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Killing vector fields on Riemannian and Lorentzian 3-manifolds

Amir Babak Aazami and Robert Ream Clark University Worcester, MA 01610 aaazami@clarku.edu, rream@clarku.edu
Abstract.

We give a complete local classification of all Riemannian 3-manifolds (M,g)(M,g) admitting a nonvanishing Killing vector field TT. We then extend this classification to timelike Killing vector fields on Lorentzian 3-manifolds, which are automatically nonvanishing. The two key ingredients needed in our classification are the scalar curvature SS of gg and the function Ric(T,T)\text{Ric}(T,T), where Ric is the Ricci tensor; in fact their sum appears as the Gaussian curvature of the quotient metric obtained from the action of TT. Our classification generalizes that of Sasakian structures, which is the special case when Ric(T,T)=2\text{Ric}(T,T)=2. We also give necessary, and separately, sufficient conditions, both expressed in terms of Ric(T,T)\text{Ric}(T,T), for gg to be locally conformally flat. We then move from the local to the global setting, and prove two results: in the event that TT has unit length and the coordinates derived in our classification are globally defined on 3\mathbb{R}^{3}, we give conditions under which SS completely determines when the metric will be geodesically complete. In the event that the 3-manifold MM is compact, we give a condition stating when it admits a metric of constant positive sectional curvature.

1. Introduction

The aim of this paper is to give a complete local classification of all Riemannian 3-manifolds (M,g)(M,g) that admit a nonvanishing Killing vector field TT. In fact this classification will also yield a related one: that of all Lorentzian 3-manifolds supporting a timelike Killing vector field. Our classification proceeds by considering the special case when TT has constant length: the general case follows from this one by applying a conformal change by the factor of g(T,Tg(T,T). But in fact there are important reasons for imposing this condition. One of them is that constant length allows us to adapt the machinery of the Newman-Penrose formalism [NP62] — a construct that originated in 4-dimensional Lorentzian geometry — to the setting of 3-dimensional Riemannian geometry; see also [SW14, NTC15, BS18], wherein similar frame techniques have been applied in dimension 3.

As shown in Section 2 below, constant length is a prerequisite for this formalism. But more importantly, our interest in constant length arises from what we regard as the “canonical” constant length Killing vector field in dimension 3: the unit length Killing vector field TT on (𝕊3,g̊)(\mathbb{S}^{3},\mathring{g}) tangent to the Hopf fibration, where g̊\mathring{g} is the standard (round) metric. Given the special geometry of (𝕊3,g̊)(\mathbb{S}^{3},\mathring{g}) as a spherical space form, and the presence of such a vector field on it, we take our motivation from the following questions:

  1. 1.

    Can one classify locally all Riemannian 3-manifolds admitting a constant length Killing vector field? Do they take on a “canonical” form?

  2. 2.

    If a Riemannian 3-manifold admits a constant length Killing vector field, then when will it be locally conformally flat, as with (𝕊3,g̊)(\mathbb{S}^{3},\mathring{g})?

(Yet another path of inquiry, which we do not pursue here, would be to examine when the circle action provided by a constant length Killing vector field is free, and the role that sectional curvature plays in this; see [BN08].) A complete answer to our first question above is provided in our first Theorem:

Theorem 1.

Let (M,g)(M,g) be a Riemannian 3-manifold that admits a unit length Killing vector field TT. Then there exist local coordinates (t,r,θ)(t,r,\theta) and a smooth function φ(r,θ)\varphi(r,\theta) such that

T=t,g=(T)2+dr2+φ2dθ2,\displaystyle T=\partial_{t}\hskip 14.45377pt,\hskip 14.45377ptg=(T^{\flat})^{2}+dr^{2}+\varphi^{2}d\theta^{2}, (1)

and where the quotient metric dr2+φ2dθ2dr^{2}+\varphi^{2}d\theta^{2} has Gaussian curvature

φrrφ=12(S+Ric(T,T)),\displaystyle-\frac{\varphi_{rr}}{\varphi}=\frac{1}{2}\big{(}S+\emph{\text{Ric}}(T,T)\big{)}, (2)

with SS and Ric the scalar curvature and Ricci tensor of gg, respectively. If (1) is given globally on M=3M=\mathbb{R}^{3} (with r,θr,\theta polar coordinates on 2\mathbb{R}^{2}), and if Ric(T,T)=0,φ(0,θ)=0\emph{\text{Ric}}(T,T)=0,\varphi(0,\theta)=0, φr(0,θ)=1\varphi_{r}(0,\theta)=1, φ(r,θ+π)=φ(r,θ)\varphi(-r,\theta+\pi)=-\varphi(r,\theta), and φ>0\varphi>0 when r>0r>0, then gg is complete if and only if

limrinf|p|rS|p0.\lim_{r\to\infty}\inf_{|p|\geq r}S\big{|}_{p}\leq 0.

The following remarks help to shed light on this result:

  1. i.

    After our first preprint appeared, we learned of the works [Man14, LM17], in which the existence of coordinates isometric to (1) are proved, as well as a result that includes (2) as a special case; these were obtained via a different method than ours, and applied to the classification of Riemannian submersions from 3-manifolds to a surface, whose fibers are the integral curves of a Killing vector field.

