Solve the equation on the round three-sphere

The round sphere has no nonzero Levi-Civita parallel spinor. Nevertheless, it supports rigid supersymmetry: an auxiliary background can compensate for the spin connection. We will choose that background, solve every component of both spinor equations, and verify that the solutions cover the whole sphere.

The preparation is the Killing spinor equations and the global geometry of the round three-sphere. We use the latter’s frame and curvature conventions, rather than changing the orientation halfway through the calculation.

The parent matter theory is three-dimensional $\mathcal N=2$ with a continuous R-symmetry: four real Poincaré supercharges before Euclidean continuation. We are constructing its background, not yet a matter action or integration contour. A gauge-neutral chiral theory will provide the later application, so no gauge rank enters this calculation. The sphere metric is real and positive definite.

Put the geometric and source data together

Write the sphere of radius $\ell>0$ using dimensionless ambient coordinates,

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

The global coframe and its orientation are

$$e^a=\ell\left(y^0dy^a-y^ady^0+\epsilon_{abc}y^bdy^c\right), \qquad e^1\wedge e^2\wedge e^3>0,\quad \epsilon_{123}=1.$$

Both $b,c$ are summed. Let $E_a$ be the dual frame. From the geometry lesson,

$$de^a=\frac1\ell\epsilon_{abc}e^b\wedge e^c,\qquad \Omega_{ab}=\frac1\ell\epsilon_{abc}e^c,\qquad [E_a,E_b]=-\frac2\ell\epsilon_{abc}E_c.$$

Use Hermitian two-by-two gamma matrices with $\gamma_a\gamma_b=\delta_{ab}\mathbf1+i\epsilon_{abc}\gamma_c$. The spinor connection in this frame is

$$\nabla_a=E_a+\mathcal A_a,\qquad \mathcal A_a=\frac14\Omega_{a\,bc}\gamma_b\gamma_c =\frac{i}{2\ell}\gamma_a.$$

Here the first index in $\Omega_{a\,bc}$ is the direction $E_a$. The calligraphic $\mathcal A_a$ is the matrix spin connection; it is different from the R-symmetry connection $A_a$.

Choose the following sources:

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

The R-symmetry bundle is trivial and $A$ is its globally zero connection. The globally zero graviphoton $C$ gives $V=-i*dC=0$. The imaginary value of $H$ is allowed by the Euclidean reality convention. It has mass dimension one, as required. We will give its two-form potential description below; a single global potential cannot represent it.

Both background gravitini vanish. The remaining question is which supersymmetry labels leave them zero.

Solve all three equations in both families

Since $A=V=0$, the two independent equations reduce to the same differential operator:

$$\nabla_a\zeta=-\frac H2\gamma_a\zeta,\qquad \nabla_a\widetilde\zeta=-\frac H2\gamma_a\widetilde\zeta.$$

The labels have R-charges $+1$ and $-1$, respectively, but that distinction does not produce an opposite sign of $H$. Insert the connection and the chosen scalar:

$$\left(E_a+\frac{i}{2\ell}\gamma_a\right)\zeta =\frac{i}{2\ell}\gamma_a\zeta.$$

Consequently $E_a\zeta=0$ for $a=1,2,3$. The identical calculation gives $E_a\widetilde\zeta=0$. Because the three frame vectors span every tangent space, these conditions mean that the component functions are constant on the connected sphere:

$$\zeta=\begin{pmatrix}c_1\\c_2\end{pmatrix}, \qquad \widetilde\zeta=\begin{pmatrix}\widetilde c_1\\\widetilde c_2\end{pmatrix}, \qquad c_1,c_2,\widetilde c_1,\widetilde c_2\in\mathbb C.$$

These are components in the global frame, not components in a singular angular frame. Since $S^3$ is simply connected and has a unique spin structure, the global oriented coframe lifts to a global spin frame. It gives a global trivialization in which the columns above are smooth sections. No extra boundary identification is imposed on the sphere, and the R-connection introduces no transition phase.

This finds every solution: any solution has zero derivative along every $E_a$, so its two components must be constant. Each family therefore has exactly two complex solution directions. The background preserves the maximal solution space of these two rigid equations. The four complex parameters are the Euclidean complexification of the parent four-real-supercharge algebra; they are not a claim of eight real Poincaré supercharges.

The two constant columns remain independent. We did not obtain $\widetilde\zeta$ by imposing pointwise complex conjugation on $\zeta$. Setting one family to zero selects a subset of the available transformations, rather than changing the solution space of the background.

