Superspace and Supertranslations

Superspace realizes supersymmetry as ordinary geometry on a space with commuting coordinates $x^\mu$ and anticommuting coordinates $\theta^\alpha,\bar\theta^{\dot\alpha}$. The mixed term in a finite supertranslation is forced by the anticommutator $\{Q_\alpha,\bar Q_{\dot\alpha}\}\propto P_\mu$: two fermionic translations differ by a spacetime translation. Deriving that group law fixes the differential generators and prevents the sign table for $Q$, $\bar Q$, $D$, and $\bar D$ from becoming a collection of unrelated conventions.

Required background. The Four-Dimensional N=1 Super-Poincaré Algebra supplies the algebra. Graded Algebra, Grassmann Variables, and Berezin Integration supplies left differentiation and the graded product rule.

Helpful background. Lie Groups, Lie Algebras, the Exponential Map, and the Adjoint Action supplies the Baker–Campbell–Hausdorff viewpoint used for the finite composition law.

Supertranslation coordinates and their dimensions

Work in four-dimensional Lorentzian spacetime with

$$\sigma^\mu=(\mathbf1,\boldsymbol\sigma), \qquad \bar\sigma^\mu=(\mathbf1,-\boldsymbol\sigma), \qquad P_\mu=i\partial_\mu.$$

The superspace point is

$$z^M=(x^\mu,\theta^\alpha,\bar\theta^{\dot\alpha}),$$

with Grassmann parities

$$\lvert x^\mu\rvert=0, \qquad \lvert\theta^\alpha\rvert =\lvert\bar\theta^{\dot\alpha}\rvert=1.$$

Because $[P_\mu]=1$ and $[Q_\alpha]=1/2$,

$$[x^\mu]=-1, \qquad [\theta^\alpha]=[\bar\theta^{\dot\alpha}]=-\frac12.$$

Barred coordinates are Lorentzian conjugates of unbarred coordinates on real Minkowski superspace. They remain independent generators while doing Grassmann algebra; conjugation reverses the order of odd factors.

Throughout this page the odd derivatives act from the left. For homogeneous superfields $F$ and $G$,

$$\begin{aligned} \partial_\alpha(FG) &=(\partial_\alpha F)G +(-1)^{\lvert F\rvert}F(\partial_\alpha G),\\ \bar\partial_{\dot\alpha}(FG) &=(\bar\partial_{\dot\alpha}F)G +(-1)^{\lvert F\rvert}F(\bar\partial_{\dot\alpha}G), \end{aligned}$$

where $\lvert F\rvert\in\{0,1\}$ is the Grassmann parity. In particular,

$$\partial_\alpha(\theta^\beta F) =\delta_\alpha{}^\beta F -\theta^\beta\partial_\alpha F.$$

The minus sign comes from moving the odd derivative past the odd factor $\theta^\beta$; it occurs whether $F$ is even or odd. By contrast, the ordinary spacetime derivative is even and obeys the ungraded product rule. Switching the side of the odd derivatives changes a package of signs, not just this example.

The finite composition law

Choose the active left action of a supertranslation $(a,\epsilon,\bar\epsilon)$ on a point to be

$$\begin{aligned} \theta'&=\theta+\epsilon,\\ \bar\theta'&=\bar\theta+\bar\epsilon,\\ x'^\mu &=x^\mu+a^\mu -i\epsilon\sigma^\mu\bar\theta +i\theta\sigma^\mu\bar\epsilon. \end{aligned}$$

The mixed shift is bilinear because an odd parameter cannot shift an even coordinate by itself. Apply first $(a_1,\epsilon_1,\bar\epsilon_1)$ and then $(a_2,\epsilon_2,\bar\epsilon_2)$. Direct substitution gives

$$\begin{aligned} \epsilon_{21}&=\epsilon_1+\epsilon_2,\\ \bar\epsilon_{21}&=\bar\epsilon_1+\bar\epsilon_2,\\ a_{21}^\mu &=a_1^\mu+a_2^\mu +i\left( \epsilon_1\sigma^\mu\bar\epsilon_2 -\epsilon_2\sigma^\mu\bar\epsilon_1 \right). \end{aligned}$$

