Symmetries and different multiplet descriptions

The same action can be described by one vector and three chiral multiplets, or by a vector and a hypermultiplet. Which symmetry is visible in each description, and how do we recover the full theory? The answer requires tracking the fields, not changing their number.

Use the construction of the N=4 action as preparation. We keep its flat Lorentzian $SU(N)$ theory, $N\geq2$, canonical coupling $g>0$, zero theta angle, and no masses. All fields are adjoint-valued; gauge rank $N-1$ is unrelated to the number of supersymmetries. The specified action has sixteen real Poincaré supercharges. Our task is to exhibit the internal symmetry relating their different presentations.

One quartet of fermions and a real six of scalars

Reserve $a,b,c$ for color-algebra indices and $A,B=1,\ldots,4$ for a new R-symmetry index. A gauge transformation acts on color; an R-symmetry rotates supercharges and their multiplets. It does not act as a color change of basis.

Combine the four left Weyl fields into

$$\chi^A=(\psi_1,\psi_2,\psi_3,i\lambda). \tag{N03.1}$$

The constant phase in the fourth entry is a basis convention matching the Yukawa coefficients of the prerequisite. It changes neither the gaugino’s kinetic norm nor its physical states. We will check it below. The quartet transforms as $\chi^A\mapsto U^A{}_B\chi^B$ for $U\in SU(4)_R$; its Lorentzian conjugate transforms in $\overline{\mathbf4}$. The gauge field is an R-symmetry singlet. These are the representations described in Minahan, section 2.

The six real scalars need a less familiar representation. An antisymmetric four-by-four tensor has $\binom42=6$ independent entries, but six arbitrary complex entries would be twice too many real fields. Define instead

$$\mathsf X^{i4}=\phi_i,\qquad \mathsf X^{ij}=\epsilon_{ijk}\phi_k^\dagger, \qquad \mathsf X^{BA}=-\mathsf X^{AB}, \qquad i,j,k=1,2,3. \tag{N03.2}$$

Thus, with $\epsilon_{123}=1$,

$$\mathsf X= \begin{pmatrix} 0&\phi_3^\dagger&-\phi_2^\dagger&\phi_1\\ -\phi_3^\dagger&0&\phi_1^\dagger&\phi_2\\ \phi_2^\dagger&-\phi_1^\dagger&0&\phi_3\\ -\phi_1&-\phi_2&-\phi_3&0 \end{pmatrix}. \tag{N03.3}$$

The entries remain color matrices. A dagger on an entry conjugates its color matrix; it is not a transpose of the displayed R-symmetry array. With $\epsilon_{1234}=1$, this construction obeys the reality condition

$$(\mathsf X^{AB})^\dagger =\frac12\epsilon_{ABCD}\mathsf X^{CD}. \tag{N03.4}$$

For example, $(\mathsf X^{14})^\dagger=\mathsf X^{23}$ and $(\mathsf X^{24})^\dagger=-\mathsf X^{13}$. Only the three complex $\phi_i$ are independent. This is the real representation $\mathbf6=\wedge^2\mathbf4$, where $\wedge^2$ means the antisymmetric two-index representation. The identification of the scalar six with antisymmetric SU(4) indices is explained in Minahan, section 4 before (4.7).

Its transformation is

$$\mathsf X'=U\mathsf XU^T. \tag{N03.5}$$

Unitarity together with $\det U=1$ preserves (N03.4): the invariant epsilon tensor converts the two uncontracted $U$ matrices into two conjugate matrices. It also preserves

$$\frac12\sum_{A,B}\operatorname{Tr} \bigl((\mathsf X^{AB})^\dagger\mathsf X^{AB}\bigr) =2\sum_i\operatorname{Tr}(\phi_i^\dagger\phi_i) =\sum_{I=1}^6\operatorname{Tr}\bigl((X^I)^2\bigr), \tag{N03.6}$$

where $\phi_i=(X^{2i-1}+iX^{2i})/\sqrt2$. Thus SU(4) induces rotations of the six real coordinates. More precisely, $SU(4)\simeq\operatorname{Spin}(6)$ is a double cover of $SO(6)$: $U$ and $-U$ act identically on $\mathsf X$, while they act oppositely on $\chi$. The distinction matters when fermions are included.

Check that the interactions respect the packaging

The six-scalar potential derived in the prerequisite, $-g^2\sum_{I<J}\operatorname{Tr}[X^I,X^J]^2$, is invariant under these real rotations. The covariant kinetic terms use invariant norms. The fermions provide a further check: scalar symmetry alone would not establish the symmetry of the full action.

