N=2 vector multiplets and hypermultiplets

Eight real supercharges can relate fields that look like separate multiplets when only four supercharges are visible. Which familiar fields belong together, and what interaction makes the grouping an actual symmetry?

We use the flat four-dimensional Lorentzian conventions of the N=1 action lesson: canonical component kinetic terms, Hermitian gauge generators, and the chiral auxiliary $F=-f_{\rm Bilal}$. The full theory will have $\mathcal N=2$ supersymmetry, while the superfield notation makes an $\mathcal N=1$ subalgebra manifest. These are eight and four real supercharges, respectively. Neither number specifies a gauge rank.

The multiplet comparison provides a compact map. Here we derive the representation assignments, construct the interactions, and test them in an Abelian example. The extended-algebra lesson supplies the distinction between massless fields and a massive BPS particle sector.

First calculation: two different pairs of N=1 multiplets

An $\mathcal N=2$ vector multiplet consists of

$$\underbrace{(A_\mu,\lambda,D)}_{\mathcal N=1\ {\rm vector}} \quad\oplus\quad \underbrace{(\phi,\psi_\Phi,F_\Phi)}_{\mathcal N=1\ {\rm chiral}}.$$

Both multiplets take values in the gauge algebra. In particular, the added chiral is adjoint, not an arbitrary charged matter field. The second supersymmetry relates its fermion and scalar to the vector fields. For a massless gauge field, each generator contributes two transverse bosonic modes. The complex scalar contributes two more; the two Weyl fields contribute two physical fermionic modes each. The result is $4$ bosonic and $4$ fermionic states per generator.

A full hypermultiplet instead consists of two chiral multiplets:

$$\mathcal H_R=Q_R\oplus\widetilde Q_{R^*}, \qquad Q=(q,\psi_Q,F_Q),\quad \widetilde Q=(\widetilde q,\psi_{\widetilde Q},F_{\widetilde Q}).$$

The tilde labels a second chiral field. It does not impose $\widetilde q=q^\dagger$, nor make $\widetilde Q$ antichiral.

Why the conjugate representation? Write $q$ as a column in a unitary representation $R$ and $\widetilde q$ as a row in its dual. Under an ordinary gauge transformation,

$$q'=U_Rq,\qquad \widetilde q'=\widetilde q\,U_R^{-1}.$$

Transposing the second equation makes its generators explicit:

$$\widetilde q^{\,T}\mapsto U_R^*\widetilde q^{\,T}, \qquad t_{R^*}^a=-(t_R^a)^T.$$

Consequently both $q$ and $\widetilde q^\dagger$ transform as columns in $R$. They can form the scalar doublet

$$h=\begin{pmatrix}q\\ \widetilde q^\dagger\end{pmatrix}$$

of the $SU(2)_R$ symmetry of the conventional theories considered here. The two components of this doublet are still independent scalar variables. Gauge transformations act on their representation indices; $SU(2)_R$ acts on the displayed two-entry index. This is the decomposition in Tachikawa, (2.1.10)–(2.1.13).

For $d_R=\dim_{\mathbb C}R$, the two complex scalars supply $4d_R$ real propagating bosonic modes and the two Weyl fields supply $4d_R$ fermionic modes. Conjugate field entries are not counted a second time:

Massless field collection	Bosonic / fermionic states
One N=2 vector for gauge group $G$	$4\dim G\,/\,4\dim G$
One full hypermultiplet in $R$	$4d_R\,/\,4d_R$

These counts include the particle and antiparticle content of the fields. A massive scalar-seed BPS sector at one fixed nonzero charge has only $2$ bosonic and $2$ fermionic states; its opposite-charge CPT partner completes the full charged hypermultiplet. The eight-state field collection therefore agrees with the previous lesson’s four-state count for one charge sector.