The components are constant, but the spinors are not Levi-Civita parallel:

$$\nabla_a\zeta=\frac{i}{2\ell}\gamma_a\zeta.$$

For a nonzero $\zeta$, the right side cannot vanish for every $a$, since each $\gamma_a$ is invertible. The auxiliary term is doing essential work. Its imaginary coefficient also makes the norm constant:

$$\partial_\mu(\zeta^\dagger\zeta) =-\frac{i}{2\ell}\zeta^\dagger\gamma_\mu\zeta +\frac{i}{2\ell}\zeta^\dagger\gamma_\mu\zeta=0.$$

The same check applies independently to $\widetilde\zeta$.

Check the result against curvature

There is a separate way to test the allowed value of $H$. For a constant scalar $H$ with $A=V=0$, define the connection

$$\mathscr K_a=\nabla_a+\frac H2\gamma_a.$$

Its curvature must annihilate a solution. In a noncoordinate frame the curvature is $\mathcal F^{\mathscr K}_{ab}=[\mathscr K_a,\mathscr K_b]-\mathscr K_{[E_a,E_b]}$. The subtraction is essential; the frame vectors do not commute.

The sphere’s spin curvature is $\mathcal F^\nabla_{ab}=[\gamma_a,\gamma_b]/(4\ell^2)$. For constant $H$, covariant constancy of the Clifford multiplication cancels the cross derivative terms. The extra matrix connection contributes $H^2[\gamma_a,\gamma_b]/4$. Thus

$$\mathcal F^{\mathscr K}_{ab} =\frac14\left(\frac1{\ell^2}+H^2\right) [\gamma_a,\gamma_b].$$

For $a=1,b=2$, the commutator is $2i\gamma_3$ and is invertible. A nonzero solution therefore requires

$$H^2=-\frac1{\ell^2}.$$

Our choice $H=-i/\ell$ passes. The other choice $H=+i/\ell$ also passes this necessary test, but has different component solutions in our fixed frame; exercise 1 constructs them.

The target paper uses $R_{ab}=-2\delta_{ab}/\ell^2$. Its equation (5.9) and the first case in section 5.2 give $R_{ab}=2H^2\delta_{ab}$ when $V=0$. Substituting our $H$ reproduces that Ricci tensor. This checks the source convention as well as the spinor equation; a positive-sphere Ricci convention cannot be inserted into this formula without translation.

Give the auxiliary scalar regular local potentials

The definition in equation (1.1) is $H=i*dB$. Therefore our background requires the globally defined three-form

$$G_B\equiv dB=-\frac1\ell\,\mathrm{vol}_g.$$

Its integral is

$$\int_{S^3}G_B=-2\pi^2\ell^2.$$

If $B$ were a single smooth two-form on the closed sphere, Stokes’ theorem would make this integral zero. The obstruction concerns the potential, not the smooth scalar $H$ or the spinors.

To exhibit local potentials, introduce polar coordinates

$$y^0=\cos\chi,\qquad (y^1,y^2,y^3)=\sin\chi\,n,\qquad 0\leq\chi\leq\pi,\quad n\in S^2.$$

With the unit two-sphere area form $\omega_2$, normalized by $\int_{S^2}\omega_2=4\pi$,

$$ds^2=\ell^2(d\chi^2+\sin^2\chi\,ds_{S^2}^2), \qquad \mathrm{vol}_g=\ell^3\sin^2\chi\,d\chi\wedge\omega_2.$$

Define

$$f(\chi)=\frac\chi2-\frac{\sin2\chi}{4}, \qquad f'(\chi)=\sin^2\chi.$$

On the sphere with the south pole removed, and the sphere with the north pole removed, respectively, take

$$B_N=-\ell^2 f(\chi)\,\omega_2,\qquad B_S=\ell^2\left(\frac\pi2-f(\chi)\right)\omega_2.$$

Direct differentiation gives $dB_N=dB_S=-\mathrm{vol}_g/\ell$. Applying the Hodge star yields $H=-i/\ell$ on both patches, including their overlap.

