The three-dimensional N=2 algebra

Supersymmetry relates bosons and fermions, but its distinguishing feature is what happens when two transformations are composed: their algebra produces a spacetime translation, sometimes accompanied by a charge transformation. We will calculate that statement on a scalar field. Along the way, we must distinguish the anticommutator of two odd operations from the commutator of two even variations.

The preparation is Spinors in three dimensions and Euclidean continuation and independent spinors. We use flat Euclidean three-space, the metric $\delta_{\mu\nu}$, and the gamma matrices $(\sigma^3,-\sigma^1,-\sigma^2)$. Spinor contractions mean $u v=u^\alpha v_\alpha$, with $\varepsilon^{12}=\varepsilon_{21}=1$. All spinor labels below are constant. There is no gauge field or curved background in this calculation.

Four real supercharges, written in a complex basis

In Lorentzian three-dimensional spacetime, a minimal Majorana spinor has two real components. The label $\mathcal N=2$ means two such spinor generators, hence four real Poincaré supercharges. Equivalently, combine them into one two-component complex spinor charge and its Lorentzian conjugate. This count is the same as for four-dimensional $\mathcal N=1$, although the spinor representations differ. Aharony and collaborators, section 2, equation (2.1).

After Euclidean continuation, write the two independent odd spinor families as $Q_\alpha$ and $\widetilde Q_\alpha$, with $\alpha=1,2$. Their independence describes the complexified algebra. It does not turn the original theory into one with eight real physical supercharges. We keep the full flat-space algebra; selecting a subset of generators for a calculation would not change that full count.

Let $P_\mu$ generate translations and let $Z$ be a central charge, meaning a bosonic generator commuting with the other generators. In a differential representation on fields, use

$$P_\mu=-i\partial_\mu.$$

The spinor part of the algebra in the three-manifold paper’s convention is

$$\{Q_\alpha,Q_\beta\}=0,\qquad \{\widetilde Q_\alpha,\widetilde Q_\beta\}=0,$$ $$\{Q_\alpha,\widetilde Q_\beta\} =2(\gamma^\mu)_{\alpha\beta}P_\mu +2i\varepsilon_{\alpha\beta}Z.$$

Here $\{A,B\}=AB+BA$, and the two-lower-index gamma matrix is defined as in the spinor lesson. Also $[P_\mu,Q_\alpha]=[P_\mu,\widetilde Q_\alpha]=0$, while $Z$ commutes with translations and supercharges. The generators rotate as spinors; $P_\mu$ is a vector and $Z$ a scalar. These relations are the flat algebra displayed in equation (1.2) of the paper, version 3.

For a field of definite central charge, write $Z\phi=z\phi$. The number $z$ has mass dimension one, like momentum. It is not automatically an electric charge. One possible origin of $Z$ is momentum along a direction removed by dimensional reduction: a mode proportional to $e^{izy}$ is an eigenvector of $-i\partial_y$ with eigenvalue $z$. Another useful representation is $Z=m\mathsf T$, where $\mathsf T$ is a dimensionless internal-symmetry generator; a field of $\mathsf T$-charge $q$ then has $z=mq$. Neither description introduces a dynamical gauge field into the present example.

First use commuting labels for odd operations

Let $\zeta_\alpha$ and $\widetilde\zeta_\alpha$ be commuting spinors. Fix the normalization of the odd differential operations by

$$\delta_\zeta=i\zeta^\alpha Q_\alpha, \qquad \delta_{\widetilde\zeta} =i\widetilde\zeta^\alpha\widetilde Q_\alpha.$$

Because the labels are even, these operations are odd. They reverse the parity of a field. Define the commuting bilinears

$$K^\mu=\zeta\gamma^\mu\widetilde\zeta, \qquad s=\zeta\widetilde\zeta.$$

The index conventions give

$$\zeta^\alpha\widetilde\zeta^\beta (\gamma^\mu)_{\alpha\beta}=-K^\mu, \qquad \zeta^\alpha\widetilde\zeta^\beta \varepsilon_{\alpha\beta}=s.$$

Thus the two factors of $i$ in the operation definitions, together with $P_\mu=-i\partial_\mu$, convert the generator algebra into

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi =-2i\bigl(K^\mu\partial_\mu\phi+s\,z\phi\bigr).$$

This is the flat, zero-background specialization of the paper’s equation (6.1). The minus sign from lowering the gamma index is essential in making the two forms agree.