  2. ii.

    An almost identical Theorem exists for unit timelike Killing vector fields on Lorentzian 3-manifolds; see Corollary 2 in Section 6 below, wherein the relevant Lorentzian terminology is also defined.

  3. iii.

    If TT does not have unit length, then (1) is scaled by g(T,T)g(T,T). Also, in the global case M=3M=\mathbb{R}^{3}, completeness can be obtained by conditions other than Ric(T,T)=0\text{Ric}(T,T)=0, using, e.g., [MR20, Theorem 6].

  4. iv.

    The “canonical form” alluded to above is manifested in (1) and (2); for the form of the metric in the coordinate basis {t,r,θ}\{\partial_{t},\partial_{r},\partial_{\theta}\}, see (36) in Section 5 below. As Theorem 1 makes clear, our classification depends entirely on two functions, the scalar curvature SS and Ric(T,T)\text{Ric}(T,T). Let us say more about the latter function, which is especially important; one way to appreciate its significance when dimM=3\text{dim}\,M=3 is as follows. If a vector field TT has constant length and geodesic flow (as does any constant length Killing vector field), then the function Ric(T,T)\text{Ric}(T,T), if nonnegative, completely governs whether its orthogonal complement TTMT^{\perp}\subseteq TM is integrable. As a consequence, it was shown in [HP16] that when such a TT satisfies Ric(T,T)>0\text{Ric}(T,T)>0 and when MM is orientable and compact, then TT^{\flat} is a contact form and TT is its Reeb vector field; if in addition TT is divergence-free and Ric(T,T)=1\text{Ric}(T,T)=1, then T(T)T_{*}(T^{\perp}) is JJ-invariant, where JJ is the Levi-Civita almost-complex structure on TTMTTM (in fact these two conditions are necessary and sufficient).

  5. v.

    The previous remark did not assume that TT is a unit length Killing vector field. Imposing this condition — as well as the condition Ric(T,T)=2\text{Ric}(T,T)=2, so that the endomorphism in (5) below defines an almost complex structure on TT^{\perp} — would make (M,g,T)(M,g,T) a Sasakian structure. In dimension 3, a classification of these on closed manifolds was obtained in [Gei97], up to diffeomorphism; an explicit metric classification was then given in [Bel01, Bel03], which also established a one-to-one correspondence between Sasakian and normal CR structures, and also classified the latter. For an application to monopole fields, see [DH19].

Given the importance of the quantities SS and Ric(T,T)\text{Ric}(T,T), it is worthwhile to mark the following relationship between unit length Killing vector fields and constant curvature in the compact setting, a result which is essentially a corollary of a well known result in [Ham82]:

Theorem 2.

Let (M,g)(M,g) be a compact Riemannian 3-manifold and TT a globally defined, unit length Killing vector field. If Ric(T,T)0\emph{\text{Ric}}(T,T)\neq 0 and

S>2|Ric(T)|g2Ric(T,T)Ric(T,T),\displaystyle S>2\frac{|\emph{\text{Ric}}(T)|_{g}^{2}}{\emph{\text{Ric}}(T,T)}-\emph{\text{Ric}}(T,T), (3)

where Ric(T)\emph{\text{Ric}}(T) is the Ricci operator, then MM admits a metric of constant positive sectional curvature.

Finally, our answer to the second question is also given in terms of Ric(T,T)\text{Ric}(T,T):

Theorem 3.

Let (M,g)(M,g) be a Riemannian 3-manifold that admits a unit length Killing vector field TT. If gg is locally conformally flat, then

4|Ric(T)|g2=3Ric(T,T)22BRic(T,T)+C\displaystyle 4|\emph{\text{Ric}}(T)|_{g}^{2}=3\emph{\text{Ric}}(T,T)^{2}-2B\emph{\text{Ric}}(T,T)+C (4)

for some constants B,CB,C, where Ric(T)\emph{\text{Ric}}(T) is the Ricci operator. Conversely, given (4), there exist coordinates (r,θ)(r,\theta) on the quotient metric in (1) with respect to which gg is conformally flat when

ωθ=0,ωr2+14(ω2+2B)2=C+B2,φ=h(θ)ωr,\omega_{\theta}=0\hskip 14.45377pt,\hskip 14.45377pt\omega_{r}^{2}+\frac{1}{4}(\omega^{2}+2B)^{2}=C+B^{2}\hskip 14.45377pt,\hskip 14.45377pt\varphi=h(\theta)\omega_{r},

where ω2=2Ric(T,T)\omega^{2}=2\emph{\text{Ric}}(T,T), φ\varphi is as in Theorem 1, and h(θ)h(\theta) is a smooth function. If Ric(T,T)\emph{\text{Ric}}(T,T) is constant, then gg is locally conformally flat if and only if S=3Ric(T,T)S=3\emph{\text{Ric}}(T,T).