Reversing the two transformations changes the sign of the bilinear term. Subtracting the 1-then-2 ordering from the 2-then-1 ordering, as required by the field convention $[\delta_1,\delta_2]=\delta_1\delta_2-\delta_2\delta_1$, gives the translation parameter

$$\xi^\mu =2i\left( \epsilon_2\sigma^\mu\bar\epsilon_1 -\epsilon_1\sigma^\mu\bar\epsilon_2 \right).$$

This is the finite-coordinate form of

$$\{Q_\alpha,\bar Q_{\dot\alpha}\} =2\sigma^\mu_{\alpha\dot\alpha}P_\mu.$$

The truncation of the Baker–Campbell–Hausdorff series is exact: the odd-odd bracket is the central translation within the supertranslation subgroup, and further nested brackets vanish. Salam and Strathdee introduced this systematic superfield realization in Salam and Strathdee 1974, pp. 477–482. In four-component notation, Weinberg 2000, § 26.2, pp. 60–61 constructs the corresponding infinitesimal superspace differential generator and checks its algebra; the finite law above follows by composing the declared coordinate action.

Differential supercharges from the coordinate action

Let $\mathscr F(x,\theta,\bar\theta)$ be a scalar superfield. Expanding its pullback under the infinitesimal transformation at fixed original coordinates yields

$$\delta\mathscr F = \left( \epsilon^\alpha Q_\alpha +\bar\epsilon_{\dot\alpha}\bar Q^{\dot\alpha} +a^\mu\partial_\mu \right)\mathscr F,$$

with

$$\begin{aligned} Q_\alpha &= \frac{\partial}{\partial\theta^\alpha} -i\sigma^\mu_{\alpha\dot\alpha} \bar\theta^{\dot\alpha}\partial_\mu,\\ \bar Q_{\dot\alpha} &= -\frac{\partial}{\partial\bar\theta^{\dot\alpha}} +i\theta^\alpha\sigma^\mu_{\alpha\dot\alpha}\partial_\mu. \end{aligned}$$

The minus sign in the leading term of $\bar Q$ implements Lorentzian conjugation together with left differentiation. It does not reverse the desired shift of the barred coordinate. Raising the operator index with $\epsilon^{\dot\alpha\dot\beta}$ gives

$$\bar\epsilon_{\dot\alpha} \bar Q^{\dot\alpha}\bar\theta^{\dot\beta} =-\bar\epsilon_{\dot\alpha} \epsilon^{\dot\alpha\dot\beta} =+\bar\epsilon^{\dot\beta}.$$

The final plus sign uses $\bar\epsilon_{\dot\alpha}\epsilon^{\dot\alpha\dot\beta} =-\bar\epsilon^{\dot\beta}$. Thus the leading minus in $\bar Q_{\dot\alpha}$ and the antisymmetry of the spinor metric are both required by the active coordinate shift $\delta\bar\theta=\bar\epsilon$. Acting on all three coordinates checks the construction:

$$\begin{array}{c|ccc} &x^\mu&\theta^\beta&\bar\theta^{\dot\beta}\\ \hline Q_\alpha& -i\sigma^\mu_{\alpha\dot\gamma}\bar\theta^{\dot\gamma}& \delta_\alpha{}^\beta&0\\ \bar Q_{\dot\alpha}& +i\theta^\gamma\sigma^\mu_{\gamma\dot\alpha}& 0&-\delta_{\dot\alpha}{}^{\dot\beta} \end{array}$$

Using the graded product rule gives

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

Terms containing $\theta\bar\theta\,\partial_\mu\partial_\nu$ cancel because their Grassmann coefficient is antisymmetric while the spacetime derivatives commute. This explicit check is more reliable than importing one generator from a source with a different $P_\mu$ convention.

Left and right actions

The supertranslation group acts on itself from both sides. Infinitesimal left-invariant vector fields generate one copy of the algebra; right-invariant vector fields commute with that action and become the supercovariant derivatives on the next page. Calling both sets “fermionic translations” without specifying the side obscures their crucial relative sign.

