Squashed spheres and their full background data

“Put the theory on a squashed sphere” does not yet specify a supersymmetric problem. Which directions are stretched? Which auxiliary sources change? Which spinors remain global? We will answer all three questions for a complete family, then compare it with another geometry also commonly denoted by $S^3_b$.

Prepare with the round-sphere solution and global spinor and R-bundle patching. We retain their orientation, spinor contractions and the target paper’s negative round-sphere scalar curvature.

Stretch one global direction

The parent is three-dimensional $\mathcal N=2$, with four real Lorentzian Poincaré supercharges and an R symmetry. The metric is Euclidean and positive definite. There is no choice of dynamical gauge rank in this background calculation.

Use the dimensionless ambient coordinates $\sum_{I=0}^3(y^I)^2=1$ and the global forms of the round-sphere lesson:

$$\eta^a=y^0dy^a-y^ady^0+\epsilon_{abc}y^bdy^c, \qquad d\eta^1=2\eta^2\wedge\eta^3 \quad\text{and cyclically}. \tag{M10.1}$$

The orientation is $\eta^1\wedge\eta^2\wedge\eta^3>0$. These three forms are independent everywhere, including the poles of angular coordinates. Choose $v_0=\eta^3$ as the Hopf direction. For radius scale $\ell>0$ and real parameter $b>0$, define $u=\log b$, $h=\cosh u=(b+b^{-1})/2$ and $d=2\sinh u=b-b^{-1}$. Thus $h^2-d^2/4=1$.

$$\begin{aligned} g_b&=\ell^2\bigl[(\eta^1)^2+(\eta^2)^2+h^2(\eta^3)^2\bigr],\\ A&=V=dh\,\eta^3, & C&=\frac{i\ell d}{2}\eta^3, & H&=\frac{ih}{\ell}. \end{aligned} \tag{M10.2}$$

Products in the metric are symmetric tensor products. The symbol $V$ in this display is the one-form dual to $V^\mu$. $A$ is the background R connection; $C$ is the graviphoton potential. They are fixed sources, not gauge fields to be integrated over. Their definitions are $V=-i*_b dC$ and $H=i*_b dB$, where $B$ is a two-form potential described locally below.

This is the real-$A,V$ branch of P (5.11)–(5.12), (8.4)–(8.5). The metric has $SU(2)\times U(1)$ isometry for $h\ne1$. It stretches the Hopf direction by $h\geq1$; the paper also discusses other real or complex source branches, which we are not identifying with real positive $b$ here.

To compare directly with P (5.11), locally take $\eta^3=\tfrac12[d\psi+(1-\cos\theta)d\varphi]$ and $(\eta^1)^2+(\eta^2)^2=\tfrac14(d\theta^2+\sin^2\theta\,d\varphi^2)$. The paper’s radius $r$ is our $\ell$, and its Hopf coordinate has period $4\pi$. The ambient construction fixes the global sphere behind these singular angular coordinates; no quotient has been introduced.

The radius factors can be checked without supersymmetry. An orthonormal coframe is $e^1=\ell\eta^1$, $e^2=\ell\eta^2$, $e^3=\ell h\eta^3$. Consequently,

$$*_b d\eta^3=\frac{2h}{\ell}\eta^3,\qquad -i*_b dC=dh\,\eta^3,\qquad V_a=\frac d\ell\delta_{a3},\qquad \operatorname{Vol}(g_b)=2\pi^2\ell^3h . \tag{M10.3}$$

In particular $A_a,V_a,H$ have mass dimension one. The one-form $C$ has length units in these dimensionless coordinates, as its dual-derivative definition requires.

At $b=1$, $h=1$ and $d=0$: the metric is round, $A=V=C=0$, and $H=+i/\ell$. This is the positive auxiliary branch. The earlier constant-component round solution used $H=-i/\ell$; its first exercise constructed the positive branch. Keeping the old negative value while varying (M10.2) would give a different problem.

Solve both spinor equations

Let $(\gamma_1,\gamma_2,\gamma_3)=(\sigma_3,-\sigma_1,-\sigma_2)$. Cartan’s equation for the stretched frame gives

$$\Omega_{12}=\frac{2/h-h}{\ell}e^3,\qquad \Omega_{23}=\frac h\ell e^1,\qquad \Omega_{31}=\frac h\ell e^2 . \tag{M10.4}$$

For example $de^3=(2h/\ell)e^1\wedge e^2$, while $de^1=(2/\ell h)e^2\wedge e^3$ and cyclically for $de^2$. These derivatives check every coefficient in (M10.4). The spin matrix is $\Gamma=\tfrac14\Omega_{ab}\gamma_a\gamma_b$, summing both antisymmetric index orders.

