Almost contact metric structures

Before this lesson: Construct geometry from one spinor.

The word “almost” marks a distinction between algebra at one point and differential geometry across neighboring points. An almost contact metric structure supplies a unit direction and a complex rotation in its perpendicular plane. It does not say that those planes fit together into surfaces. Nor does it say that they satisfy the different condition called contact.

Use the geometry constructed from one spinor as preparation. We will keep its orientation and sign of $\Phi$ , but work directly with tensors. The physical setting remains three-dimensional $\mathcal N=2$ with four real parent supercharges; this lesson does not choose background sources, count preserved solutions, or select a dynamical gauge group. We are learning to distinguish geometries before solving a supersymmetry equation.

What the algebra specifies

On an oriented Riemannian three-manifold, the data are a real one-form $\eta$ , a vector $\xi$ , a linear map $\Phi:TM\to TM$ , and a positive metric $g$ . The almost contact identities are

\eta(\xi)=1,\qquad \Phi^2=-1+\xi\otimes\eta,\qquad \Phi\xi=0,\qquad \eta\circ\Phi=0. \tag{M02.1}

Metric compatibility adds

g(\xi,X)=\eta(X),\qquad g(\Phi X,\Phi Y)=g(X,Y)-\eta(X)\eta(Y). \tag{M02.2}

Here $\xi\otimes\eta$ sends $X$ to $\eta(X)\xi$ . In particular $\xi$ is unit. On the plane $\mathcal D=\ker\eta$ , the restriction $J=\Phi|_{\mathcal D}$ obeys $J^2=-1$ . This is the definition used in P, (B.1)–(B.2).

Choose a local oriented orthonormal coframe $(e^1,e^2,e^3)$ with $e^1=\eta$ , and let $(E_1,E_2,E_3)$ be its dual vectors. Our sign convention is

\xi=E_1,\qquad \Phi E_1=0,\quad \Phi E_2=-E_3,\quad \Phi E_3=E_2. \tag{M02.3}

This is $\Phi X=X\times\xi$ . Rotating the oriented pair $E_2,E_3$ by $SO(2)\simeq U(1)$ preserves the data. That is the concrete meaning of the reduction to $U(1)$ mentioned in Appendix B; it does not promise one global choice of the pair. The preceding lesson derived this structure from a spinor. Here we will test it without repeating that construction.

A family with an adjustable twist

Work on $\mathbb R^3$ with coordinates $(t,x,y)$ and orientation $dt\wedge dx\wedge dy>0$ . For these geometric examples, measure coordinates in a fixed length unit. Let $f(x)$ be any smooth real function and set

\eta=dt+f(x)\,dy,\qquad g=(dt+f\,dy)^2+dx^2+dy^2. \tag{M02.4}

The coframe $(\eta,dx,dy)$ is nonsingular for every $f$ , and the metric is positive because it is a sum of squares. Its dual frame is

E_1=\partial_t,\qquad E_2=\partial_x,\qquad E_3=\partial_y-f\partial_t. \tag{M02.5}

The last vector is horizontal: $\eta(E_3)=f-f=0$ . Equation (M02.3) therefore defines an almost contact metric structure. We can check the assertion in the coordinate basis $(\partial_t,\partial_x,\partial_y)$ :

g_{\mu\nu}= \begin{pmatrix}1&0&f\\0&1&0\\f&0&1+f^2\end{pmatrix}, \qquad \xi=\begin{pmatrix}1\\0\\0\end{pmatrix}, \qquad \Phi^\mu{}_\nu= \begin{pmatrix}0&f&0\\0&0&1\\0&-1&0\end{pmatrix}. \tag{M02.6}

For example $\partial_y=E_3+fE_1$ , so $\Phi\partial_y=E_2$ , while $\Phi\partial_x=-E_3=f\partial_t-\partial_y$ . Multiplying the displayed matrix gives

\Phi^2= \begin{pmatrix}0&0&f\\0&-1&0\\0&0&-1\end{pmatrix} =-1+\xi\otimes\eta.

Also $g\xi=\eta^T$ and $\Phi^Tg\Phi=g-\eta^T\eta$ . The determinant of $g$ is one, hence $\operatorname{vol}_g=dt\wedge dx\wedge dy$ . No derivative of $f$ was needed for any of these checks.

Ask whether the one-form is contact

A contact form on a three-manifold is a smooth one-form for which the three-form $\eta\wedge d\eta$ never vanishes. Its kernel is called the contact plane field. The definition is differential, unlike (M02.1)–(M02.2).

