Paper guide: Section 5, additional supercharges

This is a reading guide to Section 5 of Supersymmetric Field Theories on Three-Manifolds, by Cyril Closset, Thomas T. Dumitrescu, Guido Festuccia and Zohar Komargodski. Use arXiv:1212.3388v3, 22 October 2013; the page numbers below distinguish the printed number from the PDF viewer’s count.

Your task is to keep track of which spinor equation has how many solutions, then separate the resulting local geometry from global boundary conditions. Read two supercharges and their Killing vector and the curvature obstruction first. The worked geometric examples are Seifert fibrations, four supercharges on product geometries and the complete squashed-sphere background.

The setting is three-dimensional N=2 with four real Poincaré supercharges before Euclidean continuation. Here the two charge sectors are complexified separately. A gauge-group rank is not part of this geometric count.

Read in two passes

First follow the opposite-charge pair through the vector construction. Then restart with two independent columns in each charge sector. A pair of different R charges does not itself imply the second, stronger condition. The real-vector restriction in Section 5.1 is also additional; Section 5.2 includes product examples with genuinely complex mixed vectors. P, Section 5.

Read in P v3	Question for your notes	Canonical calculation
(5.1)–(5.2), printed 17/PDF 18	Which terms cancel in the derivative?	M07.2–M07.4: mixed vector
(5.3)–(5.5), printed 17/PDF 18	Where is reality imposed? Which vector is normalized?	M07.7–M07.9: real direction
(5.6)–(5.7), printed 17–18/PDF 18–19	Do both labels use the same sources?	M07.11–M07.12: a complete pair
After (5.7), printed 18/PDF 19	Does the chosen flow close?	M08.1: exceptional-fiber model
(5.8), printed 18/PDF 19	Is curvature annihilating a line or a basis?	R09.4: full obstruction
(5.9)–(5.10), printed 19/PDF 20	What is local, and what remains global?	M09.3: enhancement conditions
(5.11)–(5.12), printed 20/PDF 21	Are radius and squashing conventions matched?	M10.2: homogeneous family
(5.13)–(5.14), printed 20/PDF 21	Which base curvature does each case require?	M09.3 and its local cases

The last two geometries are further local examples, not a classification of compact quotients. The site lessons supply explicit checks and global examples; they are teaching reconstructions, not extra results attributed to the original paper.

Translate notation before differentiating

Paper notation	Column notation used in the lessons
Opposite-charge labels $\zeta,\widetilde\zeta$	Independent charge-$+1,-1$ columns solving the same background
$\zeta\gamma_a\widetilde\zeta$	$\zeta^T C\gamma_a\widetilde\zeta$, where $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$
Index-raised conjugate in (5.3)	$C\overline\zeta$; ordinary entrywise conjugation alone is insufficient
$\eta^\mu$, $\Omega$ in (5.4)	The unit vector $\xi^\mu$ and positive length $\Omega$, so $K=\Omega\xi$
The function $s$ in (5.7)	The local spinor-amplitude function; it is not M07’s mixed scalar $\zeta^TC\widetilde\zeta$
$R_{ab}$	The source convention, negative on a round sphere; Gaussian curvature below has the ordinary positive-sphere sign

For (5.6), the two individually allowed source shifts must be compatible: $U=\widetilde U=0$, $\widetilde\kappa=-\kappa$, and $K^\mu\partial_\mu\kappa=0$. The freedom belongs to a common background, not two independently chosen backgrounds. P (5.6).

Restore the derivative behind the Killing equation

Use $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$, $\epsilon_{123}=1$ and commuting labels. Put $s_{\rm mix}=\zeta^TC\widetilde\zeta$. Starting from the two equations in M07.1, recover the antisymmetry of $\nabla_aK_b$. Explain why the connection $A$ drops out even when it is not flat.

Hint 1

Differentiate both columns. Move the matrix acting on the first column through $C$ with $\gamma_a^TC=-C\gamma_a$.

Hint 2

The $H$ terms multiply a commutator. The $V$ terms multiply an anticommutator. The two identity-matrix terms have opposite signs.

Solution

The identity contributions are $+i(A_a-V_a)K_b$ and $-i(A_a-V_a)K_b$, so they cancel pointwise; no derivative of $A$ is involved. The remaining expression is

$$\begin{aligned} \nabla_aK_b &=\frac H2\zeta^TC[\gamma_a,\gamma_b]\widetilde\zeta +\frac12\epsilon_{acd}V_c\zeta^TC\{\gamma_d,\gamma_b\}\widetilde\zeta\\ &=\epsilon_{abc}(iHK_c-s_{\rm mix}V_c). \end{aligned} \tag{PC04.1}$$

Here $[\gamma_a,\gamma_b]=2i\epsilon_{abc}\gamma_c$ and $\{\gamma_d,\gamma_b\}=2\delta_{db}\mathbf1$. In the second term, $\epsilon_{acb}=-\epsilon_{abc}$ fixes the sign. The symmetric derivative is therefore zero. Opposite R charges caused the cancellation; taking two solutions of the same charge equation would not reproduce this calculation.

