Integrability and transverse holomorphic geometry

Before this lesson: Almost contact metric structures.

An almost contact metric structure selects a direction and a complex structure on the perpendicular plane at every point. What makes those complex structures compatible as we move along the direction? A preserved supercharge supplies precisely such a restriction. We will derive it from two components of the Killing equation, then see how it produces complex coordinates transverse to a flow.

The required preparation is almost contact metric structures. In particular, a contact plane need not be tangent to a surface, and the transverse space is a quotient by the flow direction. Keep those two facts separate throughout this lesson.

The tensor whose derivative matters

Work on a smooth oriented Riemannian three-manifold with positive metric. The physical setting is three-dimensional $\mathcal N=2$ , with four real parent Poincaré supercharges. We assume one nowhere-zero commuting charge- $+1$ Killing spinor $\zeta$ ; no second solution, reality relation to $\widetilde\zeta$ , or dynamical gauge group is assumed.

Recall the neutral tensors

n=\zeta^\dagger\zeta,\qquad \eta_\mu=\frac{\zeta^\dagger\gamma_\mu\zeta}{n},\qquad \xi=\eta^\sharp,\qquad \Pi^\mu{}_\nu=\delta^\mu{}_\nu-\xi^\mu\eta_\nu,\qquad \Phi^\mu{}_\nu=\epsilon^\mu{}_{\nu\rho}\xi^\rho. \tag{M03.1}

They obey $\eta(\xi)=1$ and $\Phi^2=-\Pi$ . Our sign is $\Phi X=X\times\xi$ , so in an oriented frame with $\xi=E_1$ , $\Phi E_2=-E_3$ and $\Phi E_3=E_2$ .

The integral curves of the nowhere-zero vector $\xi$ form a one-dimensional foliation: locally they are the curves along a coordinate $\tau$ . The normal bundle $\mathcal N=TM/\mathbb R\xi$ identifies vectors differing by a multiple of $\xi$ . The map $\Phi$ induces a complex structure on this quotient. Its preservation along the flow is a differential statement, not a consequence of $\Phi^2=-\Pi$ .

For example, an isotropically growing transverse metric $e^{2s\tau}(dx^2+dy^2)$ preserves angles and its complex structure. The metric $e^{2s\tau}dx^2+e^{-2s\tau}dy^2$ changes transverse shapes. Here $s$ has dimension inverse length. The Killing equation will allow the first behavior and rule out the second for the specified direction.

Extract the restriction directly from the spinor equation

The charge-one equation in the paper’s conventions is

\nabla_\mu\zeta =i(A_\mu-V_\mu)\zeta -\frac H2\gamma_\mu\zeta -\frac12\epsilon_{\mu\nu\rho}V^\nu\gamma^\rho\zeta. \tag{M03.2}

The sources may be complex. In an orthonormal frame $A_a,V_a,H$ have dimension inverse length. A dagger in the following calculation conjugates these ordinary background values; it does not introduce a second supersymmetry parameter.

At one point choose a normal orthonormal frame and rotate the nonzero spinor to $\zeta=\sqrt n(1,0)^T$ . This fixes its value at that point, not its derivatives in a neighborhood. Write $\nabla_a\zeta=\sqrt n(\alpha_a,b_a)^T$ there. With

\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},

differentiating the normalized bilinear gives

\nabla_a\eta_1=0,\qquad \nabla_a\eta_2=-2\operatorname{Re}b_a,\qquad \nabla_a\eta_3=-2\operatorname{Im}b_a. \tag{M03.3}

The derivative of $n$ cancels the contribution of $\alpha_a$ . The scalar matrix $i(A_a-V_a)$ in (M03.2) also has no lower component on this column. The remaining lower components are

b_1=\frac{iV_2-V_3}{2},\qquad b_2=\frac{H-iV_1}{2},\qquad b_3=\frac{iH+V_1}{2}=i b_2. \tag{M03.4}

For instance, the $a=2$ epsilon term uses $\epsilon_{213}=-1$ and $(\gamma_3\zeta)_2=-i\sqrt n$ , giving $-iV_1/2$ . This is where an orientation error would change the conclusion.

Write $B_{ab}=\nabla_a\eta_b$ , with the derivative index first. Equations (M03.3)–(M03.4) imply the real matrix

B= \begin{pmatrix} 0&a&b\\ 0&c&d\\ 0&-d&c \end{pmatrix}. \tag{M03.5}

The transverse block has one isotropic expansion and one antisymmetric rotation; its symmetric traceless part vanishes. This is the absence of transverse shear. The letters $a,b,c,d$ here are real local coefficients, not the source $A$ .

