Work in adapted coordinates

Before this lesson: Integrability and transverse holomorphic geometry.

A coordinate system adapted to a foliation separates motion along a flow from motion transverse to it. It does not generally diagonalize the metric. The cross terms contain useful information: they tell us which transverse coordinate changes remain perpendicular to the chosen direction.

The prerequisite is integrability and transverse holomorphic geometry. We take its local complex-coordinate existence result as input and learn to calculate in those coordinates. The aim is to keep the same geometry through a change of chart, before reconstructing any supersymmetric background fields.

State what an adapted chart preserves

Let $\xi$ be a specified smooth unit vector on an oriented Riemannian three-manifold, and let $\eta=g(\xi,\cdot)$ . Assume the integrability condition from the preceding lesson. In an adapted chart,

\xi=\partial_\tau,\qquad \xi(z)=0,\qquad \Omega=\omega(\tau,z,\bar z)\,dz. \tag{M04.1}

Here $\tau$ is real, $z$ is a complex transverse coordinate, and $\Omega$ is any nonzero holomorphic covector, satisfying $\Omega\Phi=i\Omega$ . The coefficient $\omega$ is not assumed holomorphic; it records the choice of a local section of the holomorphic covector line. It is the line, rather than every section, that the flow preserves.

For the supersymmetry application the physical parent is three-dimensional $\mathcal N=2$ , with four real Poincaré supercharges; one preserved commuting spinor supplied this structure. The coordinate calculation itself needs neither a second spinor nor a dynamical gauge group. An adapted chart does not yet specify $A,V,H$ , and its existence does not solve a Killing equation for arbitrary fixed sources.

Use real transverse coordinates $x,y$ with $z=x-iy$ . Thus $\partial_z=(\partial_x+i\partial_y)/2$ and $\partial_{\bar z}=(\partial_x-i\partial_y)/2$ . This choice matches the oriented frame convention $e^2-ie^3=c\,dz$ in P (4.20). It will fix the signs of the complex volume form and holomorphic vectors below.

Derive the metric rather than guessing its components

Because $\eta(\partial_\tau)=1$ and $\eta$ is real, write

\eta=d\tau+h\,dz+\bar h\,d\bar z =d\tau+p\,dx+q\,dy, \qquad p=2\operatorname{Re}h,\quad q=2\operatorname{Im}h.

The symbol $h$ is a metric coefficient; it is not the supergravity scalar $H$ . The positive metric on the transverse quotient is Hermitian in its one complex dimension. It therefore has one real positive scale $c$ :

g=\eta^2+c^2dz\,d\bar z =\eta^2+c^2(dx^2+dy^2),\qquad c>0. \tag{M04.2}

The functions $h,c$ may depend on all three coordinates. In particular, this form does not require $\xi$ to be Killing. The product $dz\,d\bar z$ means the symmetric tensor $(dz\otimes d\bar z+d\bar z\otimes dz)/2$ . Consequently the transverse contribution to $g_{z\bar z}$ is $c^2/2$ , not $c^2$ . Equations (M04.1)–(M04.2) are the chart data in P (4.9)–(4.12).

An oriented orthonormal coframe and its dual frame are

\begin{aligned} &e^1=\eta,\quad e^2=c\,dx,\quad e^3=c\,dy,\\ &E_1=\partial_\tau,\quad E_2=c^{-1}(\partial_x-p\partial_\tau),\quad E_3=c^{-1}(\partial_y-q\partial_\tau). \end{aligned} \tag{M04.3}

The terms proportional to $\partial_\tau$ make $E_2,E_3$ perpendicular to $\xi$ . Dropping them would confuse the coordinate plane with the horizontal plane $\ker\eta$ .

Define the horizontal lifts of the complex coordinate vectors by $Z=\partial_z-h\partial_\tau$ and $\bar Z=\partial_{\bar z}-\bar h\partial_\tau$ . Then $Z=c(E_2+iE_3)/2$ and $\Phi Z=iZ$ . In the coordinate order $(\tau,z,\bar z)$ ,

\Phi=\begin{pmatrix} 0&-ih&i\bar h\\0&i&0\\0&0&-i \end{pmatrix},\qquad \Phi^2=-1+\xi\otimes\eta. \tag{M04.4}

For example, $\Phi\partial_z=i\partial_z-ih\partial_\tau$ . The coordinate vector $\partial_z$ is an eigenvector only when $h=0$ , whereas its quotient class is always holomorphic.

The real metric determinant is $c^4$ . With our orientation,

\operatorname{vol}_g=c^2d\tau\wedge dx\wedge dy =-\frac{i c^2}{2}\,d\tau\wedge dz\wedge d\bar z. \tag{M04.5}

The minus sign follows from $dz\wedge d\bar z=2i\,dx\wedge dy$ . In complex components this agrees with P’s convention $\epsilon^{\tau z\bar z}=2i/c^2$ , stated in Appendix B’s footnote 14.