For our family,

d\eta=f'(x)\,dx\wedge dy,\qquad \boxed{\eta\wedge d\eta=f'(x)\,\operatorname{vol}_g.} \tag{M02.7}

Thus $f=0$ gives an almost contact metric structure whose one-form is not contact. The choice $f=\lambda x$ , with constant $\lambda\ne0$ , is contact everywhere. The choice $f=x^2$ remains a smooth almost contact metric structure everywhere, but is contact only on $x\ne0$ . Nothing singular happens to its metric or $\Phi$ at $x=0$ . The positive or negative sign of $\eta\wedge d\eta$ compares the contact orientation with the fixed orientation; nonvanishing is the contact test.

Why does this test measure the failure of planes to form surfaces? For horizontal vector fields $X,Y$ , the definition of the exterior derivative gives

d\eta(X,Y) =X[\eta(Y)]-Y[\eta(X)]-\eta([X,Y]) =-\eta([X,Y]). \tag{M02.8}

Vector fields tangent to a surface have a bracket tangent to that surface: in surface coordinates their differentiations cannot create a normal component. Consequently, if $\mathcal D$ were tangent to surfaces, the right-hand side would vanish for every horizontal pair. Contact violates this requirement everywhere. In our frame,

[E_2,E_3]=-f'(x)E_1. \tag{M02.9}

The bracket has precisely the vertical component predicted by (M02.8). For $f=0$ , the planes actually are tangent to $t=\text{constant}$ surfaces. This contrasts two-plane integrability with contact using the same coordinates and algebra.

Another useful two-form is $\omega(X,Y)=g(X,\Phi Y)$ . Here $\omega=dx\wedge dy$ , whereas $d\eta=f'\omega$ . The almost contact metric identities do not fix that coefficient or require it to be nonzero. Additional definitions in contact metric geometry impose differential normalizations; none is implicit in the word “almost.”

Measure the twist with a small loop

Take $f=\lambda x$ . A horizontal path satisfies

\eta(\dot\gamma)=0 \quad\Longrightarrow\quad \dot t=-\lambda x\,\dot y.

Start at $(t_0,0,0)$ and traverse the rectangle $(0,0)\to(X,0)\to(X,Y)\to(0,Y)\to(0,0)$ in the $(x,y)$ plane, lifting every segment horizontally. Only the second side changes $t$ : on it $x=X$ and $\Delta y=Y$ . The final point is

(t_0-\lambda XY,0,0),\qquad \Delta t=-\oint\lambda x\,dy =-\int_{\text{rectangle}}\lambda\,dx\wedge dy. \tag{M02.10}

The projected path closes, but its horizontal lift does not. For $\lambda=2$ , $X=1/2$ and $Y=1/3$ , the displacement is $-1/3$ . The bracket predicts the same signed area dependence. The last equality is Stokes’ theorem and gives a second way to calculate the result. This is ordinary geometric transport along $\ker\eta$ , with no assumed gauge field or supersymmetric observable.

The unit vector and the Reeb vector

Given a contact form, its Reeb vector $R_\eta$ is defined by

\eta(R_\eta)=1,\qquad \iota_{R_\eta}d\eta=0. \tag{M02.11}

The symbol $\iota_R$ means insertion of $R$ into the first argument of a form. A nonzero two-form in three dimensions has a one-dimensional kernel. The contact condition says that $\eta$ does not vanish on that kernel, so the two conditions select one vector uniquely.

By contrast, the almost contact metric vector is defined by $\xi=\eta^\sharp$ and unit normalization. It need not obey the second condition in (M02.11). For the contact members of (M02.4), $d\eta$ contains no $dt$ , so both vectors happen to be $\partial_t$ . Exercise 1 changes the normalization and compatible metric, and separates them explicitly. Calling every unit almost contact vector “the Reeb vector” would hide that extra requirement.

Transverse complex geometry is a different question

A nowhere-zero vector has local integral curves. They form a one-dimensional foliation: locally, points on the same curve share two transverse labels. The plane $\mathcal D$ supplies representatives for the quotient bundle $TM/\mathbb R\xi$ , even when $\mathcal D$ is not tangent to surfaces.

