= Solution
Choose a local <orthonormal coframe> $e^A$ with $g=\eta_{AB}e^Ae^B$. The <Levi-Civita connection> is represented by the <connection 1 forms> satisfying
$$
de^A+\omega^A{}_B\wedge e^B=0,\qquad \omega_{AB}=\eta_{AC}\omega^C{}_B=-\omega_{BA}.
$$
The <gamma matrices> satisfy the <Clifford algebra> relations $\{\gamma^A,\gamma^B\}=2\eta^{AB}I$. The matrices $\tfrac14[\gamma^A,\gamma^B]$ represent the infinitesimal orthogonal transformations on <spinor fields>; lifting the frame connection to this representation defines the <spinor covariant derivative>
$$
\boxed{\nabla_\mu\epsilon=\partial_\mu\epsilon+\frac14\omega_{\mu AB}\gamma^A\gamma^B\epsilon
=\partial_\mu\epsilon+\frac18\omega_{\mu AB}[\gamma^A,\gamma^B]\epsilon.}
$$
Both frame indices are summed, so each antisymmetric pair occurs twice. Under a <local Lorentz transformation> with spin lift $S$, the <spin connection> transforms as $\Omega_\mu\mapsto S\Omega_\mu S^{-1}-(\partial_\mu S)S^{-1}$ and $\epsilon\mapsto S\epsilon$; consequently $\nabla_\mu\epsilon\mapsto S\nabla_\mu\epsilon$. This establishes covariance. Global construction requires a compatible <spin structure> (and the appropriate <orientation> and <time orientation>); the formulas themselves work on a local <orthonormal frame> patch.
The supplied <gamma matrices> have squares $-I,I,I$ and anticommute pairwise, so $\eta=\operatorname{diag}(-1,1,1)$. Work on a connected regular patch with $r>0$ and $V\ne0$. Since the metric depends only on $V^2$, we may use its positive smooth square root $V>0$. Take
$$
e^0=V\,dt,\qquad e^1=dr/V,\qquad e^2=r\,d\theta.
$$
Their <exterior derivatives> are $de^0=-V'e^0\wedge e^1$, $de^1=0$, and $de^2=(V/r)e^1\wedge e^2$. Solving <Cartan's first structure equation> and imposing <Lorentzian connection-form antisymmetry> gives
$$
\omega^0{}_1=\omega^1{}_0=VV'\,dt,\qquad \omega^2{}_1=V\,d\theta,\qquad \omega^1{}_2=-V\,d\theta.
$$
In particular $\omega_{01}=-VV'\,dt$ and $\omega_{12}=-V\,d\theta$. The <parallel spinor> equations are therefore
$$
\partial_t\epsilon-\frac{VV'}2\gamma^0\gamma^1\epsilon=0,\qquad
\partial_r\epsilon=0,\qquad
\partial_\theta\epsilon-\frac V2\gamma^1\gamma^2\epsilon=0.
$$
Differentiate the angular equation with respect to $r$ and use $\partial_r\epsilon=0$. This gives $V'\gamma^1\gamma^2\epsilon=0$. Since $(\gamma^1\gamma^2)^2=-I$, this matrix is invertible. A <parallel spinor> that vanishes at any point vanishes everywhere on the connected domain, by uniqueness of <parallel transport>. Thus a nonzero <parallel spinor> implies $V'=0$ throughout the patch. \b[There are no nonzero parallel spinors unless $V$ is constant.] The zero <spinor field> is always a solution, so “nonzero” is essential in interpreting the printed assertion. A smooth original $V$ of either sign is constant on such a patch exactly when $|V|$ is constant; $V=0$ is excluded because the given metric is not defined there.
Write the constant as $c=|V|>0$, and introduce $T=ct$, $\rho=r/c$, and $\phi=c\theta$. The metric becomes
$$
ds^2=-dT^2+d\rho^2+\rho^2d\phi^2.
$$
A circle at small $\rho$ has circumference $cP\rho$ if $\theta$ has period $P$, whereas its proper radius is $\rho$. Smoothness requires circumference divided by radius to tend to $2\pi$, not a multiple of $2\pi$. Conversely, when $cP=2\pi$, the coordinates $X=\rho\cos\phi$, $Y=\rho\sin\phi$ exhibit a smooth flat spatial disk. Hence
$$
\boxed{\theta\ \text{must have period }2\pi/c.}
$$
For a prescribed period $2\pi$, this requires $c=1$; otherwise there is a <conical singularity> at the apex.
For constant $V=c$, the time and radial equations make the components independent of $t,r$. The angular equation has all solutions
$$
\boxed{\epsilon(t,r,\theta)=\exp\!\left(\frac{c\theta}2\gamma^1\gamma^2\right)\epsilon_0
=\left[\cos\frac{c\theta}2\,I-\sin\frac{c\theta}2\,\gamma^0\right]\epsilon_0,}
$$
where $\epsilon_0$ is an arbitrary constant two-component <spinor field> and $\gamma^1\gamma^2=-\gamma^0$. There are two independent real solutions for real <spinor fields>, or two independent complex solutions if complex spinors are used. At the smooth angular period, these polar-frame components change sign. This is exactly the <polar-frame parallel spinor> transformation under the spin lift of a full rotation: the rotating polar <orthonormal frame> does not extend through the origin. The <spin structure> extending over the disk uses this antiperiodic polar trivialization; in a Cartesian <orthonormal frame> the same <parallel spinors> have constant components and extend smoothly. Requiring periodic components in the rotating frame would select the other spin structure on the punctured disk, which does not extend over its center.
Back to article page
