrix kinetic terms and the cubic have a common coefficient $1/g^2$ at zero theta angle. Setting $(\Phi_{\rm T},Z_{\rm T},\widetilde Z_{\rm T})=\sqrt2g(\Phi_1,\Phi_2,\Phi_3)$ gives kinetic coefficient $2\operatorname{Tr}_N$ and cubic coefficient $2\sqrt2g$. This specifies the scalar-superfield normalization; source fermions must also be translated according to their superfield expansion.

The complete superspace action is

$$\begin{aligned} S={}&\int d^4x\,d^4\theta\, 2\sum_{i=1}^3\operatorname{Tr}_N \bigl(\Phi_i^\dagger e^{2gV}\Phi_i e^{-2gV}\bigr)\\ &+\int d^4x\left[ \int d^2\theta\left( \frac14\sum_a\mathcal W^{a\alpha}\mathcal W^a_\alpha+W \right)+\text{h.c.}\right]. \end{aligned} \tag{N02.2}$$

$V$ is the vector superfield, and $\mathcal W_\alpha$ is its chiral field strength, not the superpotential. Its canonical definition is $\mathcal W_\alpha=-\bar{\mathbb D}^{\,2}(e^{-2gV}\mathbb D_\alpha e^{2gV})/(8g)$, as in the prerequisite. The gauge connection in the covariant derivative is $gA_\mu$: $\mathscr D_\mu Y=\partial_\mu Y-ig[A_\mu,Y]$ for any adjoint $Y$. Consequently $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu-ig[A_\mu,A_\nu]$. Minahan’s connection in (4.2) is instead $A_{\mu,\mathrm{MIN}}=igA_\mu$; its coupling is absorbed.

This action fixes both scalar and fermion interactions. Engineering dimensions provide an immediate check: $[\phi]=[A]=1$, $[\psi]=[\lambda]=3/2$, $[F_i]=[D]=2$, and $[g]=0$. Hence $[W]=3$, appropriate to a four-dimensional chiral integral.

Differentiate before taking a matrix norm

Let $C_1=[\phi_2,\phi_3]$, with $C_2,C_3$ obtained cyclically. Cyclicity of the ordinary trace gives

$$\begin{aligned} \delta W&=2\sqrt2g\,\operatorname{Tr}_N (\delta\phi_1C_1+\delta\phi_2C_2+\delta\phi_3C_3),\\ W_{ia}&\equiv\frac{\partial W}{\partial\phi_i^a} =2\sqrt2g\,\operatorname{Tr}_N(t^aC_i). \end{aligned} \tag{N02.3}$$

For example, $\operatorname{Tr}(\phi_1[\delta\phi_2,\phi_3]) =\operatorname{Tr}(\delta\phi_2[\phi_3,\phi_1])$. This step determines the ordering and sign of the derivative.

The Lorentzian auxiliary terms are

$$\begin{aligned} \mathcal L_{\rm aux} &=\sum_{i,a}\bigl(|F_i^a|^2+W_{ia}F_i^a+\overline{W_{ia}}\,\overline{F_i^a}\bigr)\\ &\quad+\frac12\sum_a(D^a)^2+g\sum_aD^a\mu^a,\\ \mu^a&=2\operatorname{Tr}_N(t^a\mathcal C), \qquad \mathcal C=\sum_i[\phi_i,\phi_i^\dagger]. \end{aligned} \tag{N02.4}$$

The symbol $\mathcal C$ is a Hermitian, traceless matrix. It is the adjoint charge density entering the D coupling; it is distinct from the three holomorphic commutators $C_i$. Equation (N02.4) follows from the canonical component action in Bilal, (5.27)–(5.29), with our $F=-f_{\rm Bilal}$.

Completing each square gives $F_i^a=-\overline{W_{ia}}$ and $D^a=-g\mu^a$. The constant part of the Lorentzian Lagrangian is minus the potential, so $U=\sum_{i,a}|W_{ia}|^2+g^2\sum_a(\mu^a)^2/2$.