As a check of the last statement, note that for (𝕊3,g̊)(\mathbb{S}^{3},\mathring{g}) with Hopf Killing vector field TT and radius RR,

Ric(T,T)=2R2,S=6R2\text{Ric}(T,T)=\frac{2}{R^{2}}\hskip 14.45377pt,\hskip 14.45377ptS=\frac{6}{R^{2}}\cdot

2. Divergence, Twist, Shear

A Killing vector field TT on a Riemannian manifold (M,g)(M,g) is defined by the condition 𝔏Tg=0,\mathfrak{L}_{T}g=0, where 𝔏\mathfrak{L} is the Lie derivative. However, when TT has unit length and when dimM=3\text{dim}\,M=3, there is an equivalent formulation, given by Lemma 2 below, which plays a crucial role in our classification. This formulation also involves the Lie derivative, but owing to the low dimension, only certain components of it, which components carry geometric properties of the flow of TT. These properties are the divergence, twist, and shear; as the latter two are not as well known as the former, we now digress to define them explicitly. Thus, let TT be a smooth unit length vector field defined in an open subset of a Riemannian 3-manifold (M,g)(M,g), so that vTT\nabla_{\!v}{T}\perp T for all vectors vv (\nabla is the Levi-Civita connection). Let XX and YY be two smooth vector fields such that {T,X,Y}\{T,X,Y\} is a local orthonormal frame. Now define the following endomorphism DD of the orthogonal complement TTMT^{\perp}\subseteq TM,

D:TT,vvT,\displaystyle D\colon T^{\perp}\longrightarrow T^{\perp}\hskip 14.45377pt,\hskip 14.45377ptv\ \mapsto\ \nabla_{\!v}{T}, (5)

and observe that its matrix with respect to the frame {T,X,Y}\{T,X,Y\} is

D=(g(XT,X)g(YT,X)g(XT,Y)g(YT,Y))\displaystyle D=\begin{pmatrix}g({\nabla_{\!X}{T}},{X})&g({\nabla_{\!Y}{T}},{X})\\ g({\nabla_{\!X}{T}},{Y})&g({\nabla_{\!Y}{T}},{Y})\end{pmatrix}\cdot

Contained within this matrix are three geometric properties associated to the flow of TT:

  1. 1.

    The divergence of TT, denoted divT\text{div}\,T, is simply the trace of DD.

  2. 2.

    By Frobenius’s theorem, TT^{\perp} is integrable if and only if the anti-symmetric part of DD vanishes; as seen in (8) below, this vanishing is governed by the following function, which comprises the off-diagonal elements of the anti-symmetric part of DD:

    ω=g(T,[X,Y])=g(YT,X)g(XT,Y).\displaystyle\omega\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=g({T},{[X,Y]})=g({\nabla_{\!Y}{T}},{X})-g({\nabla_{\!X}{T}},{Y}). (6)

    Since ω2\omega^{2} equals the determinant of the anti-symmetric part of DD, it is a frame independent quantity. We call ω2\omega^{2} the twist function of TT and say that the flow of TT is twist-free if ω2=0\omega^{2}=0.

  3. 3.

    The third piece of information is the shear σ\sigma of TT; it is given by the trace-free symmetric part of DD, whose components σ1,σ2\sigma_{1},\sigma_{2} we combine here into a complex-valued quantity:

    σ\displaystyle\hskip 18.7898pt\sigma\!\! =\displaystyle\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}= 12(g(YT,Y)g(XT,X))σ1+i12(g(YT,X)+g(XT,Y))σ2.\displaystyle\!\!\underbrace{\,\frac{1}{2}\Big{(}g({\nabla_{\!Y}{T}},{Y})-g({\nabla_{\!X}{T}},{X})\Big{)}\,}_{\sigma_{1}}+\ i\underbrace{\,\frac{1}{2}\Big{(}g({\nabla_{\!Y}{T}},{X})+g({\nabla_{\!X}{T}},{Y})\Big{)}\,}_{\sigma_{2}}.

    Although σ\sigma itself is not frame independent, its magnitude |σ2||\sigma^{2}| is: by (8) below, it is minus the determinant of the trace-free symmetric part of DD. We say that the flow of TT is shear-free if σ=0\sigma=0. As with being twist-free, being shear-free is a frame independent statement. In terms of divT\text{div}\,T, ω\omega, and σ\sigma, DD takes the form

    D=(12divT0012divT)+(σ1σ2σ2σ1)trace-free symmetric+(0ω2ω20)anti-symmetric\displaystyle D=\begin{pmatrix}\frac{1}{2}\text{div}\,T&0\\ 0&\frac{1}{2}\text{div}\,T\end{pmatrix}\ +\!\!\!\underbrace{\,\begin{pmatrix}-\sigma_{1}&\sigma_{2}\\ \sigma_{2}&\sigma_{1}\end{pmatrix}\,}_{\text{trace-free symmetric}}\!\!\!+\,\underbrace{\,\begin{pmatrix}0&\frac{\omega}{2}\\ -\frac{\omega}{2}&0\end{pmatrix}\,}_{\text{anti-symmetric}}\cdot (8)

We record here the well known fact that, while being divergence-free is not a conformal invariant, being shear-free or twist-free is:

Lemma 1.

