Extended algebras, R-symmetry, and BPS representations

The mass of an ordinary particle is a parameter of its representation. Supersymmetry can constrain it more strongly: in a sector carrying a central charge, positivity places a lower bound on the mass. At the bound, some supercharges act as zero and the particle has fewer superpartners. We will derive both statements by diagonalizing a small matrix.

Continue from four-dimensional interactions and vacua. We use the Lorentzian spinors of the four-dimensional algebra lesson: metric $\eta=\operatorname{diag}(+---)$, $\sigma^\mu=(I,-\tau_i)$, $\bar\sigma^\mu=(I,+\tau_i)$, and $\epsilon_{12}=-1$. Here $\tau_i$ are the Pauli matrices. The calculation concerns unitary one-particle representations in flat space, with unbroken supersymmetry in the vacuum. No particular gauge group, gauge rank, or Lagrangian is assumed.

More supercharges, with a possible new charge

Extended supersymmetry has several Weyl generators $Q^I_\alpha$, where $I=1,\ldots,\mathcal N$ labels supersymmetries and $\alpha=1,2$ is the spinor index. Their adjoints are $\bar Q_{\dot\alpha I}=(Q^I_\alpha)^\dagger$. The full algebra has $4\mathcal N$ real Poincare supercharges: eight for $\mathcal N=2$ and sixteen for $\mathcal N=4$. Choosing $I=1$ selects a manifest $\mathcal N=1$ subalgebra; it does not remove the other generators.

For particle representations with Lorentz-scalar central charges, write

$$\begin{aligned} \{Q^I_\alpha,\bar Q_{\dot\beta J}\} &=2\delta^I{}_J\sigma^\mu_{\alpha\dot\beta}P_\mu,\\ \{Q^I_\alpha,Q^J_\beta\} &=2\epsilon_{\alpha\beta}\mathcal Z^{IJ}, \qquad \mathcal Z^{IJ}=-\mathcal Z^{JI}. \end{aligned}$$

The adjoint relation fixes the barred-barred anticommutator. The antisymmetry in $I,J$ is required: exchanging the two complete charge labels leaves an anticommutator unchanged but reverses $\epsilon_{\alpha\beta}$. Thus no such scalar central charge exists for $\mathcal N=1$. Tensorial extensions associated with extended objects, such as walls, are outside this particle algebra.

“Central” means that $\mathcal Z^{IJ}$ commutes with translations, Lorentz generators, and supercharges. In an irreducible sector it is represented by a fixed complex antisymmetric matrix. It can encode conserved electric, magnetic, or flavor charges multiplied by vacuum-dependent coefficients. It is not another momentum component in this four-dimensional calculation. The dimensions are $[Q]=1/2$, $[P]=1$, and $[\mathcal Z]=1$.

This is Bilal’s algebra (3.2)–(3.3), with $Z_{\rm BIL}^{IJ}=2\mathcal Z^{IJ}$. The factor of two will matter in the bound.

R-symmetry and changes of basis

A unitary change $Q^I\mapsto U^I{}_JQ^J$ preserves the first anticommutator. The second transforms by

$$\mathcal Z\longmapsto U\mathcal Z U^T.$$

These $U(\mathcal N)$ transformations describe automorphisms of the algebra, allowing the central generators to transform too. A fixed central-charge matrix is preserved only by the subgroup satisfying $U\mathcal ZU^T=\mathcal Z$. This distinction prevents a basis choice from being mistaken for a symmetry that acts within one charge sector.

For $\mathcal N=2$, introduce a separate internal antisymmetric tensor

$$J=\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad \mathcal Z^{IJ}=J^{IJ}\mathcal Z.$$

The repeated symbol $\mathcal Z$ without indices now denotes one complex number. Since $UJU^T=(\det U)J$, the determinant-one rotations preserve a fixed nonzero $\mathcal Z$. A common phase of the supercharges instead rotates $\mathcal Z$ by twice that phase. The adjoint charges must transform by the conjugate phase.

An R-symmetry of a theory is an actual internal symmetry that rotates its supercharges. It must preserve the dynamics and survive any relevant quantum anomaly. The algebra’s possible automorphisms alone do not prove that every theory realizes them. In particular, the usual $SU(2)_R$ structure of many $\mathcal N=2$ models is additional information about the model; Tachikawa explicitly distinguishes it from the mere existence of two supersymmetries. A flavor symmetry instead commutes with the supercharges.

