The Four-Dimensional N=1 Super-Poincaré Algebra

In four-dimensional Lorentzian spacetime, the minimal super-Poincaré algebra has one left-handed Weyl charge $Q_\alpha$ and its Hermitian adjoint $\bar Q_{\dot\alpha}$. Their only nonzero odd bracket is $\{Q_\alpha,\bar Q_{\dot\beta}\}=2\sigma^\mu_{\alpha\dot\beta}P_\mu$. This normalization makes positive energy, the massless rank reduction, and closure onto translations immediately checkable.

Required background. Graded spacetime symmetry supplies the classification hypotheses and graded Jacobi identity. The dimension-by-dimension reality map explains why one Weyl charge plus its adjoint means four real supercharges.

Helpful background. Grassmann variables and Berezin integration supplies odd parameters, and continuous symmetries, generators, and charges supplies the operator convention for transformations.

Two-component convention

We use the inherited $(+---)$ metric in Lorentzian signature. Undotted indices $\alpha,\beta=1,2$ transform in $(\tfrac12,0)$; dotted indices $\dot\alpha,\dot\beta=1,2$ transform in $(0,\tfrac12)$. Set

$$\begin{aligned} \epsilon^{12}&=+1, & \epsilon_{12}&=-1, & \epsilon^{\dot1\dot2}&=+1, & \epsilon_{\dot1\dot2}&=-1,\\ \psi^\alpha&=\epsilon^{\alpha\beta}\psi_\beta, & \psi_\alpha&=\epsilon_{\alpha\beta}\psi^\beta,\\ \bar\psi^{\dot\alpha}&=\epsilon^{\dot\alpha\dot\beta}\bar\psi_{\dot\beta}, & \bar\psi_{\dot\alpha}&=\epsilon_{\dot\alpha\dot\beta}\bar\psi^{\dot\beta}. \end{aligned}$$

The sigma matrices are

$$\sigma^\mu_{\alpha\dot\alpha}=(\mathbf 1,\boldsymbol\sigma), \qquad \bar\sigma^{\mu\dot\alpha\alpha}=(\mathbf 1,-\boldsymbol\sigma),$$

so that

$$\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu =2\eta^{\mu\nu}\mathbf 1.$$

Spacetime indices are lowered with the metric:

$$\sigma_\mu=\eta_{\mu\nu}\sigma^\nu, \qquad \bar\sigma_\mu=\eta_{\mu\nu}\bar\sigma^\nu.$$

Useful Lorentz generators are

$$\begin{aligned} (\sigma^{\mu\nu})_\alpha{}^\beta &=\frac14(\sigma^\mu\bar\sigma^\nu-\sigma^\nu\bar\sigma^\mu)_\alpha{}^\beta,\\ (\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\beta} &=\frac14(\bar\sigma^\mu\sigma^\nu-\bar\sigma^\nu\sigma^\mu)^{\dot\alpha}{}_{\dot\beta}. \end{aligned}$$

We similarly define $\sigma_{\mu\nu}=\eta_{\mu\rho}\eta_{\nu\sigma}\sigma^{\rho\sigma}$ and $\bar\sigma_{\mu\nu}=\eta_{\mu\rho}\eta_{\nu\sigma}\bar\sigma^{\rho\sigma}$.

No explicit matrix representation beyond the Pauli matrices is used. Complex conjugation maps an undotted spinor to a dotted one; it does not raise an undotted index. The detailed two-component algebra in this convention is the site translation of Weinberg 2000, § 25.2, pp. 29–40, whose book convention uses the opposite metric signature. The invariant round-trip checks are the rest-frame positive anticommutator and closure onto the same translation generator below.

The algebra and adjoints

Take $P_\mu$ and $M_{\mu\nu}=-M_{\nu\mu}$ Hermitian, with

$$\begin{aligned} [M_{\mu\nu},P_\rho] &=i(\eta_{\nu\rho}P_\mu-\eta_{\mu\rho}P_\nu),\\ [M_{\mu\nu},M_{\rho\sigma}] &=i(\eta_{\mu\sigma}M_{\nu\rho} +\eta_{\nu\rho}M_{\mu\sigma} -\eta_{\mu\rho}M_{\nu\sigma} -\eta_{\nu\sigma}M_{\mu\rho}). \end{aligned}$$

