Four supercharges and product geometries

The same product metric can support different numbers of supercharges when its fixed sources change. To establish enhancement, we must find enough independent solutions of the same equations, then keep them through global transport.

Prepare with the mixed Killing vector and integrability and background verification. We will derive the maximal local conditions and construct a complete solution basis on $S^2\times S^1$. Its mixed vectors will also explain why this background differs from a topological twist on the same manifold.

Count the solution spaces

The parent is three-dimensional $\mathcal N=2$, with four real Lorentzian Poincaré supercharges. In Euclidean signature the commuting labels $\zeta$ and $\widetilde\zeta$, of R charges $+1$ and $-1$, are independent complex spinors. Write their full equations as

$$\begin{aligned} \mathscr D_\mu^+\zeta&=0,& \mathscr D_\mu^+&=\nabla_\mu-i(A_\mu-V_\mu) +\frac H2\gamma_\mu+\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho,\\ \mathscr D_\mu^-\widetilde\zeta&=0,& \mathscr D_\mu^-&=\nabla_\mu+i(A_\mu-V_\mu) +\frac H2\gamma_\mu-\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho . \end{aligned} \tag{M09.1}$$

The metric is real and positive, and $\nabla\cdot V=0$. The sources may be complex. Each label has two complex components. First-order uniqueness makes evaluation at a point injective on the solution space: a solution that vanishes there is zero everywhere on the connected domain. Thus each charge sector has at most two complex independent solutions.

If a parallel basis exists, every solution is a constant linear combination of its two columns. Writing $\zeta,2\zeta,3\zeta$ has not produced three generators. Following P, section 5.2, “four supercharges” here means the maximal pair of complex dimensions $(2,2)$ in the two sectors. This is the Euclidean complexification of the parent symmetry, not a change to a three-dimensional $\mathcal N=4$ theory.

Derive the local enhancement conditions

Two independent solutions in one sector force its curvature to annihilate a basis, hence to vanish as a matrix. For enhancement both curvatures must vanish. Conversely, two flat connections give two parallel bases on a contractible patch.

Use the curvature operator derived in the prerequisite, with $\mathscr I_a^\pm=\epsilon_{abc}\mathscr F_{bc}^\pm$. Set $W=A-V$. The identity coefficient gives $dW=0$. To expose the gamma coefficients, define the symmetric tensor

$$T_{ab}=\frac12(2R_{ab}-R\delta_{ab}) +H^2\delta_{ab}+V_aV_b .$$

The source Ricci tensor is negative on a sphere. The sum and difference of the two vanishing gamma coefficients are

$$iT_{ab}-\epsilon_{abc}\partial_cH=0,\qquad \nabla_bV_a-iH\epsilon_{abc}V_c=0 . \tag{M09.2}$$

The antisymmetric part of the first equation forces $dH=0$; its symmetric part gives $T_{ab}=0$. Taking the trace yields $R=6H^2+2V^2$. Substituting back and exchanging the indices in the second equation gives

$$\boxed{\begin{aligned} dH&=0,&d(A-V)&=0,\\ \nabla_aV_b&=-iH\epsilon_{abc}V_c,& R_{ab}&=-V_aV_b+\delta_{ab}(V^2+2H^2). \end{aligned}} \tag{M09.3}$$

These are P (5.9). They are locally sufficient as well as necessary: substitution makes both complete curvature matrices zero. They do not force the metric’s spin curvature to vanish; the source contributions cancel it. Nor do they fix the holonomy of the remaining flat connection $A-V$.

In particular, $V$ is Killing and $V^2$ is constant. The source’s local cases can be summarized using ordinary geometric sectional curvature, which has the opposite sphere sign to its Ricci convention:

Source case	Local geometric implication
$V=0$	Constant sectional curvature $-H^2$: spherical for imaginary $H$, flat for $H=0$, hyperbolic for real nonzero $H$.
$H=0$, $V\ne0$	A local product $\mathbb R\times\Sigma$, with Gaussian curvature $-V^2$ on $\Sigma$.
$H\ne0$, $V\ne0$	The constant-length Killing direction gives a local fibration over a constant-curvature surface; its Gaussian curvature is $-(V^2+4H^2)$.

For the second case, the Ricci equation and metric reality force $V$ to be real or purely imaginary; it is parallel, so its underlying real direction splits off locally. For the third case, the source analysis has imaginary $H$ and real or imaginary $V$. The explicit fibrations in P (5.11)–(5.14) are further source-supported cases. This table specifies local geometry, not all possible compact quotients or their surviving spinors.

Choose the full sphere-product background

Let $\tau$ have length units and period $L$, and let $r>0$ be the sphere radius. Use

$$g=d\tau^2+r^2(d\theta^2+\sin^2\theta\,d\varphi^2),\qquad H=0,\qquad V=\frac{i}{r}d\tau,\qquad A=V+\nu\,d\tau . \tag{M09.4}$$

Here $\nu$ has inverse-length units and is initially complex. All sources are fixed; no dynamical auxiliary equation is imposed. The source Ricci tensor in the oriented frame $e^1=d\tau,e^2=r\,d\theta,e^3=r\sin\theta\,d\varphi$ is $R_{ab}=\operatorname{diag}(0,-r^{-2},-r^{-2})$. Since $V^2=-r^{-2}$ and $\nabla V=0$, (M09.3) holds. This is the spherical product branch discussed after P (5.9). The imaginary $V$ is essential to its positive sphere curvature.