Diagonalize the positive anticommutator

Take a massive particle at rest, $P_\mu=(M,0,0,0)$ with $M>0$, in a fixed $\mathcal N=2$ central-charge sector. Dotted and undotted components can be compared using the rest-frame matrix $\sigma^0=I$. Set

$$X_\alpha=Q^1_\alpha,\qquad Y_\alpha=\sum_{\beta=1}^2\epsilon_{\alpha\beta} (Q^2_\beta)^\dagger, \qquad \epsilon=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

This is an explicit matrix acting on the column of adjoint lower-component operators; no unshown index raising is intended. Because $\epsilon\epsilon^T=I$, direct substitution gives

$$\{X_\alpha,X_\beta^\dagger\} =\{Y_\alpha,Y_\beta^\dagger\}=2M\delta_{\alpha\beta}, \qquad \{X_\alpha,Y_\beta^\dagger\}=2\mathcal Z\delta_{\alpha\beta}.$$

For each spinor component, the positive-norm matrix is therefore

$$\mathsf G=2 \begin{pmatrix} M&\mathcal Z\\ \mathcal Z^*&M \end{pmatrix}.$$

Why must it be positive? For any operator $A$ that is a linear combination of the charges and any state $|v\rangle$,

$$\langle v|\{A,A^\dagger\}|v\rangle =\|A|v\rangle\|^2+\|A^\dagger|v\rangle\|^2\ge0.$$

Write $\mathcal Z=\rho e^{i\varphi}$ with $\rho=|\mathcal Z|$. For $\rho=0$ the phase is arbitrary. The combinations

$$S_\alpha=\frac{X_\alpha-e^{i\varphi}Y_\alpha}{\sqrt2}, \qquad L_\alpha=\frac{X_\alpha+e^{i\varphi}Y_\alpha}{\sqrt2}$$

give

$$\begin{aligned} \{S_\alpha,S_\beta^\dagger\} &=2(M-\rho)\delta_{\alpha\beta},\\ \{L_\alpha,L_\beta^\dagger\} &=2(M+\rho)\delta_{\alpha\beta},\\ \{S_\alpha,L_\beta^\dagger\}&=0. \end{aligned}$$

The unbarred oscillator anticommutators also vanish. For example, $\{X,Y\}=0$ follows from the different internal indices in $\{Q,\bar Q\}$. Positivity of the first diagonal entry requires

$$\boxed{M\ge|\mathcal Z|.}$$

This is the particle Bogomolny–Prasad–Sommerfield bound, or BPS bound. We obtained it without a soliton ansatz or an equation of motion.

For a concrete check, take $\mathcal Z=(3+4i)\mu$ with $\mu>0$. Its magnitude is $5\mu$. At $M=6\mu$ the eigenvalues are $2\mu$ and $22\mu$, each repeated for the two spinor components. At $M=5\mu$ they become $0$ and $20\mu$. At $M=4\mu$ one eigenvalue would be $-2\mu$; no positive Hilbert-space representation of this algebra can have that mass.

What disappears at saturation

When $M>\rho$, normalize

$$a_\alpha=\frac{S_\alpha}{\sqrt{2(M-\rho)}}, \qquad b_\alpha=\frac{L_\alpha}{\sqrt{2(M+\rho)}}.$$

There are four independent fermionic annihilation operators, with $\{a_\alpha,a_\beta^\dagger\}=\{b_\alpha,b_\beta^\dagger\}=\delta_{\alpha\beta}$. Each creation operator can be used zero or one time.

Start with a state annihilated by all four. This is a Clifford vacuum, the seed for a particle multiplet; it is not the zero-energy vacuum of the field theory. For a spin-zero bosonic seed, occupation numbers give $2^4=16$ states. Eight have even occupation number and eight have odd occupation number.

At $M=\rho>0$, division by $\sqrt{M-\rho}$ is invalid. Instead apply the norm identity directly:

$$0=\|S_\alpha|v\rangle\|^2+ \|S_\alpha^\dagger|v\rangle\|^2.$$

Both operators act as zero on every state in this irreducible sector. Only the two $b_\alpha$ oscillators survive. The resulting four states are

$$|\Omega\rangle,\qquad b_1^\dagger|\Omega\rangle,\qquad b_2^\dagger|\Omega\rangle,\qquad b_1^\dagger b_2^\dagger|\Omega\rangle.$$

The creation operators form a spin-$1/2$ doublet under rest-frame rotations. Their antisymmetric two-particle product is a spin singlet. Thus a bosonic spin-zero seed gives two spin-zero bosonic states and one spin-$1/2$ doublet: two bosonic and two fermionic states at fixed charge and momentum.