Let (M,g)(M,g) be a Riemannian 3-manifold and TT a unit length vector field. Given a conformal metric g~=e2fg\tilde{g}=e^{2f}g, the vector field T~=efT\widetilde{T}=e^{-f}T is shear-free with respect to g~\tilde{g} if and only if TT is shear-free with respect to gg. Likewise if shear-free is replaced by twist-free.

Proof.

Note that g~(T~,T~)=1\tilde{g}(\widetilde{T},\widetilde{T})=1; given a gg-orthonormal frame {T,X,Y}\{T,X,Y\}, form the g~\tilde{g}-orthonormal frame {T~,X~,Y~}\{\widetilde{T},\widetilde{X},\widetilde{Y}\}, with X~=efX\widetilde{X}=e^{-f}X and Y~=efY\widetilde{Y}=e^{-f}Y. Denoting by ~\widetilde{\nabla} the Levi-Civita connection of g~\tilde{g}, the shear of T~\widetilde{T} with respect to g~\tilde{g} is

σ~\displaystyle\tilde{\sigma}\!\! =\displaystyle= 12(g~(~Y~T~,Y~)g~(~X~T~,X~))+i2(g~(~Y~T~,X~)+g~(~X~T~,Y~))\displaystyle\!\!\frac{1}{2}\Big{(}\tilde{g}({\widetilde{\nabla}_{\!\widetilde{Y}}{\widetilde{T}}},{\widetilde{Y}})-\tilde{g}({\widetilde{\nabla}_{\!\widetilde{X}}{\widetilde{T}}},{\widetilde{X}})\Big{)}+\frac{i}{2}\Big{(}\tilde{g}({\widetilde{\nabla}_{\!\widetilde{Y}}{\widetilde{T}}},{\widetilde{X}})+\tilde{g}({\widetilde{\nabla}_{\!\widetilde{X}}{\widetilde{T}}},{\widetilde{Y}})\Big{)}
=\displaystyle= ef2(g(~YT,Y)g(~XT,X))+ief2(g(~YT,X)+g(~XT,Y))\displaystyle\!\!\frac{e^{-f}}{2}\Big{(}g({\widetilde{\nabla}_{\!Y}{T}},{Y})-g({\widetilde{\nabla}_{\!X}{T}},{X})\Big{)}+i\frac{e^{-f}}{2}\Big{(}g({\widetilde{\nabla}_{\!Y}{T}},{X})+g({\widetilde{\nabla}_{\!X}{T}},{Y})\Big{)}
=\displaystyle= efσ,\displaystyle\!\!e^{-f}\sigma,\phantom{\frac{1}{2}}

where in the last step we have used standard formulae relating Levi-Civita connections of conformal metrics, e.g.,

~YT=YT+Y(f)T+T(f)Y.\widetilde{\nabla}_{\!Y}{T}=\nabla_{\!Y}{T}+Y(f)T+T(f)Y.

Likewise for the twist:

ω~=g~(T~,[X~,Y~])=efg(T,[X,Y])=efω.\tilde{\omega}=\tilde{g}(\widetilde{T},[\widetilde{X},\widetilde{Y}])=e^{-f}g(T,[X,Y])=e^{-f}\omega.

Thus not only twist-free, but shear-free as well, is a conformal property: |σ~|2=0|σ|2=0|\tilde{\sigma}|^{2}=0\Leftrightarrow|\sigma|^{2}=0 and ω~2=0ω2=0\tilde{\omega}^{2}=0\Leftrightarrow\omega^{2}=0. ∎

When dimM4\text{dim}\,M\geq 4, divergence and shear alone are not enough to characterize unit length Killing vector fields, but they do when dimM=3\text{dim}\,M=3:

Lemma 2.

A unit length vector field TT on a Riemannian 3-manifold (M,g)(M,g) is a Killing vector field if and only if its flow is geodesic, divergence-free, and shear-free.

Proof.

The Killing condition is equivalent to

g(vT,w)+g(wT,v)=0for allv,wTM,\displaystyle g(\nabla_{\!v}{T},w)+g(\nabla_{\!w}{T},v)=0\hskip 14.45377pt\text{for all}~v,w\in TM, (9)

from which it follows that any Killing vector field TT is divergence-free and shear-free, via (LABEL:eqn:shear). Finally, (9) also implies that any unit length Killing vector field must have geodesic flow:

TT=0.\nabla_{\!T}{T}=0.