For a field transformation there is also an active-versus-passive choice. This page uses an active coordinate action and the induced pullback written above. A passive coordinate change reverses the sign in $\delta\mathscr F$ while leaving the invariant algebra unchanged. A convention translation must change the coordinate shift and differential operators together.

Invariant one-forms

Exterior calculus on a supermanifold needs one more sign convention. We use a bigrading by form degree $p$ and Grassmann parity $\lvert\cdot\rvert$, with $\mathrm d$ raising form degree and preserving Grassmann parity. Homogeneous forms obey

$$\alpha_p\wedge\beta_q =(-1)^{pq+\lvert\alpha\rvert\lvert\beta\rvert} \beta_q\wedge\alpha_p, \qquad \mathrm d(\alpha_p\wedge\beta) =\mathrm d\alpha_p\wedge\beta +(-1)^p\alpha_p\wedge\mathrm d\beta.$$

Consequently,

$$\mathrm dz^M\wedge\mathrm dz^N =-(-1)^{\lvert z^M\rvert\lvert z^N\rvert} \mathrm dz^N\wedge\mathrm dz^M.$$

In particular, two $\mathrm dx$ forms anticommute, whereas $\mathrm d\theta$ and $\mathrm d\bar\theta$ commute under the wedge product. This last fact prevents the flat-superspace torsion below from canceling by mistake.

The one-forms

$$E^\alpha=\mathrm d\theta^\alpha, \qquad \bar E^{\dot\alpha}=\mathrm d\bar\theta^{\dot\alpha},$$

are manifestly invariant under constant supertranslations. The invariant bosonic form is

$$E^\mu = \mathrm d x^\mu +i\theta\sigma^\mu\mathrm d\bar\theta -i\mathrm d\theta\sigma^\mu\bar\theta.$$

Indeed,

$$\delta(\mathrm d x^\mu) =-i\epsilon\sigma^\mu\mathrm d\bar\theta +i\mathrm d\theta\sigma^\mu\bar\epsilon,$$

which is canceled by the variation of the remaining two terms. The exterior derivative is not merely quoted: the form-degree Leibniz rule gives

$$\begin{aligned} \mathrm dE^\mu &=i\,\mathrm d(\theta\sigma^\mu\mathrm d\bar\theta) -i\,\mathrm d(\mathrm d\theta\sigma^\mu\bar\theta)\\ &=i\,\mathrm d\theta\sigma^\mu\wedge\mathrm d\bar\theta +i\,\mathrm d\theta\sigma^\mu\wedge\mathrm d\bar\theta\\ &=2i\,\mathrm d\theta\sigma^\mu\wedge\mathrm d\bar\theta. \end{aligned}$$

In the second term, $\mathrm d(\mathrm d\theta\,\bar\theta)=-\mathrm d\theta\wedge\mathrm d\bar\theta$ because $\mathrm d\theta$ has form degree one. The resulting torsion form encodes the odd-odd translation bracket. Thus flat superspace is not an ordinary direct product with trivial graded geometry, even though its underlying coordinates look like $\mathbb R^{4|4}$.

The geometric construction and the relation between generators, covariant derivatives, and flat supervielbeins are developed in Gates, Grisaru, Roček, and Siegel 1983, §§ 3.2–3.4, pp. 62–88.

The component-constraint atlas follows this geometry through covariant derivatives, standard constraints, gauge equivalences, and component content.

Chiral coordinates

Define

$$y^\mu=x^\mu+i\theta\sigma^\mu\bar\theta, \qquad \bar y^\mu=x^\mu-i\theta\sigma^\mu\bar\theta.$$

Under the finite supertranslation,

$$\begin{aligned} y'^\mu &=y^\mu+a^\mu+2i\theta\sigma^\mu\bar\epsilon +i\epsilon\sigma^\mu\bar\epsilon,\\ \bar y'^\mu &=\bar y^\mu+a^\mu-2i\epsilon\sigma^\mu\bar\theta -i\epsilon\sigma^\mu\bar\epsilon. \end{aligned}$$

The parameter-only quadratic term is required for a finite transformation; at infinitesimal order it drops out. Most importantly, the transformed $(y,\theta)$ depend on $(y,\theta)$ and the parameters but not on the old $\bar\theta$ as an independent coordinate. This makes the chiral subspace stable under supertranslations and anticipates the solution of $\bar D_{\dot\alpha}\Phi=0$.

