Two supercharges and their Killing vector

One preserved spinor selects a direction and a transverse complex structure, but that direction need not generate an isometry. What changes when a second supercharge survives? For two labels of opposite R charge, a mixed bilinear obeys the Killing equation. The result is stronger than a direction field, but it is generally complex.

Prepare with global spinor patching and verification of background solutions. We will prove the bilinear statement, isolate its reality assumption, and check both spinor equations in a background with a nonconstant isometry length.

Keep the two charge equations distinct

The parent is three-dimensional $\mathcal N=2$, with four real Poincaré supercharges. In Euclidean signature we use independent commuting labels $\zeta$ and $\widetilde\zeta$, of R charges $+1$ and $-1$. Both must solve their equations for the same metric and sources:

$$\begin{aligned} \nabla_a\zeta &=i(A_a-V_a)\zeta-\frac H2\gamma_a\zeta -\frac12\epsilon_{abc}V_b\gamma_c\zeta,\\ \nabla_a\widetilde\zeta &=-i(A_a-V_a)\widetilde\zeta-\frac H2\gamma_a\widetilde\zeta +\frac12\epsilon_{abc}V_b\gamma_c\widetilde\zeta. \end{aligned} \tag{M07.1}$$

The metric is real and positive, $\nabla\cdot V=0$, and the sources may be complex. We retain $\epsilon_{123}=1$, $\gamma=(\sigma_3,-\sigma_1,-\sigma_2)$, and the contraction matrix $C=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$. No conjugation between the labels has been imposed.

Different R charges distinguish the two supersymmetry generators even if their component columns happen to be proportional at a point. This differs from two linearly independent solutions of a single charge equation, which was a stronger condition in the preceding verification lesson.

Define a scalar and a vector by

$$s=\zeta^TC\widetilde\zeta,\qquad K_a=\zeta^TC\gamma_a\widetilde\zeta . \tag{M07.2}$$

Both have R charge zero. On an overlap the phases of the two labels cancel, while the spin matrices obey $S^TCS=C$. Thus $s$ is an ordinary scalar and $K$ an ordinary complex vector, provided the original labels are globally defined. With the usual label dimension $-1/2$, both bilinears have dimension $-1$; constant label normalizations may instead be factored out when describing the geometric flow.

Differentiate the mixed vector

The transpose identity $\gamma_a^TC=-C\gamma_a$ is the useful algebraic step. Differentiate (M07.2), substitute (M07.1), and use covariant constancy of the gamma matrices. The two identity terms involving $A-V$ cancel. The remaining terms organize as

$$\begin{aligned} \nabla_aK_b ={}&\frac H2\zeta^TC[\gamma_a,\gamma_b]\widetilde\zeta\\ &+\frac12\epsilon_{acd}V_c \zeta^TC\{\gamma_d,\gamma_b\}\widetilde\zeta\\ ={}&\epsilon_{abc}\bigl(iH K_c-sV_c\bigr). \end{aligned} \tag{M07.3}$$

The commutator is $2i\epsilon_{abc}\gamma_c$, whereas the anticommutator is $2\delta_{db}\mathbf1$. Their different roles fix the relative sign. The right side is antisymmetric in $a,b$, so

$$\boxed{\nabla_aK_b+\nabla_bK_a=0.} \tag{M07.4}$$

This is the statement of P (5.1)–(5.2). It uses both full equations on the same sources, not only their algebraic integrability conditions. Replacing the second label by a second charge-$+1$ solution would change the cancellation and would not give this neutral vector.