To convert to matrices, use the $SU(N)$ completeness identity

$$\sum_a\operatorname{Tr}(t^aM)\operatorname{Tr}(t^aN) =\frac12\left(\operatorname{Tr}(MN) -\frac1N\operatorname{Tr}M\,\operatorname{Tr}N\right). \tag{N02.5}$$

Every commutator has zero trace, so the last term vanishes here. It must not be dropped for an arbitrary matrix source. Substituting (N02.3) and (N02.4) now yields

$$\boxed{ U_F=4g^2\sum_{i<j}\operatorname{Tr} \bigl([\phi_i,\phi_j]^\dagger[\phi_i,\phi_j]\bigr), \qquad U_D=g^2\operatorname{Tr}(\mathcal C^2). } \tag{N02.6}$$

Both terms are nonnegative on the Lorentzian physical slice. The same normalization gives the matrix equations $F_i=-\sqrt2g\,C_i^\dagger$ and $D=-g\mathcal C$. The factors differ from simply treating the entries of a traceless matrix as independent canonical scalars.

Recover the six real scalars

Set $\phi_i=(X^{2i-1}+iX^{2i})/\sqrt2$, with all $X^I$ Hermitian. Their kinetic term is $2\sum_i\operatorname{Tr}|\mathscr D\phi_i|^2 =\sum_I\operatorname{Tr}(\mathscr D X^I)^2$, so each real component has coefficient $1/2$.

The cancellation needed to combine (N02.6) is worth seeing. For two different complex-field labels, temporarily write $\phi_i=(a+ib)/\sqrt2$ and $\phi_j=(c+id)/\sqrt2$. Define $P=[a,b]$, $Q=[c,d]$, $A=[a,c]$, $B=[b,d]$, $C=[a,d]$, and $E=[b,c]$. All six are anti-Hermitian. Then

$$[\phi_i,\phi_i^\dagger]=-iP,\qquad [\phi_i,\phi_j]=\frac12(A-B+iC+iE),$$

and cyclicity removes the trace of the mixed commutator in the norm:

$$4\operatorname{Tr}|[\phi_i,\phi_j]|^2 =-\operatorname{Tr}(A^2+B^2+C^2+E^2) +2\operatorname{Tr}(AB-CE). \tag{N02.7}$$

The cross term from $\operatorname{Tr}\mathcal C^2$ is $-2\operatorname{Tr}(PQ)$. These cancel because cyclicity and the Jacobi identity give

$$\begin{aligned} \operatorname{Tr}(AB-CE) &=\operatorname{Tr}\bigl(a([c,[b,d]]-[d,[b,c]])\bigr)\\ &=\operatorname{Tr}\bigl(a[b,[c,d]]\bigr) =\operatorname{Tr}([a,b][c,d]). \end{aligned} \tag{N02.8}$$

The three same-pair terms supply the remaining commutator squares. Thus

$$\boxed{ U=-g^2\sum_{I<J}\operatorname{Tr}[X^I,X^J]^2 =g^2\sum_{I<J}\operatorname{Tr} \bigl([X^I,X^J]^\dagger[X^I,X^J]\bigr). } \tag{N02.9}$$

If the sum is written over all ordered $I,J$, its coefficient is $g^2/2$. The positive second form fixes the sign unambiguously. It also shows that zero potential requires every real-scalar commutator to vanish. Solving these equations modulo gauge transformations is a further vacuum-space calculation.

The equal coefficient for all fifteen scalar pairs is a strong check on (N02.1). It establishes the rotationally invariant scalar interaction, not by itself the complete extended supersymmetry algebra. The source construction uses the full action, including the fermions; Tachikawa, (2.1.17)–(2.1.18) explains that enhancement. The detailed symmetry organization can now build on the derived action.

The fermion couplings are fixed too

Use lower-index Weyl fields with $\psi\chi=\psi^\alpha\chi_\alpha$. For odd fields this scalar contraction is symmetric: $\psi\chi=\chi\psi$. In component indices the superpotential Hessian is particularly transparent:

$$\begin{aligned} W_{ia,jb}&=i\sqrt2g\,\epsilon_{ijk}f^{abc}\phi_k^c, \qquad \epsilon_{123}=1,\\ \mathcal L_{Y,W}&=-\frac12W_{ia,jb}\psi_i^a\psi_j^b+\text{h.c.} \end{aligned} \tag{N02.10}$$

Repeated flavor and gauge indices are summed. Interchanging $(i,a)$ with $(j,b)$ reverses both antisymmetric tensors, making the Hessian symmetric, as it must be. This avoids assigning ordinary commuting-matrix identities to Grassmann-valued matrix products.

