The three-dimensional A-twist

A generic curved sphere has no nonzero parallel spinor. A charged spinor can nevertheless be parallel if an R-symmetry connection cancels its spin connection. This cancellation is the A-twist. On a sphere times a circle it leaves a small supersymmetric quantum-mechanical problem: most field modes pair, while a finite set of unpaired modes determines a protected partition function.

We will construct that background globally and reproduce the free-chiral factor $Z=(1-y)^{-m+r-1}$. The integers $m,r$ will acquire geometric meanings before we use this expression. The project is an established reproduction with a controlled training variation, not a proposed new twist.

Prepare with the reproduction method, Seifert manifolds, gauge–matter actions, Chern–Simons and background contact terms, and field-theory localization. The line-bundle counting needed below is derived here. No Bethe equations are prerequisites.

Fix the theory and the global background

Take one free three-dimensional N=2 chiral multiplet, with no superpotential and no dynamical gauge field: gauge rank zero, flavor charge $+1$ under compact $U(1)_F$. The flat theory has four real supercharges. We use its full N=2 transformation rules and retain one solution in each of the two opposite-R-charge Killing-spinor families. In Euclidean signature these labels are independent; they are not related by imposing a new reality condition.

The scalar has integer trial R-charge $r$. Our main example uses $r=0$, a permissible conserved R-symmetry of this free massive problem. It is not the superconformal assignment $r=1/2$ on flat massless space. The global bundle specified here requires integer charges.

Let $\tau\sim\tau+2\pi$, $0\leq\theta\leq\pi$, and $\phi\sim\phi+2\pi$. For positive lengths $\ell,\beta$, use

$$ds^2=\beta^2d\tau^2+\ell^2(d\theta^2+\sin^2\theta\,d\phi^2), \qquad (e^1,e^2,e^3)=(\beta d\tau,\ell d\theta,\ell\sin\theta\,d\phi). \tag{D02.1}$$

The orientation is $e^1\wedge e^2\wedge e^3>0$. The circle has periodic spin structure. Set the rigid sources $H=V=C=0$, with vanishing flat two-form source, and choose zero R holonomy around the circle. This is the trivial circle bundle over $S^2$, so its Seifert degree is zero. It has ordinary integer magnetic fluxes; the torsion identifications of a nontrivial circle bundle must not be imported into this example.

Use the site’s matrices $\gamma^1=\sigma_3$, $\gamma^2=-\sigma_1$, $\gamma^3=-\sigma_2$ and $C_{\rm spin}=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. The polar frame has $\omega^{23}=-\cos\theta\,d\phi$, hence spinor connection $\Gamma_\phi=-i\cos\theta\,\gamma^1/2$. With $u_+=(1,0)^T$ and $u_-=(0,1)^T$, define north and south R potentials and spinor components by

$$\begin{aligned} A_N&=\tfrac12(\cos\theta-1)d\phi,& \zeta_N&=e^{-i\phi/2}u_-,&\widetilde\zeta_N&=e^{i\phi/2}u_+,\\ A_S&=\tfrac12(\cos\theta+1)d\phi,& \zeta_S&=e^{i\phi/2}u_-,&\widetilde\zeta_S&=e^{-i\phi/2}u_+. \end{aligned} \tag{D02.2}$$

For example, the north $\phi$ component of $(\nabla-iA)\zeta=0$ reduces to $-i/2+i\cos\theta/2-i(\cos\theta-1)/2=0$. The opposite family obeys $(\nabla+iA)\widetilde\zeta=0$. All other components vanish. Their bilinear is $\zeta^TC_{\rm spin}\gamma^a\widetilde\zeta=\delta^{a1}$: the mixed supersymmetry algebra follows the circle, with its appropriate flavor transformation.

These formulas also pass a global test. The overlap has $A_N-A_S=-d\phi$, $\zeta_N=e^{-i\phi}\zeta_S$, and the inverse transition for $\widetilde\zeta$. The half-angle signs of their polar-frame components are the spin lift of a rotating frame; in frames regular at each pole they are smooth. They do not mean that a physical spinor is multivalued. Around $\tau$ the labels are constant and periodic. Curvature integrability selects $\gamma^1\zeta=-\zeta$ and $\gamma^1\widetilde\zeta=\widetilde\zeta$, so no second component survives in either family.

The R flux is $\int_{S^2}dA/(2\pi)=-1$. Thus the R line bundle $L_R$ has degree $-1$, and $L_R^2$ is the canonical bundle $K_{S^2}$ of degree $-2$. This is the global content of the twist, consistent with A17 (3.14)–(3.18). In the central paper’s curvature convention the scalar curvature is $R_P=-2/\ell^2$; the usual GR scalar curvature has the opposite sign.