The odd generators obey

$$(Q_\alpha)^\dagger=\bar Q_{\dot\alpha}, \qquad (Q^\alpha)^\dagger=\bar Q^{\dot\alpha}, \qquad (\bar Q_{\dot\alpha})^\dagger=Q_\alpha, \qquad (\bar Q^{\dot\alpha})^\dagger=Q^\alpha,$$

so Hermitian conjugation changes chirality while epsilon tensors raise or lower within one chirality. Their brackets are

$$\begin{aligned} \{Q_\alpha,\bar Q_{\dot\beta}\} &=2\sigma^\mu_{\alpha\dot\beta}P_\mu,\\ \{Q_\alpha,Q_\beta\} &=0, & \{\bar Q_{\dot\alpha},\bar Q_{\dot\beta}\} &=0,\\ [P_\mu,Q_\alpha]&=0, & [P_\mu,\bar Q_{\dot\alpha}]&=0, \end{aligned}$$

and Lorentz covariance is expressed by

$$\begin{aligned} [M_{\mu\nu},Q_\alpha] &=i(\sigma_{\mu\nu})_\alpha{}^\beta Q_\beta,\\ [M_{\mu\nu},\bar Q^{\dot\alpha}] &=i(\bar\sigma_{\mu\nu})^{\dot\alpha}{}_{\dot\beta} \bar Q^{\dot\beta},\\ [M_{\mu\nu},\bar Q_{\dot\alpha}] &=-i\bar Q_{\dot\beta} (\bar\sigma_{\mu\nu})^{\dot\beta}{}_{\dot\alpha}. \end{aligned}$$

For $\mathcal N=1$, a scalar central term in $\{Q_\alpha,Q_\beta\}$ would have the form $\epsilon_{\alpha\beta}Z$. The anticommutator is symmetric under exchanging the complete labels, whereas $\epsilon_{\alpha\beta}$ is antisymmetric; with no second supersymmetry index to supply another antisymmetry, $Z$ must vanish. This conclusion is for the minimal point-particle algebra: Lorentz-tensor surface charges in wall or boundary sectors are excluded here. Extended supersymmetry and those tensorial extensions are separated on the next page.

Three Jacobi and automorphism checks

The algebra is compact, but its factors are not decorative.

Why translations commute with $Q$. Before imposing $[P,Q]=0$, Lorentz covariance and the HLS finite-generator setting allow the minimal ansatz

$$[P_\mu,Q_\alpha] =c(\sigma_\mu)_{\alpha\dot\alpha}\bar Q^{\dot\alpha}.$$

Its adjoint fixes the corresponding $[P_\mu,\bar Q]$ term. The ordinary Jacobi identity for $(P_\mu,P_\nu,Q_\alpha)$ then contains

$$|c|^2 (\sigma_\mu\bar\sigma_\nu-\sigma_\nu\bar\sigma_\mu)_\alpha{}^\beta Q_\beta.$$

Translations commute, while this Lorentz tensor is not identically zero, so a nonzero supercharge requires $c=0$. Once $[P,Q]=0$, the $(P_\rho,Q_\alpha,\bar Q_{\dot\beta})$ identity gives

$$[P_\rho,\{Q_\alpha,\bar Q_{\dot\beta}\}]=0.$$

Momentum and scalar central charges pass this test; an $M_{\mu\nu}$ term does not because $[P_\rho,M_{\mu\nu}]\neq0$. The same derivation sequence is presented pedagogically in Tong, n.d., Supersymmetric Field Theory, § 2.2, pp. 23–25, PDF.

Lorentz check. Acting with $M_{\mu\nu}$ on $\{Q_\alpha,\bar Q_{\dot\beta}\}$ uses the undotted action and the lower-dotted action with its minus sign. The required sigma identity is

$$\sigma_{\mu\nu}\sigma_\rho -\sigma_\rho\bar\sigma_{\mu\nu} =\eta_{\nu\rho}\sigma_\mu-\eta_{\mu\rho}\sigma_\nu.$$

For example, $(\mu,\nu,\rho)=(0,1,0)$ gives $\sigma_{01}=\tfrac12\sigma^1$, $\bar\sigma_{01}=-\tfrac12\sigma^1$, and both sides equal $\sigma^1$. Replacing the minus by a plus would make the left side vanish. The complete identity turns the two spinor actions into the vector transformation of $P_\rho$. Same-signature two-component identities and convention translations are cataloged in Dreiner, Haber, and Martin 2022, § 2, Eqs. (2.69)–(2.104).

Internal-symmetry check. The abstract algebra admits the automorphism

$$Q_\alpha\longmapsto e^{-i\varphi}Q_\alpha, \qquad \bar Q_{\dot\alpha}\longmapsto e^{+i\varphi}\bar Q_{\dot\alpha}.$$

With a Hermitian generator $R$ this convention is $[R,Q_\alpha]=-Q_\alpha$ and $[R,\bar Q_{\dot\alpha}]=+\bar Q_{\dot\alpha}$. Its Jacobi check is explicit:

$$[R,\{Q_\alpha,\bar Q_{\dot\beta}\}] =\{[R,Q_\alpha],\bar Q_{\dot\beta}\} +\{Q_\alpha,[R,\bar Q_{\dot\beta}]\}=0.$$

An ordinary flavor or internal generator $T$ instead satisfies $[T,Q_\alpha]=0$; a generator acting nontrivially on $Q$ is an R-generator. Whether the $U(1)_R$ automorphism is a symmetry of a Lagrangian, survives anomalies, or is preserved by the vacuum is a separate dynamical question.

