Compute the geometry of the round three-sphere

A sphere has no single nonsingular angular coordinate chart. That need not prevent us from choosing a frame that is smooth everywhere. On the three-sphere we can construct such a frame explicitly, calculate its connection, and check the curvature without concealing a coordinate singularity.

Use Covariant derivatives of spinors for the frame and spin-connection definitions. Our background here is purely geometric: the round, positive-definite three-sphere of radius $\ell>0$. No gauge field, matter action, or supersymmetry-preserving background sources have yet been chosen. Its curvature alone does not specify a number of preserved supercharges.

Describe the sphere without angular poles

Write a point of the sphere as four dimensionless ambient coordinates $y^I$, with $I=0,1,2,3$, obeying

$$(y^0)^2+(y^1)^2+(y^2)^2+(y^3)^2=1, \qquad ds^2=\ell^2\sum_{I=0}^3dy^I\,dy^I.$$

The differentials in the metric are restricted to tangent vectors, for which $\sum_I y^I dy^I=0$. The physical ambient coordinates would be $X^I=\ell y^I$. Although there are four $y^I$, the constraint leaves a three-dimensional manifold.

Define three one-forms, with $\epsilon_{123}=1$,

$$\eta^a=y^0dy^a-y^ady^0+\epsilon_{abc}y^bdy^c, \qquad e^a=\ell\eta^a,\qquad a=1,2,3.$$

Both $b,c$ are summed. For example, $\eta^1=y^0dy^1-y^1dy^0+y^2dy^3-y^3dy^2$. Every coefficient is a polynomial in the ambient coordinates, so these forms are smooth even where an angular chart fails. Smoothness alone does not prove that they form a basis: we must also check their rank.

Let $M$ be the three-by-four matrix such that $\eta^a=M_{aI}dy^I$:

$$M= \begin{pmatrix} -y^1&y^0&-y^3&y^2\\ -y^2&y^3&y^0&-y^1\\ -y^3&-y^2&y^1&y^0 \end{pmatrix}.$$

Matrix multiplication gives, on the unit sphere,

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

The last matrix projects onto the tangent space orthogonal to $y$. Consequently, for any tangent displacement,

$$\sum_a\eta^a\eta^a =dy^T(\mathbf1_4-yy^T)dy =\sum_I dy^I dy^I.$$

Thus $ds^2=\sum_a e^a e^a$, and the three rows are linearly independent everywhere. We have a global orthonormal coframe, not merely a formula on one patch. Its dual vector fields are

$$E_a=\frac1\ell M_{aI}\frac{\partial}{\partial y^I}.$$

They are tangent because $My=0$, and $e^a(E_b)=\delta^a{}_b$. Their action on a function restricted to the sphere is independent of how that function is extended away from the sphere. Choose the orientation $e^1\wedge e^2\wedge e^3>0$; at $y=(1,0,0,0)$ it agrees with $dy^1\wedge dy^2\wedge dy^3>0$.

Differentiate the coframe

The inverse tangent relations follow from $M^TM$:

$$dy^0=-y^a\eta^a,\qquad dy^a=y^0\eta^a-\epsilon_{abc}y^b\eta^c.$$

Differentiate the explicit expression for $\eta^1$:

$$d\eta^1=2\,dy^0\wedge dy^1+2\,dy^2\wedge dy^3.$$

Substitution of the inverse relations cancels the mixed coefficients and gives $d\eta^1=2\eta^2\wedge\eta^3$; the factor multiplying this last wedge is $2\sum_I(y^I)^2=2$. Cyclically, the complete result is

$$de^a=\frac1\ell\epsilon_{abc}e^b\wedge e^c, \qquad de^1=\frac2\ell e^2\wedge e^3.$$

The compact expression sums both $(b,c)=(2,3)$ and $(3,2)$. Forgetting that factor of two changes the radius inferred from the curvature.

Our frame is not a coordinate basis. Applying the definition of an exterior derivative to $e^c(E_a)=\delta^c{}_a$ gives

$$de^c(E_a,E_b)=-e^c([E_a,E_b]), \qquad [E_a,E_b]=-\frac2\ell\epsilon_{abc}E_c.$$

The nonzero bracket measures this failure to be a coordinate basis. It does not obstruct global smoothness.

