Fierz identities from explicit matrices

Two calculations can produce different-looking products of four spinors and still describe the same interaction. A Fierz rearrangement changes which spinors are paired. Its numerical coefficients come from completeness of the spinor matrices; its exchange signs come from the statistics of the entries. Keeping those two operations separate is the reliable way to use the identity.

This lesson builds on spinors in three dimensions. We use three-dimensional Euclidean space, metric $\delta_{\mu\nu}$ and orientation $\varepsilon_{123}=+1$. The argument is local algebra: no action, gauge group or preserved supersymmetry is assumed. It applies both to odd fermion fields and to commuting spinor parameters, with different exchange signs.

Fix the coefficients before moving any fields

Retain the mixed-index matrices and scalar contraction

$$\gamma^1=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad \gamma^2=\begin{pmatrix}0&-1\\-1&0\end{pmatrix},\quad \gamma^3=\begin{pmatrix}0&i\\-i&0\end{pmatrix},\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}, \tag{T09.1}$$ $$uv=u^TCv=u_2v_1-u_1v_2,\qquad u\gamma^\mu v=u^TC\gamma^\mu v. \tag{T09.2}$$

Transpose does not conjugate. Raising a spinor index uses $\varepsilon^{\alpha\beta}=-C_{\alpha\beta}$ on the left. The scalar matrix $C$ is antisymmetric; each $C\gamma^\mu$ is symmetric. For homogeneous spinors with parities $p_u,p_v\in\{0,1\}$, therefore,

$$uv=-(-1)^{p_up_v}vu,\qquad u\gamma^\mu v=(-1)^{p_up_v}v\gamma^\mu u. \tag{T09.3}$$

Parity zero means commuting and parity one means odd. If a spinor contains both parities, separate its homogeneous pieces before applying this rule. These definitions agree with the paper’s Appendix A.1, (A.1)–(A.4).

The preceding spinor lesson derived

$$M=\frac12\operatorname{tr}(M)1 +\frac12\operatorname{tr}(\gamma^\mu M)\gamma_\mu \tag{T09.4}$$

for every two-by-two matrix. The traces project onto the four basis matrices $1,\gamma^1,\gamma^2,\gamma^3$; their orthogonality fixes the factors of two. For instance, $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ gives coefficients $(a+d)/2$, $(a-d)/2$, $-(b+c)/2$, $i(c-b)/2$. Substituting them reconstructs each of the four entries without exchanging $a,b,c,d$.

For the ordered outer product $M_\alpha{}^\beta=u_\alpha v^\beta$, the same expansion is

$$u_\alpha v^\beta =-\frac12(uv)\delta_\alpha{}^\beta +\frac12(u\gamma^\mu v)(\gamma_\mu)_\alpha{}^\beta. \tag{T09.5}$$

Its coefficients retain the order $u$ before $v$, so (T09.5) itself needs no parity-dependent prefactor. A trace against an ordinary numerical gamma matrix introduces no Grassmann exchange. One must not extend that observation to arbitrary cyclic rearrangements of matrices with odd entries.

Four spinors and two pairing channels

Let $u,v,w,x$ have definite parities, and define the two contracted expressions

$$S=(uv)(wx),\qquad V=(u\gamma^\mu v)(w\gamma_\mu x). \tag{T09.6}$$

A channel here means a choice of which spinors are paired. We want the crossed pairings

$$S'=(ux)(wv),\qquad V'=(u\gamma^\mu x)(w\gamma_\mu v). \tag{T09.7}$$

The displayed order matters. Introduce the permutation sign

$$\epsilon_{\rm ex} =(-1)^{p_vp_w+p_x(p_v+p_w)}. \tag{T09.8}$$

It records changing the entry order $u,v,w,x$ into $u,x,w,v$: moving $x$ across $w,v$, then interchanging $v,w$. The antisymmetry of $C$ will be accounted for separately.