For the charge-$+1$ label the full equation is $(\nabla_a-iA_a)\zeta=-H\gamma_a\zeta/2-iV_a\zeta-\epsilon_{abc}V_b\gamma_c\zeta/2$. Since $A=V$, the identity-matrix terms cancel. Combining the remaining sources with (M10.4) yields

$$d\zeta+\mathcal B\zeta=0,\qquad \mathcal B= \left(ih\gamma_1-\frac d2\gamma_2\right)\eta^1+ \left(ih\gamma_2+\frac d2\gamma_1\right)\eta^2+ i\gamma_3\eta^3 . \tag{M10.5}$$

This intermediate equation is a useful sign check. Reversing the R charge changes the sign of the two $d$ terms, not the sign of $H$.

Define two invertible matrices,

$$U(y)=y^0\mathbf1+i y^a\gamma_a,\qquad D=\exp\!\left(\frac u2\gamma_3\right). \tag{M10.6}$$

$U$ is unitary with determinant one and obeys $U^{-1}dU=i\gamma_a\eta^a$. The constant matrix $D$ is a positive Hermitian matrix for real $u$; it is an algebraic similarity transformation, not a unitary change of spin frame. The Clifford identity gives $D(i\gamma_1)D^{-1}=ih\gamma_1-(d/2)\gamma_2$ and $D(i\gamma_2)D^{-1}=ih\gamma_2+(d/2)\gamma_1$. Therefore $\mathcal B=D(U^{-1}dU)D^{-1}$.

$$\boxed{\ \zeta(y)=D\,U(y)^{-1}\zeta_0,\qquad \widetilde\zeta(y)=D^{-1}U(y)^{-1}\widetilde\zeta_0\ }, \qquad \zeta_0,\widetilde\zeta_0\in\mathbb C^2 . \tag{M10.7}$$

To verify the first expression, differentiate $U^{-1}$: $dU^{-1}=-U^{-1}dU\,U^{-1}$. Inserting it in (M10.5) cancels the two terms. For the opposite-charge equation the corresponding matrix is $D^{-1}(U^{-1}dU)D$, giving the second expression by the same differentiation. This verifies all components of both equations, without imposing pointwise conjugacy between the labels.

Every solution has this form: multiplication of the first equation by $UD^{-1}$ gives $d(UD^{-1}\zeta)=0$, and similarly for the second. There are exactly two complex directions in each family. These four Euclidean directions are the complexification of the parent four-real-supercharge algebra, not eight real Poincaré charges. At $b=1$, $D=\mathbf1$, and (M10.7) becomes the positive-branch round solution.

Check curvature and global data

The construction uses the ordinary unquotiented $S^3$, its unique spin structure and the trivial R line bundle. The global coframe has a global spin lift; $U$ and $D$ are smooth and invertible everywhere. Thus a nonzero constant column in (M10.7) gives a nowhere-zero global spinor. There are no additional circle identifications or flat R twists to choose on this simply connected sphere.

The one-forms $A$ and $C$ in (M10.2) are global. Although $F_A=dA$ need not vanish, it is exact and has zero flux through every closed two-cycle. The auxiliary scalar is also global. Its potential is more subtle:

$$dB=-iH\,\mathrm{vol}_{g_b} =\ell^2h^2\,\eta^1\wedge\eta^2\wedge\eta^3, \qquad \int_{S^3}dB=2\pi^2\ell^2h^2 . \tag{M10.8}$$

As on the round sphere, no single smooth two-form $B$ has this exterior derivative. Local potentials and their overlap gauge data suffice; exercise 2 constructs them. We impose no additional compact two-form flux quantization, since no such charge lattice or normalization is part of this rigid-background choice.

There is a separate geometric check. Computing the curvature of (M10.4) and using the paper’s Ricci contraction gives the frame components

$$R_{ab}=\frac1{\ell^2} \operatorname{diag}(2h^2-4,\ 2h^2-4,\ -2h^2), \qquad R=\frac{2h^2-8}{\ell^2}. \tag{M10.9}$$

Substituting $V_a=d\,\delta_{a3}/\ell$ and $H=ih/\ell$ into $R_{ab}=-V_aV_b+\delta_{ab}(V^2+2H^2)$ reproduces this answer. Also $\nabla_aV_b=(hd/\ell^2)\epsilon_{ab3}=-iH\epsilon_{abc}V_c$. These are checks against P (5.9); the explicit global solutions, rather than a curvature test alone, established global existence.