The same substitution in the scalar gives a useful check:

$$\partial_a s=\epsilon_{abc}V_bK_c,\qquad K_aK_a=s^2. \tag{M07.5}$$

The second identity is the commuting-spinor completeness identity. One can verify its sign without remembering a rearrangement formula. At a point rotate the nonzero $\zeta$ to $(\alpha,0)^T$ and write $\widetilde\zeta=(u,v)^T$. Then

$$s=-\alpha v,\qquad K=(\alpha v,\alpha u,i\alpha u). \tag{M07.6}$$

These components give $K^2=s^2$ directly. They also show that $K$ cannot vanish if both labels are nonzero. First-order uniqueness makes a nontrivial label nowhere zero on the connected domain, so the mixed complex vector is nowhere zero.

There is no complex conjugation in $K^2$. A nonzero complex vector can be null even in a positive-definite metric; Exercise 2 gives an explicit example. Differentiating $K^2=s^2$ using (M07.3) agrees with the first identity in (M07.5), checking the signs independently without dividing by a possibly vanishing $s$.

Identify the real-vector case

Because the metric is real, the real and imaginary parts of $K$ separately satisfy the real Killing equation. They can generate two independent real isometries. It is an additional restriction to assume that a constant complex rescaling makes $K$ real.

Why specify constant? Multiplying either solution by a variable function generally destroys its differential equation. Also, if two nonzero real Killing fields obey $Y=fX$ on an open set, then $0=\mathcal L_{fX}g=df\otimes X^\flat+X^\flat\otimes df$ forces $df=0$. Thus a single real Killing direction cannot acquire an arbitrary position-dependent phase while remaining Killing.

Suppose we are in this real-vector case. Choose its overall sign so that it points along the normalized Hermitian direction of $\zeta$. Define $\rho=\zeta^\dagger\zeta>0$ and $\Omega=\sqrt{K^\mu K_\mu}>0$. Equation (M07.6), followed by a change back to the original frame, gives

$$\widetilde\zeta=\frac{\Omega}{\rho}C\overline\zeta,\qquad s=-\Omega,\qquad K=\Omega\xi,\qquad \xi^\flat=\frac{\zeta^\dagger\gamma_a\zeta}{\rho}\,e^a . \tag{M07.7}$$

The column $C\overline\zeta$ is the index-correct conjugate spinor in this notation. This relation describes the chosen pair of solutions with real $K$; it is not permission to manufacture the second solution by conjugating the first for arbitrary sources. It is the column version of P (5.3)–(5.4). The normalized Hermitian direction of $\widetilde\zeta$ is $-\xi$.

The length is constant along its own flow: $K(\Omega)=0$, because contracting the Killing equation with $K^aK^b$ gives $K(K^2)=0$. It may vary in transverse directions. In fact the unit vector need not itself be Killing:

$$\mathcal L_\xi g =-d\log\Omega\otimes\xi^\flat -\xi^\flat\otimes d\log\Omega,\qquad \nabla_\xi\xi^\flat=-d\log\Omega . \tag{M07.8}$$

The second equation follows from $K^a\nabla_aK_b=-\frac12\partial_bK^2$ and $K(\Omega)=0$. It explains how a genuine isometry can have nongeodesic unit-speed orbits.

Locally choose a flow coordinate $\psi$ with $K=\partial_\psi$. All metric coefficients are then independent of $\psi$. Completing the square and choosing isothermal coordinates on the two-dimensional local quotient gives

$$g=\Omega^2(x,y)(d\psi+\alpha_xdx+\alpha_y dy)^2 +c^2(x,y)(dx^2+dy^2). \tag{M07.9}$$

Here $\alpha_x,\alpha_y$ are ordinary real functions, unrelated to the R connection $A$. Isothermal means that the transverse metric is conformal to the Euclidean two-metric. This is the local content of P (5.5). A real Killing vector alone does not say that every orbit closes, identify a global circle action, or classify the manifold.