For a pseudoreal representation, an invariant antisymmetric pairing can permit a half-hypermultiplet. This is an additional representation-compatible constraint, not permission to delete one chiral in an arbitrary complex representation. Gauge-anomaly consistency must also be checked. We use full hypers throughout; see Tachikawa, (2.1.19).

The interactions that make the grouping work

Fix one full hypermultiplet, zero theta angle and no Fayet–Iliopoulos term. For $SU(N)$ retain the ordinary fundamental trace $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$ and write $\Phi=\Phi^at^a$. Its canonically normalized Kähler term is

$$K_{\rm vector} =2\operatorname{Tr}_N \left(\Phi^\dagger e^{2gV}\Phi e^{-2gV}\right).$$

The gauge action is $\frac14\int d^2\theta\,\sum_a\mathcal W^{a\alpha}\mathcal W^a_\alpha+\text{h.c.}$, with $\mathcal W_\alpha$ defined in the preceding action lesson. For the hypermultiplet add

$$\begin{aligned} K_{\rm hyper} &=Q^\dagger e^{2gV_R}Q +\widetilde Q e^{-2gV_R}\widetilde Q^\dagger,\\ W&=\sqrt2\,g\,\widetilde Q\Phi_R Q +m\,\widetilde Q Q, \qquad \Phi_R=\Phi^at_R^a. \end{aligned}$$

Here $g>0$ is dimensionless and $m$ is a complex mass of dimension one. For $U(1)$ use the component Kähler term $\Phi^\dagger\Phi$ for the neutral adjoint chiral; a unit-charge hyper has $V_R=V$ and $\Phi_R=\Phi$.

The action is $\int d^4x\,d^4\theta(K_{\rm vector}+K_{\rm hyper})$ plus the gauge and superpotential chiral integrals and their Hermitian conjugates. It contains the canonical covariant derivative $\mathscr D_\mu=\partial_\mu-igA_\mu^at_R^a$. In $\widetilde Q\Phi_RQ$, the transformations cancel: $\widetilde Q U_R^{-1}(U_R\Phi_RU_R^{-1})U_RQ=\widetilde Q\Phi_RQ$.

This is the standard N=2 gauge–hyper interaction of Tachikawa, (2.1.11), (2.1.14)–(2.1.15), expressed in our canonical normalization. His scalar kinetic coefficient is $\operatorname{Tr}|\partial\Phi_T|^2/g^2$; setting $\Phi_T=\sqrt2g\,\phi$ produces $2\operatorname{Tr}|\partial\phi|^2$ and the displayed cubic coefficient. His superfield fermion expansion also has an unconventional factor two in (2.1.5); component fermions should not be identified without that separate translation.

Check the relative coefficient

A field list alone cannot fix the symmetry of an action. Replace $\sqrt2g$ temporarily by an arbitrary cubic coefficient $\kappa$ in the unit-charge Abelian model, set $m=\phi=0$, and eliminate the auxiliaries using the N=1 rules. The potential is

$$U_\kappa=|\kappa|^2|\widetilde q q|^2 +\frac{g^2}{2}(|q|^2-|\widetilde q|^2)^2.$$

An $SU(2)_R$ rotation can take $h=(v,0)$ to $h=(v/\sqrt2,v/\sqrt2)$ for real $v$. The first configuration has potential $g^2v^4/2$ and the second $|\kappa|^2v^4/4$. Equality requires $|\kappa|^2=2g^2$. A constant field-phase choice gives the positive coefficient used above. With it,

$$U_{\rm quartic} =\frac{g^2}{2}(|q|^2+|\widetilde q|^2)^2,$$

which is invariant under the full rotation of $h$.

