What is N=4 super Yang–Mills?

How much information does the name “four-dimensional N=4 super Yang–Mills” specify? It identifies more than a gauge field accompanied by several scalars and fermions. It also fixes how those fields interact so that sixteen real supersymmetry generators act on the theory.

Begin with N=2 vector multiplets and hypermultiplets. We will use its adjoint-hypermultiplet construction, count the resulting physical polarizations, and identify what must be retained when writing the theory with fewer supercharges visible.

Our setting is flat four-dimensional Lorentzian spacetime with signature $(+---)$, gauge group $SU(N)$ for $N\geq2$, and the vacuum with all scalar expectation values zero. The full action has $\mathcal N=4$ supersymmetry; this background and vacuum preserve all sixteen real Poincaré supercharges. The $\mathcal N=1$ notation below displays four of them. The gauge rank is $N-1$, a different number from either supersymmetry count.

First calculation: eight bosonic and eight fermionic polarizations

Use Hermitian generators $t^a$ with ordinary fundamental trace $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$, where $a=1,\ldots,N^2-1$. The field content is

$$A_\mu=A_\mu^at^a,\qquad X^I=X^{Ia}t^a\quad (I=1,\ldots,6),\qquad \lambda^A_\alpha=\lambda^{Aa}_\alpha t^a\quad (A=1,\ldots,4).$$

There is one gauge connection, six real adjoint scalars, and four adjoint Weyl fermions. “Real scalar” means that $X^{Ia}$ are real Lorentzian fields, so $X^I$ is Hermitian. It does not mean that every entry of its matrix is real. The fermion label $A$ counts four species; the spinor label $\alpha=1,2$ does not count two Weyl species. This field content is stated in Minahan, section 2.

Count the modes of the quadratic theory about the stated vacuum. For a nonzero null momentum $p^\mu=E(1,0,0,1)$, a gauge polarization obeys $p\cdot\epsilon=0$ and is equivalent to $\epsilon^\mu+c\,p^\mu$. The first condition sets $\epsilon^0=\epsilon^3$; the gauge equivalence removes their common value. Only $\epsilon^1,\epsilon^2$ remain. A massless gauge field thus contributes two, not four, physical polarizations per generator.

Each real scalar contributes one. For a Weyl field, the plane-wave equation $\bar\sigma^\mu p_\mu\lambda=0$ has a one-complex-dimensional kernel: lowering the stated momentum gives $p_\mu=E(1,0,0,-1)$, and our $\bar\sigma^\mu=(1,\boldsymbol\sigma)$ convention gives $\bar\sigma^\mu p_\mu=E(1-\sigma_3)$. There is one particle helicity, and the conjugate mode supplies the opposite-helicity antiparticle. For the real adjoint representation, these are the two helicities of the corresponding Majorana field. Counting a Weyl field and its Lorentzian conjugate as two independent Weyl species would duplicate those states.

Consequently,

$$n_B=2+6=8,\qquad n_F=4\times2=8 \quad\hbox{per gauge generator}. \tag{N01.1}$$

For $SU(2)$ there are three generators, even though the rank is one. The gauge polarizations contribute $2\times3=6$ bosonic states, the scalars $6\times3=18$, and the fermions $4\times2\times3=24$. The total is $24/24$. In general it is $8(N^2-1)$ of each.

These are elementary physical polarizations in the linearized gauge theory, with color labels retained. They are not a count of gauge-invariant composite operators or a claim about asymptotic colored particles at arbitrary coupling. A scalar expectation value can also reorganize the spectrum into massive and massless sectors; that is a different vacuum calculation.

Check the helicities using the supersymmetry algebra

The extended-algebra lesson derives the massless oscillator construction. Apply it here at one fixed nonzero null momentum, with vanishing central charge. One normalized creation operator survives for each of the four supersymmetry labels. Each raises helicity by $1/2$, and each can be used only once.

Starting with helicity $-1$, choose $k$ of the four operators. The resulting helicity is $-1+k/2$, and the number of choices is $\binom4k$:

Number of creation operators	Helicity	Multiplicity	Fields supplying these modes
0	$-1$	1	Gauge field
1	$-1/2$	4	Four Weyl fields
2	$0$	6	Six real scalars
3	$+1/2$	4	Conjugate fermion modes
4	$+1$	1	Gauge field

The two bosonic ends and six scalar states give eight; the two fermionic levels give eight. This independent count matches (N01.1). CPT exchanges the two ends and the two fermionic levels. With the adjoint reality condition, the multiplet is already complete: no second copy is required. See Bilal, (3.7)–(3.11) and the N=4 example.

Sixteen physical states at fixed momentum and sixteen real Poincaré supercharges happen to be equal here. They are different counts. The first counts states on which the second act. Nor does “N=4” mean four real generators: four-dimensional $\mathcal N$ counts Weyl supersymmetry labels, each giving four real generators together with its adjoint.