For a concrete point in the family, $b=2$ gives $h=5/4$, $d=3/2$, $\operatorname{Vol}=5\pi^2\ell^3/2$ and $R=-39/(8\ell^2)$. The spinor similarity matrix is $D=(3\mathbf1+\gamma_3)/(2\sqrt2)$. Both $D$ and $U$ are invertible, so the spinors remain regular although their frame components vary with position.

Compare squashing parameters through their definitions

The metric depends only on $h=\cosh u$, whereas $A,V,C$ also depend on $\sinh u$. Thus $b$ and $b^{-1}$ have the same metric and $H$, but opposite $A,V,C$. Moreover $h'(1)=0$ while $d'(1)=2$. The complete family has a first-order source deformation even though its metric has none. No partition-function derivative has yet been computed.

The normalization of $v_0=\eta^3$ agrees with the response calculation: under $y^0=(1-|x|^2)/(1+|x|^2)$, $y^a=2x^a/(1+|x|^2)$, it becomes $v_0=\Omega^2[-x^2dx^1+x^1dx^2+(1-|x|^2)dx^3/2+x^3x^a dx^a]$ with $\Omega=2/(1+|x|^2)$. At unit radius, (M10.2) is exactly the complete source family in P (8.4)–(8.5).

A different common representative is the toric metric in W16 (2.33)–(2.34):

$$g_{\rm E}=f(\chi)^2d\chi^2+ \ell_1^2\cos^2\chi\,d\varphi^2+ \ell_2^2\sin^2\chi\,d\vartheta^2,\qquad b_{\rm E}=\sqrt{\ell_1/\ell_2}. \tag{M10.10}$$

Here $0\leq\chi\leq\pi/2$, and both angles have period $2\pi$. A concrete smooth choice is the induced ellipsoid: $f^2=\ell_1^2\sin^2\chi+\ell_2^2\cos^2\chi$. Near $\chi=0$ the $\vartheta$ circle collapses, requiring $f(0)=\ell_2$; near $\pi/2$ the $\varphi$ circle collapses, requiring $f(\pi/2)=\ell_1$. The endpoint and collapsing-circle labels in the accompanying prose of W16 v2 are interchanged; these conditions follow directly from its displayed metric. An arbitrary interpolation also needs the appropriate smooth even behavior at the endpoints, not just the endpoint values.

W16 (2.41) constructs this representative with $V=0$, $H=i/f$ and $A^{(R)}=\tfrac12(1-\ell_1/f)d\varphi+\tfrac12(1-\ell_2/f)d\vartheta$. It exhibits one label of each R charge in (2.42), with further solutions in the round limit. These sources differ from (M10.2). W16 also uses the opposite sign for the Ricci scalar; this page retains P’s convention.

Data	The homogeneous family above	The toric ellipsoid representative
Parameter definition	$h=(b+b^{-1})/2$	$b_{\rm E}^2=\ell_1/\ell_2$
Generic metric isometry	$SU(2)\times U(1)$	$U(1)\times U(1)$
Auxiliary one-form	$V=dh\,\eta^3$	$V=0$
Auxiliary scalar	$H=ih/\ell$, constant	$H=i/f(\chi)$
Displayed rigid solutions	Two per charge family	One per charge family

Equal numerical parameter values do not identify these metrics and source fields. Comparing protected partition functions requires the preserved supersymmetry, the deformation argument and the contact-term prescription as well. The geometric dictionary above supplies the background data for that subsequent question.

Exercises

1. Reverse the signed deformation

Compare $b=2$ and $b=1/2$ at fixed $\ell$ and fixed ambient coordinates. Are the backgrounds related by an ordinary R gauge transformation? What happens to the two solution families?

Hint 1

The metric sees $\cosh u$. Inspect which sources see $\sinh u$.

Hint 2

An R gauge transformation leaves $dA$ invariant. In the spinor formula, $D(-u)=D(u)^{-1}$.

Solution

Both have $h=5/4$, hence the same metric, volume, curvature and $H$. The value of $d$ changes from $3/2$ to $-3/2$, so $A,V,C$ all reverse sign. Since $dA=2dh\,\eta^1\wedge\eta^2$ is nonzero, $A$ and $-A$ cannot be connected by $A\mapsto A+d\lambda$ at fixed coordinates. In addition, an R transformation cannot change the global tensor $V$.

