Spinors in three dimensions

The expression $\epsilon^\alpha\psi_\alpha$ contracts a spinor parameter with a spinor field. To calculate with it, we need to know exactly what moving the index does. An unnoticed sign at this stage will later look like a failed supersymmetry transformation.

We will build the conventions from two-by-two matrices, check their products, and derive a useful identity that replaces a product of two spinors by a scalar and a vector. The preparation is Grassmann calculus and ordinary matrix multiplication. All calculations here take place in flat three-dimensional Euclidean space, with metric $\delta_{\mu\nu}$, coordinates labeled $\mu=1,2,3$, and orientation $\varepsilon_{123}=+1$. No gauge field, curved background, or particular action is involved.

These spinor conventions are the ones used in Appendix A.1 of Supersymmetric Field Theories on Three-Manifolds, v3. Choosing them is a choice of notation and signature, not a claim that a given background preserves supersymmetry.

A spinor is a two-component rotation representation

A Euclidean three-dimensional spinor is a column

$$\psi=\begin{pmatrix}\psi_1\\\psi_2\end{pmatrix}$$

on which $\operatorname{Spin}(3)=SU(2)$ acts. Here $SU(2)$ means the two-by-two unitary matrices of determinant one. It is the double cover of the ordinary rotation group: two matrices, $U$ and $-U$, produce the same rotation of vectors.

For example, a spinor rotation through angle $\vartheta$ around the first axis is represented in our basis by

$$U(\vartheta)= \begin{pmatrix} e^{-i\vartheta/2}&0\\ 0&e^{i\vartheta/2} \end{pmatrix}.$$

At $\vartheta=2\pi$, this is $-I$, while at $4\pi$ it is $I$. A bilinear made from two spinors returns to itself after $2\pi$ because the two minus signs multiply to plus.

The components may be ordinary commuting complex numbers, as for a spinor labeling a supersymmetry generator, or Grassmann-odd quantities, as for a fermion field. Their rotation law is the same. Their exchange signs differ.

Multiply the gamma matrices first

The gamma matrices connect vector directions with operators on spinors. With row index $\alpha$ and column index $\beta$, fix

$$(\gamma^\mu)_\alpha{}^\beta: \qquad \gamma^1= \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad \gamma^2= \begin{pmatrix}0&-1\\-1&0\end{pmatrix}, \quad \gamma^3= \begin{pmatrix}0&i\\-i&0\end{pmatrix}.$$

Equivalently, these are $(\sigma^3,-\sigma^1,-\sigma^2)$ in terms of the Pauli matrices. They act on lower-index columns: $(\gamma^\mu\psi)_\alpha=(\gamma^\mu)_\alpha{}^\beta\psi_\beta$.

Each matrix squares to $I$. For two different directions, explicitly,

$$\gamma^1\gamma^2 =\begin{pmatrix}0&-1\\1&0\end{pmatrix} =i\gamma^3, \qquad \gamma^2\gamma^1=-i\gamma^3.$$

The other cyclic products work the same way. Together these calculations give

$$\gamma^\mu\gamma^\nu =\delta^{\mu\nu}I+i\varepsilon^{\mu\nu\rho}\gamma_\rho.$$

Adding this formula to the one with $\mu,\nu$ interchanged yields the Clifford algebra,

$$\{\gamma^\mu,\gamma^\nu\}=2\delta^{\mu\nu}I.$$

The antisymmetric part fixes the orientation-sensitive sign:

$$[\gamma^\mu,\gamma^\nu] =2i\varepsilon^{\mu\nu\rho}\gamma_\rho, \qquad \gamma^1\gamma^2\gamma^3=iI.$$

Checking only the anticommutator would miss an error in that last sign. These matrices and product identities agree with equations (A.3)–(A.4) of the source paper.