We must check the polar endpoints, where $\omega_2$ alone is not a smooth coordinate expression. Near the north pole use Cartesian normal coordinates $u=\chi n$ and $\rho=|u|=\chi$. Then

$$\omega_2=\frac{ u^1du^2\wedge du^3+ u^2du^3\wedge du^1+ u^3du^1\wedge du^2}{\rho^3}.$$

The coefficient multiplying the smooth numerator in $B_N$ is $-\ell^2 f(\rho)/\rho^3$. It extends smoothly to the origin because

$$\frac{f(\rho)}{\rho^3} =\frac13-\frac{\rho^2}{15} +\frac{2\rho^4}{315}+\cdots$$

is analytic in $\rho^2=(u^1)^2+(u^2)^2+(u^3)^2$. This proves smoothness of the two-form coefficients in Cartesian coordinates, rather than relying on its angular prefactor alone. At the south pole put $\rho=\pi-\chi$. The identity $\pi/2-f(\pi-\rho)=f(\rho)$ gives the same regularity check for $B_S$.

Their difference on the overlap is

$$B_S-B_N=k\,\omega_2,\qquad k=\frac{\pi\ell^2}{2}.$$

It has no single global one-form primitive on that overlap, since its integral over an $S^2$ slice is $4\pi k\ne0$. Refine the overlap using angular coordinates $n=(\sin\theta\cos\varphi,\sin\theta\sin\varphi,\cos\theta)$. On the patches excluding the south and north angular poles, respectively,

$$\Lambda_+=k(1-\cos\theta)\,d\varphi,\qquad \Lambda_-=-k(1+\cos\theta)\,d\varphi$$

satisfy $d\Lambda_\pm=k\omega_2$. Thus locally $B_S=B_N+d\Lambda_\pm$. The apparent angular singularities are removable at the included poles: for example, $(1-\cos\theta)d\varphi=(n^1dn^2-n^2dn^1)/(1+n^3)$. On a further azimuthal patch where $\varphi$ is a single-valued coordinate,

$$\Lambda_+-\Lambda_-=d(\pi\ell^2\varphi).$$

This supplies the local overlap data rather than pretending that one gauge parameter covers the whole overlap.

Our rigid background is specified globally by $H$ and $G_B$, with these local potential representatives. The local chiral couplings to be used later depend on this two-form background through $H$, so the representatives give the same couplings. No compact two-form gauge group or normalization of charged string operators has been specified here. Their large gauge transformations and possible flux conditions would be additional global data; the local supergravity definitions and this spinor solution do not quantize the radius.

Check the large-radius limit locally

The scalar source and spin-connection coefficients both scale as $1/\ell$. Their cancellation therefore approaches the ordinary flat-space equation if the radius grows while the sources are retuned to $H=-i/\ell$.

The comparison must hold physical distances fixed. Near the north pole set $X^a=\ell y^a$ and $y^0=\sqrt{1-|X|^2/\ell^2}$. The coframe is

$$e^a=y^0dX^a+ \frac{X^aX^b\,dX^b}{\ell^2y^0} +\frac1\ell\epsilon_{abc}X^b dX^c.$$

At fixed bounded $X$, it tends to $dX^a$. The metric tends to the Cartesian metric, $H$ tends to zero, and the constant frame components become the familiar constant flat-space labels. This is a local limit on regions of fixed physical size, not a statement that the entire compact sphere becomes one finite coordinate patch.

Practice

1. Change the sign of the auxiliary scalar

Keep the same oriented frame and $A=V=C=0$, but choose $H=+i/\ell$. Define

$$q(y)=y^0\mathbf1+i y^a\gamma_a.$$

Use the Clifford identity and the unit-sphere constraint to prove $q^{-1}=y^0\mathbf1-i y^a\gamma_a$ and $q^{-1}dq=i e^a\gamma_a/\ell$. Then find the global solutions. Does the opposite-charge spinor have to use the opposite sign of $H$?

First hint

The scalar part of $q^{-1}dq$ is $\sum_I y^I dy^I=0$. Its gamma-matrix part is the global coframe.

Second hint

The equation is now $(E_a+i\gamma_a/\ell)\varepsilon=0$ for either family. Differentiate $q^{-1}$ and test $q^{-1}\xi$ with $\xi$ constant.