Add a mass without breaking the retained pair

The flavor source is a fixed vector multiplet, not an integration variable. Specify

$$\begin{aligned} a_{F,N}&=\tfrac m2(1-\cos\theta)d\phi+h\,d\tau,& a_{F,S}&=-\tfrac m2(1+\cos\theta)d\phi+h\,d\tau,\\ \sigma_F&=M>0,&D_F&=-\frac{m}{2\ell^2}. \end{aligned} \tag{D02.3}$$

Here $m\in\mathbb Z$ is the flavor flux, and $h\sim h+1$ specifies holonomy $2\pi h$. The scalar $M$ has mass dimension one, $D_F$ dimension two; $m,h,r$ are dimensionless. In an orthonormal frame the magnetic field is $b^1=m/(2\ell^2)$. The two source-gaugino variations reduce to $i(D_F+b^1)\zeta$ and $-i(D_F+b^1)\widetilde\zeta$, both zero. These are supersymmetry conditions on an external source; a Maxwell equation is not an additional requirement.

For the circle determinant it is convenient to use

$$\nu=-h+i\beta M,\qquad y=e^{2\pi i\nu}=e^{-2\pi\beta M-2\pi ih}. \tag{D02.4}$$

The minus sign of $h$ follows from our charge-$+1$ derivative. Acting on $e^{in\tau}$, the unpaired first-order operator is $-M+D_1=(i/\beta)(n+\nu)$, where $D_1=\beta^{-1}(\partial_\tau-ih)$. It is not a redefinition of the flavor charge. These are the product-background specialization of A17 (3.20)–(3.25).

Count modes by gluing two polynomials

The scalar is a section of $L=L_F\otimes L_R^r$. Its degree is $d=m-r$. On $S^2\simeq\mathbb{CP}^1$, write this line bundle as $\mathcal O(d)$. The covariant operator $\bar\partial_L$ differentiates a section in the antiholomorphic direction; it maps scalar sections to sections carrying one antiholomorphic differential. Supersymmetry pairs its nonzero modes. The unmatched bosons lie in its kernel; the unmatched fermions lie in its cokernel, equivalently the kernel of its adjoint. Denote their numbers by $h^0,h^1$.

We can find both without assuming an index theorem. On the north complex coordinate $z=\tan(\theta/2)e^{i\phi}$, a holomorphic section is a polynomial $P_N(z)$. At the south pole, $w=1/z$, its expression is $P_S(w)=w^dP_N(1/w)$. Regularity at both poles permits precisely $1,z,\ldots,z^d$ when $d\geq0$, and no nonzero section when $d<0$.

For the cokernel, gluing local solutions of $\bar\partial_L f=g$ can leave a holomorphic function on the overlap. Expand it in Laurent powers $z^k$. Powers $k\geq0$ can be removed on the north patch; powers $k\leq d$ can be removed on the south patch. The obstruction consists of the remaining integers $d<k<0$. Consequently

$$h^0=\max(d+1,0),\qquad h^1=\max(-d-1,0),\qquad \chi=h^0-h^1=m-r+1. \tag{D02.5}$$

For $d=-1$ neither kind survives. For $d=-2$ there is one unpaired fermion; a negative index does not mean a negative number of bosons. The pairing statement is the $p=0$ case of A17 (4.35)–(4.40). This bundle count supplies the exponent, but not the phase convention of an infinite circle product.

Fix the determinant before evaluating it

The remaining formal ratio is $\prod_{n\in\mathbb Z}(n+\nu)^{-\chi}$. Its constant factors and phase require a regulator. We use the source’s $U(1)_{-1/2}$ quantization for a unit-charge chiral, with its associated flavor–R, R–R and gravitational contact terms and no added integer Chern–Simons terms. In CK19’s notation the contact coefficients are $\kappa_{FF}=-1/2$, $\kappa_{FR}=-(r-1)/2$, $\kappa_{RR}=-(r-1)^2/2$, $\kappa_g=-1$. These describe the regulated fermion and its contact prescription together; they do not license an isolated arbitrary half-integer classical counterterm. See CK19 (2.28)–(2.34).