Solve for the connection and its curvature

Use the convention from the frame lesson, $\nabla E_b=\Omega^a{}_b E_a$. Cartan’s torsion equation is $de^a+\Omega^a{}_b\wedge e^b=0$. The antisymmetric connection

$$\Omega_{ab}=\frac1\ell\epsilon_{abc}e^c$$

solves it: exchanging $e^c\wedge e^b$ into $e^b\wedge e^c$ supplies the minus sign needed to cancel $de^a$. For example, $\Omega^1{}_2=e^3/\ell$ and $\Omega^1{}_3=-e^2/\ell$. Metric compatibility and zero torsion identify this as the Levi-Civita connection.

The curvature of this connection is the matrix of two-forms

$$\mathcal R^a{}_b=d\Omega^a{}_b+ \Omega^a{}_c\wedge\Omega^c{}_b.$$

Compute an entry rather than infer its sign from a memorized sphere formula:

$$\begin{aligned} d\Omega^1{}_2&=\frac1\ell de^3 =\frac2{\ell^2}e^1\wedge e^2,\\ \Omega^1{}_3\wedge\Omega^3{}_2 &=\left(-\frac{e^2}\ell\right)\wedge \left(-\frac{e^1}\ell\right) =-\frac1{\ell^2}e^1\wedge e^2. \end{aligned}$$

Their sum is $e^1\wedge e^2/\ell^2$. The cyclic entries give

$$\mathcal R^a{}_b=\frac1{\ell^2}e^a\wedge e^b.$$

The radius appears with the required inverse-length-squared coefficient in an orthonormal basis. Unlike the cylindrical example, this curvature cannot be removed by rotating the frame.

Take the paper’s curvature trace explicitly

The target paper orders the input frame index first: $\omega_{\mu a}{}^b=\Omega_\mu{}^b{}_a$. Its equation (A.9), version 3, defines

$$R_{\mu\nu a}{}^b= \partial_\mu\omega_{\nu a}{}^b-\partial_\nu\omega_{\mu a}{}^b +\omega_{\nu a}{}^c\omega_{\mu c}{}^b -\omega_{\mu a}{}^c\omega_{\nu c}{}^b.$$

Substituting $\omega_a{}^b=\Omega^b{}_a$ identifies this with the output-$b$, input-$a$ entry of the curvature two-form. In orthonormal components,

$$\mathcal R^b{}_a=\frac12R_{cd a}{}^b e^c\wedge e^d, \qquad R_{cd a}{}^b= \frac1{\ell^2} \left(\delta_c^b\delta_{da}-\delta_d^b\delta_{ca}\right).$$

The paper then declares $R_{\mu\nu}=R_{\mu\rho\nu}{}^\rho$. The same contraction in the orthonormal frame is

$$\begin{aligned} R_{ab} &=\sum_{c=1}^3R_{ac b}{}^c\\ &=\frac1{\ell^2}\sum_c \left(\delta_a^c\delta_{cb}-\delta_c^c\delta_{ab}\right) =-\frac2{\ell^2}\delta_{ab}. \end{aligned}$$

In the summed expression each fixed $\delta_c^c$ is one. Therefore

$$\boxed{R_{\mu\nu}=-\frac2{\ell^2}g_{\mu\nu}, \qquad R=-\frac6{\ell^2}.}$$

This is the negative round-sphere value stated immediately after (A.9). A common GR Ricci convention instead contracts the commutator tensor as $R_{\rho\mu\nu}{}^\rho$, with the first two indices reversed, and gives the opposite sign. The geometry has not changed. When importing a curvature coupling into an action, translating the Ricci definition is essential; simply replacing this result by a familiar positive sphere value would change that action.