For a real vector $v_\mu$, define $v\cdot\gamma=v_\mu\gamma^\mu$. Its square is

$$(v\cdot\gamma)^2 =v_\mu v_\nu\delta^{\mu\nu}I +i v_\mu v_\nu\varepsilon^{\mu\nu\rho}\gamma_\rho =|v|^2I.$$

The second term vanishes because a symmetric product is contracted with an antisymmetric tensor. This gives a useful check with a physical interpretation: the matrix associated with a unit direction has eigenvalues $+1$ and $-1$.

Raising an index is an antisymmetric contraction

Use spinor indices $\alpha,\beta=1,2$, and define

$$\varepsilon^{\alpha\beta} =\begin{pmatrix}0&1\\-1&0\end{pmatrix}, \qquad \varepsilon_{\alpha\beta} =\begin{pmatrix}0&-1\\1&0\end{pmatrix}.$$

Thus $\varepsilon^{12}=\varepsilon_{21}=1$. The two matrices are inverses. They act on the left:

$$\psi^\alpha=\varepsilon^{\alpha\beta}\psi_\beta, \qquad \psi_\alpha=\varepsilon_{\alpha\beta}\psi^\beta.$$

In components,

$$\psi^1=\psi_2,\qquad \psi^2=-\psi_1.$$

Lowering again recovers $\psi_1=-\psi^2$ and $\psi_2=\psi^1$. No complex conjugation has taken place.

The scalar contraction is defined by

$$\psi\chi:=\psi^\alpha\chi_\alpha =\psi_2\chi_1-\psi_1\chi_2.$$

If $C$ denotes the lower-index epsilon matrix, the same expression is $\psi^T C\chi$, where $T$ is transpose without conjugation. The identity $U^T C U=C$ for $U\in SU(2)$ makes this contraction invariant under simultaneous spinor rotations.

For commuting columns $\zeta=(1,0)^T$ and $\eta=(0,1)^T$, one finds

$$\zeta\eta=-1,\qquad \eta\zeta=1.$$

The sign comes from the antisymmetric contraction, even though the entries themselves commute.

The Hermitian adjoint is a different operation. In the source convention,

$$(\psi^\dagger)^\alpha=\overline{\psi_\alpha},$$

where the bar complex conjugates the component. For $\psi=(1,i)^T$, epsilon raising gives components $(i,-1)$, whereas the adjoint has upper-index components $(1,-i)$. For commuting columns, $\psi^\dagger\psi=|1|^2+|i|^2=2$. For Grassmann-valued fields, adjunction reverses product order, and a formal even bilinear is not a positive real number to which a numerical inequality can automatically be applied.

Scalar and vector bilinears have different exchange signs

Define the vector bilinear by

$$\psi\gamma^\mu\chi :=\psi^\alpha(\gamma^\mu)_\alpha{}^\beta\chi_\beta.$$

Its three components follow by multiplication:

$$\begin{aligned} \psi\gamma^1\chi&=\psi_1\chi_2+\psi_2\chi_1,\\ \psi\gamma^2\chi&=\psi_1\chi_1-\psi_2\chi_2,\\ \psi\gamma^3\chi&=i(\psi_1\chi_1+\psi_2\chi_2). \end{aligned}$$

Unlike the scalar contraction, the coefficient matrices $C\gamma^\mu$ are symmetric. If the spinors have definite parities $p_\psi,p_\chi\in\{0,1\}$, exchanging their entries therefore gives

$$\psi\chi=-(-1)^{p_\psi p_\chi}\chi\psi, \qquad \psi\gamma^\mu\chi =(-1)^{p_\psi p_\chi}\chi\gamma^\mu\psi.$$

There are two potential signs: one from the contraction matrix and one from moving odd components past each other. For two odd spinors, the scalar bilinear is symmetric under exchanging the spinors, while the vector bilinear is antisymmetric. For two commuting spinors, those statements are reversed. With one commuting and one odd spinor, no Grassmann exchange sign appears.