There are two complex null $S_\alpha$, or four real combinations of supercharges that annihilate the multiplet. Four of the original eight are preserved: this is a half-BPS massive representation. The vacuum still preserves the full algebra; the fraction refers to the charged particle state.

The contrast is visible in the long multiplet’s rotation content. Its four creation operators are two spinor doublets. At occupation two their exterior square is

$$\bigwedge\nolimits^2(\mathbf2\oplus\mathbf2) =\mathbf1\oplus(\mathbf1\oplus\mathbf3)\oplus\mathbf1.$$

Including occupation zero and four gives five scalar states and one spin-one triplet, totaling eight bosonic states. Occupations one and three give four spinor doublets, totaling eight fermionic states. “Sixteen states” counts polarizations, not sixteen independent fields.

Charge sectors and antiparticles

The counting above concerns one irreducible sector. If it carries nonzero electric, magnetic, or flavor charge, a CPT-invariant theory also contains the antiparticle sector with opposite charges. In the usual additive-charge convention its central charge changes sign. The two sectors are distinct even though their masses agree.

Consequently the four-state short multiplet with a scalar seed and its charged CPT partner together have four bosonic and four fermionic states. This is the familiar count for a massive charged full hypermultiplet. It does not contradict the four-state answer obtained before CPT completion. A field and its antiparticle were already included in the propagating-field counts of the multiplet comparison.

More generally a Clifford seed of spin $j_0$ supplies $2j_0+1$ states before the creation operators act. Multiply the oscillator count by that number. Spin addition, rather than just the total number, determines whether the result contains a scalar or a vector; exercise 2 makes this distinction explicit.

Translate the normalization before using a mass formula

Tachikawa’s (2.3.1)–(2.3.8) uses $\{Q_{\rm T},Q_{\rm T}^\dagger\}=\sigma\cdot P$, without our factor two. After aligning the antisymmetric-tensor orientations, the translation is

$$Q_{\rm here}=\sqrt2\,Q_{\rm T},\qquad \mathcal Z=Z_{\rm T}.$$

Both anticommutators scale together, so both conventions give $M\ge|Z_{\rm T}|$. Bilal instead keeps our momentum normalization and calls the coefficient of $\epsilon_{\alpha\beta}$ itself $Z_{\rm BIL}^{12}=2\mathcal Z$; his (3.16) reads $2M\ge|Z_{\rm BIL}^{12}|$. These are the same inequality.

The later Seiberg–Witten period convention on this site uses a scalar $Z_\gamma$ with $M\ge\sqrt2|Z_\gamma|$. In that convention $\mathcal Z=\sqrt2 Z_\gamma$, after matching the overall phase. This statement alone does not identify either source’s periods or integer electric-charge unit. Those need their own basis and action normalization. The mass inequality is a useful check on that dictionary.

More central-charge blocks

For general $\mathcal N$, a unitary change of supercharge basis puts the complex antisymmetric matrix $\mathcal Z^{IJ}$ into two-by-two blocks $\rho_r J$ with $\rho_r\ge0$, and a zero row and column if $\mathcal N$ is odd. These $\rho_r$ are its skew singular values; each is the magnitude appearing in one copy of the calculation above. This standard matrix normal form is used in Bilal, (3.13)–(3.16).

Every block requires $M\ge\rho_r$, hence

$$M\ge\max_r\rho_r.$$

If $k$ blocks saturate at positive $M$, two complex oscillators per saturated block vanish. There remain $2\mathcal N-2k$ annihilation operators. For a spin-zero seed,

$$\text{states}=2^{\,2\mathcal N-2k}, \qquad \text{preserved real supercharges}=4k.$$

For $\mathcal N=4$ this gives $256$ states when no block saturates, $64$ for one saturated block, and $16$ for two. The latter fractions are respectively one-quarter and one-half of its sixteen real supercharges. These are massive particle representations; they are not counts of the elementary massless $\mathcal N=4$ vector fields. The rest-frame derivation does not apply at $M=0$.