Two useful normalization conditions are flavor large-gauge invariance, $\nu\to\nu+1$, and decoupling to one as $M\to+\infty$. A symmetric pairing of positive and negative circle integers has logarithmic derivative $-\pi\cot\pi\nu$. It misses the contact contribution $-i\pi$. Adding it and fixing the decoupling constant gives

$$\frac{d}{d\nu}\log\Pi(\nu)=-\pi\cot(\pi\nu)-i\pi, \qquad \Pi(\nu)=\frac1{1-e^{2\pi i\nu}}, \qquad Z_{S^2\times S^1}=\Pi(\nu)^\chi. \tag{D02.6}$$

Indeed $1-e^{2\pi i\nu}=-2i e^{i\pi\nu}\sin\pi\nu$. The factor $e^{-i\pi\nu}$ corrects the sign under $\nu\to\nu+1$ and the behavior at large positive mass. The adopted result is A17 (4.45)–(4.48); its absolute normalization is a physical convention, not something an unregulated product can determine.

For $|y|<1$, each unpaired bosonic oscillator contributes $\sum_{q\geq0}y^q=(1-y)^{-1}$, and each unpaired fermionic oscillator contributes the graded two-state trace $1-y$. This is the supersymmetric twisted trace, not an ordinary thermal partition function or the untwisted superconformal index. Away from this convergence domain one uses meromorphic continuation. At $y=1$ a positive $\chi$ gives a pole from unsuppressed bosonic states; it is not a numerical error.

Reproduce a number and change one input

Take $r=0,m=2,h=0$ and $\beta M=\log3/(2\pi)$. The three unpaired sections are $1,z,z^2$, $y=1/3$, and $Z=27/8$. At total occupation $q$, distributing quanta among three oscillators gives $\binom{q+2}{2}$ states. The first degeneracies are $1,3,6,10$, so direct state counting reproduces $(1-y)^{-3}$.

There is a check that uses the complete spatial kinetic spectrum rather than just holomorphic sections. The scalar curvature and flavor auxiliary terms in the action combine to $-d/(2\ell^2)$. The spatial operator is therefore $L_d=-D^2-d/(2\ell^2)$. Its monopole spherical harmonics have

$$\ell^2\lambda_j=j(j+1)-\frac{d(d+2)}4,\qquad j=\frac{|d|}{2}+n,\quad n=0,1,\ldots,\qquad \text{multiplicity}=2j+1. \tag{D02.7}$$

For a direct differential check set $x=\cos\theta$. The radial part at azimuthal integer $l$ obeys $\ell^2 L_df=-(1-x^2)f''+2xf'+[l-d(1-x)/2]^2f/(1-x^2)-df/2$. At $d=2$, the functions $(1-x)^{l/2}(1+x)^{(2-l)/2}$ for $l=0,1,2$ have eigenvalue zero. Excited functions acquire Jacobi polynomials. On the fermionic side the adjoint-pair operator has the spectrum of $L_{-d-2}$. The positive eigenvalues match: for $d=2$, the bosonic sequence starts at $j=1$ and the fermionic sequence at $j=2$. Only the three zero modes remain. Thus, for every dimensionless $s>0$, $\operatorname{Tr}e^{-s\ell^2L_d}-\operatorname{Tr}e^{-s\ell^2L_{-d-2}}=d+1$.

This suggests a useful numerical diagnosis. Sum both heat traces separately and bound their omitted positive tails before subtracting them. For this example at $s=0.3$, including $n=0,\ldots,65$ in each leaves a combined tail below $7\times10^{-592}$; ordinary rounding will dominate long before that bound matters. Agreement with $3$ checks spectral multiplicities and the auxiliary-field sign independently of the polynomial count.

Now vary $\ell$ while holding $\beta M,h,m,r$ fixed. The nonzero spatial eigenvalues move as $\ell^{-2}$, but their paired determinant cancels and $Z$ stays $27/8$. This is a controlled metric variation of this specified free system. Increasing the discrete flux from $m=2$ to $m=3$ instead adds one unpaired boson and multiplies $Z$ by $3/2$. Flux sectors cannot be connected by a smooth variation of one globally defined connection. A failed radius check suggests an incomplete pairing or contact prescription; a changed flux answer is expected physics.

Practice with changed assumptions

1. A negative exponent

Keep $r=0$ and $y=1/3$, but set $m=-2$. Find the two unpaired multiplicities and the partition function. Repeat for $m=-1$. Explain why neither case requires negative occupation numbers.

Hint 1

Count regular polynomial sections and the Laurent powers which neither patch can remove separately.

Hint 2

The graded trace of one fermionic oscillator has only the empty and occupied states.