The gauge D-term supplies the other Yukawa coupling. With the Hermitian adjoint generators $(t_{\rm ad}^b)_{ac}=-if^{bac}$, it is

$$\mathcal L_{Y,g} =i\sqrt2g\sum_i \phi_i^{*a}(t_{\rm ad}^b)_{ac}\lambda^b\psi_i^c +\text{h.c.} \tag{N02.11}$$

This is the adjoint specialization of Bilal, (5.27); its matrix ordering is discussed in (8.5). The component formula fixes the phases without an ambiguous matrix anticommutator.

Together with (N02.9), these terms give the eliminated component action:

$$\begin{aligned} \mathcal L={}&-\frac14F^a_{\mu\nu}F^{a\mu\nu} -i\lambda^a\sigma^\mu\mathscr D_\mu\bar\lambda^a\\ &+\sum_{i,a}\left[ (\mathscr D_\mu\phi_i^a)^*\mathscr D^\mu\phi_i^a -i\psi_i^a\sigma^\mu\mathscr D_\mu\bar\psi_i^a \right]-U+\mathcal L_{Y,g}+\mathcal L_{Y,W}. \end{aligned} \tag{N02.12}$$

The sigma matrices and barred Lorentzian conjugates are those of the four-dimensional prerequisite. The auxiliaries made the chosen $\mathcal N=1$ subalgebra work off shell. Eliminating them does not demonstrate a finite-auxiliary off-shell realization of all sixteen charges. No three-dimensional spinor or auxiliary convention enters these formulas.

Worked example: a scalar mass checks the coefficient

Take $SU(2)$ with $t^a=\tau_a/2$, and consider $X^1=u\,t^3$, $X^3=v\,t^1$, all other real scalars zero, with real $u,v$. The only nonzero commutator is $[X^1,X^3]=iuv\,t^2$. Equation (N02.9) gives

$$U=\frac{g^2}{2}u^2v^2. \tag{N02.13}$$

In the complex description, $\phi_1=u t^3/\sqrt2$ and $\phi_2=v t^1/\sqrt2$ are Hermitian, so $\mathcal C=0$ and only $C_3=iuv\,t^2/2$ contributes. Equation (N02.6) independently gives $4g^2(u^2v^2/4)\operatorname{Tr}(t^2)^2=g^2u^2v^2/2$.

Hold $u$ constant and expand about $v=0$. The kinetic term is $(\partial v)^2/2$, so this scalar fluctuation has mass $g|u|$. The background is a zero-potential configuration, and the chiral fermions provide a further test. The quadratic superpotential in their scalar partners is

$$W_{\rm quadratic}=igu (\phi_2^1\phi_3^2-\phi_2^2\phi_3^1),$$

which gives fermion pair masses of magnitude $g|u|$ through (N02.10). The phases in the two pairings differ, but their singular values agree. This is a check of the action’s local normalization, not a classification of the whole broken-phase spectrum or of gauge-inequivalent vacua.

In a $U(N)$ extension, a central component commutes with every field and drops out of these interactions. Its free multiplet must still be retained when that group is chosen. Our default action uses the traceless $SU(N)$ fields, not a normalized trace pretending to remove a central sector.

Exercises

1. Differentiate within the traceless subspace

Let a general matrix $M$ couple to a traceless adjoint auxiliary through $W_a=2c\,\operatorname{Tr}(t^aM)$, with complex constant $c$. Derive $\sum_a|W_a|^2$ without assuming $\operatorname{Tr}M=0$. Evaluate it for $SU(2)$ and $M=\operatorname{diag}(m_1,m_2)$. Explain why no trace subtraction survives in (N02.6).

Hint 1

Apply (N02.5) to $M$ and $M^\dagger$, not to $M$ twice.

