Geometry of the round sphere

How can a computer distinguish a negative sphere-curvature convention from a wrong connection? Reproduce the connection from its defining equations, keep the tensor indices in a declared order, and compare with an experiment that does not use the same formula. Here that experiment is parallel transport: a tangent vector returns rotated after traveling around a closed curve.

Prepare with the geometry of the round three-sphere. That lesson supplies the analytic derivation. This laboratory implements it, deliberately breaks one sign, and tests finite transport using the sphere’s embedding in four-dimensional Euclidean space.

Specify the geometry and the output

The manifold is the round, positive-definite three-sphere of physical radius $\ell>0$. Its dimensionless ambient coordinates satisfy $y^Ty=1$, and its metric is the restriction of $\ell^2dy^Tdy$ to tangent vectors. This is a geometric calculation: no gauge field, matter theory or supersymmetry-preserving background has been specified. A sphere metric alone does not determine preserved supercharges.

Use the orientation $e^1\wedge e^2\wedge e^3>0$ and the global coframe

$$e^a=\ell M_{aI}dy^I,\qquad M=\begin{pmatrix} -y^1&y^0&-y^3&y^2\\ -y^2&y^3&y^0&-y^1\\ -y^3&-y^2&y^1&y^0 \end{pmatrix}. \tag{LAB03.1}$$

The three frame indices run over $1,2,3$; the four ambient indices run over $0,1,2,3$. The code uses zero-based indices for both. Before differentiating anything, it verifies the polynomial identities

$$My=0,\qquad MM^T=(y^Ty)\mathbf1_3,\qquad M^TM=(y^Ty)\mathbf1_4-yy^T.$$

On $y^Ty=1$ these show that the coframe has rank three everywhere and reproduces the tangent metric. Testing them at a few nonsingular points would not establish global rank. The code verifies the polynomials before restricting to the sphere, so the check also covers points where an angular chart degenerates.

The target output is

$$\Omega_{ab}=\frac1\ell\epsilon_{abc}e^c,\qquad \mathcal R^a{}_b=\frac1{\ell^2}e^a\wedge e^b,\qquad R_{ab}=-\frac2{\ell^2}\delta_{ab},\qquad R=-\frac6{\ell^2}. \tag{LAB03.2}$$

Here $\nabla E_b=\Omega^a{}_b E_a$ defines our connection, and $\mathcal R=d\Omega+\Omega\wedge\Omega$. The last two expressions use the target paper’s Ricci convention, which we will contract explicitly. The curvature two-form is not itself a scalar curvature.

Derive rather than insert the connection

Let $E_a=M_{aI}\partial_{y^I}/\ell$ be the dual tangent vectors. Differentiate the coefficients of (LAB03.1) in the ambient space and evaluate on $E_b,E_c$. The result is

$$D_{abc}:=de^a(E_b,E_c)=\frac2\ell\epsilon_{abc},\qquad [E_b,E_c]=-\frac2\ell\epsilon_{bca}E_a.$$

The factor two is essential: $de^1=2e^2\wedge e^3/\ell$. In the compact expression $de^a=\epsilon_{abc}e^b\wedge e^c/\ell$, both ordered pairs are summed.

Write $w_{abc}=\Omega_{ab}(E_c)$. Metric compatibility means $w_{abc}=-w_{bac}$, leaving nine unknown coefficients at each point. The torsion equation gives nine linear equations,

$$D_{abc}+w_{acb}-w_{abc}=0\qquad(b<c). \tag{LAB03.3}$$

The script solves this system and obtains $w_{abc}=\epsilon_{abc}/\ell$. Constancy of these coefficients is a result of the calculation; it does not mean that the connection is flat. The basis one-forms still have nonzero exterior derivatives.

