Construct geometry from one spinor

A spinor has components that depend on the chosen spin frame. How can a preserved supersymmetry nevertheless select an ordinary direction on the manifold? The answer is to form a bilinear whose frame dependence is exactly that of a tensor. One commuting spinor will determine a unit vector, its perpendicular two-plane, and a ninety-degree rotation within that plane.

The preparation is the Killing-spinor equation, spinor bundles and the commuting-spinor projector. We will apply that projector rather than repeat the general Fierz derivation. No complex geometry is assumed.

Specify the spinor and its norm

Let $M$ be a smooth oriented Riemannian three-manifold with positive metric $g$. Fix a spin structure and a Hermitian R-symmetry line bundle $L$. A charge-$+1$ spinor is a section of $L\otimes\mathscr S$, where $\mathscr S$ is the two-component spin bundle. We use conventional unitary R-gauge transitions; the connection $A$ may still have a complex, gauge-invariant imaginary part.

The physical application is three-dimensional $\mathcal N=2$, with four real parent Poincaré supercharges. We will use one charge-one Killing-spinor solution; the existence of additional preserved solutions is a separate question. No dynamical gauge group or gauge rank is chosen for this geometry problem.

Our spinor $\zeta$ is commuting, smooth and nowhere zero. For now it need not solve any differential equation. Its Hermitian norm is

$$n=\zeta^\dagger\zeta>0. \tag{M01.1}$$

The dagger uses the spin and line-bundle Hermitian metrics. It is not a second independent supersymmetry label $\widetilde\zeta$. An odd matter fermion cannot be substituted here: its bilinear is not a positive ordinary number that we may divide by.

Use an oriented orthonormal coframe $e^a$, dual vectors $E_a$, and the established matrices

$$\gamma^1=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\quad \gamma^2=\begin{pmatrix}0&-1\\-1&0\end{pmatrix},\quad \gamma^3=\begin{pmatrix}0&i\\-i&0\end{pmatrix},\qquad C=\begin{pmatrix}0&-1\\1&0\end{pmatrix}. \tag{M01.2}$$

Thus $\gamma^a\gamma^b=\delta^{ab}+i\epsilon^{abc}\gamma_c$, $\epsilon^{123}=1$, and $\zeta\gamma^a\zeta$ means $\zeta^TC\gamma^a\zeta$. These are the source’s Appendix A conventions. Coordinate matrices are $\gamma_\mu=e_\mu{}^a\gamma_a$.

As a first example, choose $\zeta=(1,0)^T$ in a local trivialization. Its three Hermitian bilinears are $(1,0,0)$. The distinguished direction is $E_1$, and the perpendicular plane is spanned by $E_2,E_3$. The construction below makes this statement independent of that particular column and frame.

Extract a unit direction

Define a real one-form and its metric-dual vector:

$$\eta_\mu=\frac{\zeta^\dagger\gamma_\mu\zeta}{n}, \qquad \xi^\mu=g^{\mu\nu}\eta_\nu. \tag{M01.3}$$

The source writes the vector as $\eta^\mu$; the separate symbol $\xi$ here keeps vectors and one-forms easy to distinguish. Hermiticity of $\gamma_a$ makes the components real.

The normalized rank-one spinor matrix is

$$\rho=\frac{\zeta\zeta^\dagger}{n} =\frac12(1+\eta_a\gamma^a),\qquad \rho^2=\rho.$$

The scalar coefficient in $\rho^2=\rho$ gives $\eta_a\eta^a=1$: the antisymmetric Clifford term vanishes when contracted with $\eta_a\eta_b$. Applying $\rho$ to $\zeta$ also gives

$$\boxed{\eta_\mu\xi^\mu=1,\qquad (\eta_\mu\gamma^\mu)\zeta=\zeta.} \tag{M01.4}$$

The normalization removed the arbitrary size of the label. If its frame components have engineering dimension $-1/2$, then $n$ has dimension $-1$, while the frame components of $\eta$ are dimensionless. A Hermitian contraction and division by $n$, not the Killing equation, produced the unit vector. This is the construction of equations (3.3)–(3.4).