As a separate pointwise check, take $\zeta=(\alpha,0)^T$ and $\widetilde\zeta=(u,v)^T$. Then $s_{\rm mix}=-\alpha v$ and $K=(\alpha v,\alpha u,i\alpha u)$, giving $K\cdot K=s_{\rm mix}^2$. This square has no complex conjugation and need not be positive. A real-vector assumption cannot be inferred from it.

Check the adapted coordinate carefully

There is a useful coordinate check when passing from a Killing flow to its unit-speed parametrization. Use the already verified background of M07,

$$g=e^{2kx}d\psi^2+dx^2+dy^2,\qquad K=\partial_\psi, \qquad \xi=e^{-kx}\partial_\psi,$$

with real constant $k$ of inverse-length dimension. Along each orbit, arc length obeys $d\tau=e^{kx}d\psi$. Across a coordinate patch the full differential is different. Choosing $\tau=e^{kx}\psi$ gives

$$d\tau=e^{kx}d\psi+k\tau\,dx,\qquad g=(d\tau-k\tau\,dx)^2+dx^2+dy^2. \tag{PC04.2}$$

Thus $\xi=\partial_\tau$ has unit norm, yet the metric depends on $\tau$. For example $(\mathcal L_\xi g)_{\tau x}=-k$ and $(\mathcal L_\xi g)_{xx}=2k^2\tau$. This is a complete coordinate calculation, not a new background assumption. It explains how to use the along-flow coordinate relation after P (5.5): a literal equality of one-forms $d\tau=\Omega d\psi$ throughout a patch would require $d\Omega\wedge d\psi=0$. The changing transverse length in this example violates that requirement. The canonical geometric derivation is M07.8.

Exit calculation: change only the flat holonomy

Start with the fully patched product solution of M09. Fix the sphere radius $r>0$ and circle period $L>0$, with length coordinate $\tau\sim\tau+L$:

$$g=d\tau^2+r^2d\Omega_2^2,\qquad H=0,\quad V=\frac{i}{r}d\tau,\quad A=V+\nu\,d\tau. \tag{PC04.3}$$

The parameter $\nu$ may be complex and has inverse-length dimension. Keep the same sphere patching and choose circle spin sign $\varepsilon=\pm1$, with no additional R transition. For which $\nu$ do two global solutions in each charge sector survive? Compare $\nu=0$, $\pi/L$, and $i/L$ for both spin structures. Then explain why the local conditions cannot distinguish these cases.

Hint 1

In the complete circle operator the two source terms combine to $A-V$. Solve the first-order equation in each charge sector.

Hint 2

The sphere solution matrices are invertible. The circle factor therefore acts on both columns of a sector by the same scalar. Impose the boundary condition on that factor, including its absolute value.

Solution

The full circle operators are $\mathscr D_\tau^\pm=\partial_\tau\mp i\nu$, so the factors are $e^{\pm i\nu\tau}$. Consequently,

$$e^{i\nu L}=\varepsilon=e^{-i\nu L},\qquad \begin{cases} \nu L=2\pi n,&\varepsilon=+1,\\ \nu L=(2n+1)\pi,&\varepsilon=-1, \end{cases} \quad n\in\mathbb Z. \tag{PC04.4}$$

In particular $\operatorname{Im}\nu=0$ is necessary: a spin sign has modulus one. For periodic spin structure, $\nu=0$ preserves both columns of both sectors and $\nu=\pi/L$ preserves none. For antiperiodic spin structure the answers reverse. The choice $\nu=i/L$ preserves none for either spin structure, because its two factors around the circle are $e^{-1}$ and $e^{+1}$.

Every case has $d(A-V)=0$ and the same $H,V$ and metric. All therefore satisfy the same local curvature equations. Global transport depends on the flat connection’s period, not just its curvature. Using $e^{i\oint A}=e^{-L/r}e^{i\nu L}$ instead of transport for the complete spinor operator would incorrectly reject even the periodic $\nu=0$ example. Nothing here projects out just one of the two columns within a charge sector, since the circle transport is scalar there.

One final local check completes the reading map. If $v^2=V_\mu V^\mu$, M09’s enhancement equations give the ordinary Gaussian curvature of the transverse base as $-(v^2+4H^2)$. Insert the data of P (5.11), (5.13) and (5.14): with $H=ih/r$ and respective $v^2=4(h^2-1)/r^2$, $4h^2/r^2$, $4(h^2+1)/r^2$, the answers are $4/r^2$, $0$ and $-4/r^2$. This checks the radius $r/2$ and all three local signs, without asserting that any chosen quotient preserves the spinors.

Continue to the curved supersymmetry algebra for P Section 6: the vector just constructed acts on matter together with source-dependent internal transformations. To follow the sphere application instead, use the full squashing family before differentiating an observable with respect to its parameter.

For the full component and action map, continue with the Section 6 guide. Its changed-source example shows why shifting a scalar background also requires its auxiliary completion.