Verify a Killing spinor

A spinor that has constant components in one frame usually has nonconstant components in another. Can a calculation recognize the same supersymmetry in both descriptions? Here you will evaluate every component of the equation, deliberately omit a term in the frame transformation, and check what happens at a coordinate chart’s missing point.

Prepare with the complete round-sphere solution. That lesson supplies the analytic geometry and global existence proof. This laboratory implements its equation and tests a second description; it does not replace a global argument by sampling points.

Specify the background and the residual

The setting is three-dimensional Euclidean rigid $\mathcal N=2$ supersymmetry, with a Lorentzian parent containing four real Poincaré supercharges. The metric is the round sphere of radius $\ell>0$, the spin structure is the unique one on $S^3$, and the R-symmetry bundle is trivial. Take

$$A=V=C=0,\qquad H=-\frac{i}{\ell}. \tag{LAB04.1}$$

These are fixed sources, not fields to be integrated over. The smooth scalar $H$ has the local two-form potentials already constructed in the sphere lesson; it does not come from a single global two-form. No compact two-form charge lattice is added here. There is no dynamical gauge group or gauge rank in this background calculation.

Use the explicit Hermitian matrices

$$\gamma_1=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad \gamma_2=\begin{pmatrix}0&-1\\-1&0\end{pmatrix},\quad \gamma_3=\begin{pmatrix}0&i\\-i&0\end{pmatrix},\qquad \gamma_a\gamma_b=\delta_{ab}\mathbf1+i\epsilon_{abc}\gamma_c. \tag{LAB04.2}$$

For a coframe $e^a=e^a{}_{i}dx^i$, define the coordinate Clifford matrix $G_i=e^a{}_{i}\gamma_a$ and matrix spin connection $\Gamma_i$. Our residual is the left side minus the right side of P (3.1):

$$\mathcal R_i[\Psi]=\partial_i\Psi+\Gamma_i\Psi+ \frac H2G_i\Psi. \tag{LAB04.3}$$

The columns of the two-by-two matrix $\Psi$ are two separate trial spinors. Testing its two columns tests arbitrary constant linear combinations. There are three coordinate directions and two spinor components, hence twelve scalar equations for this basis.

The opposite R-charge family obeys the same operator because $A=V=0$. Its two coefficients are independent; it is not imposed to be the complex conjugate of the first family. Thus the calculation finds two complex solution directions in each family, the Euclidean complexification of the parent algebra. It does not double the number of real parent supercharges. P (3.2) leaves $H$ unchanged when the R-charge changes.

Use one chart without forgetting the rest of the sphere

Let $x=(x^1,x^2,x^3)\in\mathbb R^3$, $d=1+|x|^2$, and set

$$y^0=\frac{1-|x|^2}{d},\qquad y^a=\frac{2x^a}{d}, \qquad (y^0)^2+y^ay^a=1. \tag{LAB04.4}$$

This chart misses the south pole $(y^0,y^a)=(-1,0)$. Its coordinates are dimensionless; the physical metric is $4\ell^2 dx\cdot dx/d^2$. Coordinate connection matrices are dimensionless, whereas orthonormal-frame coefficients have dimension inverse length. A constant normalization supplies the spinor label’s mass dimension $-1/2$; the basis matrices below describe its dimensionless component functions.

Define the smooth ambient matrix

$$q(y)=y^0\mathbf1+i y^a\gamma_a,\qquad q^{-1}=q^\dagger=y^0\mathbf1-i y^a\gamma_a. \tag{LAB04.5}$$

The unit-sphere constraint gives $q^\dagger q=\mathbf1$ and $\det q=1$. The following left and right coframes are global:

$$e_{L,R}^a=\ell\left(y^0dy^a-y^ady^0 \mathbin{\pm}\epsilon_{abc}y^bdy^c\right), \qquad q^{-1}dq=\frac{i}{\ell}e_L^a\gamma_a, \quad dq\,q^{-1}=\frac{i}{\ell}e_R^a\gamma_a. \tag{LAB04.6}$$

Both ordered indices $b,c$ are summed. The label “right” changes the coframe, not the orientation of the sphere: conjugation by $q$ rotates the left coframe by an element of $SO(3)$. In the chart, both coframe determinants are $(2\ell/d)^3>0$.

The candidate Levi-Civita connections, in the ordinary Cartan convention $de^a+\Omega^a{}_b\wedge e^b=0$, are

$$\Omega_{L\,ab}=\frac1\ell\epsilon_{abc}e_L^c, \qquad \Omega_{R\,ab}=-\frac1\ell\epsilon_{abc}e_R^c, \qquad \Gamma_{L\,i}=\frac{i}{2\ell}G_{L\,i}, \quad\Gamma_{R\,i}=-\frac{i}{2\ell}G_{R\,i}. \tag{LAB04.7}$$

The program checks the induced metric and the torsion-free equation for these inputs; metric compatibility follows from the displayed antisymmetry of $\Omega_{ab}$. The sign is consistent with P (A.8)–(A.11): its input-first connection has all-lowered components $\omega_{iab}=-\Omega_{iab}$, so its $-i\omega_{iab}\epsilon^{abc}\gamma_c/4$ equals our $\Omega_{iab}\gamma_a\gamma_b/4$.