We can verify the result without assuming it. Introduce a complex scalar $\phi$, an odd two-component field $\psi_\alpha$, and an even scalar $F_\Phi$. The last field will be explained as an auxiliary field in the chiral-multiplet lesson. For now, only the following transformation rules are needed:

$$\begin{aligned} \delta_\zeta\phi&=\sqrt2\,\zeta\psi, &\delta_{\widetilde\zeta}\phi&=0,\\ \delta_\zeta\psi&=\sqrt2\,\zeta F_\Phi, &\delta_{\widetilde\zeta}\psi &=-\sqrt2\,i\left( z\widetilde\zeta\,\phi+ \gamma^\mu\widetilde\zeta\,\partial_\mu\phi \right). \end{aligned}$$

These are the first two lines of equation (6.8), separated into the two parameter families and specialized to vanishing background fields. The central eigenvalue $z$ is retained. We are testing scalar closure, so the transformation of $F_\Phi$ and the equations of motion are not needed.

Composition acts from the rightmost operation first. Since $\delta_{\widetilde\zeta}\phi=0$,

$$\delta_\zeta\delta_{\widetilde\zeta}\phi=0.$$

In the other order, the commuting label passes through the odd operation without a sign:

$$\begin{aligned} \delta_{\widetilde\zeta}\delta_\zeta\phi &=\sqrt2\,\zeta \bigl(\delta_{\widetilde\zeta}\psi\bigr)\\ &=-2i\left( z\,\zeta\widetilde\zeta\,\phi +\zeta\gamma^\mu\widetilde\zeta\,\partial_\mu\phi \right). \end{aligned}$$

Adding the two orders reproduces the algebra. Two operations of the first family also close correctly on the scalar:

$$\{\delta_\zeta,\delta_\eta\}\phi =2(\eta\zeta+\zeta\eta)F_\Phi=0,$$

because commuting-spinor scalar contractions are antisymmetric. Two operations of the tilde family annihilate $\phi$ directly. This checks the scalar representation of the stated algebra, rather than merely inspecting its symbols.

Now attach odd parameters and take a commutator

An ordinary infinitesimal variation preserves the parity of the field being varied. Obtain such an even operation by multiplying an odd operation by an odd parameter. For example, let $\lambda_1,\lambda_2$ be independent constant Grassmann generators and define

$$\Delta_1=\lambda_1\delta_\zeta, \qquad \Delta_2=\lambda_2\delta_{\widetilde\zeta}.$$

An odd differential passes an external odd parameter with a minus sign. Therefore

$$\begin{aligned} \Delta_1\Delta_2 &=-\lambda_1\lambda_2 \delta_\zeta\delta_{\widetilde\zeta},\\ \Delta_2\Delta_1 &=+\lambda_1\lambda_2 \delta_{\widetilde\zeta}\delta_\zeta. \end{aligned}$$

Subtracting gives the translation between conventions:

$$[\Delta_1,\Delta_2] =-\lambda_1\lambda_2 \{\delta_\zeta,\delta_{\widetilde\zeta}\}.$$

Hence

$$[\Delta_1,\Delta_2]\phi =2i\lambda_1\lambda_2 \bigl(K^\mu\partial_\mu\phi+s\,z\phi\bigr).$$

The overall sign changed for a reason: the odd operation crossed an odd parameter. Replacing an anticommutator by a commutator while retaining every other sign would miss this step.