In particular, a commuting spinor $\zeta$ satisfies $\zeta\zeta=0$, but its vector bilinear need not vanish. An odd spinor $\psi$ instead satisfies

$$\psi\psi=-2\psi_1\psi_2,\qquad \psi\gamma^\mu\psi=0.$$

These are statements about the parity of the components, not different choices of gamma matrices.

Derive the two-spinor rearrangement

A two-by-two matrix has four independent entries. The four matrices $I,\gamma^1,\gamma^2,\gamma^3$ form a basis: they obey

$$\operatorname{tr}\gamma^\mu=0, \qquad \operatorname{tr}(\gamma^\mu\gamma^\nu)=2\delta^{\mu\nu}.$$

Write $M=aI+b_\mu\gamma^\mu$. Taking a trace fixes $a$, while multiplying by $\gamma^\nu$ and tracing fixes $b_\nu$. Thus

$$M=\frac12\operatorname{tr}(M)I +\frac12\operatorname{tr}(\gamma^\mu M)\gamma_\mu.$$

This formula also works with Grassmann-valued entries: the basis matrices have ordinary numerical entries, so no odd factors have been exchanged.

Now choose an outer product

$$M_\alpha{}^\beta=u_\alpha v^\beta.$$

Keep the order of coefficients $u$ then $v$ throughout. Using the component rules above gives

$$\operatorname{tr}M=u_1v_2-u_2v_1=-uv, \qquad \operatorname{tr}(\gamma^\mu M)=u\gamma^\mu v.$$

Substitute these coefficients into the matrix expansion:

$$u_\alpha v^\beta =-\frac12(uv)\delta_\alpha{}^\beta +\frac12(u\gamma^\mu v)(\gamma_\mu)_\alpha{}^\beta.$$

This is a Fierz identity: a rearrangement following from completeness of the spinor matrices. It is valid for commuting or odd $u,v$ with the displayed ordering. Any subsequent exchange of spinors must still use their parities.

For an immediate check, let $u=(1,0)^T$, $v=(0,1)^T$. The left side is $\operatorname{diag}(1,0)$, while $uv=-1$ and $u\gamma^\mu v=(1,0,0)$. The right side is therefore $(I+\gamma^1)/2$, the same matrix.

Acting on a third spinor $w$ gives a particularly useful version,

$$u_\alpha(vw) =-\frac12(uv)w_\alpha +\frac12(u\gamma^\mu v)(\gamma_\mu w)_\alpha.$$

No reordering was needed here. This identity turns products of spinor parameters and fermion fields into the scalar and vector structures needed when composing supersymmetry transformations.

To compare directly with the source’s all-lower-index equation, lower the second gamma index by

$$(\gamma^\mu)_{\alpha\beta} =\varepsilon_{\beta\lambda} (\gamma^\mu)_\alpha{}^\lambda.$$

The resulting symmetric matrices are $(\sigma^1,\sigma^3,-iI)$. The same completeness relation gives

$$(\gamma^\mu)_{\alpha\beta} (\gamma_\mu)_{\kappa\lambda} =-\bigl( \varepsilon_{\alpha\kappa}\varepsilon_{\beta\lambda} +\varepsilon_{\alpha\lambda}\varepsilon_{\beta\kappa} \bigr).$$

This is the second identity in the source’s equation (A.4). As a component check, set $\alpha=\beta=1$ and $\kappa=\lambda=2$: both sides give $-2$. The coefficient arrays for this lowered gamma matrix and for the bilinear $C\gamma^\mu$ occupy different index positions; do not replace one by the other without doing the contractions.

Exercises

1. Invert a vector matrix

For a nonzero real vector $v$, use the Clifford algebra to find $(v\cdot\gamma)^{-1}$. Check it explicitly for $v=(1,2,0)$, and state what happens as $v\to0$.

First hint

The lesson already computed $(v\cdot\gamma)^2$.

Second hint

For the specified vector, $v\cdot\gamma=\begin{pmatrix}1&-2\\-2&-1\end{pmatrix}$.