In (M02.4), the curves of $\xi$ keep $x,y$ fixed. Their labels can be combined into

z=x-iy,\qquad dz=dx-i\,dy,\qquad dz\circ\Phi=i\,dz. \tag{M02.12}

The sign agrees with the covector convention in the preceding lesson. Translation along $t$ preserves $\Phi$ , and this same complex label works on every cross-section $t=\text{constant}$ . Thus this explicit foliation has a transverse complex structure for every $f(x)$ , including its contact and noncontact cases.

There is no conflict with (M02.9). A cross-section to the curves need not be tangent to $\mathcal D$ : on $t=\text{constant}$ , the vector $\partial_y$ has $\eta(\partial_y)=f$ . “Transverse” describes geometry across the one-dimensional leaves, not surfaces whose tangent planes are $\ker\eta$ .

In general, a transversely holomorphic foliation requires compatible complex labels in overlapping charts, with holomorphic changes of the transverse label. Pointwise $J^2=-1$ alone does not ensure this. The source derives the additional condition in (4.6)–(4.12) and explains its chart meaning in Appendix B.3. Here we have constructed one transparent example; the general criterion and its derivation from the Killing-spinor equation are the next geometric step. Exercise 3 shows why an additional condition is needed.

A Seifert fibration asks a further, global question: do the leaves form circle fibers, with the permitted exceptional circle fibers? No pointwise identity or contact test ensures closed orbits. The torus exercise below makes that issue visible without assuming the classification of Seifert manifolds.

Structure	What must be supplied or checked
Almost contact metric structure	The algebra of $\eta,\xi,\Phi$ and compatibility with $g$
Contact form	The differential test $\eta\wedge d\eta\ne0$
Transversely holomorphic foliation	Complex geometry preserved across the one-dimensional leaves and compatible chart changes
Seifert fibration	Global circle fibers, including the allowed exceptional fibers

These are distinct statements about specified data. A manifold may admit more than one choice of these data. Failure for a chosen vector or plane field is not a proof that every possible choice on that manifold fails.

Exercises

1. Preserve a contact plane field but change its Reeb vector

On $\mathbb R^3$ , start with $\eta=dt+x\,dy$ and set $\alpha=e^{cx}\eta$ , for a real constant $c$ . Use the compatible metric $g_\alpha=\alpha^2+dx^2+dy^2$ . Show that $\ker\alpha=\ker\eta$ and that $\alpha$ is contact. Find its unit metric-dual vector $\xi_\alpha$ and its Reeb vector, and explain why they need not coincide.

Hint 1

In $\alpha\wedge d\alpha$ , the term containing two copies of $\eta$ vanishes. The dual vector to the first member of the coframe $(\alpha,dx,dy)$ has no $x$ or $y$ component.

Hint 2

Write $R=a\partial_t+b\partial_x+d\partial_y$ . Use $d\alpha=e^{cx}dx\wedge[c\,dt+(1+cx)dy]$ before imposing $\alpha(R)=1$ .

Solution

The multiplier is everywhere positive, so the kernel is unchanged. For any positive $F$ , $(F\eta)\wedge d(F\eta)=F^2\eta\wedge d\eta$ . Here it gives

\alpha\wedge d\alpha=e^{2cx}dt\wedge dx\wedge dy\ne0.

The dual coframe conditions give $\xi_\alpha=e^{-cx}\partial_t$ . For the Reeb vector, insertion into the two-form gives

\iota_Rd\alpha=e^{cx} \left\{b[c\,dt+(1+cx)dy]-[ca+(1+cx)d]dx\right\}.

Since $c$ and $1+cx$ cannot both vanish, $b=0$ . The remaining equations are $ca+(1+cx)d=0$ and $a+xd=e^{-cx}$ . Solving them,

R_\alpha=e^{-cx}\big[(1+cx)\partial_t-c\partial_y\big].

Substitution verifies both defining equations. In contrast $\iota_{\xi_\alpha}d\alpha=-c\,dx$ , so the unit vector is Reeb only when $c=0$ . The metric has changed with $\alpha$ ; no claim is made that an arbitrary multiple of the old one-form stays unit in the old metric. The horizontal plane field survives the rescaling, whereas the normalized contact dynamics changes.

2. Test contact and orbit closure on a compact manifold

Let $(x,y,z)$ be angular coordinates of period $2\pi$ on $T^3$ , with flat metric and orientation $dx\wedge dy\wedge dz>0$ . Set

\eta=\cos z\,dx+\sin z\,dy,\qquad \beta=-\sin z\,dx+\cos z\,dy.

