Orbifolds and spindle backgrounds

A fractional magnetic flux need not violate charge quantization. On an orbifold, the missing information can be the action of a finite rotation group on a charged field. The useful question is therefore more precise than “can we put supersymmetry on a cone?”: can the metric, spinors and gauge bundles all be made regular on the covering disks, with compatible finite group actions?

Prepare with the reproduction method, Seifert fibers, complete squashed backgrounds, and flux and holonomy. Our first reproduction is a twisted spindle times a circle. We then change its interior metric while holding its orbifold orders fixed. This is a geometry and bundle calculation; a regulated interacting partition function is a later project.

Specify the quotient charts

A two-dimensional orbifold chart consists of a smooth disk, a finite group acting on it, and an identification of the quotient with a neighborhood in the space. Near a point of order $n$, use a disk coordinate $w$ with

$$w\sim e^{2\pi i/n}w.$$

A field must also carry a representation of this group. Regularity means smoothness and the specified transformation law on the disk, not just boundedness in a singular polar coordinate.

A spindle has the topology of a sphere with two such points. We use coprime positive orders $(n_+,n_-)$. Unlike an exceptional fiber in the Seifert example, the product of this spindle with a circle has genuine orbifold loci: the finite rotation does not also translate the circle to make the total action free. We do not assume that the whole spindle is one finite quotient of a smooth sphere.

Let $0\leq\theta\leq\pi$, with $\varphi$ and $\psi$ both of period $2\pi$. The circle radius is $\beta>0$ and the equatorial radius is $\ell>0$. Choose

$$\begin{aligned} ds^2&=\beta^2d\psi^2+ \ell^2\bigl[F(\theta)^2d\theta^2+\sin^2\theta\,d\varphi^2\bigr],\\ F(\theta)&=a+b\cos\theta,\qquad a=\frac{n_++n_-}{2},\qquad b=\frac{n_+-n_-}{2}. \end{aligned} \tag{D07.1}$$

Thus $F(0)=n_+$, $F(\pi)=n_-$ and $F>0$. Near the north pole, $\rho_+=\ell n_+\theta$ gives $ds^2_\Sigma=d\rho_+^2+\rho_+^2d\varphi^2/n_+^2+O(\rho_+^2)d\rho_+^2+O(\rho_+^4)d\varphi^2$. The covering angle is $\varphi/n_+$ and the cone angle downstairs is $2\pi/n_+$. At the south pole use $\rho_-=\ell n_-(\pi-\theta)$ and covering angle $-\varphi/n_-$; that minus sign preserves the orientation of the surface. All radial coefficients have even Taylor series, so the lifted metrics are smooth at their disk origins.

This specializes Inglese–Martelli–Pittelli, O23 (2.30)–(2.37): set their $x=\cos\theta$, $\Omega=0$, $f(x)=F(\theta)/\sin\theta$, and their dimensionless circle radius to $\beta/\ell$, then restore the overall $\ell^2$. Their twist and anti-twist are distinct backgrounds. We construct the twist here.

Solve the local equations before checking the poles

The parent is a three-dimensional $\mathcal N=2$ theory with four real flat-space supercharges. A concrete probe is one free chiral multiplet of flavor charge $+1$, scalar R charge zero and no superpotential. There is no dynamical gauge group. We seek one solution in each of the two independent Euclidean R-charge families; that does not enlarge the parent supersymmetry.

Orient the coframe as $e^1=\beta d\psi$, $e^2=\ell Fd\theta$, $e^3=\ell\sin\theta d\varphi$, and use $(\gamma_1,\gamma_2,\gamma_3)=(\sigma_3,-\sigma_1,-\sigma_2)$. Set $c=\cos\theta/F$ and $H=V=C=0$. Cartan’s equation gives $\omega_{23}=-c\,d\varphi$, hence $\Gamma_\varphi=-ic\gamma_1/2$ and $\Gamma_\theta=\Gamma_\psi=0$. The Killing-spinor equations reduce to $(\nabla-iA_R)\zeta=0$ and $(\nabla+iA_R)\widetilde\zeta=0$.