Separate the transverse plane

At each point, let $\mathcal D$ denote the real plane perpendicular to $\xi$: $\mathcal D=\{X:\eta_\mu X^\mu=0\}$. The tensor

$$\Pi^\mu{}_\nu=\delta^\mu{}_\nu-\xi^\mu\eta_\nu \tag{M01.5}$$

projects any vector onto this plane. Indeed, $\Pi X=X-\eta(X)\xi$, so $\eta(\Pi X)=0$, $\Pi^2=\Pi$ and $\Pi\xi=0$. This three-by-three tangent-space projector is different from the two-by-two spinor projector $\rho$.

Orientation supplies another tensor:

$$\Phi^\mu{}_\nu=\epsilon^\mu{}_{\nu\rho}\xi^\rho. \tag{M01.6}$$

Here $\epsilon$ is the metric volume tensor, not an unnormalized coordinate symbol. In an orthonormal frame $\Phi X=X\times\xi$. In particular it is the negative of $\xi\times X$; this order fixes a useful sign.

The vector triple-product identity gives

$$\Phi^2X=(X\times\xi)\times\xi =\eta(X)\xi-X=-\Pi X. \tag{M01.7}$$

Also $\Phi\xi=0$ and $\eta(\Phi X)=0$. Since $\Phi$ is skew-adjoint with respect to $g$, we obtain

$$g(\Phi X,\Phi Y) =-g(X,\Phi^2Y) =g(X,Y)-\eta(X)\eta(Y). \tag{M01.8}$$

Consequently the restriction $J=\Phi|_{\mathcal D}$ satisfies $J^2=-1$ and preserves lengths. This is what a complex structure on a real two-dimensional vector space means: multiplication by $i$ is represented by a ninety-degree rotation.

For the first example, $\xi=E_1$, and

$$\Phi= \begin{pmatrix}0&0&0\\0&0&1\\0&-1&0\end{pmatrix}, \qquad \Phi E_2=-E_3,\quad \Phi E_3=E_2.$$

The metric and tensors satisfying (M01.4)–(M01.8) form an almost contact metric structure. At this stage that name records algebraic identities. It does not say that the transverse planes are tangent to surfaces, or that the geometry is contact or transversely holomorphic. Those are additional questions about how the tensors vary between points. The algebra agrees with the source’s (3.5)–(3.6) and Appendix B.1.

Complex transverse directions and an index trap

Allow complex coefficients in the transverse plane. Since $J^2=-1$, its eigenvalues are $+i$ and $-i$. The corresponding tangent-space projectors are

$$\Pi_+=\frac12(\Pi-i\Phi),\qquad \Pi_-=\frac12(\Pi+i\Phi). \tag{M01.9}$$

Using $\Phi^2=-\Pi$ and $\Pi\Phi=\Phi\Pi=\Phi$, multiplication gives $\Pi_\pm^2=\Pi_\pm$, $\Pi_+\Pi_-=0$ and $\Pi_++\Pi_-=\Pi$. Each image has one complex dimension.

Following the source’s terminology, call a vector $X$ holomorphic when $\Phi X=iX$. At this point “holomorphic” labels an eigenspace; no holomorphic coordinate chart has yet been constructed. A useful spinor test is

$$\Phi X=iX \quad\Longleftrightarrow\quad X^\mu\gamma_\mu\zeta=0. \tag{M01.10}$$

To see the forward implication, compute an ordinary positive norm:

$$\|X^a\gamma_a\zeta\|^2 =n\,\bar X^a(\delta_{ab}+i\Phi_{ab})X^b=0.$$

For the reverse implication, multiply $X^a\gamma_a\zeta=0$ first by $\zeta^\dagger$ and then by $\zeta^\dagger\gamma_b$. These give $\eta_aX^a=0$ and $X_b+i\Phi_{ba}X^a=0$, hence $\Phi X=iX$. The positive Hermitian norm is essential in the forward step.