Conversely, suppose that a unit length vector field TT is geodesic, divergence-free, and shear-free, and consider (9). Writing v,wv,w with respect to an orthonormal frame {T,X,Y}\{T,X,Y\} as

v=a0T+a1X+a2Y,w=b0T+b1X+b2Y,v=a_{0}T+a_{1}X+a_{2}Y\hskip 14.45377pt,\hskip 14.45377ptw=b_{0}T+b_{1}X+b_{2}Y,

we have

g(vT,w)+g(wT,v)\displaystyle g(\nabla_{\!v}{T},w)+g(\nabla_{\!w}{T},v)\!\! =\displaystyle= a1g(XT,w)+a2g(YT,w)\displaystyle\!\!a_{1}g(\nabla_{\!X}{T},w)+a_{2}g(\nabla_{\!Y}{T},w)
+a1g(wT,X)+a2g(wT,Y)\displaystyle\hskip 14.45377pt+\ a_{1}g(\nabla_{\!w}{T},X)+a_{2}g(\nabla_{\!w}{T},Y)
=\displaystyle= 2a1b1g(XT,X)+(a1b2+b1a2)g(XT,Y)\displaystyle\!\!2a_{1}b_{1}g(\nabla_{\!X}{T},X)+(a_{1}b_{2}+b_{1}a_{2})g(\nabla_{\!X}{T},Y)
+(a1b2+b1a2)g(YT,X)+2a2b2g(YT,Y)g(XT,X)\displaystyle\hskip 14.45377pt+\ (a_{1}b_{2}+b_{1}a_{2})g(\nabla_{\!Y}{T},X)+2a_{2}b_{2}\underbrace{\,g(\nabla_{\!Y}{T},Y)\,}_{g(\nabla_{\!X}{T},X)}
=\displaystyle= 2(a1b1+a2b2)g(XT,X)12divT\displaystyle\!\!2(a_{1}b_{1}+a_{2}b_{2})\underbrace{\,g(\nabla_{\!X}{T},X)\,}_{\frac{1}{2}\text{div}\,T}
+(a1b2+b1a2)(g(XT,Y)+g(YT,X))2σ2.\displaystyle\hskip 14.45377pt+\ (a_{1}b_{2}+b_{1}a_{2})\underbrace{\,\big{(}g(\nabla_{\!X}{T},Y)+g(\nabla_{\!Y}{T},X)\big{)}\,}_{2\sigma_{2}}.

This vanishes by our assumptions, completing the proof. ∎

We can now state our plan of attack: divergence, geodesic flow, twist, and shear all involve first derivatives of TT, whereas curvature involves second derivatives. Our plan of attack, therefore, is to express the components of the Riemann curvature tensor in terms of the divergence, twist, and shear of TT, thereby reducing second-order equations to first-order ones — indeed, further encouraged by the fact that, as we have just seen, if TT is a unit length Killing vector field, then divT,σ\text{div}\,T,\sigma, and TT\nabla_{\!T}{T} all vanish, so that only TT’s twist function ω2\omega^{2} is unknown. The hope is that this will simplify things enough to allow a full determination of the metric. And it will — after we express the curvature in terms of the divergence, twist, and shear, which we now proceed to do.

3. The Newman-Penrose Formalism for Riemannian 3-manifolds

In what follows we present the Newman-Penrose formalism for Riemannian 3-manifolds, presenting here only the resulting equations; complete derivations can be found in [Aaz15]. Let {T,X,Y}\{T,X,Y\} be an orthonormal frame — with TT not necessarily a Killing vector field — and form the complex-valued quantities

𝒎=12(XiY),𝒎¯=12(X+iY).\displaystyle{\boldsymbol{m}}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\frac{1}{\sqrt{2}}(X-iY)\hskip 14.45377pt,\hskip 14.45377pt\overline{\boldsymbol{m}}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\frac{1}{\sqrt{2}}(X+iY).

Henceforth we work with the complex frame {T,𝒎,𝒎¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\}, for which only

g(T,T)=1,g(𝒎,𝒎¯)=1g({T},{T})=1\hskip 14.45377pt,\hskip 14.45377ptg({{\boldsymbol{m}}},{\overline{\boldsymbol{m}}})=1

are nonzero. The following quantities associated to this complex frame play a central role in all that follows.

Definition 1.

The spin coefficients of the complex frame {T,𝐦,𝐦¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\} are the complex-valued functions

κ\displaystyle\kappa\!\!\! =\displaystyle= g(TT,𝒎),ρ=g(𝒎¯T,𝒎),σ=g(𝒎T,𝒎),\displaystyle\!\!\!-g({\nabla_{\!T}{T}},{{\boldsymbol{m}}})\hskip 14.45377pt,\hskip 14.45377pt\rho=-g({\nabla_{\!\overline{\boldsymbol{m}}}{T}},{{\boldsymbol{m}}})\hskip 14.45377pt,\hskip 14.45377pt\sigma=-g({\nabla_{\!{\boldsymbol{m}}}{T}},{{\boldsymbol{m}}}),
ε=g(T𝒎,𝒎¯),β=g(𝒎𝒎,𝒎¯).\displaystyle\hskip 32.52127pt\varepsilon=g({\nabla_{\!T}{{\boldsymbol{m}}}},{\overline{\boldsymbol{m}}})\hskip 14.45377pt,\hskip 14.45377pt\beta=g({\nabla_{\!{\boldsymbol{m}}}{{\boldsymbol{m}}}},{\overline{\boldsymbol{m}}}).