Complete solution

The product of the proposed inverse with $q$ is $[(y^0)^2+y^ay^a]\mathbf1=\mathbf1$; the antisymmetric gamma term vanishes against $y^ay^b$. Expanding its differential gives

$$q^{-1}dq =i\left(y^0dy^a-y^ady^0+ \epsilon_{abc}y^bdy^c\right)\gamma_a =\frac i\ell e^a\gamma_a.$$

Therefore

$$d(q^{-1})=-\frac i\ell e^a\gamma_a q^{-1}, \qquad E_a(q^{-1})=-\frac i\ell\gamma_a q^{-1}.$$

It follows that

$$\zeta=q^{-1}\xi,\qquad \widetilde\zeta=q^{-1}\widetilde\xi$$

solve all components, with $\xi,\widetilde\xi$ arbitrary independent constant columns. These formulas are global: $q^{-1}$ is a smooth polynomial matrix in the ambient coordinates and is invertible everywhere. Conversely, the equation implies $d(q\varepsilon)=0$, so these are all the solutions.

The two-form potentials may be negated to produce the new $H$. Both charge sectors use that same new scalar. Changing the sign of $H$ selects another background; it is not the operation of changing a label’s R-charge. The number of solutions is unchanged, but their components in the original frame are no longer constant.

2. Compare parallel transport with rigid transport

Return to $H=-i/\ell$. Follow the unit-speed $E_3$ orbit through the north pole:

$$y(t)=\bigl(\cos(t/\ell),\,0,\,0,\,\sin(t/\ell)\bigr), \qquad 0\leq t\leq2\pi\ell.$$

The curve is closed and the global frame is the same at its beginning and end. Find the Levi-Civita parallel transport of a spinor column along it. Compare this with a rigid Killing spinor. Does a minus sign in the first transport make our globally constant Killing spinors inconsistent?

First hint

Along $E_3$, the Levi-Civita connection matrix is $i\gamma_3/(2\ell)$. A parallel spinor obeys a first-order matrix equation.

Second hint

Use $\gamma_3^2=\mathbf1$ to exponentiate the matrix. In the Killing equation the $H\gamma_3/2$ term cancels that matrix.

Complete solution

For a Levi-Civita parallel column $\psi(t)$,

$$\frac{d\psi}{dt}+\frac{i}{2\ell}\gamma_3\psi=0, \qquad \psi(t)= \left[\cos\frac{t}{2\ell}\,\mathbf1 -i\sin\frac{t}{2\ell}\,\gamma_3\right]\psi(0).$$

After one circuit, $\psi(2\pi\ell)=-\psi(0)$. This is the holonomy of the Levi-Civita spin connection along the curve. Curvature allows nontrivial parallel transport around a closed contractible loop.

A rigid Killing spinor instead obeys transport by $\mathscr K=\nabla+H\gamma/2$. Its matrix connection is zero in the global frame, so $\zeta(t)=\zeta(0)$ and it returns to itself. These are different transport equations. A globally defined Killing spinor is not required to be Levi-Civita parallel; the minus sign therefore creates no contradiction.

3. Change the radius while holding the source fixed

Start with radius $\ell_0$ and $H=-i/\ell_0$. Change the metric to a round sphere of radius $L>0$ while keeping $A=V=C=0$ and keeping that same value of $H$. Can a nonconstant spinor restore supersymmetry when $L\ne\ell_0$? What must change to obtain a supersymmetric large-radius limit?

First hint

Failure of a constant trial spinor would not exclude all other solutions. Use the curvature condition on any solution.

Second hint

The coefficient multiplying $[\gamma_1,\gamma_2]$ becomes $L^{-2}-\ell_0^{-2}$.

Complete solution

The curvature condition is

$$\frac14\left(\frac1{L^2}-\frac1{\ell_0^2}\right) [\gamma_1,\gamma_2]\varepsilon=0.$$

The commutator is invertible. For positive radii this forces $\varepsilon=0$ unless $L=\ell_0$. A nonconstant ansatz cannot evade the obstruction.

To keep the chosen family supersymmetric, retune $H=-i/L$ as the radius changes. Then both the spin connection and auxiliary source tend to zero in the local flat limit. Sending $L$ to infinity at fixed nonzero $H$ would leave a different background, with the same obstruction as the flat-space constant-$H$ example.

Checkpoint

The complete background data now include the round metric, a global spin frame, trivial R and graviphoton connections, the smooth scalar $H=-i/\ell$, and regular local representatives of its three-form field strength. Both independent spinor families have two global solution directions, verified component by component and against curvature.

The next task is to put matter on this background. The curved chiral multiplet supplies the transformations, and the complete sphere action checks their invariance term by term. Substituting the sphere metric into the flat action alone would discard the compensation we have just established.

Evaluate every spinor component and test the change of frame in the Killing-spinor laboratory.