A one-form $\omega$ instead acts on $\Phi$ from the left: it is called holomorphic when $\omega_\mu\Phi^\mu{}_\nu=i\omega_\nu$. Because $\Phi$ is skew-adjoint, lowering a holomorphic vector’s index gives an antiholomorphic one-form:

$$\Phi X=iX \quad\Longrightarrow\quad X_\mu\Phi^\mu{}_\nu=-iX_\nu. \tag{M01.11}$$

This is a consequence of the metric pairing, not a disagreement between conventions. Complex conjugation also exchanges the two eigenspaces, but lowering an index is not complex conjugation.

The bilinear that retains a phase

Besides the neutral Hermitian direction, form

$$P_\mu=\zeta^TC\gamma_\mu\zeta. \tag{M01.12}$$

There is no dagger. This is a complex one-form of R-charge $+2$, rather than an ordinary real direction. Its algebraic properties are

$$\eta^\mu P_\mu=0,\qquad P_\mu P^\mu=0,\qquad \bar P_\mu P^\mu=2n^2. \tag{M01.13}$$

Here the bar uses the Hermitian line-bundle pairing when the two factors are contracted. In particular $P$ never vanishes when $\zeta$ is nonzero. Its complex null norm is compatible with the positive Euclidean metric because $P_\mu P^\mu$ has no conjugation.

For a direct check, write $\zeta=(a,b)^T$, with commuting complex entries. Matrix multiplication gives

$$\begin{aligned} n&=|a|^2+|b|^2,\\ n\eta_a&=(|a|^2-|b|^2,\,-\bar a b-\bar b a,\, i\bar a b-i\bar b a),\\ P_a&=(2ab,\,a^2-b^2,\,i(a^2+b^2)). \end{aligned} \tag{M01.14}$$

The first contraction in (M01.13) cancels term by term. The null contraction reduces to $4a^2b^2+(a^2-b^2)^2-(a^2+b^2)^2=0$. For the Hermitian norm, the two last components contribute $2(|a|^4+|b|^4)$, and the first contributes $4|a|^2|b|^2$. Their sum is $2n^2$.

The same components give $P^a\gamma_a\zeta=0$. Thus $P^\sharp$, the metric-dual vector, is holomorphic by (M01.10), while

$$\boxed{P_\mu\Phi^\mu{}_\nu=-iP_\nu.} \tag{M01.15}$$

This explains the source’s description of $P$ as an antiholomorphic one-form in (3.7)–(3.8).

Locally write $P/n=u+iv$, taking the real and imaginary parts of its frame components. Equations (M01.13) imply $u^2=v^2=1$, $u\cdot v=0$, and both are orthogonal to $\xi$. Equation (M01.15), after raising the index, gives $\Phi u=-v,\ \Phi v=u$. With our orientation, $u\times v=\xi$. A local spinor therefore supplies an oriented orthonormal triad; the neutral tensors do not retain its transverse phase.

Work through a direction away from the axes

On flat Cartesian space choose the constant column $\zeta=(1,1+i)^T$, in an arbitrary fixed overall normalization. Then

$$n=3,\qquad \xi=\frac13(-1,-2,-2),\qquad \frac{P}{n}=\frac13(2+2i,\,1-2i,\,-2+i). \tag{M01.16}$$

The real transverse vectors are $u=(2,1,-2)/3$ and $v=(2,-2,1)/3$. Their dot products give an orthonormal triad, and their cross product is $\xi$. For an explicit matrix check,

$$\Phi=\frac13 \begin{pmatrix}0&-2&2\\2&0&-1\\-2&1&0\end{pmatrix}, \qquad \Phi u=-v,\quad \Phi v=u.$$

There is a supersymmetric background realizing this simple example: the Cartesian spin connection and $A,V,H$ all vanish, so the constant $\zeta$ solves every component of the charge-$+1$ Killing equation.