To derive the scalar rearrangement, insert (T09.5) into the middle two spinors:

$$\begin{aligned} (uv)(wx) &=u^\alpha\bigl(v_\alpha w^\beta\bigr)x_\beta\\ &=-\frac12u^\alpha(vw)x_\alpha +\frac12u^\alpha(v\gamma^\mu w)(\gamma_\mu x)_\alpha. \end{aligned} \tag{T09.9}$$

Move each middle bilinear to the right of $x$. Both moves give $(-1)^{p_x(p_v+p_w)}$. Then use $vw=-(-1)^{p_vp_w}wv$ and $v\gamma^\mu w=(-1)^{p_vp_w}w\gamma^\mu v$. The minus sign of the scalar contraction cancels the first minus in (T09.9). Hence

$$S=\frac{\epsilon_{\rm ex}}2(S'+V'). \tag{T09.10}$$

For the vector channel, insert the same outer product between the two gamma matrices. The only additional matrix identities needed are

$$\sum_{\mu=1}^3\gamma^\mu\gamma_\mu=3\,1,\qquad \sum_{\mu=1}^3\gamma^\mu\gamma^\nu\gamma_\mu=-\gamma^\nu. \tag{T09.11}$$

For fixed $\nu$, the term $\mu=\nu$ contributes $+\gamma^\nu$ and the other two contribute $-\gamma^\nu$ each. The same entry permutation as before then gives

$$\boxed{\begin{pmatrix}S\\V\end{pmatrix} =\epsilon_{\rm ex} \begin{pmatrix}\tfrac12&\tfrac12\\[2pt]\tfrac32&-\tfrac12\end{pmatrix} \begin{pmatrix}S'\\V'\end{pmatrix}.} \tag{T09.12}$$

This is the full change of basis between these two scalar and vector contraction channels. The numerical matrix squares to the identity. The inverse permutation has the same $\epsilon_{\rm ex}$, so applying the rearrangement twice recovers the original expression. This catches a sign or a factor-of-three mistake without choosing spinor components.

For four odd spinors $\epsilon_{\rm ex}=-1$; for four commuting spinors it is $+1$. With mixed statistics calculate (T09.8). The number of spinors appearing in the expression is not enough to determine the sign.

Worked example: one independent quartic coupling

Take two independent odd spinors $\psi=(\theta_1,\theta_2)^T$ and $\chi=(\theta_3,\theta_4)^T$. All four $\theta_i$ anticommute and square to zero. Define

$$s=\psi\chi,\qquad j^\mu=\psi\gamma^\mu\chi.$$

Set $u=w=\psi$ and $v=x=\chi$ in (T09.12). The crossed expressions are now the original ones. Its scalar row becomes $s^2=-(s^2+j_\mu j^\mu)/2$, or

$$\boxed{j_\mu j^\mu=-3s^2 =\frac32(\psi\psi)(\chi\chi).} \tag{T09.13}$$

The second equality deserves an independent check. Put every Grassmann product in the order $\theta_1\theta_2\theta_3\theta_4$:

$$\begin{aligned} s&=\theta_2\theta_3-\theta_1\theta_4, &s^2&=-2\theta_1\theta_2\theta_3\theta_4,\\ j^1&=\theta_1\theta_4+\theta_2\theta_3, &j^2&=\theta_1\theta_3-\theta_2\theta_4,\\ j^3&=i(\theta_1\theta_3+\theta_2\theta_4). \end{aligned} \tag{T09.14}$$

Each $(j^\mu)^2$ is $2\theta_1\theta_2\theta_3\theta_4$. Also $\psi\psi=-2\theta_1\theta_2$ and $\chi\chi=-2\theta_3\theta_4$. Thus both equalities in (T09.13) are verified by actual anticommuting entries. A computation with four commuting symbols would test a different identity.

Consequently an interaction written as $g_s s^2+g_vj_\mu j^\mu$ is $(g_s-3g_v)s^2$: these two quartic structures are not independent for this field content and contraction. The spinors have independent components; no Euclidean conjugation condition was imposed. Nor is $j_\mu j^\mu$ a positive number: it is a Grassmann polynomial, and no dagger occurs in its definition.