Replacing $D$ with $D^{-1}$ exchanges the two formulas (M10.7), consistently with exchanging the charge-$+1$ and charge-$-1$ equations when $A,V$ reverse sign. This is a relation between the background equations. It does not by itself prove a charge-conjugation symmetry of an arbitrary matter theory or an equality of its partition functions in arbitrary contact-term schemes.

2. Supply local two-form potentials

Write $y^0=\cos\rho$, $y^a=\sin\rho\,n^a$, with $n\in S^2$ and $0\leq\rho\leq\pi$. Let $\omega_2$ be the oriented unit-sphere area form, $\int\omega_2=4\pi$, so $\eta^1\wedge\eta^2\wedge\eta^3=\sin^2\rho\,d\rho\wedge\omega_2$. Use $F(\rho)=\rho/2-\sin(2\rho)/4$ to find potentials regular near the north and south poles. Check their flux and explain why their difference need not have a single global one-form primitive on the overlap.

Hint 1

$F'(\rho)=\sin^2\rho$. Choose its additive constant separately at each pole.

Hint 2

Near the north pole, $F(\rho)=\rho^3/3+O(\rho^5)$. On a sphere inside the overlap, integrate the difference before attempting to write it as an exact two-form.

Solution

Take $B_N=\ell^2h^2 F(\rho)\omega_2$ and $B_S=\ell^2h^2[F(\rho)-\pi/2]\omega_2$. Both differentiate to (M10.8). The cubic vanishing of the north coefficient cancels the apparent singularity of $\omega_2$ in Cartesian normal coordinates; the south coefficient vanishes cubically in $\pi-\rho$ in the same way.

Their difference is $\pi\ell^2h^2\omega_2/2$, with integral $2\pi^2\ell^2h^2$ over an overlap sphere. It is closed but cannot be exact on that entire overlap, since an exact form has zero integral there. Refining the overlap to contractible patches provides local one-form gauge primitives, just as in the round construction. This is compatible with smooth $H=ih/\ell$; it forbids only the proposed single global $B$ or global overlap primitive. No new flux quantization follows without specifying an additional compact two-form gauge theory and its units.

3. Derive the ellipsoid’s endpoint conditions

Embed a three-dimensional ellipsoid in $\mathbb R^4$ by $X^1+iX^2=\ell_1\cos\chi\,e^{i\varphi}$ and $X^3+iX^4=\ell_2\sin\chi\,e^{i\vartheta}$. Find its induced metric, test absence of a conical singularity at each endpoint, and evaluate the circumference ratio of its two noncollapsing endpoint circles. Is this ratio $h$ for the homogeneous family?

Hint 1

Differentiate both complex coordinates and sum their absolute squares. Terms mixing the radial and angular differentials vanish.

Hint 2

At $\chi=0$, proper radial distance is approximately $f(0)\chi$, while the small $\vartheta$ circle has circumference $2\pi\ell_2\chi$. Reverse the roles at the other endpoint.

Solution

The induced metric is (M10.10) with $f^2=\ell_1^2\sin^2\chi+\ell_2^2\cos^2\chi$. At $\chi=0$ the ratio of the small circumference to proper radius is $2\pi\ell_2/f(0)$, which equals $2\pi$ because $f(0)=\ell_2$. Writing $\rho=\pi/2-\chi$ at the other endpoint gives $2\pi\ell_1/f(\pi/2)$, hence $f(\pi/2)=\ell_1$. This explicit $f$ also has the required smooth even expansion at both ends.

The noncollapsing circles have circumferences $2\pi\ell_1$ at $\chi=0$ and $2\pi\ell_2$ at $\pi/2$, with ratio $\ell_1/\ell_2=b_{\rm E}^2$. By contrast, $h=(b+b^{-1})/2$ stretches a homogeneous Hopf direction relative to the transverse round frame. It is not this endpoint-circle ratio. For example, assigning the same value $b_{\rm E}=b=2$ gives a toric ratio four, but a homogeneous stretch $5/4$.

The background needed for a response calculation is now fully specified: metric, sources, global choices and preserved spinors. What squashing the sphere measures differentiates the partition function along this family, treating current insertions and local subtractions with the same conventions.

Use the Section 5 paper guide to locate this calculation in the original argument and compare its local and global assumptions.

Orbifolds and spindle backgrounds replaces smooth polar caps by specified finite quotients. Construct the spin and R transitions before counting regular charged sections.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 5, additional supercharges