Use the coframe $(\eta,\beta,dz)$ to define $\Phi$ by (M02.3). Verify the almost contact metric and contact statements, find the Reeb flow, and decide whether this flow gives circle fibers everywhere.

Hint 1

The two-by-two rotation taking $(dx,dy)$ to $(\eta,\beta)$ is orthogonal with determinant one. Compute $d\eta$ in that coframe.

Hint 2

Along the candidate flow, $z$ is constant. A trajectory closes only if both angular displacements return to multiples of $2\pi$ after the same nonzero time.

Solution

All forms are globally defined: although the angles themselves are local coordinates, $dx,dy,dz$ and the displayed periodic coefficients patch on the torus. The coframe is oriented and orthonormal. Its dual first vector is $\xi=\cos z\,\partial_x+\sin z\,\partial_y$ , and the prescribed frame action of $\Phi$ verifies (M02.1)–(M02.2).

Since $d\eta=dz\wedge\beta$ ,

\eta\wedge d\eta=-\eta\wedge\beta\wedge dz =-dx\wedge dy\wedge dz.

The form is contact everywhere, with contact volume $\int_{T^3}\eta\wedge d\eta=-(2\pi)^3$ . Both $\beta(\xi)$ and $dz(\xi)$ vanish, so $\xi$ is also the Reeb vector. The flow through $(x_0,y_0,z_0)$ is

(x(s),y(s),z(s))= (x_0+s\cos z_0,\ y_0+s\sin z_0,\ z_0)\pmod{2\pi}.

It closes exactly when the direction $(\cos z_0,\sin z_0)$ is proportional to an integer pair. For finite slope this means $\tan z_0$ is rational; the vertical direction also closes. For irrational slope, no common return time exists, so these orbits are not circles. Thus this particular flow is not a circle fibration. The conclusion concerns the specified flow, not every possible fibration of $T^3$ .

3. Keep contact while changing the transverse complex rotation

Return to $\eta=dt+x\,dy$ , $\xi=\partial_t$ , and replace the metric by

g_h=\eta^2+e^{2h(t)}dx^2+e^{-2h(t)}dy^2,

where $h$ is any smooth real function. Define $\Phi_h$ in the oriented orthonormal coframe $(\eta,e^h dx,e^{-h}dy)$ . Does contact survive? On the transverse quotient, use the classes of $\partial_x,\partial_y$ as a fixed basis and find $J_h$ . For $h(t)=ct$ with $c\ne0$ , does the flow of $\xi$ preserve this chosen transverse complex structure?

Hint 1

The dual horizontal vectors have factors $e^{-h}$ and $e^h$ . The contact test uses $\eta$ , which has not changed.

Hint 2

The coordinate flow sends $(t,x,y)$ to $(t+s,x,y)$ , so it induces the identity on the chosen transverse coordinate basis. Compare the matrices at $t$ and $t+s$ , not only their squares at one point.

Solution

The metric stays positive and $\det g_h=1$ . Also $d\eta=dx\wedge dy$ and $\eta\wedge d\eta=dt\wedge dx\wedge dy$ , independent of $h$ . The coframe construction verifies the almost contact metric identities, and $\xi$ remains both unit and Reeb.

The horizontal orthonormal vectors are $e^{-h}\partial_x$ and $e^h(\partial_y-x\partial_t)$ . Consequently the induced matrix on the transverse quotient is

J_h= \begin{pmatrix}0&e^{-2h}\\-e^{2h}&0\end{pmatrix}, \qquad J_h^2=-1,\qquad \partial_tJ_h= \begin{pmatrix}0&-2h'e^{-2h}\\-2h'e^{2h}&0\end{pmatrix}.

For $h'=c\ne0$ the flow does not preserve this matrix. Equivalently the holomorphic covector $\Omega_h=e^h dx-i e^{-h}dy$ obeys $\partial_t\Omega_h=h'\overline{\Omega_h}$ . Its change has an antiholomorphic component, rather than only rescaling the same complex line. This is an explicit obstruction for the chosen transverse structure, despite the contact test and the pointwise identity $J_h^2=-1$ . It is precisely the sort of differential issue that the general integrability condition must detect.

Continue with integrability and transverse holomorphic geometry to determine when a pointwise transverse complex structure can be transported consistently along the distinguished curves. Derive the differential restriction from the Killing-spinor equation, then distinguish local complex coordinates from a global foliation.