Hint 2

Only the traceless projection of a matrix can pair with every $t^a$. For a two-by-two diagonal matrix, that projection depends on $m_1-m_2$.

Solution

Completeness gives

$$\sum_a|W_a|^2 =2|c|^2\left( \operatorname{Tr}M^\dagger M-\frac1N|\operatorname{Tr}M|^2 \right) =2|c|^2\operatorname{Tr}(M_0^\dagger M_0), \quad M_0=M-\frac{\operatorname{Tr}M}{N}\boldsymbol1.$$

For $SU(2)$, $M_0=\operatorname{diag}(m_1-m_2,m_2-m_1)/2$, so the answer is $|c|^2|m_1-m_2|^2$. If $m_1=m_2$, the coupling vanishes, a useful check. In the theory above, every $M=C_i$ is a commutator, and $\operatorname{Tr}C_i=0$ exactly; the projection therefore changes nothing. Treating all matrix entries as independent would miss the projection for the general source.

2. Vanishing F terms are not sufficient

In $SU(2)$ set $\phi_1=v\,t^+$, $\phi_2=\phi_3=0$, where $t^+=t^1+it^2=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $v$ is complex. Calculate every $F_i$, the raw vector auxiliary $D$, and the potential. For which $v$ is the configuration a supersymmetric constant vacuum with zero gauge field?

Hint 1

The superpotential derivatives require two different chiral fields, but the moment matrix contains a field and its own conjugate.

Hint 2

Use $[t^+,t^-]=2t^3$ and $\operatorname{Tr}(t^3)^2=1/2$.

Solution

All three $C_i$ vanish, hence $F_i=0$. However, $\mathcal C=2|v|^2t^3$ and $D=-2g|v|^2t^3$. Therefore

$$U=g^2\operatorname{Tr}\mathcal C^2=2g^2|v|^4.$$

It vanishes only at $v=0$. The raw $D$ also vanishes only there, agreeing with the gaugino supersymmetry variation for a constant zero-field-strength background. A complex scalar matrix may commute with the other two while failing to commute with its own conjugate. Holomorphic F equations alone therefore do not solve the physical vacuum problem.

3. Change the superpotential coefficient

Replace $g$ in (N02.1) by a complex parameter $h$, leaving the gauge and kinetic terms fixed. Compare two $SU(2)$ configurations: first $X^1=u t^1$, $X^2=v t^2$; second $X^1=u t^1$, $X^3=v t^2$, with all other scalars zero. They are related by a rotation in the real-scalar $2$–$3$ plane. Determine the condition for equal potentials. Does the test alone prove sixteen supercharges?

Hint 1

The first configuration lies in a single complex chiral and tests the D term. The second has two Hermitian complex chirals and tests the F term.

Hint 2

Changing the phase of $h$ does not change its F-term potential. The magnitude is what this scalar test can determine.

Solution

For the first configuration, $\phi_1=(ut^1+ivt^2)/\sqrt2$ and $\mathcal C=uv\,t^3$. Thus $U_1=g^2u^2v^2/2$. For the second, $[\phi_1,\phi_2]=iuv\,t^3/2$ and $\mathcal C=0$, so $U_2=|h|^2u^2v^2/2$. Equality for arbitrary nonzero $u,v$ requires $|h|=g$.

At the classical level, a fixed chiral-field phase can bring such an $h$ to the displayed positive convention, with the fermion interactions and transformations translated consistently. In the quantum theory its anomalous fermion measure can also shift the theta angle, so this is not a fixed-theta equivalence claim. If $h=0$, the second configuration has zero potential while the first does not, immediately exposing the lost scalar rotation symmetry.

This is a necessary coefficient test, not a proof of all supersymmetry transformations or quantum conformality. The full source construction supplies the extended symmetry; the derived fermion couplings and the worked mass comparison test additional consequences. The same field list at arbitrary $h$ is not enough.

The action is now explicit: the gauge coupling fixes the gauge Yukawa interaction, the cubic fixes the chiral Yukawa interaction, and their auxiliary equations give a common commutator potential for all six real scalars. Continue with Symmetries and different multiplet descriptions to organize these interactions under SU(4) and compare the N=1 and N=2 descriptions.