Positivity and the spectrum condition

For any commuting test spinor $z^\alpha$, define $A=z^\alpha Q_\alpha$. Then

$$\{A,A^\dagger\} =2z^\alpha\sigma^\mu_{\alpha\dot\beta} \bar z^{\dot\beta}P_\mu.$$

On a momentum eigenstate this is nonnegative because $\|A|\Psi\rangle\|^2+\|A^\dagger|\Psi\rangle\|^2\geq0$. Since this holds for every $z$, the Hermitian matrix $\sigma^\mu p_\mu$ is positive semidefinite. For $p^\mu=(p^0,\mathbf p)$ its eigenvalues are

$$p^0+|\mathbf p|, \qquad p^0-|\mathbf p|.$$

Both must be nonnegative, so the momentum lies in the future causal cone:

$$p^0\geq|\mathbf p|\geq0.$$

This derives the spectrum condition from positivity of the superalgebra within a unitary representation. In a massive rest frame $P_\mu=(m,\mathbf0)$,

$$\{Q_\alpha,Q_\beta^\dagger\}=2m\,\delta_{\alpha\beta}.$$

For a nonzero future-null momentum the $2\times2$ matrix $\sigma\cdot P$ has rank one, so half the complex charge components act trivially in an irreducible massless representation. At $p^\mu=0$ its rank is zero. These are the seeds of the massive and massless oscillator constructions on unitary supermultiplets.

Taking the trace at arbitrary momentum gives the operator identity

$$H=P_0=\frac14\sum_{\alpha=1}^2 \{Q_\alpha,Q_\alpha^\dagger\}\geq0.$$

If a normalizable state has zero energy, every term in this sum has zero expectation value, hence both $Q_\alpha$ and $Q_\alpha^\dagger$ annihilate it. The converse is immediate. This is an algebraic statement about a positive representation; it does not by itself decide whether a theory possesses such a vacuum.

Closure onto a spacetime translation

A quick field-level round trip checks the normalization. Fix

$$\delta_\xi=\xi Q+\bar\xi\bar Q, \qquad P_\mu=i\partial_\mu, \qquad [\delta_\xi,\delta_\eta] =\delta_\xi\delta_\eta-\delta_\eta\delta_\xi.$$

For a chiral multiplet $(A,\psi_\alpha,F)$, choose

$$\begin{aligned} \delta_\xi A&=\sqrt2\,\xi\psi,\\ \delta_\xi\psi_\alpha &=i\sqrt2(\sigma^\mu\bar\xi)_\alpha\partial_\mu A +\sqrt2\,\xi_\alpha F. \end{aligned}$$

For anticommuting parameters, the two ordered actions are

$$\begin{aligned} \delta_\xi\delta_\eta A &=2i\,\eta\sigma^\mu\bar\xi\,\partial_\mu A +2\eta\xi F,\\ \delta_\eta\delta_\xi A &=2i\,\xi\sigma^\mu\bar\eta\,\partial_\mu A +2\xi\eta F. \end{aligned}$$

The contracted odd spinors satisfy $\eta\xi=\xi\eta$, so the auxiliary terms cancel and

$$[\delta_\xi,\delta_\eta]A =a^\mu\partial_\mu A, \qquad a^\mu=2i(\eta\sigma^\mu\bar\xi-\xi\sigma^\mu\bar\eta).$$

Reversing the definition of the transformation commutator reverses $a^\mu$; the ordered calculation is what fixes the sign. Closure on every component, including the distinction between off-shell and on-shell closure, is developed on component multiplets and closure. Here the calculation verifies that the operator algebra and field convention generate the same translation. Wess and Zumino developed detailed linearly realized four-dimensional multiplets in Wess and Zumino 1974, pp. 39–50 and a renormalizable interacting model in Wess and Zumino 1974, pp. 52–54.

Common pitfalls

Mixing metric conventions inside a sigma identity. With $(+---)$, the site uses $\sigma^\mu=(1,\boldsymbol\sigma)$ and $\bar\sigma^\mu=(1,-\boldsymbol\sigma)$ so their symmetrized product is $+2\eta^{\mu\nu}$. Importing a $(-+++)$ identity without its compensating sign breaks positivity or Lorentz closure.