Now test

$$\Psi_L=\mathbf1,\qquad \Psi_R=q. \tag{LAB04.8}$$

The left residual cancels between its two connection terms. For the right one, $\partial_iq=iG_{R\,i}q/\ell$, so its three terms have coefficients $i/\ell$, $-i/(2\ell)$ and $-i/(2\ell)$. Their sum is zero. The script evaluates the literal coordinate derivatives and all matrix entries before comparison with zero.

Globality is a separate fact: the ambient expressions for $q$ and the two coframes are smooth on the entire sphere. In particular, $q=-\mathbf1$ at the missing south pole, with no directional ambiguity. The global spin frame argument supplies the lift of the left coframe, and $q$ is the explicit lift relating it to the right one. An exact identity in $x$ alone would not have supplied that extension.

Transform the connection, not just its matrix indices

In a fixed coordinate chart the change is

$$G_{R\,i}=qG_{L\,i}q^{-1},\qquad \Gamma_{R\,i}=q\Gamma_{L\,i}q^{-1} -(\partial_iq)q^{-1}. \tag{LAB04.9}$$

To see the derivative term, require $(\partial_i+\Gamma_{R\,i})(q\psi)=q(\partial_i+\Gamma_{L\,i})\psi$ and apply the product rule. The script checks (LAB04.9) independently of the residual. Merely conjugating $\Gamma_L$ gives the wrong connection even though it preserves matrix traces.

At the north pole $x=0$, $q=\mathbf1$, $\partial_iq=2i\gamma_i$, and $G_{L\,i}=G_{R\,i}=2\ell\gamma_i$. The correct right-frame residual is $2i\gamma_i-i\gamma_i-i\gamma_i=0$. With only conjugation, it becomes $2i\gamma_i$. This is a hand check of a sign and factor that the computer should detect.

Download and reproduce

Download the complete lab (ZIP) with the program, pinned dependencies, reference output and run instructions.

Save the editable calculation, pinned requirements, and complete expected output in one folder. The tested environment is Python 3.13.7, SymPy 1.13.3 and mpmath 1.3.0. Only SymPy is used directly; its supporting mpmath dependency is pinned too.

In Windows PowerShell, from that folder:

python -m venv .venv
.venv/Scripts/python.exe -m pip install -r requirements.txt
.venv/Scripts/python.exe killing_spinor.py

On macOS or Linux, create the environment with python3 -m venv .venv, then use .venv/bin/python for the remaining commands. Installation needs package-registry access; execution needs no network or repository files.

Mathematical object	Code name
Dimensionless coordinate and positive radius	x, ell
Smooth ambient lift	q
Coframe matrices, row $a$ and column $i$	frames
Coordinate Clifford matrices and spin connections	curved, connection
Three complete matrix equations	residual
Exact rational-function reduction	clean, is_zero

The default output includes

{
  "total": 64,
  "scalar_equations_per_frame_per_charge_family": 12,
  "left_Killing_residual_zero": true,
  "right_Killing_residual_zero": true,
  "profile_pole_sample_zero": true,
  "profile_global_residual_zero": false
}

The 64 grouped assertions include matrix identities, the input geometry and negative controls; they are not 64 separate supersymmetry results. The two charge families have identical operators here, so duplicating their identical arithmetic would not add an independent check. No numerical tolerance, grid or truncation is used: denominators are powers of $d>0$ and $\ell>0$, and exact rational reduction decides the local identities. This does not make a wrongly specified connection correct.

An independent chart-and-seam benchmark

Instead of starting with Maurer forms, one can compute Christoffel symbols from the diagonal metric and insert them into the tetrad postulate. This gives a separate check of the spin connection and its signs. The two oriented diagonal coframes and solutions are

$$\begin{aligned} f_N^a&=\frac{2\ell\,dx^a}{1+|x|^2},& U_N(x)&=\frac{\mathbf1+i x\cdot\gamma}{\sqrt{1+|x|^2}},\\ f_S^a&=-\frac{2\ell\,ds^a}{1+|s|^2},& U_S(s)&=\frac{\mathbf1-i s\cdot\gamma}{\sqrt{1+|s|^2}}. \end{aligned} \tag{LAB04.10}$$

The south chart uses $y^0=(|s|^2-1)/(1+|s|^2)$, $y^a=2s^a/(1+|s|^2)$ and misses the north pole. On the overlap $x=s/|s|^2$. The minus sign in $f_S$ compensates the orientation-reversing coordinate inversion. For either chart, with its own coordinate $u$, the directly computed connection is

$$\Gamma_i^{\rm diag}=-\frac{[\gamma_i,u\cdot\gamma]}{2(1+|u|^2)}. \tag{LAB04.11}$$

Substitution of $U_N$ or $U_S$, its own coframe and the same $H$ gives zero for all three residual matrices. On the punctured overlap, write $r=|s|>0$, $n=s/r$. Their spin-frame transition is

$$T(s)=-i n\cdot\gamma,\qquad U_S(s)=T(s)U_N(x(s)). \tag{LAB04.12}$$

In comparing connections, first pull the north one-form back with $\partial x^j/\partial s^i$; then include $-(\partial_iT)T^{-1}$. The independent check derives both connections from their metrics and tests that full equality. It also follows a closed great circle through both charts. The two seam matrices restore the identity return map; treating a half-angle component formula as a globally fixed frame would give a spurious minus sign. These are exact coordinate and transition checks, not a numerical claim about additional backgrounds.