For example, calculate the $(1,2)$ curvature entry:

$$d\Omega^1{}_2=\frac2{\ell^2}e^1\wedge e^2,\qquad \Omega^1{}_3\wedge\Omega^3{}_2=-\frac1{\ell^2}e^1\wedge e^2.$$

Their sum has coefficient $1/\ell^2$, not $2/\ell^2$. The implementation evaluates every entry with

F[a, b, c, d] = sum(
    omega[a, b, f] * de[f, c, d]
    + omega[a, f, c] * omega[f, b, d]
    - omega[a, f, d] * omega[f, b, c]
    for f in range(3)
)

Thus F[a,b,c,d] means $\mathcal R^a{}_b(E_c,E_d)$. No extra factor $1/2$ belongs in this evaluated component: the $1/2$ appears when reconstructing a two-form from an ordered-pair sum.

Contract the indices in the paper’s order

The paper writes its connection with the input frame index first: $\omega_a{}^b=\Omega^b{}_a$. Its commutator tensor and Ricci contraction are consequently

$$R_{cd a}{}^b=F^b{}_{a cd},\qquad R_{ac}=\sum_b R_{ab c}{}^b=\sum_b F^b{}_{c ab}. \tag{LAB03.4}$$

These are the definitions in Appendix A.2, (A.8)–(A.9) and the following paragraph. Inserting the computed curvature gives

$$R_{ac}=\frac1{\ell^2}\sum_b \left(\delta_a^b\delta_{bc}-\delta_b^b\delta_{ac}\right) =-\frac2{\ell^2}\delta_{ac}.$$

Each fixed $\delta_b^b$ in the sum is one. The final scalar is $-6/\ell^2$. A common GR convention reverses the first two indices in this Ricci contraction and gives $+6/\ell^2$ for the same connection and geometry. That legitimate convention change leaves torsion zero. Negating the connection while keeping the coframe fixed is a different operation.

The script also starts afresh from the coordinate metric

$$g_{\mu\nu}=\ell^2\operatorname{diag} \bigl(1,\cos^2\chi,\sin^2\chi\bigr),\qquad 0<\chi<\frac\pi2,$$

with both angular coordinates of period $2\pi$. Direct Christoffel differentiation gives $R_{\mu\nu}=-2g_{\mu\nu}/\ell^2$. Integrating its volume density yields $2\pi^2\ell^3$. This second calculation uses neither the solved frame connection nor its structure coefficients. The coordinate chart is local; global coverage came from (LAB03.1).

Run the editable calculation

Save these three files together:

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

• sphere_geometry.py, the complete calculation;
• requirements.txt, the exact dependency versions;
• expected-output.json, the complete default output.

The tested environment is Python 3.13.7, SymPy 1.13.3 and mpmath 1.3.0 in an isolated environment. From the download directory, create and activate a virtual environment using your platform’s usual commands, then run:

python -m pip install -r requirements.txt
python sphere_geometry.py
python sphere_geometry.py --wrong-connection-sign

No repository checkout, network request during execution or external data file is needed. To inspect just the symbolic result interactively, this complete small entry point uses the downloaded module:

from sphere_geometry import exact_geometry

result = exact_geometry()
print(result["paper_Ricci_diagonal"])
print(result["paper_scalar"])
print(result["volume"])

It prints ['-2/ell**2', '-2/ell**2', '-2/ell**2'], -6/ell**2, and 2*pi**2*ell**3. The default run passes 163 grouped exact assertions, covering the global frame, structure equations, solved connection, curvature, coordinate comparison and spin lift. Some assertions compare whole matrices and many curvature components vanish; this count is not 163 independent physical results.

Mathematical object	Code representation
Global coframe coefficients	M, a $3\times4$ symbolic matrix
$de^a(E_b,E_c)$	de[a,b,c]
$\Omega_{ab}(E_c)$	omega[a,b,c], solved pointwise
$\mathcal R^a{}_b(E_c,E_d)$	F[a,b,c,d]
Paper Ricci contraction	paper_ricci(F)
Three transported ambient vectors	V, a $4\times3$ numerical matrix

The deliberate-failure mode exits successfully only after detecting the specified wrong result. Its output includes torsion component 4/ell and scalar 18/ell**2. It does not label this as an alternative sphere convention.