The polar spin connection is $\Gamma=-i\cos\theta\,\gamma_1\,d\varphi/2$. With $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$, the two sphere operators become

$$\begin{aligned} \mathscr D_\theta^\pm&=\partial_\theta\mp\frac i2\gamma_3,\\ \mathscr D_\varphi^\pm&=\partial_\varphi -\frac i2\cos\theta\,\gamma_1 \ \pm\frac i2\sin\theta\,\gamma_2,\\ \mathscr D_\tau^\pm&=\partial_\tau\mp i\nu . \end{aligned} \tag{M09.5}$$

The radial scale has cancelled from the angular-coordinate operators, as it must; their physical frame coefficients are proportional to $1/r$. Define the two invertible matrices

$$U_\pm(\theta,\varphi) =e^{\pm i\theta\gamma_3/2}e^{i\varphi\gamma_1/2}. \tag{M09.6}$$

Differentiating the first exponential checks the $\theta$ equation. For the other equation use

$$e^{\pm i\theta\gamma_3/2}\gamma_1e^{\mp i\theta\gamma_3/2} =\cos\theta\,\gamma_1\mp\sin\theta\,\gamma_2.$$

Thus the complete local solution spaces are

$$\zeta=e^{i\nu\tau}U_+c,\qquad \widetilde\zeta=e^{-i\nu\tau}U_-\widetilde c,\qquad c,\widetilde c\in\mathbb C^2\ \hbox{constant}. \tag{M09.7}$$

Both matrices have determinant one. They supply two independent columns in each charge sector, rather than just one convenient pair.

Check the sphere patches and the circle

The polar frame is singular at the poles, so the matrices in (M09.6) are not yet a global check. Rotate to regular northern and southern frames with spin matrices

$$S_N=e^{-i\varphi\gamma_1/2},\qquad S_S=e^{i\varphi\gamma_1/2},\qquad U_{\pm,N}=S_NU_\pm,\quad U_{\pm,S}=S_SU_\pm . \tag{M09.8}$$

At the north pole $U_{\pm,N}=\mathbf1$. Near it the nonconstant terms contain $\sin(\theta/2)\cos\varphi$ and $\sin(\theta/2)\sin\varphi$, which are smooth local Cartesian functions times a smooth radial factor. At the south pole $U_{\pm,S}=\pm i\gamma_3$, with the corresponding smooth southern expansion. Both regular matrices are periodic in $\varphi$. On the overlap, $U_{\pm,N}=e^{-i\varphi\gamma_1}U_{\pm,S}$, the required spin transition. There is no R monopole in this example: $A$ is globally defined along the circle.

Choose a circle spin structure with sign $\varepsilon=\pm1$ and no extra R transition. Equation (M09.7) then descends precisely when

$$e^{i\nu L}=\varepsilon,\qquad e^{-i\nu L}=\varepsilon . \tag{M09.9}$$

These conditions are identical because $\varepsilon^{-1}=\varepsilon$. If they hold, all four complexified solution directions survive; if not, neither sector has a nonzero global column. For example, $\nu=0$ with periodic spin structure gives the promised global $(2,2)$ background. The R connection alone has holonomy $e^{i\oint A}=e^{-L/r}e^{i\nu L}$. It is the full connection, including $V$, that determines (M09.9).

The other auxiliary potential is also worth specifying. With $V=-i*dC$ and the declared orientation, regular sphere potentials are

$$C_N=-r(1-\cos\theta)d\varphi,\qquad C_S=r(1+\cos\theta)d\varphi,\qquad C_N-C_S=-2r\,d\varphi . \tag{M09.10}$$

They give $dC=-r\sin\theta\,d\theta\wedge d\varphi$, hence the stated $V$, and $\int_{S^2}dC=-4\pi r$. Set $B=0$, consistent with $H=0$. The one-form $C$ has length units, unlike an ordinary dimensionless normalized circle connection.

For a fully specified compact central-source model, one possible choice is $a_C=C/(2r)$ as a $U(1)$ connection whose unit-charge gauge parameter has period $2\pi$. Its flux number is $-1$. A field of dimensionful central charge $z$ has dimensionless charge $Q_Z=2rz$, and its overlap phase is $e^{-iQ_Z\varphi}$; this model therefore requires $Q_Z\in\mathbb Z$. This is an explicit choice of global source normalization, not a universal quantization law derived from supersymmetry or from P. An uncharged matter sector with $z=0$ imposes no such nonzero-charge restriction. For a theory with a preassigned central-charge lattice, that lattice and the radius must instead be compatible with its specified source group.