Note that, because TT has unit length, its flow is geodesic, TT=0\nabla_{\!T}{T}=0, if and only if κ=0\kappa=0; that σ\sigma, when written out in terms of its real and imaginary parts, is precisely the complex shear (LABEL:eqn:shear); and that the spin coefficient ρ\rho has real and imaginary parts given by

ρ=divT2iω2\displaystyle\rho=-\frac{\text{div}\,T}{2}-i\,\frac{\omega}{2}\cdot (10)

In other words, the first three spin coefficients κ,ρ,σ\kappa,\rho,\sigma stand in for the geometric properties of the flow of TT discussed above. In terms of all five spin coefficients, the Lie brackets are

[T,𝒎]\displaystyle[T,{\boldsymbol{m}}]\!\! =\displaystyle= κT+(ε+ρ¯)𝒎+σ𝒎¯,\displaystyle\!\!\kappa\,T+(\varepsilon+\bar{\rho})\,{\boldsymbol{m}}+\sigma\,\overline{\boldsymbol{m}}, (11)
[𝒎,𝒎¯]\displaystyle\,[{\boldsymbol{m}},\overline{\boldsymbol{m}}]\!\! =\displaystyle= (ρ¯ρ)T+β¯𝒎β𝒎¯.\displaystyle\!\!(\bar{\rho}-\rho)\,T+\bar{\beta}\,{\boldsymbol{m}}-\beta\,\overline{\boldsymbol{m}}. (12)

(The remaining Lie bracket [T,𝒎¯][T,\overline{\boldsymbol{m}}] is obtained by complex conjugation.) Now to the curvature; to begin with, our sign convention for the Riemann curvature tensor is

R(X,Y)Z=XYZYXZ[X,Y]Z,R(X,Y)Z=\nabla_{\!X}{\nabla_{\!Y}{Z}}-\nabla_{\!Y}{\nabla_{\!X}{Z}}-\nabla_{\![X,Y]}{Z},\\

in which case the Ricci tensor with respect to the complex frame {T,𝒎,𝒎¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\} is

Ric(v,w)=R(T,v,w,T)+R(𝒎,v,w,𝒎¯)+R(𝒎¯,v,w,𝒎).\text{Ric}(v,w)=R(T,v,w,T)+R({\boldsymbol{m}},v,w,\overline{\boldsymbol{m}})+R(\overline{\boldsymbol{m}},v,w,{\boldsymbol{m}}).

The following identities satisfied by the Ricci tensor in the complex frame {T,𝒎,𝒎¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\} will appear in formulae below:

{Ric(𝒎,𝒎)=R(T,𝒎,𝒎,T),Ric(T,T)=2R(𝒎,T,T,𝒎¯),Ric(T,𝒎)=R(𝒎,T,𝒎,𝒎¯),Ric(𝒎,𝒎¯)=12Ric(T,T)+R(𝒎¯,𝒎,𝒎¯,𝒎).\left\{\begin{array}[]{rcl}\text{Ric}({\boldsymbol{m}},{\boldsymbol{m}})\!\!&=&\!\!R(T,{\boldsymbol{m}},{\boldsymbol{m}},T),\\ \text{Ric}(T,T)\!\!&=&\!\!2R({\boldsymbol{m}},T,T,\overline{\boldsymbol{m}}),\\ \text{Ric}(T,{\boldsymbol{m}})\!\!&=&\!\!R({\boldsymbol{m}},T,{\boldsymbol{m}},\overline{\boldsymbol{m}}),\\ \text{Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})\!\!&=&\!\!\frac{1}{2}\text{Ric}(T,T)+R(\overline{\boldsymbol{m}},{\boldsymbol{m}},\overline{\boldsymbol{m}},{\boldsymbol{m}}).\end{array}\right.

The Newman-Penrose begins by expressing the Lie brackets in terms of spin coefficients, as we saw in (11) and (12) above. It then moves down to the level of curvature, by expressing the following components of the curvature tensor,

R(T,𝒎¯,T,𝒎)\displaystyle R(T,\overline{\boldsymbol{m}},T,{\boldsymbol{m}}) ,\displaystyle\hskip 7.22743pt,\hskip 7.22743pt R(T,𝒎,T,𝒎),R(𝒎¯,𝒎,T,𝒎)\displaystyle R(T,{\boldsymbol{m}},T,{\boldsymbol{m}})\hskip 14.45377pt,\hskip 14.45377ptR(\overline{\boldsymbol{m}},{\boldsymbol{m}},T,{\boldsymbol{m}})
R(T,𝒎,𝒎,𝒎¯),R(𝒎¯,𝒎,𝒎,𝒎¯),\displaystyle\hskip-54.2025ptR(T,{\boldsymbol{m}},{\boldsymbol{m}},\overline{\boldsymbol{m}})\hskip 14.45377pt,\hskip 14.45377ptR(\overline{\boldsymbol{m}},{\boldsymbol{m}},{\boldsymbol{m}},\overline{\boldsymbol{m}}),

in terms of the Ricci tensor and the spin coefficients. Doing so, the following (first-order) equations arise; they play the driving role in our classification:

T(ρ)𝒎¯(κ)\displaystyle T(\rho)-\overline{\boldsymbol{m}}(\kappa)\!\! =\displaystyle= |κ|2+|σ|2+ρ2+κβ¯+12Ric(T,T),\displaystyle\!\!|\kappa|^{2}+|\sigma|^{2}+\rho^{2}+\kappa\bar{\beta}+\frac{1}{2}{\rm Ric}(T,T), (13)
T(σ)𝒎(κ)\displaystyle T(\sigma)-{\boldsymbol{m}}(\kappa)\!\! =\displaystyle= κ2+2σε+σ(ρ+ρ¯)κβ+Ric(𝒎,𝒎),\displaystyle\!\!\kappa^{2}+2\sigma\varepsilon+\sigma(\rho+\bar{\rho})-\kappa\beta+{\rm Ric}({\boldsymbol{m}},{\boldsymbol{m}}),\phantom{\frac{1}{2}} (14)
𝒎(ρ)𝒎¯(σ)\displaystyle{\boldsymbol{m}}(\rho)-\overline{\boldsymbol{m}}(\sigma)\!\! =\displaystyle= 2σβ¯+(ρ¯ρ)κ+Ric(T,𝒎),\displaystyle\!\!2\sigma\bar{\beta}+(\bar{\rho}-\rho)\kappa+{\rm Ric}(T,{\boldsymbol{m}}),\phantom{\frac{1}{2}} (15)
T(β)𝒎(ε)\displaystyle T(\beta)-{\boldsymbol{m}}(\varepsilon)\!\! =\displaystyle= σ(κ¯β¯)+κ(ερ¯)+β(ε+ρ¯)Ric(T,𝒎),\displaystyle\!\!\sigma(\bar{\kappa}-\bar{\beta})+\kappa(\varepsilon-\bar{\rho})+\beta(\varepsilon+\bar{\rho})-{\rm Ric}(T,{\boldsymbol{m}}),\phantom{\frac{1}{2}} (16)
𝒎(β¯)+𝒎¯(β)\displaystyle{\boldsymbol{m}}(\bar{\beta})+\overline{\boldsymbol{m}}(\beta)\!\! =\displaystyle= |σ|2|ρ|22|β|2+(ρρ¯)εRic(𝒎,𝒎¯)+12Ric(T,T).\displaystyle\!\!|\sigma|^{2}-|\rho|^{2}-2|\beta|^{2}+(\rho-\bar{\rho})\varepsilon-{\rm Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})+\frac{1}{2}{\rm Ric}(T,T).

Finally, up to complex conjugation, there are two nontrivial differential Bianchi identities:

T(Ric(T,𝒎))12𝒎(Ric(T,T))+𝒎¯(Ric(𝒎,𝒎))=\displaystyle\hskip-28.90755ptT({\rm Ric}(T,{\boldsymbol{m}}))\,-\,\frac{1}{2}{\boldsymbol{m}}({\rm Ric}(T,T))+\overline{\boldsymbol{m}}({\rm Ric}({\boldsymbol{m}},{\boldsymbol{m}}))=
κ(Ric(T,T)Ric(𝒎,𝒎¯))+(ε+2ρ+ρ¯)Ric(T,𝒎)\displaystyle\hskip-7.22743pt\kappa\,\big{(}{\rm Ric}(T,T)-\text{Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})\big{)}+\big{(}\varepsilon+2\rho+\bar{\rho}\big{)}{\rm Ric}(T,{\boldsymbol{m}})\phantom{\frac{1}{2}}
+σRic(T,𝒎¯)(κ¯+2β¯)Ric(𝒎,𝒎)\displaystyle\hskip 93.95122pt+\,\sigma\,{\rm Ric}(T,\overline{\boldsymbol{m}})-\big{(}\bar{\kappa}+2\bar{\beta}\big{)}{\rm Ric}({\boldsymbol{m}},{\boldsymbol{m}})\phantom{\frac{1}{2}}
and
𝒎(Ric(T,𝒎¯))+𝒎¯(Ric(T,𝒎))T(Ric(𝒎,𝒎¯)(1/2)Ric(T,T))=\displaystyle\hskip-28.90755pt{\boldsymbol{m}}({\rm Ric}(T,\overline{\boldsymbol{m}}))+\overline{\boldsymbol{m}}({\rm Ric}(T,{\boldsymbol{m}}))-T\big{(}{\rm Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})-(1/2){\rm Ric}(T,T)\big{)}=\phantom{\frac{1}{2}}
(ρ+ρ¯)(Ric(T,T)Ric(𝒎,𝒎¯))σ¯Ric(𝒎,𝒎)σRic(𝒎¯,𝒎¯)\displaystyle\hskip-7.22743pt(\rho+\bar{\rho})\big{(}{\rm Ric}(T,T)-{\rm Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})\big{)}-\bar{\sigma}{\rm Ric}({\boldsymbol{m}},{\boldsymbol{m}})-\sigma{\rm Ric}(\overline{\boldsymbol{m}},\overline{\boldsymbol{m}})\phantom{\frac{1}{2}}
(2κ¯+β¯)Ric(T,𝒎)(2κ+β)Ric(T,𝒎¯).\displaystyle\hskip 86.72377pt-\,\big{(}2\bar{\kappa}+\bar{\beta}\big{)}{\rm Ric}(T,{\boldsymbol{m}})-\big{(}2\kappa+\beta\big{)}{\rm Ric}(T,\overline{\boldsymbol{m}}).\phantom{\frac{1}{2}}

We now immediately specialize to the case when TT is a Killing vector field:

Lemma 3.