Change one input

1. Omit the frame-derivative term

Run

.venv/Scripts/python.exe killing_spinor.py --omit-frame-term

Predict which frame fails and calculate the sum $\sum_i\operatorname{tr}(\mathcal R_i^\dagger\mathcal R_i)$ at $x=0$ for the two basis columns. Is this sum an invariant squared norm of the residual one-form?

Hint 1

Only the right connection was changed. The residual left over from (LAB04.9) is $\partial_iq$.

Hint 2

Each $\gamma_i^2=\mathbf1$ and $\operatorname{tr}\mathbf1=2$. To turn a coordinate sum into a metric norm, insert the inverse metric at the pole.

Solution

The left test passes and the right test fails. At the pole each residual is $2i\gamma_i$, so the displayed coordinate sum is $3\times4\times2=24$, exactly the diagnostic output. The physical contraction instead uses $g^{ij}=\delta^{ij}/(4\ell^2)$ there and equals $6/\ell^2$ for the two normalized basis columns. It is $3/\ell^2$ for either single normalized column.

The command exits successfully because it detects this intended error. It does not certify the altered connection. Changing only a spinor’s frame while keeping the old source scalar is allowed; changing only part of the connection transformation is not a consistent frame change.

2. A point test that falsely passes

In the left frame replace the first constant column $c=(1,0)^T$ by $\psi=y^0c$, keeping the background fixed. Find its full residual norm, identify where a point test misses it, and calculate its mean over the round sphere. Could multiplication by an arbitrary nonconstant smooth scalar give another solution?

Hint 1

The two matrix connection terms still cancel. The only residual is $d y^0$ times $c$; use the inverse stereographic metric.

Hint 2

The restriction of the ambient coordinate has $|d y^0|_g^2=(1-(y^0)^2)/\ell^2$. Rotational symmetry makes the mean of each of the four squared ambient coordinates $1/4$.

Solution

Since $\partial_i y^0=-4x^i/d^2$,

$$\mathcal R_i=-\frac{4x^i}{d^2}c,\qquad g^{ij}\mathcal R_i^\dagger\mathcal R_j =\frac{1-(y^0)^2}{\ell^2},\qquad \left\langle|\mathcal R|^2\right\rangle=\frac{3}{4\ell^2}.$$

The norm vanishes at both poles but is positive elsewhere. Testing only $x=0$ therefore gives a false impression. The average also follows from the direct ratio $\ell^{-2}\int_0^\pi\sin^4\chi\,d\chi/\int_0^\pi\sin^2\chi\,d\chi$.

For a general smooth multiplier $f$, the residual is $(df)c$. A nonzero constant column has at least one nonzero component, so the full residual vanishes only if $df=0$. Connectedness then makes $f$ constant. The test rejects this particular class of nonconstant trials; the complete solution count is supplied by the analytic argument in the sphere lesson.

3. A direction-dependent limit in the wrong frame

As $x=R n$ with $R\to\infty$, $U_N(x)\to i n\cdot\gamma$. Does that rule out a smooth spinor at the south pole? Use the overlap formula to decide, then compare with the global right-frame trial $q$.

Hint 1

The diagonal north coframe is defined only on its chart. Its components need not have a direction-independent limit in a frame that fails at the missing point.

Hint 2

On the overlap $s=n/R$ and $T=-i n\cdot\gamma$. Multiply before taking the limit; $(n\cdot\gamma)^2=\mathbf1$.

Solution

The south components are $T U_N=U_S(s)$, whose limit at $s=0$ is $\mathbf1$, independent of $n$. The full south expression is smooth there. The apparent directional limit belonged to a missing-point frame, not to an obstruction in the spinor section. On the overlap the Clifford matrices and the spin connection transform too, including the coordinate Jacobian and inhomogeneous term.

The right global frame provides another description: $q(x)\to-\mathbf1$ as $R\to\infty$, independent of direction, and its ambient expression extends polynomially to the pole. Constant left-frame components, nonconstant right-frame components and the two diagonal-chart expressions describe the same two-dimensional solution space. A limit of components becomes a global test only after the frame and transition data are specified.

The laboratory checks the stated classical background equation and its global realization. It does not test a matter action, choose its Euclidean integration contour, determine a partition function or establish a spinor on a different quotient of the sphere. Those changes need their own source and global data.