On the cylinder excluding the poles, choose

$$A_R=\frac{1-c}{2}d\varphi,\qquad \zeta=e^{i\varphi/2}u_+,\qquad \widetilde\zeta=e^{-i\varphi/2}u_-,\qquad u_+=\binom{1}{0},\quad u_-=\binom{0}{1}. \tag{D07.2}$$

For example, $(\partial_\varphi+\Gamma_\varphi-iA_{R\varphi})\zeta =[i/2-ic/2-i(1-c)/2]\zeta=0$. The opposite-charge equation has the opposite three signs and also vanishes. The other two coordinate equations vanish separately. With the spinor contraction matrix $\varepsilon_{\alpha\beta}=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$, their bilinear is $K=\beta^{-1}\partial_\psi$.

The curvature of the first spinor equation is $-ic'(\gamma_1-1)d\theta\wedge d\varphi/2$. Here $c'=-a\sin\theta/F^2$ is nonzero in the interior. It excludes the other spinor component; the opposite equation excludes the opposite component. Take periodic spin structure along $\psi$, with $A_{R\psi}=0$. The pair is then global in the circle direction. The half-angle signs in (D07.2) concern the polar frame around the spindle, not an antiperiodic thermal circle.

The connection agrees with the $H=\Omega=0$ twist in O23 (2.40)–(2.41), with its cylinder phase chosen as $\varphi/2$. We derived our spinor columns directly in the stated frame; columns written in another gamma-matrix basis cannot simply be copied into it.

Make the spin and R actions global

Take the specific orders $(3,5)$, so $F=4-\cos\theta$. The cylinder R connection has endpoint values $1/3$ and $3/5$. Its regular covering-disk representatives and labels are

$$\begin{array}{c|c|c} &A_R\text{ on the cover}&\zeta\text{ in the polar frame}\\\hline N&A_R-\tfrac13d\varphi&e^{i\varphi/6}u_+\\ S&A_R-\tfrac35d\varphi&e^{-i\varphi/10}u_+ \end{array} \tag{D07.3}$$

The tilded labels have opposite phases. Both connection coefficients vanish quadratically at their corresponding pole. After changing from the polar spin frame to a smooth Cartesian spin frame on each covering disk, these labels have constant components.

These are covering-chart gauges, not ordinary gauges on two coarse disks. Their comparison phase $e^{4i\varphi/15}$ has integer winding four on the common fifteen-fold angular cover, where $\varphi$ has period $30\pi$. The finite deck actions below specify how to descend this comparison.

There is still a finite-group test. For an odd order $n$, choose the spin lift of a positive disk rotation by $2\pi/n$ to be

$$S_n=-\exp\!\left(-\frac{i\pi}{n}\gamma_1\right), \qquad S_n^n=1. \tag{D07.4}$$

The minus sign matters: the lift without it has $n$th power $-1$. On $u_+$, this lift has character $\exp[2\pi i(n-1)/(2n)]$. Choose the charge-one R line to have the inverse character, with integer weight $(n+1)/2$ modulo $n$. The combined spin-times-R action on the constant $\zeta$ is then the identity. The inverse R line similarly cancels the action on $u_-$.

For $(3,5)$ the R weights are $2$ modulo $3$ and $3$ modulo $5$. They also follow from (D07.3): a positive north deck rotation increases $\varphi$ by $2\pi$, producing the R transition $e^{-2\pi i/3}$; a positive south deck rotation decreases $\varphi$ by $2\pi$, producing $e^{+6\pi i/5}$. These are exactly the two required characters.

This is an honest choice of separate spin and R orbibundles. It is why we selected odd orders. More general spin-and-R structures can work when a separate spin lift does not; the even-order exercise identifies that boundary without denying the paper’s broader constructions.