Check in a coordinate patch

As an independent calculation, introduce

$$\begin{aligned} y^0&=\cos\chi\cos\alpha,& y^1&=\cos\chi\sin\alpha,\\ y^2&=\sin\chi\cos\beta,& y^3&=\sin\chi\sin\beta. \end{aligned}$$

For $0<\chi<\pi/2$ and angles $\alpha,\beta$ of period $2\pi$, the embedded metric becomes

$$ds^2=\ell^2\left(d\chi^2+\cos^2\chi\,d\alpha^2 +\sin^2\chi\,d\beta^2\right).$$

The nonzero Christoffel symbols, including symmetric lower-index partners, are

$$\Gamma^\chi_{\alpha\alpha}=\sin\chi\cos\chi,\quad \Gamma^\chi_{\beta\beta}=-\sin\chi\cos\chi,\quad \Gamma^\alpha_{\chi\alpha}=-\tan\chi,\quad \Gamma^\beta_{\chi\beta}=\cot\chi.$$

Use the coordinate commutator tensor with the same derivative-index order as the paper. For example,

$$R_{\chi\alpha\chi}{}^\alpha =\partial_\chi(-\tan\chi)+(-\tan\chi)^2=-1.$$

The $\beta$ contribution to $R_{\chi\chi}$ is likewise $\partial_\chi(\cot\chi)+\cot^2\chi=-1$. The other contractions give

$$R_{\chi\chi}=-2,\qquad R_{\alpha\alpha}=-2\cos^2\chi,\qquad R_{\beta\beta}=-2\sin^2\chi,$$

with off-diagonal entries zero. Contracting with the inverse metric again gives $-6/\ell^2$.

The angular chart degenerates at its endpoints because one of its circles shrinks. The ambient coframe remains smooth and of rank three there, as the earlier matrix identities establish. The chart also supplies a normalization check independent of the connection:

$$\operatorname{Vol}(S^3_\ell) =\ell^3(2\pi)^2 \int_0^{\pi/2}\sin\chi\cos\chi\,d\chi =2\pi^2\ell^3.$$

Thus the parameter $\ell$ is the metric radius, rather than a radius hidden behind a factor in the frame normalization.

The connection acting on spinors

The generic construction from the preceding lesson gives

$$\nabla_{E_a}\psi=E_a\psi+\mathcal A_a\psi,\qquad \mathcal A_a=\frac1{4\ell}\epsilon_{bca}\gamma^b\gamma^c.$$

For the three-dimensional Clifford orientation $\gamma^a\gamma^b=\delta^{ab}\mathbf1+i\epsilon^{abc}\gamma_c$, the ordered-pair sum reduces to

$$\mathcal A_a=\frac{i}{2\ell}\gamma_a, \qquad \nabla_{E_a}\psi=E_a\psi+\frac{i}{2\ell}\gamma_a\psi.$$

This identity can be used abstractly; it does not require choosing explicit matrices. It also agrees with the paper’s (A.11), since its all-lower connection is $\omega_{ab}=-\Omega_{ab}$. The derivative is now ready to act on a proposed spinor field. Deciding which spinor equation describes a supersymmetric background requires the background-source construction still to come.

Exercises

1. Double the radius

Keep the coordinates $y^I$ fixed and replace $\ell$ by $2\ell$. How do the coframe, the connection one-form, its orthonormal components, the scalar curvature, and the volume change? Explain why an unchanged connection one-form does not imply unchanged scalar curvature.

First hint

Write $\Omega_{ab}=\epsilon_{abc}\eta^c$ before changing the radius.

Second hint

Distinguish the one-form from its coefficient in the rescaled basis $e'^a=2e^a$.

Complete solution

The coframe doubles. The connection one-form $\Omega_{ab}=\epsilon_{abc}\eta^c$ is unchanged as a form in the dimensionless coordinates. Its orthonormal coefficient halves:

$$\Omega_{ab}=\frac1{2\ell}\epsilon_{abc}e'^c.$$

The curvature two-form is also unchanged in the coordinate basis, but its coefficient in $e'^a\wedge e'^b$ is one-quarter as large. Metric contraction therefore gives $R'=-6/(2\ell)^2=-3/(2\ell^2)$. The three-dimensional volume grows by $2^3=8$. Curvature in inverse-length-squared units and volume in length-cubed units provide independent dimensional checks. The inverse metric used in the trace changes when the metric is rescaled.