If $\chi=\psi$, there are only two independent generators in this algebraic example, and the quartic expression vanishes. That observation does not set every quartic fermion interaction to zero. Extra flavor or gauge indices, or differentiated fields, supply further independent entries. Exercise 2 shows exactly where a tempting generalization fails.

These are identities of ordered field polynomials. For coincident quantum composite operators one must use a consistent regulator and operator definition; a classical rearrangement alone does not calculate contact terms or operator mixing.

A commuting spinor gives a projector

For a geometric application, take an ordinary nonzero complex column $\zeta$. Now the Hermitian norm $n=\zeta^\dagger\zeta$ is a positive number. The matrix

$$\rho=\frac{\zeta\zeta^\dagger}{n}$$

is Hermitian, has trace one, and obeys $\rho^2=\rho$. Applying (T09.4) yields

$$\rho=\frac12(1+\eta_\mu\gamma^\mu),\qquad \eta_\mu=\frac{\zeta^\dagger\gamma_\mu\zeta}{n}. \tag{T09.15}$$

Hermiticity makes $\eta_\mu$ real. Substituting this expansion into $\rho^2=\rho$ and using $(\eta\cdot\gamma)^2=\eta_\mu\eta^\mu1$ gives

$$\eta_\mu\eta^\mu=1,\qquad (\eta\cdot\gamma)\zeta=\zeta. \tag{T09.16}$$

Thus a nonzero commuting spinor determines a unit direction at a point. This is the algebra behind the paper’s (3.3)–(3.4). It does not say that an arbitrary spinor field solves a Killing-spinor equation or that its direction has any particular differential or global property.

Compare this Hermitian construction with $P^\mu=\zeta\gamma^\mu\zeta$. Since $\zeta\zeta=0$, the all-commuting version of (T09.10) implies $P_\mu P^\mu=0$. There is no contradiction with Euclidean positivity: $P$ is complex and that contraction has no conjugation. For $\zeta=(1,0)^T$, $\eta=(1,0,0)$ while $P=(0,1,i)$; $P\cdot P=0$ but $\bar P\cdot P=2$. These two sorts of bilinear answer different geometric questions.

Exercises

1. Two parameters and two fields

Let $\psi,\chi$ be odd, and $\zeta,\eta$ commuting. Rearrange $(\psi\zeta)(\eta\chi)$ into the pairings $(\psi\chi)(\eta\zeta)$ and $(\psi\gamma^\mu\chi)(\eta\gamma_\mu\zeta)$. Check it for $\zeta=(1,0)^T$, $\eta=(0,1)^T$ and arbitrary odd components of $\psi,\chi$.

Hint 1

Use $u=\psi$, $v=\zeta$, $w=\eta$, $x=\chi$. The two middle spinors have parity zero.

Hint 2

In the component check $\eta\zeta=1$ and $\eta\gamma^\mu\zeta=(1,0,0)$.

Solution

Equation (T09.8) gives $\epsilon_{\rm ex}=1$, so

$$(\psi\zeta)(\eta\chi) =\frac12(\psi\chi)(\eta\zeta) +\frac12(\psi\gamma^\mu\chi)(\eta\gamma_\mu\zeta).$$

Write $\psi=(p,q)^T$ and $\chi=(r,t)^T$, with odd entries. The left side is $qr$. The right side is $[(qr-pt)+(pt+qr)]/2=qr$. No odd entries were exchanged in that last addition. Using the all-odd minus sign would give $-qr$, an error for independent $q,r$. The parity of the spinor parameters matters even though they carry the same rotation indices as the fields.

2. A spectator index changes the question

For two flavors define $S=\sum_{i=1}^2\psi_i\chi_i$ and $J^\mu=\sum_{i=1}^2\psi_i\gamma^\mu\chi_i$. Must $J_\mu J^\mu=-3S^2$ still hold? Test the claim with four independent odd generators and

$$\psi_1=(\theta_1,0)^T,\quad \chi_1=(0,\theta_2)^T,\quad \psi_2=(0,\theta_3)^T,\quad \chi_2=(\theta_4,0)^T.$$

Then write the correctly rearranged flavor sums.