The fermions give a further check. The N=1 gauge terms and the Hessian of $W$ contain the following unbarred Yukawa terms, with their Hermitian conjugates understood:

$$\begin{aligned} \mathcal L_{\rm Yukawa}={}& i\sqrt2g(q^*\lambda\psi_Q -\widetilde q^*\lambda\psi_{\widetilde Q})\\ &-\sqrt2g(\widetilde q\,\psi_\Phi\psi_Q +q\,\psi_{\widetilde Q}\psi_\Phi). \end{aligned}$$

Use the fixed phase convention $\Lambda=(\lambda,i\psi_\Phi)^T$ and $J=\left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$ for the internal doublet. The same terms become

$$i\sqrt2g\left[(h^\dagger\Lambda)\psi_Q +(h^T J\Lambda)\psi_{\widetilde Q}\right].$$

Both contractions are invariant when $h,\Lambda\mapsto Uh,U\Lambda$ with $U\in SU(2)_R$; the hypermultiplet fermions are singlets. Reordering a contracted pair of odd Weyl fields uses the symmetric scalar contraction already established in the spinor lesson.

These checks explain why the two interaction strengths belong together. The source’s construction obtains the second supersymmetry by rotating the manifest one with this R-symmetry. They do not imply that every abstract N=2 theory must possess an unbroken $SU(2)_R$ symmetry, or that arbitrary interactions of the same fields preserve eight supercharges.

Eliminate the auxiliaries and test a mass

The non-Abelian calculation is compact once each derivative of $W$ is visible. Put $\mathsf M=m\mathbf1+\sqrt2g\,\phi_R$. Then

$$W_{\Phi^a}=\sqrt2g\,\widetilde q\,t_R^a q,\qquad W_{Q}=\widetilde q\,\mathsf M,\qquad W_{\widetilde Q}=\mathsf Mq.$$

The real moment map includes the adjoint and both matter representations:

$$\mu^a=2\operatorname{Tr}_N(t^a[\phi,\phi^\dagger]) +q^\dagger t_R^a q -\widetilde q\,t_R^a\widetilde q^\dagger.$$

The minus sign is precisely the dual-representation generator derived above. With the canonical norms,

$$U=\|\mathsf Mq\|^2+\|\widetilde q\mathsf M\|^2 +2g^2\sum_a|\widetilde q t_R^a q|^2 +\frac{g^2}{2}\sum_a(\mu^a)^2.$$

This follows from $F_I=-\overline{W_I}$ and $D^a=-g\mu^a$, including the adjoint chiral’s $F_\Phi$. Norms use the positive Lorentzian Hermitian metric. The gauge-index sum is explicit; no normalized trace has been silently substituted.

For a worked example, take $G=U(1)$ with unit charges $+1,-1$. Its adjoint scalar is neutral and its commutator vanishes. The complete classical potential reduces to

$$U=|m+\sqrt2g\,\phi|^2(|q|^2+|\widetilde q|^2) +\frac{g^2}{2}(|q|^2+|\widetilde q|^2)^2.$$

Every constant $\phi=a$ with $q=\widetilde q=0$ is a supersymmetric vacuum. Expanding about it gives the same mass to both complex hypermultiplet scalars:

$$M_{\rm hyper}=|m+\sqrt2g\,a|.$$

The fermion term is $-(m+\sqrt2g\,a)\psi_{\widetilde Q}\psi_Q+\text{h.c.}$. The two Weyl fields form a charged Dirac fermion with that mass. Thus the particle/antiparticle collection has four bosonic and four fermionic states. At a nonzero mass each fixed-charge sector is the short multiplet described in the previous lesson. With a common central-charge phase convention, its electric/flavor central charge is $\mathcal Z=m+\sqrt2g\,a$ and it saturates $M=|\mathcal Z|$. Here $a$ is a canonical elementary scalar expectation value; no Seiberg–Witten period normalization is being identified with it. The electric/flavor interpretation is the weak-coupling one of Tachikawa, (2.3.9) and (2.4.10)–(2.4.11).