When the superconformal algebra is discussed, there are also special conformal supercharges. They are not included in this sixteen-charge Poincaré count; Minahan, (3.5)–(3.7) distinguishes the two sets.

One visible supersymmetry organizes all the fields

Choose one $\mathcal N=1$ subalgebra and label its gaugino $\lambda$. Pair the six real scalars into three complex adjoint matrices,

$$\phi_i=\frac{X^{2i-1}+iX^{2i}}{\sqrt2}, \qquad i=1,2,3.$$

The field organization is then

N=1 multiplet	Physical fields	Auxiliary fields
Vector $V$	$A_\mu,\lambda$	Real $D$
Chiral $\Phi_1$	$\phi_1,\psi_1$	Complex $F_1$
Chiral $\Phi_2$	$\phi_2,\psi_2$	Complex $F_2$
Chiral $\Phi_3$	$\phi_3,\psi_3$	Complex $F_3$

Every entry is adjoint-valued. The complex matrices $\phi_i$ need not be Hermitian; their real and imaginary Hermitian parts reconstruct the six $X^I$. Their kinetic normalization is consistent with the pairing:

$$2\sum_{i=1}^3\operatorname{Tr} (\partial_\mu\phi_i^\dagger\partial^\mu\phi_i) =\sum_{I=1}^6\operatorname{Tr} (\partial_\mu X^I\partial^\mu X^I).$$

The component scalar norm on the right is $\frac12\sum_{I,a}(\partial X^{Ia})^2$.

The vector supplies $2/2$ physical states and each chiral $2/2$, giving $8/8$ once more. Auxiliary fields have no wave equation carrying an additional particle polarization. They make the chosen $\mathcal N=1$ description work off shell. Their presence does not establish off-shell closure of all sixteen supersymmetries.

One can also group $(V,\Phi_1)$ into an $\mathcal N=2$ vector and $(\Phi_2,\Phi_3)$ into a full adjoint hypermultiplet, as in the prerequisite. That notation can display eight charges. Changing the grouping changes which transformations are manifest; it does not remove a physical symmetry of the specified action.

A field list is not the complete theory

Keep the canonical gauge normalization and coupling $g=g_{\rm YM}>0$ of the prerequisite: the component covariant derivative on an adjoint field is $\mathscr D_\mu\phi=\partial_\mu\phi-ig[A_\mu,\phi]$. For simplicity set the theta angle to zero. There are no scalar or fermion masses.

The defining $\mathcal N=1$ interactions use the canonical gauge action, the covariant chiral kinetic term, and the particular cubic superpotential

$$\begin{aligned} K&=2\sum_{i=1}^3\operatorname{Tr}_N \big(\Phi_i^\dagger e^{2gV}\Phi_i e^{-2gV}\big),\\ W&=2\sqrt2\,g\,\operatorname{Tr}_N \big(\Phi_3[\Phi_1,\Phi_2]\big). \end{aligned} \tag{N01.2}$$

The factor two converts the trace pairing to canonical adjoint component pairing. The relative coefficient is the massless adjoint-hyper interaction already obtained in the prerequisite’s third exercise. A fixed field phase can change the displayed phase of $W$ together with the corresponding transformation conventions.

Equation (N01.2) specifies the theory; deriving its component Yukawa and scalar interactions is the next action calculation. The classical engineering dimensions are $[\phi_i]=1$, $[\psi_i]=3/2$, $[A_\mu]=1$ and $[g]=0$, so the cubic superpotential has dimension three, as required by a chiral superspace integral. Gauge invariance follows from conjugating every matrix inside the trace.

Neither this dimensional check nor equal bosonic and fermionic counts fixes the cubic coefficient. Replacing $g$ in $W$ by an unrelated parameter keeps an $\mathcal N=1$ gauge theory with the same field list, but generically loses the additional supersymmetries. Adding a mass is another change of theory. Equal counts are a necessary consistency check, not a substitute for invariant interactions.

What changes for U(N)?

An adjoint $SU(N)$ field is traceless. For $U(N)$ it also contains a central component. Choose $t^0=I_N/\sqrt{2N}$, so $\operatorname{Tr}(t^0t^0)=1/2$, and write $Y=Y^0t^0+\widehat Y$ with $\operatorname{Tr}\widehat Y=0$. Then

$$[t^0,\widehat Y]=0,\qquad 2\operatorname{Tr}(Y^\dagger Y) =|Y^0|^2+2\operatorname{Tr}(\widehat Y^\dagger\widehat Y).$$

In the ordinary flat local action (N01.2), the central fields drop out of every commutator. A central entry in the outer factor of $W$ also vanishes because the trace of a commutator is zero. The central vector and its partners therefore form a decoupled free N=4 multiplet, while the traceless fields form the interacting sector.