Derive the allowed changes of chart

Keep the same normalized vector $\xi$ . Another adapted chart must satisfy $\xi(\tau')=1$ and $\xi(z')=0$ . Preserving the same transverse holomorphic structure then gives

\tau'=\tau+t(z,\bar z),\qquad z'=f(z), \qquad t\in\mathbb R,\quad f'(z)\ne0. \tag{M04.6}

This is a local invertible holomorphic change of transverse coordinate and a change of origin along each flow line. It is the overlap rule in P (4.8). A generic three-dimensional diffeomorphism is still legitimate, but it need not leave the chart adapted to this specified $\xi$ .

Substitute $d\tau=d\tau'-t_zdz-t_{\bar z}d\bar z$ and $dz=dz'/f'(z)$ into the same one-form and metric. The result is

\boxed{\quad h'(\tau',z',\bar z')=\frac{h(\tau,z,\bar z)-t_z}{f'(z)},\qquad c'(\tau',z',\bar z')=\frac{c(\tau,z,\bar z)}{|f'(z)|}. \quad} \tag{M04.7}

Every unprimed function on the right is evaluated at the inverse point: $z=f^{-1}(z')$ on the chosen patch and $\tau=\tau'-t(z,\bar z)$ . In particular, a $\tau$ -dependent $c$ must not simply be copied with $\tau$ replaced by $\tau'$ . The complex conjugate equation gives $\bar h'$ .

A holomorphic covector transforms as $\omega'=\omega/f'$ because $\Omega$ itself is unchanged. The horizontal vector transforms as $Z'=Z/f'$ , which follows by combining the transformed coordinate derivative with $h'$ . This is a change of coordinate basis, not an R-gauge transformation. P (4.11) records the covector transition; charged-spinor and background-bundle patching are further questions.

Finally, the real transverse Jacobian is $|f'|^2$ . The extra terms in $d\tau'=d\tau+dt$ vanish when wedged with the transverse area form. Thus $c'^2d\tau'\wedge dx'\wedge dy'=c^2d\tau\wedge dx\wedge dy$ , as a volume form must.

Work through a nonlinear overlap

Consider

\eta=d\tau+\lambda x\,dy,\qquad c=e^{s\tau},\qquad h=\frac{i\lambda x}{2}. \tag{M04.8}

The real constants $\lambda,s$ have inverse-length units when $\tau,x,y$ have length units. The horizontal complex structure is independent of $\tau$ , although its metric scale changes. The vector $\xi$ is generally not Killing if $s\ne0$ .

On a patch with $z\ne0$ , choose a branch of the invertible map $z'=z^2/L$ , with fixed positive length $L$ , and let $t=a|z|^2$ , where $a$ has inverse-length units. Then

h'=\frac{L}{2z}\left(\frac{i\lambda x}{2}-a\bar z\right),\qquad c'=\frac{L}{2|z|}\exp\!\left[s\bigl(\tau'-a|z|^2\bigr)\right], \qquad x=\frac{z+\bar z}{2}. \tag{M04.9}

To check the result, pull back $h'dz'$ : it is $(i\lambda x/2-a\bar z)dz$ . Together with its conjugate and $d\tau'=d\tau+a(\bar zdz+zd\bar z)$ , the added terms cancel. Similarly $c'^2dz'd\bar z'=e^{2s\tau}dz\,d\bar z$ . The apparent singularity of these formulas at $z=0$ says that this chart change fails there; the original metric is smooth.

The calculation also separates coordinate cross terms from curvature of the horizontal distribution. Here $d\eta=\lambda\,dx\wedge dy$ and $[\partial_x,\partial_y-\lambda x\partial_\tau]=-\lambda\partial_\tau$ . These are unchanged geometric statements. Setting $\lambda=s=0$ gives a flat product metric, even though $h'=-aL\bar z/(2z)$ can be nonzero. The cross terms produced by the overlap have not created a new background.

Check with a scalar kinetic term

For a real scalar $\varphi$ , the inverse metric follows directly from the dual frame:

|d\varphi|_g^2 =(\partial_\tau\varphi)^2+\frac1{c^2} \left[(\partial_x\varphi-p\partial_\tau\varphi)^2 +(\partial_y\varphi-q\partial_\tau\varphi)^2\right] =(\partial_\tau\varphi)^2+\frac4{c^2}(Z\varphi)(\bar Z\varphi). \tag{M04.10}

The last product is nonnegative because $\varphi$ is real. Under an overlap, $Z'\varphi'=Z\varphi/f'$ and $c'=c/|f'|$ , so the factors cancel. With the invariant volume form, the integrated kinetic term is the same in either chart. For a complex scalar use absolute squares in the real-frame expression; its transverse complex form is $2c^{-2}(|Z\varphi|^2+|\bar Z\varphi|^2)$ . Using $4c^{-2}(Z\varphi)(\bar Z\varphi)$ for an arbitrary complex scalar would be a different expression.

This check is useful even before any supersymmetric action is constructed. A missing factor of two, an omitted shift in $\tau$ , or a coordinate vector mistaken for a horizontal lift will generally change it.

Reconstruct the background fields uses the adapted data to construct compatible sources. What we have established here is the tensor and overlap calculation those formulas will need. We have not imposed a source equation of motion or a global spinor-patching condition.