Fractional flux is quantized global data

Integrating curvature over the regular part with the orbifold measure gives

$$\frac1{2\pi}\int_\Sigma dA_R =\frac12\left(\frac1{n_+}+\frac1{n_-}\right) =\frac4{15}. \tag{D07.5}$$

The connection is not an ordinary line bundle over a smooth sphere with fractional Chern number. Its disk characters are part of its definition. More generally, a compact flavor connection can have pole residues $m_+/n_+$ and $m_-/n_-$, with integers $m_\pm$. Its flux is $(n_+m_--n_-m_+)/(n_+n_-)$. Coprimality permits every integer numerator by Bézout’s identity. This is the explicit flux lattice of O23 (3.1)–(3.6), with charge-one probes allowed.

For one unit of numerator, $3(2)-5(1)=1$. A convenient flavor multiplet is

$$a_F=\frac{11-\cos\theta}{30}d\varphi+h\,d\psi, \qquad \sigma_F=M,\qquad D_F=\frac1{30\ell^2F},\qquad \frac1{2\pi}\int_\Sigma da_F=\frac1{15}. \tag{D07.6}$$

Here $M$ is a real constant of mass dimension one, $D_F$ has dimension two, and $h\sim h+1$ specifies circle holonomy $e^{2\pi ih}$. The flavor group is compact $U(1)$; these are fixed sources. Subtract $d\varphi/3$ in the north cover and $2d\varphi/5$ in the south cover. The resulting connections are smooth, with characters of weights $(2\bmod3,2\bmod5)$.

For a check of the auxiliary sign, the magnetic field is $b_1=(da_F)_{23}=1/(30\ell^2F)$. At constant $\sigma_F$, the two gaugino variations reduce to $i(D_F-b_1\gamma_1)\zeta$ and $-i(D_F+b_1\gamma_1)\widetilde\zeta$. Both vanish for (D07.6). This also reproduces O23 (3.9) after $x=\cos\theta$ and radius restoration. Turning on a flux without its auxiliary source would describe a different, generally nonsupersymmetric problem.

Check curvature and the smooth specialization

Let $R_+$ denote the positive-sphere scalar-curvature convention used in O23’s Gauss–Bonnet formula; the site’s target-paper convention is $R_P=-R_+$. Directly differentiating the spin connection gives

$$R_+=-\frac2{\ell^2F\sin\theta}\frac{d}{d\theta} \left(\frac{\cos\theta}{F}\right) =\frac{2a}{\ell^2F^3},\qquad \frac1{4\pi}\int_\Sigma R_+\,d\mathrm{vol} =\frac1{n_+}+\frac1{n_-}. \tag{D07.7}$$

The last equality is a boundary term, so it tests both cone normalizations independently of the KSE. For $(3,5)$ the area is $16\pi\ell^2$ and the orbifold Euler characteristic is $8/15$. If instead one treats the coarse space as a sphere with distributional curvature concentrated at the tips, the deficit-angle terms restore the ordinary Euler number $2$. Do not add those terms again to the orbifold integral.

Set $n_+=n_-=1$. Then $F=1$, $R_+=2/\ell^2$, the area is $4\pi\ell^2$, and (D07.2) is the smooth twisted sphere times a circle with R flux $+1$. The local finite groups disappear; the north and south R potentials now differ by $d\varphi$, an ordinary unit-winding transition. This is the smooth specialization of our general formulas. Since orbifold orders are integers, it is not a continuous deformation of the fixed $(3,5)$ orbifold.

A charged-section test that detects the orbifold

There is a useful check before attempting a quantum determinant. Define $L_k$ using the connection $k(a_F-hd\psi)$ and $D=d-ia$. It has oriented flux $k/15$. We choose the positive complex coordinates on the covering disks used above. A holomorphic section is a charged function whose covariant derivative in the antiholomorphic direction vanishes. On the cylinder that equation is

$$\left[D_\theta+\frac{iF}{\sin\theta}D_\varphi\right]s=0. \tag{D07.8}$$

The paper labels its connection bundle $\mathcal O(-\mathfrak m)$. Here $k=\mathfrak m$ matches its connection and flux data; $L_k$ and (D07.8) explicitly fix our holomorphic convention rather than importing an $\mathcal O$ label.