Define the acceleration one-form and expansion scalar by $\mathfrak a_\nu=\xi^\rho\nabla_\rho\eta_\nu$ and $\vartheta=\nabla_\rho\xi^\rho$ . In this frame $\mathfrak a=(0,a,b)$ and $\vartheta=2c$ . Restoring tensor indices in (M03.5) gives

\boxed{ \nabla_\mu\eta_\nu+\nabla_\nu\eta_\mu =(g_{\mu\nu}-\eta_\mu\eta_\nu)\vartheta +\eta_\mu\mathfrak a_\nu+\eta_\nu\mathfrak a_\mu . } \tag{M03.6}

Equivalently, $\Pi_\mu{}^\rho\Pi_\nu{}^\sigma\nabla_{(\rho}\eta_{\sigma)} =\vartheta(g_{\mu\nu}-\eta_\mu\eta_\nu)/2$ . This is the metric condition (4.6). No source equation of motion was used. Nor does this say that $\xi$ is Killing: the right-hand side can be nonzero.

Why a projector belongs in the integrability equation

For a mixed tensor the Lie derivative is

(\mathcal L_\xi\Phi)^a{}_b =\xi^c\nabla_c\Phi^a{}_b -\Phi^c{}_b\nabla_c\xi^a +\Phi^a{}_c\nabla_b\xi^c. \tag{M03.7}

We can check the required projection without guessing it. Before imposing the spinor restriction, the most general derivative of a unit vector at this point is

B=\begin{pmatrix}0&a&b\\0&c&d\\0&e&f\end{pmatrix}.

Insert $\Phi^a{}_b=\epsilon^a{}_{b1}$ and $\nabla_c\Phi^a{}_b=\epsilon^a{}_{bd}B_{cd}$ into (M03.7):

\mathcal L_\xi\Phi= \begin{pmatrix} 0&b&-a\\ 0&d+e&f-c\\ 0&f-c&-d-e \end{pmatrix},\qquad \Phi\mathcal L_\xi\Phi= \begin{pmatrix} 0&0&0\\ 0&f-c&-d-e\\ 0&-d-e&c-f \end{pmatrix}. \tag{M03.8}

Thus (M03.6) is equivalent to

\boxed{\Phi^\mu{}_\nu(\mathcal L_\xi\Phi)^\nu{}_\rho=0.} \tag{M03.9}

Since $\ker\Phi=\mathbb R\xi$ , the Lie derivative may still take values along $\xi$ . It vanishes after passage to the normal bundle. Requiring $\mathcal L_\xi\Phi=0$ itself would additionally set $a=b=0$ and impose a stronger condition. The distinction is also emphasized in Appendix B.3.

Here is an explicit example with a nonzero longitudinal remainder:

\eta=d\tau+\kappa\tau\,dx,\qquad g=\eta^2+e^{2s\tau}(dx^2+dy^2),\qquad \xi=\partial_\tau . \tag{M03.10}

The coordinates have length units and $\kappa,s$ have inverse-length units. From the oriented coframe $(\eta,e^{s\tau}dx,e^{s\tau}dy)$ one obtains, in the coordinate order $(\tau,x,y)$ ,

\Phi=\begin{pmatrix} 0&0&-\kappa\tau\\0&0&1\\0&-1&0 \end{pmatrix},\qquad \mathcal L_\xi\Phi=\begin{pmatrix} 0&0&-\kappa\\0&0&0\\0&0&0 \end{pmatrix}. \tag{M03.11}

Multiplying by $\Phi$ gives zero. Also $\mathfrak a=\iota_\xi d\eta=\kappa\,dx$ and $\vartheta=2s$ . The transverse complex structure is preserved even though $\xi$ is generally neither geodesic nor Killing. This checks the geometric condition; it does not specify a solution for any preassigned sources.

Turn the preserved line into a complex coordinate

A local holomorphic one-form $\Omega$ is a nonzero complex row satisfying $\Omega\Phi=i\Omega$ . It annihilates $\xi$ and obeys $\Omega\wedge\bar\Omega\ne0$ on the transverse plane. Taking a Lie derivative and using (M03.9) gives

(\mathcal L_\xi\Omega)\Phi =i\mathcal L_\xi\Omega-\Omega(\mathcal L_\xi\Phi) =i\mathcal L_\xi\Omega . \tag{M03.12}

The last term vanished because its vector argument lies along $\xi$ . There is only one complex holomorphic covector direction, so $\mathcal L_\xi\Omega=q\Omega$ for some local complex function $q$ .