$U(N)$ has $N^2$ generators and rank $N$, so its linearized count is $8N^2$ of each statistics. The extra $8/8$ is real field content, not a trace convention. This local decomposition does not identify the global groups $U(N)$ and $SU(N)\times U(1)$ or determine their allowed line operators. When a matrix observable uses $U(N)$, the central contribution must be retained or explicitly removed.

Lorentzian conjugation and state counts have been used throughout. In a Euclidean path integral, independent barred variables before choosing an integration contour are not extra Lorentzian particle species.

Exercises

1. Count states, then count auxiliary components

For $SU(3)$, give the gauge rank, number of generators, and physical bosonic and fermionic counts. Then count the real off-shell component functions of the $\mathcal N=1$ description per generator, subtracting one local gauge function. Does that count prove off-shell N=4 closure?

Hint 1

The rank is the number of independent traceless diagonal generators. It is not the dimension of the full algebra.

Hint 2

Before using field equations, a Weyl field has four real components. A complex scalar and its complex auxiliary together have four real components.

Solution

$SU(3)$ has rank two and eight generators. The physical count is $8\times8=64$ bosonic states and 64 fermionic states.

Off shell, the vector has $4-1$ gauge components and one real $D$, hence four bosonic functions, matching the gaugino’s four real components. Each chiral has two real scalar and two real auxiliary components, matching four real fermionic components. Per generator the total is therefore $4+3\times4=16$ of each, or $128/128$ after including all eight generators.

These are component functions before the field equations, not twice as many particles. The count is compatible with the manifest N=1 transformations. Closure of the other transformations requires their actual algebra; equality of two integers does not prove it.

2. The same fields with a different interaction

Take $SU(2)$ with $t^a=\sigma_a/2$ and replace the cubic coefficient in (N01.2) by an independent complex $\kappa$: $W_\kappa=2\sqrt2\kappa\operatorname{Tr}(\Phi_3[\Phi_1,\Phi_2])$. On scalar matrices $\phi_1=x t^1$, $\phi_2=y t^2$, $\phi_3=z t^3$, compute $W_\kappa$. Check gauge invariance and engineering dimension, and explain why these tests and the unchanged state count cannot establish N=4 at $\kappa=0$, $g\ne0$.

Hint 1

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

Hint 2

At the zero-field vacuum a cubic term does not change the quadratic spectrum. Its absence can still change the interacting theory.

Solution

The trace is $ixyz/2$, so

$$W_\kappa=i\sqrt2\,\kappa xyz.$$

Conjugating all three matrices leaves the trace unchanged, and $[x]=[y]=[z]=1$ makes $W_\kappa$ dimension three for dimensionless $\kappa$. Both checks hold for any $\kappa$, including zero.

The cubic term has no quadratic part at the origin, so deleting it leaves the $8/8$ massless count unchanged. But for nonzero $g$, the N=4 action requires its cubic interaction with the relative magnitude in (N01.2). At $\kappa=0$ the nonzero expression above is absent for generic $x,y,z$, while gauge interactions remain. The manifest N=1 construction still works; the field list alone does not supply the missing extended symmetry.

3. Is a central multiplet a second interacting copy?

Compare $SU(2)$ and $U(2)$ with the same canonical local coupling. Decompose each $U(2)$ adjoint using $t^0=I_2/2$ and count the additional physical modes. Show directly that adding arbitrary central pieces to all three $\phi_i$ leaves the cubic interaction unchanged. What remains at $N=1$ for the unitary theory?

Hint 1

The identity commutes with all matrices. A central factor outside a commutator is tested by its trace.

Hint 2

Use $\operatorname{Tr}[A,B]=0$ after expanding the outer factor. Compare the dimensions four and three, not only the ranks two and one.

Solution

The $U(2)$ count is $32/32$ and the $SU(2)$ count is $24/24$. Their difference is one gauge field with two polarizations, six real scalars and four Weyl fermions: $8/8$.

Write $\phi_i=\widehat\phi_i+c_iI_2$. The inner commutator becomes $[\widehat\phi_1,\widehat\phi_2]$. The outer central term gives $c_3\operatorname{Tr}[\widehat\phi_1,\widehat\phi_2]=0$. Thus the cubic term only involves traceless fields. The central covariant derivative is an ordinary derivative for the same reason.

At $N=1$, all unitary adjoint fields are central and the commutator interactions vanish: the local theory is the free Abelian N=4 multiplet. It is not an interacting $SU(1)$ theory. This limiting check concerns the local action and states; it does not discard global gauge or line-operator data.

The defining data now travel together: a gauge group, a field collection, an interaction strength shared by the required terms, and a specified background and vacuum. Continue with Build the action from three adjoint chiral multiplets to derive how (N01.2), with the auxiliaries eliminated, produces the scalar and fermion interactions required by all sixteen supercharges.