Hint 1

The products with two different flavors survive even though each same-flavor product contains only two generators.

Hint 2

Apply (T09.12) separately to $(\psi_i\chi_i)(\psi_j\chi_j)$. Its crossed pairings contain $\psi_i\chi_j$ and $\psi_j\chi_i$.

Solution

Here $S=-\theta_1\theta_2+\theta_3\theta_4$, $J^1=\theta_1\theta_2+\theta_3\theta_4$, and $J^2=J^3=0$. Thus

$$S^2=-2\theta_1\theta_2\theta_3\theta_4,\qquad J_\mu J^\mu=2\theta_1\theta_2\theta_3\theta_4.$$

The proposed factor $-3$ fails. It did not fail because the Fierz identity changed, but because the crossed flavor contractions are different operators. Define

$$S_{\rm cross}=\sum_{i,j}(\psi_i\chi_j)(\psi_j\chi_i),\qquad V_{\rm cross}=\sum_{i,j}(\psi_i\gamma^\mu\chi_j) (\psi_j\gamma_\mu\chi_i).$$

The correct relations are $S^2=-(S_{\rm cross}+V_{\rm cross})/2$ and $J^2=-(3S_{\rm cross}-V_{\rm cross})/2$. In the specified example, $S_{\rm cross}=0$ and $V_{\rm cross}=4\theta_1\theta_2\theta_3\theta_4$, which reproduce both results. A spinor identity does not remove an independent flavor or color contraction.

3. Recover a direction and identify the missing information

Take $\zeta=(1,z)^T/\sqrt{1+|z|^2}$ with an ordinary complex number $z$. Find $\eta$ from (T09.15), check its unit norm, and determine what happens under $\zeta\mapsto e^{i\beta}\zeta$ for real $\beta$. Can a finite $z$ represent the direction $\eta=(-1,0,0)$?

Hint 1

Evaluate the three Hermitian matrix products separately. Both off-diagonal gamma matrices carry a minus sign relative to the usual Pauli ordering.

Hint 2

Write $z=x+iy$. The unit-norm check reduces to $(1-x^2-y^2)^2+4x^2+4y^2=(1+x^2+y^2)^2$.

Solution

The result is

$$\eta=\frac{(1-|z|^2,\,-2\operatorname{Re}z,\,-2\operatorname{Im}z)} {1+|z|^2}.$$

The identity in the second hint proves $\eta^2=1$ without appealing to the abstract projector argument. The phase cancels between $\zeta$ and $\zeta^\dagger$, so $\rho$ and $\eta$ are unchanged. By contrast, the complex bilinear $P^\mu=\zeta\gamma^\mu\zeta$ acquires the phase $e^{2i\beta}$.

For finite $z$, $(1-|z|^2)/(1+|z|^2)>-1$. The missing direction is represented by the perfectly regular spinor $(0,1)^T$. It is absent from this coordinate chart because the first component was required to be nonzero. A unit direction therefore retains less information than a normalized spinor, and a chosen component ratio need not cover every direction.

Carry the method, not the coefficients

The same strategy works in other dimensions: choose a complete matrix basis, project onto it, and then make the graded permutation explicitly. The matrix basis changes. A four-component Dirac representation needs sixteen independent matrices; Weyl bilinears require their dotted and undotted index types. The two-channel matrix (T09.12) is the three-dimensional result and should not be copied into a four-dimensional calculation. The optional four-dimensional spinor lesson develops that separate package.

For the three-manifold path, the checkpoint is to explain why a commuting-spinor projector yields an ordinary unit vector while the formally similar fermion expression cannot be divided by a positive norm. Construct geometry from one spinor uses this projector to build the transverse tensors and test their index and charge assignments. Differential equations and bundle patching are additional inputs; completeness alone supplies the pointwise algebra.

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.