At $a=-m/(\sqrt2g)$ the hyper is massless, but this alone does not produce a Higgs branch. The extra equation $W_\Phi=\sqrt2g\,\widetilde q q=0$, together with $|q|^2-|\widetilde q|^2=0$, forces both scalars to vanish. The manifestly positive quartic expression confirms this. The N=1 SQED example had only the second condition; adding the adjoint chiral and the N=2 interaction changes the vacuum equations. This conclusion is classical and specific to one full Abelian hyper with zero FI parameter.

What the auxiliary fields establish

The vector has one real $D^a$ and one complex $F_\Phi^a$: three real auxiliaries per generator. Together with four gauge components minus one local gauge freedom and the complex scalar, they give eight off-shell bosonic component functions, matching two unconstrained Weyl fields. This is different from the $4/4$ propagating count.

There is a useful non-Abelian qualification. For the pure vector sector define $\mathcal C=[\phi,\phi^\dagger]$. In our trace normalization its auxiliary terms are

$$\begin{aligned} \mathcal L_{\rm aux,vector} &=\operatorname{Tr}(D^2) +2\operatorname{Tr}(F_\Phi^\dagger F_\Phi) +2g\operatorname{Tr}(D\mathcal C)\\ &=\operatorname{Tr}(D+g\mathcal C)^2 +2\operatorname{Tr}(F_\Phi^\dagger F_\Phi) -g^2\operatorname{Tr}(\mathcal C^2). \end{aligned}$$

The natural shifted real auxiliary is $\widehat D=D+g\mathcal C$. A covariant N=2 auxiliary triplet therefore requires a field-dependent change from the raw N=1 components; it is not obtained merely by renaming $D,\operatorname{Re}F_\Phi,\operatorname{Im}F_\Phi$. Its normalized real quadratic entries are $\widehat D^a,\sqrt2\operatorname{Re}F_\Phi^a,\sqrt2\operatorname{Im}F_\Phi^a$, up to a fixed choice of triplet axes. The same shift is visible in Bilal, (8.7)–(8.8), after allowing for his different generator normalization.

Completing this square does not erase the commutator potential, and setting the shifted auxiliary to zero is not a new vacuum criterion.

For a hypermultiplet, $F_Q,F_{\widetilde Q}$ complete the two N=1 chirals off shell. That finite list guarantees closure for the chosen N=1 subalgebra. A matching count of component functions is not a proof that the second supersymmetry closes without field equations. We have constructed the conventional physical N=2 hypermultiplet and its action, not a full extended off-shell auxiliary formalism. The same caution will apply when more multiplets are assembled into N=4 Yang–Mills.

The Hermitian conjugates, state counts and positive potential above refer to Lorentzian fields. Euclidean continuation treats barred and unbarred variables independently before a contour is chosen; it does not create additional physical particle states.

Exercises

1. A full fundamental hyper of SU(3)

Take $R=\mathbf3$ and one full hyper. Derive the generators acting on the column $\widetilde q^{\,T}$, verify the invariance of $\widetilde q\phi q$, and count all massless bosonic and fermionic states after including the N=2 gauge multiplet. Explain why using the gauge rank would give the wrong answer.

Hint 1

Expand $U=1+i\alpha^at^a+O(\alpha^2)$ in the row transformation before transposing.

Hint 2

The adjoint of SU(3) has dimension eight, while the fundamental has complex dimension three.

Solution

The row variation is $\delta\widetilde q=-i\alpha^a\widetilde q\,t^a$; hence the column variation uses $t_{\overline{\mathbf3}}^a=-(t^a)^T$. With $\phi'=U\phi U^{-1}$, the adjacent inverse matrices cancel in $\widetilde q'\phi' q'$. Two fundamentals without this dual assignment would not give the same contraction.

The vector has $4\times8=32$ bosonic and 32 fermionic states. The full hyper has $4\times3=12$ of each, for a total $44/44$. The rank two counts commuting Cartan generators, not all gauge generators of the unbroken theory. The tilde chiral has already been included in the factor four for the hyper; adding its conjugate as another independent chiral would double-count.