Now replace it by $\zeta'=e^{i\lambda(x)}\zeta$, with real smooth $\lambda$, and take $A'=d\lambda,\ V'=H'=0$. The derivative of its phase cancels the R-connection, so it remains a solution. Yet $n,\eta,\Pi,\Phi$ have not changed, while $P'=e^{2i\lambda}P$. The geometry extracted from the normalized Hermitian bilinear contains less information than the full charged spinor and its source.

Which statements are global or differential?

On overlapping oriented frames, use the spinor transition

$$\zeta_j=e^{i\lambda_{ji}}S_{ji}\zeta_i,\qquad S_{ji}^{-1}\gamma^aS_{ji}=R_{ji}{}^a{}_b\gamma^b.$$

Unitarity and $S_{ji}^TCS_{ji}=C$ imply

$$n_j=n_i,\qquad \eta_j^a=R_{ji}{}^a{}_b\eta_i^b,\qquad P_j^a=e^{2i\lambda_{ji}}R_{ji}{}^a{}_bP_i^b. \tag{M01.17}$$

The first two laws make $\eta,\xi,\Pi,\Phi$ ordinary global tensors. The last law makes $P$ a section of $L^2\otimes T^*M$. Its local real and imaginary parts need not join into a globally chosen pair $u,v$. A nonzero charged bilinear is not, by itself, a trivialization of each bundle factor separately.

If $\zeta$ is a nontrivial global solution of the Killing equation on a connected smooth background, its nowhere-zero property follows from uniqueness of the transport equation along every curve. This was proved in the prerequisite lesson and is the argument of the source’s section 3, footnote 5. It does not promote an arbitrary local solution to a global one.

Conversely, arbitrary pointwise algebra does not solve

$$(\nabla_\mu-iA_\mu)\zeta =-\frac H2\gamma_\mu\zeta-iV_\mu\zeta -\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\zeta. \tag{M01.18}$$

Differentiating bilinears using this equation adds restrictions. The source derives the transverse integrability condition in (4.6)–(4.7); it is the step beyond the almost contact algebra. Its geometric interpretation and the reconstruction of suitable sources require further work. Nor should $\xi$ be called a Killing vector merely because it came from one spinor: $\nabla_{(\mu}\eta_{\nu)}=0$ is another differential assertion.

The next calculations will distinguish these structures carefully. For now the exit test is to construct $\eta,\Phi,P$, check their index-sensitive algebra and overlaps, and state exactly which differential claim has not yet been established. The last exercise gives a concrete obstruction even on flat space.

Exercises

1. Change information that the unit direction forgets

Let $\zeta'=f e^{i\beta}\zeta$, with smooth $f>0$ and real $\beta$. Determine the changes in $n,\eta,\Phi,P/n$ and in the local pair $u,v$. If $\zeta$ solves the Killing equation on a fixed background, does an arbitrary such rescaling remain a solution? Separate a change of the label from an allowed R-gauge transformation.

Hint 1

A dagger conjugates the phase but a transpose does not. The norm of the rescaled spinor contains two factors of $f$.

Hint 2

For the fixed-source first-order operator $\mathscr K_\mu$, use $\mathscr K_\mu(h\zeta)=h\mathscr K_\mu\zeta+(\partial_\mu h)\zeta$, where $h=f e^{i\beta}$.

Solution

The norm becomes $n'=f^2n$, while $\eta,\xi,\Pi,\Phi$ are unchanged. The charged bilinear obeys $P'=f^2e^{2i\beta}P$, hence $P'/n'=e^{2i\beta}(u+iv)$. Separating the parts gives

$$u'=\cos(2\beta)u-\sin(2\beta)v,\qquad v'=\sin(2\beta)u+\cos(2\beta)v.$$

This rotation preserves $u'\times v'=\xi$ and the transverse metric. The doubled angle tests the R-charge two of $P$.

With all sources fixed, the residual is $(\partial_\mu h)\zeta$. Since $\zeta$ is nowhere zero, the rescaled field solves all equations only if $dh=0$ on the connected region. The pointwise tensors did not detect that failure.