Verify an isometry with varying length

Use length coordinates $(\psi,x,y)$, let $k$ be a real inverse length, and take

$$\Omega=e^{kx},\qquad g=\Omega^2d\psi^2+dx^2+dy^2,\qquad A=V=-k\,dy,\qquad H=0. \tag{M07.10}$$

Choose the orientation of $e^1=\Omega d\psi$, $e^2=dx$, $e^3=dy$. Its only independent Cartan connection is $\Omega_{12}=k e^1$. The connection one-form $\Omega_{12}$ and the positive function $\Omega$ are different objects; the indices distinguish them. Thus $\Gamma_1=ik\gamma_3/2$ and $\Gamma_2=\Gamma_3=0$.

The candidate labels are

$$\zeta=\sqrt\Omega\,u_+,\qquad \widetilde\zeta=\sqrt\Omega\,u_-, \qquad u_+=\begin{pmatrix}1\\0\end{pmatrix},\quad u_-=\begin{pmatrix}0\\1\end{pmatrix}. \tag{M07.11}$$

For the charge-one equation, the complete frame operators are

$$\mathscr D_1=E_1+\frac k2(i\gamma_3+\gamma_2),\qquad \mathscr D_2=\partial_x-\frac k2\gamma_1,\qquad \mathscr D_3=\partial_y . \tag{M07.12}$$

The matrix $i\gamma_3+\gamma_2$ annihilates $u_+$, while $\partial_x\sqrt\Omega=k\sqrt\Omega/2$ cancels the second operator. For the opposite charge the two matrices become $k(i\gamma_3-\gamma_2)/2$ and $+k\gamma_1/2$. They annihilate the lower column with the same derivative cancellation. Both full equations hold.

The source constraint also holds: $V^\sharp=-k\partial_y$ has zero divergence. A potential is $C=i\Omega\,d\psi$, since $-i*dC=V$; $H=0$ permits $B=0$ locally. Constant prefactors of the labels can restore their engineering dimensions without changing any equation.

Direct contraction now gives

$$s=-\Omega,\qquad K^a=(\Omega,0,0),\qquad K=\partial_\psi . \tag{M07.13}$$

The coordinate metric is independent of $\psi$, so $\mathcal L_Kg=0$. There is also a nontrivial component check: $K^\flat=\Omega^2d\psi$, $\nabla_xK_\psi=k\Omega^2$ and $\nabla_\psi K_x=-k\Omega^2$. In the orthonormal frame, (M07.3) predicts $\nabla_1K_2=-k\Omega$, in agreement. Equation (M07.5) predicts $\partial_xs=-k\Omega$, again agreeing.

The pair can be made global on $S^1_\psi\times\mathbb R_x\times\mathbb R_y$ with periodic spin structure: the frame, sources and labels are single-valued. This verifies the displayed pair, rather than claiming that no further solutions exist. Global orbit classification and the conditions for maximal enhancement are separate questions.

Exercises

1. Add the opposite charge to the twisted product

Use the preceding patching lesson’s $S^1_\tau\times S^2$ background: $H=V=0$, $\Gamma_N=iA_N\gamma_1$, $\Gamma_S=iA_S\gamma_1$, $A_N-A_S=d\varphi$ and $S_{NS}=e^{-i\varphi\gamma_1}$. Its charge-one solution has local column $u_+$. Construct an opposite-charge solution, check its overlap, and compute $K$. If the same flat source $\nu\,d\tau$ is added to $A$, can both labels satisfy the same spin boundary sign $\varepsilon$ around the circle?

Hint 1

The opposite-charge operator is $d+\Gamma+iA$, and $\gamma_1u_-=-u_-$.

Hint 2

Its R overlap phase is $e^{-i\varphi}$. The two circle equations have opposite signs of $\nu$.

Solution

Take $\widetilde\zeta_N=\widetilde\zeta_S=u_-$. Then $(d+\Gamma+iA)u_-=iA(\gamma_1+\mathbf1)u_-=0$. On the overlap, $e^{-i\varphi}S_{NS}u_-=u_-$, so the original spinor glues. The neutral bilinears are $s=-1$ and $K=\partial_\tau$. They are globally defined even though the individual spin and R lines have nontrivial sphere transitions.