Write $s=e^{ij\varphi}u(\theta)$ with integer $j$. Equation (D07.8) says $u'/u=F[j-k(11-\cos\theta)/30]/\sin\theta$. Integrating gives, up to an arbitrary constant,

$$u(\theta)= \sin^{p}\frac\theta2\, \cos^{q}\frac\theta2\, e^{-k\cos\theta/30},\qquad p=3j-k,\quad q=2k-5j. \tag{D07.9}$$

In a regular north gauge the angular dependence is $e^{i(j-k/3)\varphi}$: the disk angular power is $p$. At the south it is $e^{i(j-2k/5)\varphi}$, whose power in the positive covering angle $-\varphi/5$ is $q$. Smooth holomorphic sections therefore require the integers $p,q\geq0$. This is stronger than imposing square integrability alone.

For $k=1$, no integer lies between $1/3$ and $2/5$, so there is no such section despite positive flux. For $k=15$, $j=5,6$ give $(p,q)=(0,5),(3,0)$ and two sections. An independent counting description uses weighted coordinates $(Z_+,Z_-)$ identified by $(Z_+,Z_-)\sim(t^3Z_+,t^5Z_-)$ for nonzero complex $t$. A polynomial of weighted degree $k$ has monomials $Z_+^aZ_-^b$ with $3a+5b=k$. The map $(a,b)=(q,p)$ is bijective because $3q+5p=k$ and the nonnegative exponents recover $j$. At $k=15$ these monomials are $Z_+^5$ and $Z_-^3$.

As a quantitative check, the squared norm of the normalization in (D07.9) is

$$\|s\|^2=4\pi\ell^2 e^{-k/15} \int_0^1(3+2t)t^p(1-t)^q e^{2kt/15}\,dt. \tag{D07.10}$$

This follows from $t=\sin^2(\theta/2)$ and the actual metric measure. For $k=15$, the two norms divided by $4\pi\ell^2$ are approximately $0.2834583988$ and $2.2405418095$. Direct polar integration and the beta-integral expansion of the exponential agree. These are bundle and mode-regularity checks, not yet unpaired fluctuation counts or a chiral determinant: those require the relevant supersymmetry complex, kinetic operators and phase prescription.

Change the interior while keeping the global problem fixed

A bounded training extension is $F_\epsilon=4-\cos\theta+\epsilon\sin^2\theta$, with $|\epsilon|\leq1/2$. It stays positive and has the same two even endpoint expansions. Recompute $A_R$ by (D07.2) and $D_F$ by (D07.6) with $F_\epsilon$. The same KSE calculation works, the fluxes and disk characters remain fixed, but the area becomes $4\pi\ell^2(4+2\epsilon/3)$. The radial equation shows exactly how the profiles change: multiply (D07.9) by $\exp\{-\epsilon[(j-11k/30)\cos\theta+k\cos^2\theta/60]\}$. This factor is finite and nonzero on both covering disks. Thus the two endpoint exponents and the allowed integer interval do not change.

This gives three useful diagnostics: unchanged quantized data, changed ordinary metric quantities, and unchanged dimension of this holomorphic-section space. It does not prove equality of full partition functions. Such a comparison also needs a specified supersymmetric regulator, source counterterms, field contours and behavior at the orbifold loci.

Practice with changed assumptions

1. Forgetting a cone factor

Someone replaces $F$ by $F/3$ while retaining $\varphi\sim\varphi+2\pi$, claiming this merely rescales the radius. Find the two new cone angles and decide whether this is the same $(3,5)$ orbifold. Then recover the genuine smooth specialization.

Hint 1

The radial proper distance, not the coordinate $\theta$, sets the circumference-to-radius ratio.

Hint 2

An effective disk orbifold of order $n$ has cone angle $2\pi/n$ with integer $n$.