Check finite transport without the connection formula

Choose a closed latitude inside the great two-sphere $y^3=0$:

$$y(t)=(c,s\cos t,s\sin t,0),\qquad s=\sqrt{1-c^2},\quad -1<c<1,\quad 0\leq t\leq2\pi. \tag{LAB03.5}$$

Here $t$ is an angular parameter, not physical arclength. A transported vector $v(t)$ is an ambient four-vector tangent to the sphere, not a three-column of coefficients in the global frame. Intrinsic parallel transport means that its ambient derivative is normal to the sphere. Differentiating $y\cdot v=0$ then fixes the normal component:

$$\frac{dv}{dt}=-y(t)\left(\frac{dy}{dt}\cdot v(t)\right). \tag{LAB03.6}$$

This equation follows directly from the induced metric; it contains no $M$, $\Omega$ or Ricci tensor. The factor $\ell$ cancels when the physical embedding $X=\ell y$ is used. Transport therefore depends on the angular path, not its overall scale.

An exact comparison comes from a second, local tangent basis along this curve:

$$u_\theta=(-s,c\cos t,c\sin t,0),\qquad u_\varphi=(0,-\sin t,\cos t,0),\qquad u_3=(0,0,0,1).$$

Projecting their derivatives onto the tangent space gives $\nabla_tu_\theta=c\,u_\varphi$ and $\nabla_tu_\varphi=-c\,u_\theta$. For $v=a\,u_\theta+b\,u_\varphi$, parallel transport requires $a'=cb$, $b'=-ca$. Consequently, a vector starting at $u_\theta(0)$ returns as

$$v(2\pi)=\cos(2\pi c)u_\theta(0) -\sin(2\pi c)u_\varphi(0). \tag{LAB03.7}$$

The third vector is fixed. The oriented rotation is $-2\pi c$, equivalently $2\pi(1-c)$ modulo $2\pi$. The latter is the northern cap area divided by $\ell^2$, since its area is $2\pi\ell^2(1-c)$. This is ordinary sphere focusing, compatible with the paper’s negative Ricci contraction. It is not negative geometric Gaussian curvature.

The code integrates all three initial tangent vectors using fourth-order Runge–Kutta, without projecting or renormalizing after each step. Tangency and norm drift therefore remain useful error diagnostics. At the default $c=3/5$, the exact rotation is $4\pi/5$ modulo $2\pi$:

Equal steps around the circle	Largest ambient-component error	Largest inner-product error
64	$1.25840\times10^{-6}$	$1.51030\times10^{-6}$
128	$7.89533\times10^{-8}$	$9.55829\times10^{-8}$
256	$4.94297\times10^{-9}$	$6.01022\times10^{-9}$

These are actual errors against (LAB03.7), computed with 50 decimal digits of arithmetic. The component-error ratios are approximately $15.94$ and $15.97$, approaching the fourth-order expectation $2^4=16$. Using 70 digits at the same step count leaves the reported errors unchanged to twelve significant digits: integration error dominates roundoff. Neither the working precision nor the exact symbolic result makes the finite-step transport exact.

Try python sphere_geometry.py --cos-latitude 1/3 --steps 64 128 256 --digits 70. The rotation changes, but the sphere curvature does not. The diagnostic compares with the new exact rotation rather than with the old numerical answer.

Exercises

1. Interpret the deliberately wrong sign

Keep $e^a$ fixed and replace $\Omega$ by $-\Omega$. Derive the torsion and curvature returned by the failure mode. Why does the scalar $18/\ell^2$ rule out interpreting this as just the opposite Ricci convention?

First hint

The exterior derivative term changes sign; the quadratic connection term does not.

Second hint

For the original connection, $d\Omega^1{}_2$ has coefficient $2/\ell^2$, while $(\Omega\wedge\Omega)^1{}_2$ has coefficient $-1/\ell^2$.

Complete solution

Originally $\Omega^a{}_b\wedge e^b=-de^a$. The new torsion is therefore $T'^a=de^a-\Omega^a{}_b\wedge e^b=2de^a$, so $T'^1(E_2,E_3)=4/\ell$. It is not the Levi-Civita connection of the specified coframe.