A phase change accompanied by $A'=A+d\beta$ is instead the allowed R-gauge transformation and preserves the equation. A nonconstant positive rescaling alone is not such a unitary transformation. Formally it could be compensated by changing the imaginary background connection to $A'=A-i\,d\log f$, but that changes a gauge-invariant source rather than using the original background’s gauge freedom.

2. Project a vector and then lower its index

Use the reference spinor $\zeta=(1,0)^T$ and the Euclidean frame above. For a real vector $X=aE_1+bE_2+cE_3$, compute $\Pi_\pm X$. Verify that the $+i$ part annihilates $\zeta$ under Clifford multiplication. Is its metric-dual one-form holomorphic in the one-form convention?

Hint 1

First remove $aE_1$. On the remaining plane, $\Phi(bE_2+cE_3)=cE_2-bE_3$.

Hint 2

Factor $\Pi_+X$ as a complex scalar times $E_2+iE_3$. For the one-form test multiply a row by the displayed matrix $\Phi$.

Solution

The two parts are

$$\Pi_+X=\frac{b-ic}{2}(E_2+iE_3),\qquad \Pi_-X=\frac{b+ic}{2}(E_2-iE_3).$$

They are conjugate and sum to $bE_2+cE_3$, as required for a real vector. Also $(\gamma_2+i\gamma_3)(1,0)^T=(0,-1)^T+(0,1)^T=0$. This verifies the spinor characterization of the holomorphic vector line.

Lowering the index gives a multiple of $e^2+i e^3$. As a row, $(0,1,i)\Phi=(0,-i,1)=-i(0,1,i)$. It is antiholomorphic as a one-form. Treating the row and column eigenvalue equations as identical would reverse this answer. The nonzero vector has zero complex bilinear norm but positive Hermitian norm; this is the same distinction encountered for $P$.

3. Pointwise geometry without a Killing spinor

On Cartesian $\mathbb R^3$, let $\kappa\ne0$ be real with dimension inverse length, and take

$$\zeta(x)= \begin{pmatrix} \cos(\kappa x^2/2)\\-\sin(\kappa x^2/2) \end{pmatrix}.$$

Here $x^2$ is the second coordinate, not the square of $x$. Find $n$ and $\eta$. Can any choice of smooth complex $A,V,H$ make this specified field satisfy (M01.18) near the origin? Use only the lower component of its $\mu=2,3$ equations there.

Hint 1

At the origin $\zeta=(1,0)^T$. Its lower component has derivatives $\partial_2\zeta_2=-\kappa/2$ and $\partial_3\zeta_2=0$. The spin connection vanishes in the Cartesian frame.

Hint 2

The terms proportional to $A_\mu\zeta$ and $V_\mu\zeta$ have zero lower component at that point. Evaluate the remaining gamma and epsilon terms before eliminating $H,V_1$.

Solution

The norm is one everywhere. Direct Hermitian products give $\eta=(\cos(\kappa x^2),\sin(\kappa x^2),0)$. It is a smooth unit one-form and defines all the algebraic tensors of this lesson without a singularity.

At the origin the two lower-component equations would require

$$\partial_2\zeta_2=\frac H2-\frac i2V_1,\qquad \partial_3\zeta_2=\frac{iH}{2}+\frac12V_1 =i\,\partial_2\zeta_2.$$

The actual derivatives instead give $0=-i\kappa/2$, which is impossible. The failure occurs even before imposing the additional condition $\nabla_\mu V^\mu=0$; allowing arbitrary complex source values does not remove it. A nonzero local rescaling of this spinor does not help, since at the origin the lower component vanishes and both displayed derivatives simply acquire the same rescaling factor.

Thus the pointwise almost contact data are not sufficient for the rigid equation. This does not say that flat space has no Killing spinors: constant columns on its zero-source background already provide examples. It says that this particular varying spinor line fails a necessary differential compatibility test.

Continue with almost contact metric structures. Explicit frames and horizontal loops show which additional conditions turn these algebraic tensors into contact or transverse geometric data.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 3, geometry from one spinor