For general even variations, combine the odd scalar coefficients and commuting spinor directions into odd spinor parameters $\epsilon_i,\widetilde\epsilon_i$, with $i=1,2$. The same component rules then read

$$\Delta_i\phi=\sqrt2\,\epsilon_i\psi, \qquad \Delta_i\psi=\sqrt2\,\epsilon_iF_\Phi -\sqrt2\,i\left( z\widetilde\epsilon_i\phi+ \gamma^\mu\widetilde\epsilon_i\partial_\mu\phi \right).$$

Since $\Delta_i$ is even and the parameters are fixed,

$$\Delta_1\Delta_2\phi =2\epsilon_2\epsilon_1F_\Phi -2i\left( z\epsilon_2\widetilde\epsilon_1\phi+ \epsilon_2\gamma^\mu\widetilde\epsilon_1\partial_\mu\phi \right).$$

Subtract the expression with $1$ and $2$ interchanged. Odd-spinor scalar contractions are symmetric, so $\epsilon_2\epsilon_1=\epsilon_1\epsilon_2$ and the $F_\Phi$ terms cancel:

$$[\Delta_1,\Delta_2]\phi =a^\mu\partial_\mu\phi+i\Lambda_Z\,z\phi,$$ $$\begin{aligned} a^\mu&=2i\left( \epsilon_1\gamma^\mu\widetilde\epsilon_2 -\epsilon_2\gamma^\mu\widetilde\epsilon_1 \right),\\ \Lambda_Z&=2\left( \epsilon_1\widetilde\epsilon_2 -\epsilon_2\widetilde\epsilon_1 \right). \end{aligned}$$

The first term translates the scalar by the constant even vector $a^\mu$. The second is a central-charge transformation. If $Z=m\mathsf T$, it is an internal transformation with dimensionless parameter $m\Lambda_Z$. In the complexified Euclidean calculation these parameters need not be real.

A translation you can see in components

Choose a positive reference mass $M$ solely to display the dimensions of the spinor labels:

$$\zeta=M^{-1/2}\begin{pmatrix}1\\0\end{pmatrix}, \qquad \widetilde\zeta=M^{-1/2}\begin{pmatrix}0\\1\end{pmatrix}.$$

Direct multiplication gives

$$s=-\frac1M,\qquad K^\mu=\frac1M(1,0,0).$$

Test the closure on $\phi(x)=A e^{ikx^1}$, with real momentum $k$ and constant amplitude $A$. This is a test field configuration; no equation of motion is imposed. The odd-operation algebra gives

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi =-\frac{2i}{M}(\partial_1-z)\phi =\frac{2k+2iz}{M}\phi.$$

For the even variations $\Delta_1,\Delta_2$ above, the commutator instead gives

$$[\Delta_1,\Delta_2]\phi =-\lambda_1\lambda_2 \frac{2k+2iz}{M}\phi.$$

At $z=0$, only the derivative along $x^1$ remains. At $k=0$, the derivative vanishes but a charged scalar can still transform through the central term. A constant field configuration is not automatically invariant under supersymmetry.

There are also dimensional checks. The generator algebra requires $[Q]=1/2$ and $[Z]=[P]=1$ in mass units. Spinor parameters have dimension $-1/2$, so $[a^\mu]=[\Lambda_Z]=-1$. Both $a^\mu\partial_\mu\phi$ and $\Lambda_Z z\phi$ consequently have the same dimension as $\phi$. For free chiral-field engineering dimensions, $[\phi]=1/2$, $[\psi]=1$, and $[F_\Phi]=3/2$, every term in the scalar and fermion transformation rules also matches.

Exercises

1. Does a supercharge square to zero?

With commuting labels, define the odd operation $\mathscr Q=\delta_\zeta+\delta_{\widetilde\zeta}$. Calculate $\mathscr Q^2\phi$. Why does the ordinary identity $[\mathscr Q,\mathscr Q]=0$ not answer this question? If $\lambda$ is one odd constant, what is $(\lambda\mathscr Q)^2$?