2. Do constant connection matrices imply zero curvature?

Use $\mathcal A_a=i\gamma_a/(2\ell)$ and $[E_a,E_b]=-2\epsilon_{abc}E_c/\ell$. Compute the spin curvature operator

$$\mathcal F_{ab}= [\nabla_{E_a},\nabla_{E_b}] -\nabla_{[E_a,E_b]}.$$

Explain the subtraction and compare the result with $\frac14\mathcal R^c{}_d(E_a,E_b)\gamma_c\gamma^d$.

First hint

The matrices $\mathcal A_a$ are constant in this frame, but the frame vectors do not commute.

Second hint

Separate the commutator of the matrices from the connection contribution in $\nabla_{[E_a,E_b]}$.

Complete solution

The subtraction removes differentiation along the vector bracket, leaving the tensorial curvature action. Constancy of the matrices eliminates $E_a\mathcal A_b$, but not the bracket correction:

$$\begin{aligned} \mathcal F_{ab} &=[\mathcal A_a,\mathcal A_b] +\frac2\ell\epsilon_{abc}\mathcal A_c\\ &=-\frac{i}{2\ell^2}\epsilon_{abc}\gamma_c +\frac{i}{\ell^2}\epsilon_{abc}\gamma_c\\ &=\frac{i}{2\ell^2}\epsilon_{abc}\gamma_c. \end{aligned}$$

From $\mathcal R^c{}_d=e^c\wedge e^d/\ell^2$, the other expression is

$$\frac1{4\ell^2}[\gamma_a,\gamma_b] =\frac{i}{2\ell^2}\epsilon_{abc}\gamma_c.$$

They agree. Omitting the noncoordinate-frame correction would even give the opposite sign. Constant connection coefficients in a noncoordinate frame need not describe a flat connection.

3. Add a flat fourth direction

Form the product metric $ds_4^2=ds_{S^3_\ell}^2+dw^2$. Extend the stated curvature and Ricci conventions to four dimensions. Find its scalar curvature and decide whether it is an Einstein metric, meaning $R_{AB}=\Lambda g_{AB}$ for one constant $\Lambda$. What changes if the flat factor is instead a time direction with metric $-dt^2$?

First hint

The sphere frame and connection are independent of the new coordinate; all connection entries mixing it with the sphere can be zero.

Second hint

Compare the Ricci tensor in a sphere direction and in the flat direction before tracing.

Complete solution

The mixed curvature entries and flat-direction Ricci component vanish. In an orthonormal product frame,

$$R_{AB}=\operatorname{diag} \left(-\frac2{\ell^2},-\frac2{\ell^2},-\frac2{\ell^2},0\right), \qquad R=-\frac6{\ell^2}.$$

The flat component would require $\Lambda=0$, while a sphere component requires $\Lambda=-2/\ell^2$. Thus the product is not Einstein for finite $\ell$. Replacing $dw^2$ by $-dt^2$ changes the inverse metric in the flat direction, but it still multiplies a zero Ricci component. The same scalar and the same failure to be Einstein follow. A three-sphere inside a four-dimensional product is not a four-sphere; counting the ambient or total dimension is insufficient to choose a sphere-curvature formula.

Carry the geometric data into field theory

We now have a global coframe, a torsion-free connection, and a curvature trace checked in two descriptions. Before using them in an action, reproduce the ordered-pair factor in $de^1$, identify the Ricci contraction that produces the minus sign, and explain why the angular endpoints are harmless for the global frame.

The next field-theory question is how a metric and other prescribed fields act as sources for operators. That source language will tell us which additional backgrounds accompany the sphere metric when we impose supersymmetry.

For a complementary check, the Appendices A/B paper guide reconstructs the spin-curvature identity and recovers the sphere sign directly from a spinor.

Reproduce the connection and test finite parallel transport in the round-sphere geometry laboratory.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Appendices A and B, conventions and geometry