More solutions need not give a real mixed vector

Take $\nu=0$, $c=u_+=(1,0)^T$ and $\widetilde c=u_-=(0,1)^T$. Using $C_{\mathrm{spin}}=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$ for the spinor contraction, direct multiplication gives

$$s=\zeta^TC_{\mathrm{spin}}\widetilde\zeta=-\cos\theta,\qquad K=\partial_\tau+\frac{i}{r}\partial_\varphi . \tag{M09.11}$$

The real and imaginary parts generate circle translation and sphere rotation. The rotation vanishes at the poles, but $K$ itself never vanishes. Its complex square is $\cos^2\theta=s^2$ and may vanish at the equator. No real null vector has appeared.

In fact no opposite-charge pair on this background has a nonzero real $K$. If it did, $K^2=s^2>0$ would make $s$ real and nonzero. But the mixed derivative identity $\nabla_aK_b=-s\epsilon_{abc}V_c$, with $H=0$ and imaginary nonzero $V$, would equate a real tensor to a nonzero imaginary one. This is the product example singled out after P (5.7). Its metric also supports other source choices, such as the two-supercharge twisted product studied earlier, but those are different backgrounds.

Exercises

1. Vary the product source

Keep the product metric, set $H=0$ and $A=V=v\,d\tau$ with constant complex $v$. Compute the angular curvature of each full connection. Which values of $v$ admit any nonzero local solution? Compare with setting all sources to zero.

Hint 1

The spin curvature in the polar frame is $\mathscr F_{\theta\varphi}^{\mathrm{spin}}=i\sin\theta\,\gamma_1/2$. The source terms are $N_\theta^\pm=\mp rv\gamma_3/2$ and $N_\varphi^\pm=\pm rv\sin\theta\,\gamma_2/2$.

Hint 2

Include the spin-connection commutator when differentiating the source matrices. The covariant derivative terms cancel because $V$ is parallel.

Solution

The source commutator is $[N_\theta^\pm,N_\varphi^\pm]=ir^2v^2\sin\theta\,\gamma_1/2$. Together with the spin curvature this gives

$$\mathscr F_{\theta\varphi}^\pm =\frac i2(1+r^2v^2)\sin\theta\,\gamma_1 .$$

Away from the coordinate poles $\gamma_1$ is invertible, so a nonzero local solution requires $v=\pm i/r$. At either value both full connections are flat and have two local columns each. For every other $v$, including zero or any real nonzero value, the curvature has no kernel. The vanishing coordinate factor $\sin\theta$ at a pole does not evade the obstruction: the orthonormal curvature is proportional to $(r^{-2}+v^2)\gamma_1$ and is regular there. The imaginary source supplies the needed sign, with the correct inverse-length dimension.

2. Change only the global circle data

Return to (M09.4). Determine the allowed complex $\nu$ for each circle spin structure. How many global solutions remain if the condition fails? Why is testing the R holonomy alone incorrect here?

Hint 1

The same scalar circle multiplier acts on both independent columns in one sector. The opposite sector has its inverse multiplier.

Hint 2

Take the absolute value of $e^{i\nu L}=\pm1$ before solving for its phase.

Solution

The modulus condition is $e^{-L\,\operatorname{Im}\nu}=1$, hence $\operatorname{Im}\nu=0$. Periodic spin structure requires $\nu L=2\pi n$; antiperiodic spin structure requires $\nu L=(2n+1)\pi$. Both sectors then have complex dimension two. If the condition fails, neither sector has a nonzero periodic section, because no choice of a constant initial column changes its scalar multiplier. Local dimension $(2,2)$ has become global dimension $(0,0)$.

The source $A$ contains the imaginary term $i\,d\tau/r$. Its non-unit-modulus R transport is cancelled in the supersymmetry equation by the identity contribution of $V$. The relevant circle connection is $A-V=\nu\,d\tau$. The cancellation is part of the fixed background, not a real R gauge transformation that removes an imaginary connection.

3. A complex vector with a null locus

Using the pair in (M09.11), compute the three orthonormal components of $K$ directly from the columns of $U_\pm$. Check its square and its Hermitian norm. Explain what happens at the poles and at the equator, and why counting independent real Killing vectors is not the same as counting supercharges.

Hint 1

The opposite $e^{\pm i\varphi/2}$ factors cancel between these particular columns. The two polar exponentials have opposite signs.

Hint 2

Convert a frame component to a coordinate component with $E_3=(r\sin\theta)^{-1}\partial_\varphi$. Use the smooth sphere rotation field when taking a polar limit.

Solution

Matrix multiplication gives $K^a=(1,0,i\sin\theta)$ and $s=-\cos\theta$. Therefore $K\cdot K=1-\sin^2\theta=\cos^2\theta=s^2$, while $\overline K\cdot K=1+\sin^2\theta>0$. In coordinates this is exactly (M09.11).

At a pole the sphere rotation vector is zero and $K=\partial_\tau$. At the equator the complex square vanishes, but the positive Hermitian norm is two, so the vector is nonzero. On a generic patch the real and imaginary parts are independent Killing vectors. No constant complex rescaling can turn both components into a real vector.

Supercharges are independent solutions in the two spinor sectors. Their bilinears are quadratic functions of those solutions, can obey identities, and can produce several isometries. The number of real vector fields obtained from them is not the dimension of the original solution spaces.

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

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 5, additional supercharges