A round-trip check is immediate. Substitute $x=y-i\theta\sigma\bar\theta$ into the original transformation, calculate $y'$, and then reconstruct $x'=y'-i\theta'\sigma\bar\theta'$. The original $x$ shift is recovered exactly.

Common pitfalls

Odd parameters do not commute through one another. Treating $\epsilon_1\epsilon_2$ as an ordinary commuting product erases the translation in the group commutator.

The sign of one generator is not independent. Changing $P_\mu$, the active/passive convention, derivative side, or the sign in $y^\mu$ changes a package of formulas. Verify the anticommutator after every translation.

Chiral coordinates are not extra dimensions. They are an adapted coordinate system on the same superspace. A chiral superfield still has ordinary spacetime dependence.

Exercises

1. Compose two pure odd translations

Set $a_1=a_2=0$. Compute both orders and subtract the 1-then-2 shift from the 2-then-1 shift.

Solution

The shifts of $\theta$ and $\bar\theta$ cancel. The two orderings differ by

$$\Delta x^\mu =2i( \epsilon_2\sigma^\mu\bar\epsilon_1 -\epsilon_1\sigma^\mu\bar\epsilon_2),$$

so the field commutator is the translation $\xi^\mu\partial_\mu$.

2. Recover the barred-coordinate shift

Use the leading term of $\bar Q_{\dot\alpha}$ to show that $\bar\epsilon_{\dot\alpha}\bar Q^{\dot\alpha}$ shifts $\bar\theta^{\dot\beta}$ by $+\bar\epsilon^{\dot\beta}$, not by its negative.

Solution

Because $\bar Q_{\dot\gamma}\bar\theta^{\dot\beta} =-\delta_{\dot\gamma}{}^{\dot\beta}$,

$$\bar\epsilon_{\dot\alpha}\bar Q^{\dot\alpha} \bar\theta^{\dot\beta} =-\bar\epsilon_{\dot\alpha} \epsilon^{\dot\alpha\dot\beta} =+\bar\epsilon^{\dot\beta}.$$

The two minus signs have different origins: one is built into the left-derivative representation of $\bar Q$, and the other follows from reversing the indices of the antisymmetric epsilon tensor.

3. Verify the invariant form

Show directly that $\delta E^\mu=0$.

Solution

Using constant parameters,

$$\delta\mathrm d x^\mu =-i\epsilon\sigma^\mu\mathrm d\bar\theta +i\mathrm d\theta\sigma^\mu\bar\epsilon.$$

The variations of $i\theta\sigma^\mu\mathrm d\bar\theta$ and $-i\mathrm d\theta\sigma^\mu\bar\theta$ are respectively $+i\epsilon\sigma^\mu\mathrm d\bar\theta$ and $-i\mathrm d\theta\sigma^\mu\bar\epsilon$. The sum vanishes.

4. Track the finite chiral shift

Why does $y'^\mu$ contain $i\epsilon\sigma^\mu\bar\epsilon$?

Solution

It arises from substituting both shifted odd coordinates into $i\theta'\sigma^\mu\bar\theta'$. It is second order in the finite odd parameters, so it is invisible in an infinitesimal formula but necessary for exact group composition.

Continue

Supercovariant Derivatives, Chirality, and Integrability constructs the commuting right-action operators and solves the chiral constraint. Chiral, Vector, Linear, and Field-Strength Superfields then turns those constraints into standard multiplets.

References

• Gates, S. James, Jr., Marcus T. Grisaru, Martin Roček, and Warren Siegel. Superspace, or One Thousand and One Lessons in Supersymmetry. Frontiers in Physics 58. Reading, MA: Benjamin/Cummings, 1983. Corrected open edition, 2001. arXiv:hep-th/0108200.

• Salam, Abdus, and John Strathdee. “Super-Gauge Transformations.” Nuclear Physics B 76 (1974): 477–482. DOI.

• Weinberg, Steven. The Quantum Theory of Fields. Volume III: Supersymmetry. Cambridge: Cambridge University Press, 2000, § 26.2. DOI.