The new curvature entry is $(-2-1)e^1\wedge e^2/\ell^2=-3e^1\wedge e^2/\ell^2$, and the other entries follow cyclically. Taking the same paper contraction multiplies its original Ricci tensor by $-3$, giving $R'_{ab}=6\delta_{ab}/\ell^2$ and $R'=18/\ell^2$. Merely reversing the Ricci contraction would give $+6/\ell^2$ with zero torsion. The factor three and nonzero torsion diagnose a different connection, not a notation translation.

2. Hold the path or hold its physical size

Double $\ell$ while holding the angular curve (LAB03.5) fixed. What happens to curvature and holonomy? Then instead hold a small physical circumference $L$ fixed and choose the northern latitude $c>0$. Find its holonomy as a function of $L,\ell$ and its large-radius limit. Explain the difference between the two experiments.

First hint

The circumference is $L=2\pi\ell\sqrt{1-c^2}$. A fixed angular path is not a fixed-size path.

Second hint

Insert the positive square root for $c$ into $2\pi(1-c)$, then expand $\sqrt{1-u}$ for small $u$.

Complete solution

At fixed $c$, the scalar becomes one-quarter as large, $R'=-3/(2\ell^2)$, while (LAB03.6) and its holonomy are unchanged. The cap area has grown by four, compensating the curvature scale.

At fixed $L$, with $0<L<2\pi\ell$ and $c>0$, the small-cap rotation is

$$\Theta(L,\ell)=2\pi\left[1-\sqrt{1-\frac{L^2}{4\pi^2\ell^2}}\right] =\frac{L^2}{4\pi\ell^2} +O\left(\frac{L^4}{\ell^4}\right). \tag{LAB03.8}$$

Thus it tends to zero in the locally flat limit. The leading area is the planar disk value $L^2/(4\pi)$, multiplied by $1/\ell^2$. Holding the loop’s angular size fixed instead makes its physical size grow with $\ell$, so that experiment does not approach a fixed local flat-space loop. No change of supersymmetric background is involved in either comparison.

3. Repair a constant-matrix curvature calculation

The spin lift of the solved connection is $\mathcal A_a=i\gamma_a/(2\ell)$, where $\gamma_a\gamma_b=\delta_{ab}\mathbf1+i\epsilon_{abc}\gamma_c$. A program identifies spin curvature with $[\mathcal A_1,\mathcal A_2]$. Compute its answer and repair it using the actual frame bracket. Does a zero derivative of the matrix entries imply zero curvature?

First hint

For noncommuting frame vectors, the curvature is $[\nabla_{E_1},\nabla_{E_2}]-\nabla_{[E_1,E_2]}$.

Second hint

Use $[E_1,E_2]=-2E_3/\ell$. The subtraction removes the vector differentiation and also leaves a connection contribution.

Complete solution

The naive matrix commutator is $-i\gamma_3/(2\ell^2)$. Restoring the bracket term gives

$$\mathcal F_{12} =[\mathcal A_1,\mathcal A_2]+\frac2\ell\mathcal A_3 =\frac{i}{2\ell^2}\gamma_3 =\frac1{4\ell^2}[\gamma_1,\gamma_2]. \tag{LAB03.9}$$

The last expression is the spin lift of the vector curvature in (LAB03.2), in agreement with the paper’s (A.11)–(A.12). The missing term changes the sign. Constant coefficients in a noncoordinate frame neither imply zero curvature nor permit dropping the frame bracket. The downloaded script retains this correction in every spin-curvature component.

What the calculation establishes

You have a globally nonsingular coframe, a connection derived from compatibility and torsion, an explicitly translated Ricci trace, and a finite-transport test with a measured discretization error. The symbolic checks concern exact geometry; the numerical transport concerns a specified closed curve. Neither establishes a Killing-spinor solution, action invariance or a quantum result. Those require their own background data and calculations. Return to the canonical sphere calculation whenever a field-theory curvature coupling needs this sign.