Solution

The endpoint values become $1$ and $5/3$, so the cone angles are $2\pi$ and $6\pi/5$. The second does not equal $2\pi/n$ for any positive integer $n$. This changes the local quotient problem, rather than simply rescaling $\ell$. Multiplying the whole metric by a constant preserves both cone angles. To recover the smooth sphere within (D07.1), choose the different discrete data $(1,1)$ and hence $F=1$; then both angles are $2\pi$, R flux is $1$ and the integrated positive curvature is $8\pi$. The checks concern geometry and flux, not a continuous path between orbifold orders.

2. More flux can still change the mode count unevenly

Compute the regular holomorphic sections at $k=8$ and $k=16$ and check them using weighted monomials. If a periodic charged field also depends on the circle, how does the fixed flavor holonomy $h$ alter its circle momenta? Does the spatial count determine its full quantum partition function?

Hint 1

Use $k/3\leq j\leq2k/5$ before writing the wave functions.

Hint 2

The charge-one circle derivative is $D_\psi=\partial_\psi-ih$; a field of charge $k$ has $kh$ in its place if $L_k$ is realized by taking that charge.

Solution

For $k=8$, only $j=3$ is allowed, with $(p,q)=(1,1)$ and monomial $Z_+Z_-$. For $k=16$, only $j=6$ is allowed, with $(p,q)=(2,2)$ and monomial $Z_+^2Z_-^2$. Thus doubling this flux numerator does not double the section count. Their radial functions are respectively $\sin(\theta/2)\cos(\theta/2)e^{-4\cos\theta/15}$ and $\sin^2(\theta/2)\cos^2(\theta/2)e^{-8\cos\theta/15}$, regular on both covers. A periodic charge-$k$ Fourier mode $e^{in\psi}$ has physical covariant momentum $(n-kh)/\beta$. Alternatively, realize $L_k$ using a charge-one field and an independently specified degree-$k$ source, in which case its independently chosen circle holonomy must be stated. The spatial bundle count does not supply a fluctuation pairing, an infinite-product phase or a counterterm prescription, so it alone does not determine a partition function.

3. An even-order obstruction and what it does not prove

Try to use the same separate spin-bundle construction at a cone of order $2$. Show that neither lift of its generator has order $2$ on spinors. What must change before applying a general supersymmetric spindle formula there?

Hint 1

The two lifts differ by the central sign in $\mathrm{Spin}(2)$.

Hint 2

Compute the square of $\pm\exp(-i\pi\gamma_1/2)$; do not only test its action on vectors.

Solution

Both squares are $\exp(-i\pi\gamma_1)=-1$. They therefore do not define a representation of the order-two orbifold group on a separate spin fiber. The same problem occurs at any even order because the sign of the lift drops out of its even power. This rules out our particular separate spin construction, not every charged-spinor construction. One may instead need combined spin-and-R transition data, often described using a spin-c structure, in which the common central minus signs are identified. Matter charges and all transition laws must then be checked in that combined structure. Merely retaining the local differential equation or declaring a half-integer R phase is insufficient. The odd $(3,5)$ example avoids this additional choice and explicitly passes the stronger separate-lift test.

Reading and the next deliverable

Literature checked 20 October 2026: O23 v3, dated 5 August 2026, already treats twisted and anti-twisted spindle backgrounds, circle orbibundles, localization and duality tests. Read section 2.2 first for geometry, section 3.1 for source bundles, then section 3.2 for the extra mode-pairing and regularization work needed for a determinant. Appendix A is essential when translating spinor components. Our interior-profile variation and mode count are reproductions and training, not claims of a new classification or an unsolved problem.

The next bounded deliverable is to choose one of the paper’s free-chiral factors, reconstruct the relevant orbifold characters and unpaired complex, and compare its smooth specialization in a fixed contact prescription. A mismatch in the numerator of a flux, a circle period or an orbifold character should be resolved before interpreting any discrepancy as new physics.