Write $\overline{\mathsf X}_{AB}^a=(\mathsf X^{AB\,a})^*$ in a Hermitian color basis. The two Yukawa terms of the prerequisite combine into

$$\mathcal L_Y =-\frac{ig}{\sqrt2}\, f^{abc}\overline{\mathsf X}_{AB}^a \chi^{Ab}\chi^{Bc} +\text{h.c.} \tag{N03.7}$$

Both R indices and all color indices are summed. The Weyl contraction is the inherited $\psi\eta=\psi^\alpha\eta_\alpha$, symmetric under exchanging two odd Weyl fields. The terms with $A,B=i,j$ give $-ig\epsilon_{ijk}f^{abc}\phi_k^a\psi_i^b\psi_j^c/\sqrt2$. The two placements of index $4$, using $\chi^4=i\lambda$, give $-\sqrt2g f^{abc}\phi_i^{*a}\lambda^b\psi_i^c$. These are precisely the superpotential and gauge Yukawas after cyclic relabeling of the color indices. Omitting the $i$ in (N03.1) would spoil their relative phase.

Equation (N03.7) contracts a conjugate antisymmetric tensor with two fundamentals, so it is SU(4) invariant. This checks the actual interactions, rather than just an automorphism of a free supercharge algebra. Tachikawa’s enhancement argument likewise uses the specified massless adjoint-hyper action in (2.1.16)–(2.1.18).

We write the undotted supercharges as $Q_{\alpha A}$ in $\overline{\mathbf4}$ and their adjoints as $\bar Q_{\dot\alpha}^{\,A}$. This R-index placement agrees with Minahan’s discussion before (3.5). It specifies how the internal index transforms; it does not change the earlier normalization $\{Q_{\alpha A},\bar Q_{\dot\beta}^{\,B}\} =2\delta_A{}^B\sigma^\mu_{\alpha\dot\beta}P_\mu$. In particular, a charge in the conjugate representation has the opposite diagonal R weight to the corresponding $\chi$.

Choose one supersymmetry

Choose $Q_{\alpha4}$ and its adjoint as the manifest N=1 subalgebra. The SU(3) acting on indices $1,2,3$ leaves this charge fixed, so it is a flavor symmetry relative to this subalgebra. It rotates the three chiral multiplets. Their kinetic term is invariant, and the cubic $\operatorname{Tr}(\Phi_1[\Phi_2,\Phi_3])$ transforms with the determinant of this three-by-three rotation, which is one.

Normalize the N=1 R generator by $R_1(\theta)=1$ and hence $R_1(Q_{\alpha4})=-1$. The traceless generator acting on the quartet is

$$R_1=\operatorname{diag}\left(-\frac13,-\frac13,-\frac13,1\right). \tag{N03.8}$$

An entry $\mathsf X^{AB}$ has the sum of the two fundamental weights. Therefore

$$\begin{aligned} \mathbf4&\longrightarrow \mathbf3_{-1/3}\oplus\mathbf1_{1},\\ \mathbf6&\longrightarrow \mathbf3_{2/3}\oplus\overline{\mathbf3}_{-2/3}. \end{aligned} \tag{N03.9}$$

The subscripts are R weights. The three $\phi_i$ have charge $2/3$, their $\psi_i$ have charge $-1/3$, and $\lambda$ has charge one. The scalar cubic has charge two, canceling the charge $-2$ of $d^2\theta$. These are representation and action checks; no anomalous-dimension formula was needed to obtain them.

N=1 multiplet	Physical fields	Bosonic / fermionic polarizations per color generator
Vector $V$	$A_\mu,\lambda$	$2/2$
Chiral $\Phi_i$, each of three	$\phi_i,\psi_i$	$2/2$
Total	Same full N=4 field collection	$8/8$

The other twelve real charges remain symmetries. They are not all represented as translations in this one pair of superspace coordinates. Also, the three complex $F_i$ and real $D$ used in this presentation do not prove finite-auxiliary off-shell closure of all sixteen charges.

Choose two supersymmetries

Now retain $Q_{\alpha1},Q_{\alpha4}$ and their adjoints. This makes an N=2 subalgebra with eight real charges. The fields group as

N=2 multiplet	N=1 constituents	Physical fields
Vector	$V,\Phi_1$	$A_\mu,\phi_1,\lambda,\psi_1$
Full adjoint hyper	$\Phi_2,\Phi_3$	$\phi_2,\phi_3,\psi_2,\psi_3$

Each row has $4/4$ physical polarizations per generator, giving the same $8/8$ total. The real adjoint representation does not identify the two independent hyper chirals or turn this full hyper into half as many fields.