Treating dotted indices as raised undotted indices. Dotted and undotted indices label inequivalent complex Lorentz representations. Epsilon tensors raise indices within one representation; Hermitian conjugation changes undotted to dotted.

Calling the $U(1)_R$ automorphism an exact quantum symmetry. The algebra admits it. An action, regulator, anomaly, coupling, or vacuum can preserve only a subgroup or none of it.

Exercises

1. Diagnose the Lorentz sign

Evaluate the sigma identity for $(\mu,\nu,\rho)=(0,1,0)$ and derive the lower-dotted Lorentz action from the upper-dotted one.

Solution

Lowering spacetime indices gives

$$\sigma_{01}=\frac12\sigma^1, \qquad \bar\sigma_{01}=-\frac12\sigma^1, \qquad \sigma_0=\mathbf1, \qquad \sigma_1=-\sigma^1.$$

Hence

$$\sigma_{01}\sigma_0-\sigma_0\bar\sigma_{01} =\sigma^1 =\eta_{10}\sigma_0-\eta_{00}\sigma_1.$$

Lowering $\dot\alpha$ with $\epsilon_{\dot\alpha\dot\beta}$ moves the dotted representation matrix to the right and gives $[M_{\mu\nu},\bar Q_{\dot\alpha}] =-i\bar Q_{\dot\beta} (\bar\sigma_{\mu\nu})^{\dot\beta}{}_{\dot\alpha}$.

2. Derive the future cone

Diagonalize $\sigma^\mu p_\mu$, derive the spectrum condition, and identify its ranks for timelike, nonzero null, and zero momentum.

Solution

With $p_\mu=(p^0,-\mathbf p)$,

$$\sigma^\mu p_\mu=p^0\mathbf1-\mathbf p\cdot\boldsymbol\sigma$$

has eigenvalues $p^0\pm|\mathbf p|$. Positivity for every test spinor requires both to be nonnegative, hence $p^0\geq|\mathbf p|$. The matrix has rank two for future timelike momentum, rank one for nonzero future-null momentum, and rank zero at $p=0$. Finally, $4H=\sum_\alpha\{Q_\alpha,Q_\alpha^\dagger\}$, so zero energy makes every associated norm vanish.

3. Track the ordered closure sign

Compute $\delta_\xi\delta_\eta A$ and $\delta_\eta\delta_\xi A$ separately. Why does $F$ cancel, and how would reversing the commutator definition change the answer?

Solution

The two ordered actions are

$$2i\,\eta\sigma^\mu\bar\xi\,\partial_\mu A+2\eta\xi F, \qquad 2i\,\xi\sigma^\mu\bar\eta\,\partial_\mu A+2\xi\eta F.$$

The contraction of two odd Weyl spinors is symmetric, $\eta\xi=\xi\eta$, so the $F$ terms cancel in the difference. The remaining vector is $2i(\eta\sigma^\mu\bar\xi-\xi\sigma^\mu\bar\eta)$. Defining the commutator in the reverse order multiplies it by $-1$.

4. Flavor, R-symmetry, and a scalar central term

Explain why a flavor generator and an R-generator act differently on $Q$, and why $\mathcal N=1$ has no scalar point-particle central charge.

Solution

A flavor generator satisfies $[T,Q_\alpha]=0$; an R-generator rotates the supercharge, here $[R,Q_\alpha]=-Q_\alpha$. The mixed bracket remains neutral because $\bar Q$ has the opposite R-charge. A scalar same-chirality term would be $\epsilon_{\alpha\beta}Z$. Symmetry of the anticommutator conflicts with antisymmetry of $\epsilon_{\alpha\beta}$ unless a second antisymmetric supersymmetry tensor is present, so the $\mathcal N=1$ scalar coefficient vanishes.

References

• Herbi K. Dreiner, Howard E. Haber, and Stephen P. Martin, “Two-Component Spinor Techniques and Feynman Rules for Quantum Field Theory and Supersymmetry,” Physics Reports 494 (2010), 1–196, corrected version 6 (2022), DOI, arXiv.
• David Tong, Supersymmetric Field Theory, University of Cambridge Part III lecture notes, n.d., §§ 2.2–2.2.2, PDF.
• Steven Weinberg, The Quantum Theory of Fields, Volume III: Supersymmetry, Cambridge University Press (2000), § 25.2, DOI.
• Julius Wess and Bruno Zumino, “A Lagrangian Model Invariant under Supergauge Transformations,” Physics Letters B 49 (1974), 52–54, DOI.
• Julius Wess and Bruno Zumino, “Supergauge Transformations in Four Dimensions,” Nuclear Physics B 70 (1974), 39–50, DOI.