Finally, the algebra gives a bound on a state if that state exists. It does not construct a particle of every allowed charge or prove its stability. Even a BPS mass can vary as the central charge varies with the vacuum. Short multiplets can combine into longer ones, or a discrete particle can encounter a continuum. For two proposed BPS constituents, the triangle inequality gives $|\mathcal Z_1+\mathcal Z_2|\le|\mathcal Z_1|+|\mathcal Z_2|$; aligned phases make equality possible. That threshold condition by itself does not establish a bound state or a decay.

Exercises

1. Follow the central-charge phase

Take $\mathcal Z=(3+4i)\mu$, $M=5\mu$, with $\mu>0$. Write the two null operators $S_\alpha$ explicitly in terms of the lower components of $Q^1,Q^2$. Then change only the mass to $6\mu$: what norms and how many states follow for a scalar seed?

First hint

Use $e^{i\varphi}=(3+4i)/5$ and the explicit matrix $\epsilon$, rather than suppressing its indices.

Second hint

At the bound, a zero anticommutator forces both an operator and its adjoint to act as zero. Above the bound, normalize both doublets.

Solution

Writing $u=(3+4i)/5$, the two combinations are

$$S_1=\frac{Q^1_1+u(Q^2_2)^\dagger}{\sqrt2}, \qquad S_2=\frac{Q^1_2-u(Q^2_1)^\dagger}{\sqrt2}.$$

The relative signs follow from $\epsilon_{12}=-1$, $\epsilon_{21}=1$. Their anticommutators vanish at $M=5\mu$, while $\{L_\alpha,L_\beta^\dagger\}=20\mu\,\delta_{\alpha\beta}$. There are two active oscillators and four states.

At $M=6\mu$, $\{S,S^\dagger\}=2\mu I$ and $\{L,L^\dagger\}=22\mu I$. Four active oscillators give sixteen states, eight bosonic and eight fermionic, in the fixed-charge scalar-seed representation. An antiparticle sector, if distinct, is additional to these counts.

2. Change the seed spin

At a massive $\mathcal N=2$ BPS bound, use a spin-$1/2$ fermionic Clifford seed instead of a scalar seed. Find the spin and statistics content at occupation numbers zero, one, and two. Count the polarizations, first at fixed nonzero charge and then including the distinct CPT partner.

First hint

The oscillator Fock levels have spins $0,1/2,0$ and multiplicities $1,2,1$. Tensor each level with the seed doublet.

Second hint

Two spinor doublets decompose into a symmetric spin-one triplet and an antisymmetric scalar. One fermionic creation operator reverses the seed’s parity.

Solution

Occupation zero gives a fermionic spin-$1/2$ doublet. Occupation one is bosonic and gives $\mathbf2\otimes\mathbf2=\mathbf3\oplus\mathbf1$: one massive spin-one particle with three polarizations and one scalar. Occupation two is again a fermionic doublet, since the two creation operators together are a spin singlet.

Thus the fixed-charge sector contains four bosonic and four fermionic polarizations, totaling $4(2j_0+1)=8$. With a distinct oppositely charged CPT partner, it has eight bosonic and eight fermionic polarizations in total. The larger count comes from the seed spin and charge completion, not from restoring any of the null supercharges.

3. One or two saturated blocks

For a massive $\mathcal N=4$ sector let $\rho_1=5\mu$, $\rho_2=2\mu$, and $M=5\mu$. Find the preserved fraction and scalar-seed dimension. Repeat after changing only $\rho_2$ to $5\mu$. Could either sector instead have $M=4\mu$?

First hint

Test the mass against each block separately. Count saturated blocks, not just the number of nonzero central charges.

Second hint

Each saturated block removes two complex oscillators and supplies four real annihilating supercharges.

Solution

Initially only the first block saturates: $k=1$. Four of sixteen real supercharges annihilate the states, giving one-quarter BPS. Six complex oscillators survive, so the scalar-seed representation has $2^6=64$ states.

When $\rho_2=5\mu$, both blocks saturate. Then $k=2$, eight real supercharges annihilate the states, and four complex oscillators give $2^4=16$ states: one-half BPS. Both answers concern one charge sector.

At $M=4\mu$, the first block has norm eigenvalue $2(M-\rho_1)=-2\mu$. Positivity excludes that mass in either case. The stronger block determines the bound even when another block would separately permit a smaller mass.

We can now determine a particle multiplet by fixing its charge sector, diagonalizing its supercharge norms, and adding the spin of its Clifford seed. N=2 vector multiplets and hypermultiplets organizes elementary fields into the $\mathcal N=1$ packages already used for actions, while keeping those field counts distinct from this massive-particle calculation.