The selected charge plane is rotated by $SU(2)_R$. The other plane, with indices $2,3$, has an $SU(2)_F$ commuting with the selected eight charges. Its subscript means flavor, not the gauge group. Introduce a commuting R generator $r$ and the third generator $I_3$ of this $SU(2)_R$:

$$r=\operatorname{diag}(1,-1,-1,1), \qquad I_3=\operatorname{diag}\left(-\frac12,0,0,\frac12\right). \tag{N03.10}$$

Under $SU(2)_R\times SU(2)_F$ with the stated $r$ weights,

$$\begin{aligned} \mathbf4&\longrightarrow (\mathbf2,\mathbf1)_{1} \oplus(\mathbf1,\mathbf2)_{-1},\\ \mathbf6&\longrightarrow (\mathbf1,\mathbf1)_{2} \oplus(\mathbf1,\mathbf1)_{-2} \oplus(\mathbf2,\mathbf2)_{0}. \end{aligned} \tag{N03.11}$$

To derive the second line, antisymmetrize two indices. Two indices in the first plane give a singlet of weight two; two in the other give its conjugate. One index from each plane gives a $(\mathbf2,\mathbf2)$ with zero weight. Its four components obey the inherited reality condition, so they supply four real hyper scalars, not four independent complex scalars. Equations (N03.9) and (N03.11) state Lie-algebra branchings and weight normalizations; they do not assert a global direct-product decomposition of SU(4).

The vector scalar $\phi_1$ is the charge-two singlet. The vector fermions form the first doublet; the hyper fermions form the second. The scalar hyper doublet uses a conjugate, as in the N=2 multiplet construction. That fact can be checked by an actual rotation.

A rotation that mixes two descriptions

Take $U$ to be the identity on indices $2,3$, with the real block $\left(\begin{smallmatrix}c&s\\-s&c\end{smallmatrix}\right)$ on indices $1,4$, where $c=\cos\beta$, $s=\sin\beta$. Substitution in (N03.5) gives

$$\begin{aligned} \phi_1'&=\phi_1,\\ \phi_2'&=c\phi_2+s\phi_3^\dagger,\\ \phi_3'&=c\phi_3-s\phi_2^\dagger. \end{aligned} \qquad \begin{pmatrix}\lambda'\\ i\psi_1'\end{pmatrix} =\begin{pmatrix}c&s\\-s&c\end{pmatrix} \begin{pmatrix}\lambda\\i\psi_1\end{pmatrix}. \tag{N03.12}$$

Meanwhile $\psi_2,\psi_3$ stay fixed. Thus $(\phi_2,\phi_3^\dagger)$ and $(\lambda,i\psi_1)$ are doublets for this rotation. Directly, $\operatorname{Tr}(|\phi_2'|^2+|\phi_3'|^2) =\operatorname{Tr}(|\phi_2|^2+|\phi_3|^2)$: the cross terms cancel and $c^2+s^2=1$. The invariant Yukawa contraction provides a check beyond the scalar norm.

For generic $\beta$, the rotation also mixes $Q_1$ and $Q_4$. It therefore does not commute with the single chosen N=1 charge. A transformation can be an R-symmetry of the N=2 description even though it mixes a chiral scalar with an antichiral scalar in N=1 notation. This is not a conjugacy constraint on two independent chiral fields.

The SU(3) rotations visible in the N=1 description and these extra rotations together generate SU(4). For example, commuting a generator that mixes indices $2,1$ with one that mixes $1,4$ produces a generator mixing $2,4$; similarly one obtains the $3,4$ generators and the remaining diagonal generator. This explains the enhancement mechanism in the source. The field grouping changes which part is manifest; the fixed action is unchanged.

Exercises

1. Why the six is real, and why its cover matters

Starting from three arbitrary complex numbers $z_i$ in one Hermitian color-generator component, construct (N03.3), verify the three independent conjugacy relations, and evaluate $\frac12\sum_{A,B}|\mathsf X^{AB}|^2$. What do $U=-1_4$ and $U=i1_4$ do to the scalars and fermions? Both matrices have determinant one.

Hint 1

The independent complementary pairs are $(14,23)$, $(24,13)$ and $(34,12)$. The middle pair has a minus sign.