2. Does a vanishing shifted auxiliary diagnose a vacuum?

In pure SU(2), take $t^a=\tau_a/2$ and $\phi=u t^1+iw t^2$ with real $u,w$. Eliminate $D,F_\Phi$, compute $\widehat D$ and the potential, and determine which members of this family are supersymmetric vacua.

Hint 1

Use $[t^1,t^2]=it^3$ to calculate $\mathcal C=[\phi,\phi^\dagger]$.

Hint 2

The raw auxiliary equation and the remaining commutator potential are both present in the completed-square formula.

Solution

The commutator is $\mathcal C=2uw\,t^3$. There is no pure-vector superpotential, so $F_\Phi=0$, while $D=-g\mathcal C=-2guw\,t^3$. Thus $\widehat D=0$ throughout this family after elimination.

Nevertheless,

$$U=g^2\operatorname{Tr}(\mathcal C^2) =g^2(2uw)^2\operatorname{Tr}[(t^3)^2] =2g^2u^2w^2.$$

The ordinary N=1 gaugino variation still tests the raw $D$. With constant scalars and zero field strength, preserving the manifest supersymmetry requires $D=0$, hence $uw=0$. These are the supersymmetric members; when $uw\ne0$ the positive potential already excludes a supersymmetric vacuum. The shift reorganizes the off-shell variables but does not change the action or the vacuum test.

3. Add an adjoint hyper

Let $G=SU(N)$ and take a full adjoint hyper described by two independent canonical adjoint chirals $X,Y$. Translate the cubic and mass superpotentials into ordinary fundamental traces, and count the massless states when the hyper mass vanishes. What supersymmetry does a nonzero hyper mass preserve?

Hint 1

The adjoint action of $\Phi$ on $X$ is $[\Phi,X]$, and the canonical pairing of adjoint components is $2\operatorname{Tr}(YX)$.

Hint 2

The gauge multiplet and the adjoint hyper each supply $4/4$ states per generator. Their N=1 description has one vector and three chirals.

Solution

The interaction and mass are

$$W=2\sqrt2g\,\operatorname{Tr}\!\left(Y[\Phi,X]\right) +2m\,\operatorname{Tr}(YX).$$

The factor two follows from $2\operatorname{Tr}(t^at^b)=\delta^{ab}$; it is needed to match the component contraction in the general formula. All three chiral kinetic terms have coefficient $2\operatorname{Tr}$.

For $m=0$, the field collection contains one gauge field, three complex scalars and four Weyl fermions per generator. The counts are $2+6=8$ bosonic and $4\times2=8$ fermionic states, multiplied by $N^2-1$. With precisely these interactions this is N=4 Yang–Mills, with sixteen real supercharges, even though only four are manifest in the N=1 notation. A mere collection of the same fields with arbitrary couplings would not establish that enhancement.

The adjoint hyper mass preserves N=2 and reduces the full supersymmetry to eight real charges; the mass-deformed theory is commonly called N=2*. It distinguishes $X,Y$ from $\Phi$. These statements and the interaction-based enhancement are discussed in Tachikawa, (2.1.16)–(2.1.18) and the following paragraph. No claim about finite-auxiliary off-shell closure of all sixteen charges follows from the N=1 decomposition.

The representation assignments and the interactions now travel together. The vector’s adjoint scalar supplies both a mass to charged hypers and an additional F equation; an adjoint hyper, with the required cubic coupling, supplies the fields of N=4 Yang–Mills. These are the field-theory starting points for the later four-dimensional dynamics.

Continue to What is N=4 super Yang–Mills? to specify that theory, recover its physical helicities, and distinguish its interacting special-unitary sector from a unitary theory’s free center.

For the pure N=2 theory, What does the Seiberg–Witten solution solve? instead derives the charged-vector threshold and uses it to define the low-energy approximation.

From 4d N=2 to 3d N=4 reduces the Coulomb multiplet on a periodic spatial circle and derives its four real scalar coordinates, including their global angular periods.