Exercises

1. A transverse coordinate that moves along a leaf

Starting from an adapted chart, consider $\tau'=\tau$ and $z'=e^{a\tau}z$ , with a nonzero real constant $a$ . Compute the vector $\xi$ in the primed coordinates. Is the new chart adapted to the same fixed vector? Can the failure be diagnosed without transforming the metric?

Hint 1

Apply $\xi=\partial_\tau$ to each primed coordinate. A vector’s components in that chart are precisely these derivatives.

Hint 2

The transverse coordinate must be constant on each leaf. Equivalently, the one-form $dz'$ must annihilate $\xi$ .

Solution

The derivatives give

\xi=\partial_{\tau'}+a z'\partial_{z'} +a\bar z'\partial_{\bar z'},\qquad dz'(\xi)=a z'.

The transformation is a valid local diffeomorphism, but the chart is not adapted to the specified vector on an open set. The failure is already visible in $dz'(\xi)$ , regardless of the metric. Declaring $\partial_{\tau'}$ to be the preferred direction would choose a different flow. The isolated locus $z'=0$ does not repair a chart condition required in a neighborhood. A transverse-only shift of $\tau$ cannot make the nonconstant $z'$ into a first integral of the original flow.

2. A covector’s norm across an overlap

Let $\Omega=\omega\,dz$ be a nonzero holomorphic covector. Compute its Hermitian norm $|\Omega|^2=g^{\mu\nu}\Omega_\mu\bar\Omega_\nu$ . Check its invariance under (M04.6), and explain why a nonconstant $\omega$ need not violate transverse holomorphicity. Does a pole in the component $\omega'$ always mean that the tensor is singular?

Hint 1

Use $dz=(e^2-ie^3)/c$ . The two orthonormal components both contribute to the norm.

Hint 2

Both $\omega$ and $c$ change when $f'$ changes. Distinguish a legitimate overlap with $f'\ne0$ from a map that ceases to be a coordinate chart.

Solution

The orthonormal components are $(0,\omega/c,-i\omega/c)$ , so

|\Omega|^2=\frac{2|\omega|^2}{c^2},\qquad \frac{2|\omega'|^2}{c'^2} =\frac{2|\omega/f'|^2}{c^2/|f'|^2} =\frac{2|\omega|^2}{c^2}.

The eigenvalue equation $\Omega\Phi=i\Omega$ is pointwise and is unchanged by multiplication by any nonzero smooth complex function. It does not assert $\partial_{\bar z}\omega=0$ or $\partial_\tau\omega=0$ . Those would be additional conditions on a section, not the definition of this covector line.

On a legitimate smooth overlap a smooth nonzero $f'$ gives no singularity. In the map $z'=z^2/L$ , a component can diverge as $z\to0$ because that map is not a chart there. Evaluate the invariant norm in a regular chart before assigning a geometric singularity. Conversely, the calculation does not make a tensor with a divergent invariant norm regular by changing its components.

3. Fiber momentum as a transverse connection

Use the $s=0$ geometry in (M04.8) and a complex scalar mode $\varphi=e^{ik\tau}u(x,y)$ , with real $k$ . Derive its local kinetic density. Then make a transverse-dependent shift $\tau'=\tau+t(x,y)$ , leaving $x,y$ fixed, and check invariance using the transformed amplitude $u'$ . State what changes if the flow coordinate is a circle of period $2\pi R$ and the scalar is single-valued.

Hint 1

The horizontal derivatives are $\partial_x$ and $\partial_y-\lambda x\partial_\tau$ . Acting on the Fourier factor replaces $\partial_\tau$ by $ik$ .

Hint 2

Write $\eta=d\tau+\mathfrak b$ , with $\mathfrak b=\lambda x\,dy$ . Under the shift, $\mathfrak b'=\mathfrak b-dt$ and $u'=e^{-ikt}u$ . Test the derivative $d-ik\mathfrak b$ .

Solution

The real-frame expression gives

|d\varphi|_g^2=k^2|u|^2+|\partial_xu|^2 +|(\partial_y-ik\lambda x)u|^2.

The derivative on the transverse amplitude is $D_i=\partial_i-ik\mathfrak b_i$ , with $[D_x,D_y]=-ik\lambda$ . This connection comes from the metric’s horizontal lift, not from an added dynamical gauge field or an R-source.

Since $e^{ik\tau}u=e^{ik\tau'}e^{-ikt}u$ , the primed amplitude is as stated. Direct differentiation yields

(d-ik\mathfrak b')u' =(d-ik\mathfrak b+ik\,dt)(e^{-ikt}u) =e^{-ikt}(d-ik\mathfrak b)u.

All absolute squares and the longitudinal term are unchanged. Locally $k$ is any real Fourier momentum. If the specified coordinate is globally periodic with period $2\pi R$ and the scalar is single-valued with no extra twist, $e^{2\pi i kR}=1$ requires $k=n/R$ , $n\in\mathbb Z$ . Those global assumptions supply the quantization; the local adapted metric alone does not.