Complete solution

Since $|v|^2>0$ for a nonzero real Euclidean vector,

$$(v\cdot\gamma)^{-1}=\frac{v\cdot\gamma}{|v|^2}.$$

For $v=(1,2,0)$, multiplication gives

$$\begin{pmatrix}1&-2\\-2&-1\end{pmatrix}^2=5I,$$

so division by five gives the inverse. At $v=0$, the matrix is zero and has no inverse. If $v$ has units of momentum, the inverse has inverse-momentum units, as required. The restriction to real Euclidean vectors matters: a nonzero complex vector can have $v_\mu v^\mu=0$.

2. Change the statistics, not the matrices

Let $\zeta=(a,b)^T$ have commuting entries. Let $\psi=(\theta_1,\theta_2)^T$ have independent odd entries. Calculate $\zeta\zeta$, $\psi\psi$, $\zeta\psi$, and $\psi\zeta$. Explain why $\psi\psi$ need not vanish although each component squares to zero.

First hint

Use $u v=u_2v_1-u_1v_2$ before applying exchange rules.

Second hint

The product $\theta_1\theta_2$ is nonzero, and $\theta_2\theta_1=-\theta_1\theta_2$.

Complete solution

The results are

$$\begin{aligned} \zeta\zeta&=ba-ab=0,\\ \psi\psi&=\theta_2\theta_1-\theta_1\theta_2 =-2\theta_1\theta_2,\\ \zeta\psi&=b\theta_1-a\theta_2,\\ \psi\zeta&=a\theta_2-b\theta_1=-\zeta\psi. \end{aligned}$$

The self-contraction of $\psi$ uses two different generators, so nilpotency of each generator does not set their product to zero. Its square does vanish, providing another check. The mixed exchange has no extra Grassmann sign because $a,b$ commute.

3. Catch an orientation error

Replace only $\gamma^3$ by $\widehat\gamma^3=-\gamma^3$, leaving $\gamma^1,\gamma^2$ and $\varepsilon_{123}=+1$ unchanged. Does the Clifford algebra still hold? Do the orientation-sensitive product formula and triple-product check still agree with the original convention?

First hint

Squaring a matrix is insensitive to its overall sign.

Second hint

The product $\gamma^1\gamma^2$ has not changed, but the matrix on the right side of $i\widehat\gamma^3$ has.

Complete solution

The Clifford algebra still holds: the new matrix squares to $I$ and anticommutes with the other two. However,

$$\widehat\gamma^1\widehat\gamma^2 =-i\widehat\gamma^3, \qquad \widehat\gamma^1\widehat\gamma^2\widehat\gamma^3=-iI.$$

The original oriented product formula therefore fails with the stipulated $\varepsilon_{123}=+1$. This alternative matrix basis would require translating the orientation-dependent convention consistently. Passing the Clifford check alone is insufficient to justify copying formulas involving $\varepsilon^{\mu\nu\rho}$.

From Euclidean spinors to supersymmetry

Two complex components specify the spinor representation used here; they do not by themselves specify the number of supercharges in a theory. Three-dimensional $\mathcal N=2$ has four real Lorentzian Poincaré supercharges, as explained at the start of section 2 of Aharony and collaborators. A field, an odd symmetry generator, and a commuting spinor used to label that generator can all carry the same spinor index.

The next step is Euclidean continuation and independent spinors. It explains how Lorentzian conjugate fields are treated after continuation and why a Euclidean tilde label is not automatically a dagger. Before moving on, take two spinors of specified parity and explain which of three operations you are performing: raising an index, conjugating a component, or exchanging the spinors. Each has its own rule.

For a further algebraic application, Fierz identities from explicit matrices rearranges four-spinor products, checks a fermion interaction and constructs the unit direction associated with a commuting spinor.

The gamma-matrix and bilinear laboratory provides a runnable exact calculation of these identities with genuine Grassmann entries. Change the orientation or the entry statistics and compare the predicted diagnostic with the output.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Appendices A and B, conventions and geometry