First hint

The square includes the two same-family squares and the mixed anticommutator.

Second hint

When squaring $\lambda\mathscr Q$, move $\mathscr Q$ past the second $\lambda$ before using $\lambda^2=0$.

Complete solution

The same-family squares vanish on the scalar, leaving

$$\mathscr Q^2\phi =\{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi =-2i(K^\mu\partial_\mu+s z)\phi.$$

The ordinary commutator subtracts an expression from itself, so it is identically zero even when $\mathscr Q^2$ is nonzero. The relevant bracket for an odd operation is $\{\mathscr Q,\mathscr Q\}=2\mathscr Q^2$. On the other hand,

$$(\lambda\mathscr Q)^2 =-\lambda^2\mathscr Q^2=0.$$

The last equality follows from the external parameter’s nilpotency, not from vanishing of the underlying odd operation’s square. Setting $z=0$ still leaves the derivative unless the field is constant along $K$.

2. A constant charged scalar

Take $\phi=\phi_0$ constant, $\psi=F_\Phi=0$, and nonzero real $z$. Is this configuration invariant under $\delta_{\widetilde\zeta}$ for every tilde label? What does the mixed closure do to $\phi$? Then set $z=0$ and explain the change.

First hint

Invariance of a configuration requires the variations of all its fields to vanish, not only the scalar variation.

Second hint

The derivative in $\delta_{\widetilde\zeta}\psi$ vanishes, but the term proportional to $z\phi_0$ remains.

Complete solution

For nonzero $\phi_0$,

$$\delta_{\widetilde\zeta}\psi =-\sqrt2\,i z\widetilde\zeta\,\phi_0$$

is generally nonzero. Thus the displayed transformations already show that the configuration is not invariant. Correspondingly,

$$\{\delta_\zeta,\delta_{\widetilde\zeta}\}\phi_0 =-2i s z\phi_0$$

can be nonzero despite the absence of a translation contribution. If $z=0$, both this scalar closure and the displayed scalar/fermion variations vanish on the chosen constant configuration. Establishing a vacuum would additionally require a specified action and its field equations; no such claim follows from this kinematic check alone.

3. What charge does a product carry?

Let two even scalar fields have $Z\phi_1=z_1\phi_1$ and $Z\phi_2=z_2\phi_2$. Assume the symmetry operations act by the usual product rule. Find $Z(\phi_1\phi_2)$ and check that the closure on the product is consistent with the closures on its factors. What happens if $z_2=-z_1$?

First hint

A generator acts on a product as a derivation, rather than multiplying the two charge eigenvalues.

Second hint

The commutator of even derivations is again a derivation. Use the same $a^\mu,\Lambda_Z$ for both fields.

Complete solution

The product rule gives

$$Z(\phi_1\phi_2) =(Z\phi_1)\phi_2+\phi_1(Z\phi_2) =(z_1+z_2)\phi_1\phi_2.$$

Applying the even commutator to both factors yields

$$[\Delta_1,\Delta_2](\phi_1\phi_2) =a^\mu\partial_\mu(\phi_1\phi_2) +i\Lambda_Z(z_1+z_2)\phi_1\phi_2.$$

This has exactly the same form as the scalar algebra. The central contribution cancels for opposite charges, although each factor can carry a nonzero charge. Multiplying $z_1z_2$ would fail both the product rule and the mass-dimension check.

What has been established

We have verified scalar closure without using equations of motion, with both parameter conventions made explicit. This calculation does not yet prove closure on every component or invariance of an action. The next step, the chiral multiplet, supplies the complete collection of fields and transformations. Keep the distinction between an odd operation with commuting spinor labels and an even variation with odd parameters visible when making that next calculation.