Solution

At $d=-2$ there are no holomorphic sections. The only obstruction has exponent $k=-1$, so $h^0=0,h^1=1$ and $\chi=-1$. The empty fermionic state contributes $1$, the occupied state contributes $-y$, giving $Z=1-y=2/3$. At $d=-1$ no integer satisfies $-1<k<0$ and no polynomial section exists. Both multiplicities vanish and $Z=1$. All these counts are nonnegative; the minus sign is fermion parity in the trace. The negative-flux spectrum checks the same result: $L_{-2}$ has no zero mode while $L_0$ has one.

2. A suspiciously symmetric regulator

A calculation keeps only the symmetric mode product and proposes $\Pi_{\rm sym}(\nu)=c/\sin\pi\nu$, with constant $c$. Can any choice of $c$ satisfy both large-gauge invariance and positive-mass decoupling? Find the multiplicative correction needed for the chosen prescription.

Hint 1

Check $\nu\to\nu+1$ before choosing the constant.

Hint 2

At $\nu=i\beta M$ with $M\to+\infty$, the inverse sine decays exponentially. Use the identity just below (D02.6).

Solution

The sine changes sign under a unit shift, so no nonzero constant makes $\Pi_{\rm sym}$ periodic. Also $1/\sin(i\pi\beta M)=1/[i\sinh(\pi\beta M)]$ tends to zero, rather than one. Multiplication by $e^{-i\pi\nu}$ supplies both the compensating sign and exponential growth. Fixing the constant gives $\Pi=-e^{-i\pi\nu}/(2i\sin\pi\nu)=1/(1-y)$. Differentiating its logarithm restores $-i\pi$. For even $\chi$ the sign of $\Pi_{\rm sym}^\chi$ alone happens to be periodic, but its decoupling normalization still fails. A regulator must define the unit-charge chiral consistently across the allowed flux sectors, not only at one accidental even exponent.

3. Change the circle spin structure

Replace periodic by antiperiodic circle spin structure, initially keeping $A_\tau=0$. Does the local background still define global supercharges? Then add the flat R connection $a_Rd\tau$ with $a_R=1/2$ and find globally admissible labels. Explain why this operation must be recorded when comparing partition functions.

Hint 1

The circle equations are $(\partial_\tau-ia_R)\zeta=0$ and $(\partial_\tau+ia_R)\widetilde\zeta=0$.

Hint 2

A nonzero constant component cannot be antiperiodic. A half-unit R holonomy cannot be removed by a single-valued unit-charge R gauge transformation.

Solution

For $a_R=0$ the labels must be constant in $\tau$. Antiperiodicity then forces both to vanish: the same local equations have no global supersymmetry. With $a_R=1/2$, multiply (D02.2) by $e^{i\tau/2}$ for $\zeta$ and by $e^{-i\tau/2}$ for $\widetilde\zeta$. Both change sign after $2\pi$, as required. Curvature and the retained local projectors are unchanged, but the spin structure and R holonomy have changed together. Matter fields also feel their R charges around the circle. In the conventions of CK19 (6.2)–(6.5), $\nu_R=-a_R$ is half-integral and the twisted trace contains $(-1)^F e^{2\pi i\nu_R(F+R)}$; at a half unit this becomes $(-1)^R$ for integer charges. Thus comparing the two backgrounds requires tracking the trace insertion and contact terms. Particular free-field assignments can make the two answers coincide, but the global background data are not interchangeable.

A bounded next project and its literature status

The first deliverable is now concrete: reproduce the patched background, the integer $\chi$, and $27/8$ by state counting and a regulated determinant; then change either the radius or one flux integer while preserving the other input data. Success tests this reproduction and its convention dictionary. It does not establish a new metric-independence theorem or a new Seifert formula.

Literature checked 20 October 2026: Closset–Kim–Willett, A17 v2 develops the A-model organization of Seifert observables; Closset–Kim, CK19 v2 supplies the spin-sensitive twisted-index and contact-term dictionary used here. For a later focused direction, Closset–Furrer–Keyes–Khlaif, Q25 v3 studies pure Chern–Simons theories and global gauge-group quotients through generalized symmetries. Its scope must not be enlarged into an arbitrary matter-theory result.

The next calculation is to gauge a specified Abelian flavor symmetry. That introduces a dynamical holonomy, a flux sum, a contour prescription, and possible singular or nonisolated vacua. Effective twisted superpotentials and Bethe equations organize that problem, but they must first be defined and their assumptions checked. Our free factor is a building block for it; it is already a complete first A-twist computation on its own.

Seifert partition functions and fibering operators changes the circle-bundle degree and follows the resulting vacuum phases through a finite sum, residues and a Gaussian check.