With the flat term, the two columns become $e^{i\nu\tau}u_+$ and $e^{-i\nu\tau}u_-$. The conditions are $e^{i\nu L_\tau}=\varepsilon$ and $e^{-i\nu L_\tau}=\varepsilon$. Since $\varepsilon=\pm1$ equals its inverse, these impose the same condition. It forces $\nu$ real; periodic spin structure requires an integer multiple of $2\pi/L_\tau$, and antiperiodic structure an odd multiple of $\pi/L_\tau$. Their phases cancel in $K$, but checking only $K$ would miss either spinor’s individual boundary condition.

2. A nonzero null Killing vector

On flat $\mathbb R^3$ with zero sources, take $\zeta=\widetilde\zeta=u_+$ as columns, remembering their different R charges. Find $s,K$, $K\cdot K$ and $\overline K\cdot K$. Can a constant complex rescaling make $K$ real? Compare with the choice $\widetilde\zeta=u_-$.

Hint 1

Use the three components in (M07.6) with $\alpha=u=1$, $v=0$.

Hint 2

If a nonzero multiple of both $1$ and $i$ were real, their ratio would be real.

Solution

The first pair gives $s=0$ and $K=\partial_2+i\partial_3$. Thus $K\cdot K=1+i^2=0$, whereas $\overline K\cdot K=2$. The vector has not vanished and the positive metric has not developed a real null direction. Its real and imaginary parts generate independent translations. No nonzero constant rescaling can make both components real.

For $\widetilde\zeta=u_-$, the bilinears instead are $s=-1$ and $K=\partial_1$, so the real-vector assumption holds. Both pairs solve the full flat equations. The difference is a choice of preserved generators within that background, not a change in the metric or a justification for imposing Euclidean conjugation on every pair.

3. Normalize the vector and test the metric again

For (M07.10), compute the Lie derivative of the metric along $\xi=K/\Omega=e^{-kx}\partial_\psi$ and its acceleration. Compare the finite flows generated by $K$ and $\xi$. Explain why the factor $e^{-kx}$ cannot simply be absorbed into a position-dependent redefinition of one of the preserved labels.

Hint 1

The mixed component is $(\mathcal L_\xi g)_{\psi x} =g_{\psi\psi}\partial_x\xi^\psi$.

Hint 2

The unit vector’s only nonzero component is $\xi^\psi=\Omega^{-1}$, and $\Gamma^x{}_{\psi\psi}=-k\Omega^2$. Apply the full first-order equation to $f\zeta$.

Solution

For $k\ne0$, $(\mathcal L_\xi g)_{\psi x}=-k\Omega$, with the symmetric $x\psi$ component equal to it; the other components vanish. The acceleration is $\xi^\mu\nabla_\mu\xi^x=-k$, or $(\nabla_\xi\xi)^\flat=-k\,dx=-d\log\Omega$.

The flow of $K$ sends $\psi$ to $\psi+t$ and leaves the metric unchanged. The flow of $\xi$ sends $\psi$ to $\psi+t e^{-kx}$. Its differential contains the extra term $-tk e^{-kx}dx$, producing the mixed metric component above. Unit-speed parametrization has turned a rigid translation into a transverse-position-dependent shift.

Finally, $\mathscr D(f\zeta)=df\,\zeta$ when $\mathscr D\zeta=0$. Since $\zeta$ is nowhere zero, the same fixed-source equation allows this rescaling only when $df=0$. Normalizing a geometric vector is legitimate, but does not preserve either its Killing property or the original supersymmetry equations automatically.

The local Killing equation leaves the orbits’ global behavior open. Seifert fibrations and explicit examples tracks their periods, exceptional fibers and the spin/R data that let the preserved pair descend to a quotient.

Four supercharges and product geometries takes the other next step: determine when two solutions in each charge sector coexist, and why a product background can have a genuinely complex mixed Killing vector.

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