Choose flow coordinates with $\xi=\partial_\tau$ and $\Omega=\Omega_x\,dx+\Omega_y\,dy$ . The two coefficients are nonzero on a sufficiently small real coordinate patch, and their ratio $\rho=\Omega_x/\Omega_y$ has nonzero imaginary part. The proportionality just proved implies

\partial_\tau\rho=0,\qquad \Omega\wedge d\Omega =\Omega_y^2\,\partial_\tau\rho\, d\tau\wedge dx\wedge dy=0. \tag{M03.13}

This exterior-form test is invariant under $\Omega\mapsto h\Omega$ for any nonzero complex function $h$ : the three-form multiplies by $h^2$ . In a local R-bundle trivialization one may equally test the complex conjugate of the charged bilinear $P$ from the preceding lessons. Conversely, the vanishing three-form forces $\rho$ to be independent of $\tau$ . The lines spanned by $\Omega$ and $\bar\Omega$ then stay fixed in the transverse cotangent space, so their complex structure is preserved. This proves the reverse implication locally as well.

To obtain a complex coordinate $z=z(x,y)$ we seek $\Omega=\omega\,dz$ . Its equation is $\partial_xz=\rho(x,y)\partial_yz$ , with nonzero real Jacobian. The local isothermal-coordinate theorem in two dimensions supplies such coordinates for a smooth positive conformal structure. This elliptic existence theorem is the additional mathematical input; solving the algebraic identity $J^2=-1$ alone would not construct $z$ . Appendix B.3, (B.8)–(B.14) explains this local step.

The result is a transversely holomorphic foliation, or THF: flow lines have local complex transverse coordinates, and changes of transverse chart are holomorphic. For a fixed normalized generator, overlaps have $z'=f(z)$ and $\tau'=\tau+t(z,\bar z)$ , where $f'\ne0$ and $t$ is real. Work in adapted coordinates tracks the metric and tensors through such changes. The transverse coordinates need not form a single global complex-valued function or a smooth global space of leaves.

A geometric check and the scope of the result

On $\mathbb R^3\setminus\{0\}$ with its Euclidean metric, take the radial unit vector $\xi^i=X^i/r$ , $r^2=X^iX^i$ . Direct differentiation gives

\partial_i\eta_j=\frac{\delta_{ij}-\eta_i\eta_j}{r}, \qquad \vartheta=\frac2r,\qquad \mathfrak a=0 . \tag{M03.14}

Equation (M03.6) holds although radial translations are not isometries. The flow carries a sphere to a larger sphere; its transverse lengths scale equally. In spherical coordinates about the $X^1$ axis, $\Omega=d\theta-i\sin\theta\,d\varphi$ has a transverse complex coordinate $z=e^{-i\varphi}\tan(\theta/2)$ away from its coordinate poles. The pole of this expression is a chart issue, not a singularity of the radial vector. This example has $d\eta=d(dr)=0$ , so contact geometry was unnecessary.

Conversely, being contact is not enough: the first exercise supplies a contact Reeb flow with transverse shear. A Seifert fibration adds global circle-fiber information, which neither the projector equation nor a local complex chart alone supplies.

We have proved that the specified Killing spinor implies (M03.9), and explained its equivalence to a local THF condition. This is a necessary test for a chosen spinor direction on a chosen metric. Passing it does not make arbitrary fixed $A,V,H$ supersymmetric. The converse construction in the paper requires choosing compatible background fields and patching their bundles; those are further steps. Likewise, a local solution or a small numerical residual proves no global spinor existence.

The exit task is to calculate either the transverse shear tensor or $\Omega\wedge d\Omega$ , and then state which global or source question remains unanswered.

Exercises

1. A contact Reeb field that fails the spinor test

Use dimensionless angular coordinates $x,y,z$ of period $2\pi$ and the flat unit-radius metric on $T^3$ . Let $\eta=\cos z\,dx+\sin z\,dy$ and $\xi=\cos z\,\partial_x+\sin z\,\partial_y$ . Show that $\xi$ is the Reeb field of a contact form but does not satisfy (M03.6). Check the failure independently with a holomorphic one-form. What conclusion about supersymmetry is justified?

Hint 1

Use the oriented coframe $e^1=\eta$ , $e^2=-\sin z\,dx+\cos z\,dy$ , $e^3=dz$ . Only the $z$ derivative of $\eta$ is nonzero.

Hint 2

Take $\Omega=e^2-i e^3$ . Compute $d\Omega$ before forming its wedge product; do not identify contact-plane integrability with transverse integrability.

Solution

We have $d\eta=dz\wedge e^2$ and $\eta\wedge d\eta=-dx\wedge dy\wedge dz\ne0$ . Moreover $\eta(\xi)=1$ and $\iota_\xi d\eta=0$ , so $\xi$ is the Reeb field.