Hint 2

A scalar tensor has two fundamental indices; a fermion has one. Apply the central phase once for each index.

Solution

The entries are $\mathsf X^{14}=z_1$, $\mathsf X^{24}=z_2$, $\mathsf X^{34}=z_3$, $\mathsf X^{23}=\bar z_1$, $\mathsf X^{13}=-\bar z_2$, and $\mathsf X^{12}=\bar z_3$, with the reversed entries fixed by antisymmetry. They satisfy $(\mathsf X^{14})^*=\mathsf X^{23}$, $(\mathsf X^{24})^*=-\mathsf X^{13}$, and $(\mathsf X^{34})^*=\mathsf X^{12}$. There are six real independent coordinates and

$$\frac12\sum_{A,B}|\mathsf X^{AB}|^2 =2\sum_i|z_i|^2 =\sum_I(X^I)^2.$$

For $-1_4$, the two scalar factors cancel while each fermion changes sign. It acts as fermion parity on these elementary fields. For $i1_4$, the scalars change sign and each $\chi$ acquires $i$. Thus the two central transformations are different; only the first acts trivially on the scalar six. The bosonic rotation description alone loses this fermionic information.

2. Recover one R generator from the other description

Find constants $a,b$ such that $R_1=ar+bI_3$ has charge one on $\lambda$ and charge $-1/3$ on $\psi_1$. Check all three complex scalars and the selected supercharge. Explain why equal R charges of the three scalars do not make them three components of an $SU(2)_R$ triplet.

Hint 1

The two fermions have equal $r=1$ and opposite $I_3=\pm1/2$.

Hint 2

Scalar weights add the weights of their two R indices. A supercharge in the conjugate quartet reverses the fundamental weights.

Solution

The equations are $a+b/2=1$ and $a-b/2=-1/3$. Hence

$$R_1=\frac13r+\frac43I_3.$$

For $\phi_1$, the weights are $(r,I_3)=(2,0)$; for $\phi_2,\phi_3$ they are $(0,1/2)$. All give $R_1=2/3$. The conjugates have opposite weights. On $Q_{\alpha4}$, $(r,I_3)=(-1,-1/2)$, giving $R_1=-1$ as required.

Equal eigenvalues of this one generator do not determine an entire non-Abelian representation. Under the selected $SU(2)_R$, $\phi_1$ is a singlet while the hyper scalars are organized into doublets involving conjugates. They are not a triplet of three holomorphic scalars.

3. Change the action instead of its notation

Add $W_m=2m\,\operatorname{Tr}(\Phi_2\Phi_3)$ with nonzero constant $m$ of mass dimension one. Around the zero-scalar vacuum, determine the quadratic masses of the hyper scalars and fermions. Which supersymmetry does this preserve, and why is this different from merely choosing the N=2 description above?

Hint 1

The trace normalization makes $W_m=m\sum_a\Phi_2^a\Phi_3^a$. At the origin, the cubic superpotential gives no quadratic mass.

Hint 2

Use $U=\sum_I|W_I|^2$ and the Weyl mass matrix $W_{IJ}$. Compare the hyper mass with the vector’s mass before considering rotations that exchange their fields.

Solution

For every color generator,

$$U_{\rm quadratic}=|m|^2(|\phi_2^a|^2+|\phi_3^a|^2), \qquad \mathcal M_{\rm Weyl}= \begin{pmatrix}0&m\\m&0\end{pmatrix}.$$

Writing each complex scalar as two real scalars divided by $\sqrt2$ gives four canonical real masses squared $|m|^2$. Also $\mathcal M^\dagger\mathcal M=|m|^2 1_2$: the two Weyl fields have mass magnitude $|m|$. The vector multiplet remains massless at this vacuum.

This is the adjoint-hyper mass of the N=2 construction, preserving eight real supercharges; the resulting theory is called N=2*. The original SU(4) cannot mix massive hyper fields with massless vector fields while preserving the quadratic action. It is therefore a changed theory, whereas (N03.11) only reorganized an unchanged massless theory.

The original $R_1$ also assigns the scalar product charge $4/3$, rather than the charge two required for a superpotential with a fixed neutral coefficient. Assigning a formal compensating charge to $m$ can keep track of the deformation, but does not restore the broken symmetry at a fixed nonzero mass.

The full field content, its R representations, and its smaller multiplet descriptions now agree. These checks establish how the specified symmetries act; quantum scale dependence is a separate question. Continue to couplings and conformal invariance to calculate the one-loop cancellation and identify the additional reasoning needed for the exact statement.