The flat covariant derivative in the stated orthonormal frame has the sole nonzero component $B_{32}=1$ . Its expansion and acceleration vanish, but $B_{23}+B_{32}=1$ . Equation (M03.6) would require this sum to vanish. Equivalently, in the general matrix (M03.8), $e=1$ and $a=b=c=d=f=0$ , so the projected Lie derivative is nonzero.

For the independent test, $d\Omega=-dz\wedge\eta$ . Thus

\Omega\wedge d\Omega =-e^2\wedge dz\wedge\eta =-dx\wedge dy\wedge dz\ne0 .

No choice of sources can produce a Killing spinor whose normalized Hermitian direction is this specified $\xi$ . This does not forbid other supersymmetric backgrounds on the same flat torus. In particular, with a periodic spin structure and zero sources, a constant spinor supplies a different, constant direction.

2. Solve a nonconstant transverse coordinate equation

On a real coordinate patch take

\eta=d\tau+\lambda x\,dy,\qquad g=\eta^2+dx^2+e^{2\kappa x}dy^2,\qquad \xi=\partial_\tau ,

where $\kappa,\lambda$ are real constants with inverse-length units. Construct a holomorphic one-form, find a complex coordinate $z$ , and check its Jacobian. Does setting $\lambda=0$ change the transverse-integrability conclusion?

Hint 1

Use $e^2=dx$ and $e^3=e^{\kappa x}dy$ ; the holomorphic row is $e^2-i e^3$ in our orientation.

Hint 2

Divide the row by $e^{\kappa x}$ . The resulting $dx$ coefficient has a real antiderivative depending only on $x$ .

Solution

The one-form is $\Omega=dx-i e^{\kappa x}dy$ . It is independent of $\tau$ , so $\mathcal L_\xi\Omega=0$ . For $\kappa\ne0$ choose

z=\frac{1-e^{-\kappa x}}{\kappa}-iy,\qquad dz=e^{-\kappa x}dx-i\,dy,\qquad \Omega=e^{\kappa x}dz.

The real Jacobian $\partial(\operatorname{Re}z,\operatorname{Im}z)/\partial(x,y)$ is $-e^{-\kappa x}$ , everywhere nonzero on the patch. The $\kappa\to0$ limit is $z=x-iy$ , which checks the orientation and integration constant.

Here $d\eta=\lambda\,dx\wedge dy$ and $\eta\wedge d\eta=\lambda\,d\tau\wedge dx\wedge dy$ . The contact property disappears at $\lambda=0$ , while the same transverse complex coordinate remains valid. Neither case specifies the background fields needed to solve the Killing equation.

3. A local complex direction that cannot survive an identification

Let $A=\begin{pmatrix}2&1\\1&1\end{pmatrix}$ act on a two-torus with real angular column $u=(x,y)^T$ . Form a three-manifold by

(\tau,u)\sim(\tau+L,A u),\qquad L>0.

The vector $\xi=\partial_\tau$ descends to this quotient. The torus point $u=0$ gives a closed flow line. Explain why no transverse complex structure preserved by this particular flow can be smooth and well defined around that closed line. Does this obstruct every possible foliation or metric on the three-manifold?

Hint 1

A preserved complex structure at $u=0$ is constant as $\tau$ changes on the cover. The identification must carry that same linear map into itself.

Hint 2

Derive $AJ=JA$ and $J^2=-I$ on the real transverse tangent plane. The matrix $A$ has two distinct real eigenvalues, $(3\pm\sqrt5)/2$ .

Solution

At the fixed torus point, flow preservation makes the normal complex structure independent of $\tau$ . Smooth descent through the identification requires $J=AJA^{-1}$ . Since the real eigenvalues of $A$ are distinct, every real matrix commuting with it preserves each one-dimensional real eigenspace. In that eigenbasis $J$ is diagonal with real entries. Neither real entry can have square $-1$ , contradicting $J^2=-I$ .

This is a global holonomy obstruction for the specified flow, even though a small flow box always admits a chosen complex coordinate. Local choices cannot be made flow invariant and compatible with this identification simultaneously.

There is no obstruction to a global almost contact metric structure with this direction. For example, let $S=\log A$ , which is real symmetric because $A$ is positive definite, and take on the cover

\eta=d\tau,\qquad g=d\tau^2+du^T e^{-2\tau S/L}du .

Under $(\tau,u)\mapsto(\tau+L,Au)$ , the transverse metric pulls back to itself. It is positive and makes $\xi$ unit. Its transverse complex structure changes along $\tau$ , which is precisely the failed integrability requirement. The argument makes no assertion about another flow, another choice of background sources, or the existence of a THF with different global data.