Let (M,g)(M,g) be a Riemannian 3-manifold admitting a unit length Killing vector field TT with twist function ω2\omega^{2}. With respect to any complex frame {T,𝐦,𝐦¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\}, the Ricci tensor Ric and scalar curvature SS satisfy

T(ω)\displaystyle T(\omega)\!\!\! =\displaystyle= 0,Ric(T,T)=ω22,Ric(𝒎,𝒎)=0,\displaystyle\!\!\!0\hskip 14.45377pt,\hskip 14.45377pt\emph{\text{Ric}}(T,T)=\frac{\omega^{2}}{2}\hskip 14.45377pt,\hskip 14.45377pt\emph{\text{Ric}}({\boldsymbol{m}},{\boldsymbol{m}})=0, (20)
𝒎(β¯)+𝒎¯(β)=2|β|2iωε12(Sω22)\displaystyle\hskip-21.68121pt{\boldsymbol{m}}(\bar{\beta})+\overline{\boldsymbol{m}}(\beta)=-2|\beta|^{2}-i\omega\varepsilon-\frac{1}{2}\Big{(}S-\frac{\omega^{2}}{2}\Big{)}\cdot

When (20) is written in terms of the underlying orthonormal frame {T,X,Y}\{T,X,Y\} of the complex frame, it is

X(divX)+Y(divY)=(divX)2(divY)2iωε12(Sω22)\displaystyle X(\emph{\text{div}}\,X)+Y(\emph{\text{div}}\,Y)=-(\emph{\text{div}}\,X)^{2}-(\emph{\text{div}}\,Y)^{2}-i\omega\varepsilon-\frac{1}{2}\Big{(}S-\frac{\omega^{2}}{2}\Big{)}\cdot (21)
Proof.

By Lemma 2, we know that

κ=σ=ρ+ρ¯=0;\kappa=\sigma=\rho+\bar{\rho}=0;

inserting these into (13) and (14) directly yields the first line of equations in the statement of Lemma 3; e.g.,

Ric(T,T)=ω22,T(ω)=0,\text{Ric}(T,T)=\frac{\omega^{2}}{2}\hskip 14.45377pt,\hskip 14.45377ptT(\omega)=0,

are, respectively, the real and imaginary parts of (13). Meanwhile, (20) follows from (LABEL:eqn:S4), which has no imaginary part, and the fact that the scalar curvature SS in terms of the complex frame {T,𝒎,𝒎¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\} is

S=Ric(T,T)+2Ric(𝒎,𝒎¯)Ric(𝒎,𝒎¯)=S2ω24\displaystyle S=\text{Ric}(T,T)+2\text{Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})\hskip 14.45377pt\Rightarrow\hskip 14.45377pt\text{Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})=\frac{S}{2}-\frac{\omega^{2}}{4}\cdot (22)

Finally, (21) follows from the fact that, when TT=0\nabla_{\!T}{T}=0,

β=12(g(YX,Y)+ig(XX,Y))=12(divXidivY),\displaystyle\beta=\frac{1}{\sqrt{2}}\big{(}g({\nabla_{\!Y}{X}},{Y})+ig({\nabla_{\!X}{X}},{Y})\big{)}=\frac{1}{\sqrt{2}}(\text{div}\,X-i\,\text{div}\,Y), (23)

which completes the proof. ∎

We have not yet considered the differential Bianchi identities; let us do so now. Inserting the contents of Lemma 3 into (3) and (3), as well as ρ¯=ρ\bar{\rho}=-\rho, yields

T(𝒎(ρ))=(ε+ρ¯)𝒎(ρ)T({\boldsymbol{m}}(\rho))=(\varepsilon+\bar{\rho}){\boldsymbol{m}}(\rho)

for (3); but this is precisely the Lie bracket (11) applied to ρ\rho (bearing in mind that T(ρ)=0T(\rho)=0), and therefore carries no new information. As for (3), it yields

𝒎(𝒎¯(ρ))+𝒎¯(𝒎(ρ))=β¯𝒎(ρ)+β𝒎¯(ρ),-{\boldsymbol{m}}(\overline{\boldsymbol{m}}(\rho))+\overline{\boldsymbol{m}}({\boldsymbol{m}}(\rho))=-\bar{\beta}\,{\boldsymbol{m}}(\rho)+\beta\,\overline{\boldsymbol{m}}(\rho),

where T(S)=0T(S)=0 and ρ¯=ρ\bar{\rho}=-\rho have been used. But this is precisely the Lie bracket (12) applied to ρ\rho, so that (3) also yields no new information.

4. Local Coordinates

The goal of this section is to establish the “right” local coordinates in which to prove Theorem 1 in the next section. To begin with, recall that because

κ=σ=ρ+ρ¯=0,\kappa=\sigma=\rho+\bar{\rho}=0,

the only spin coefficients remaining are ε\varepsilon and β\beta. Observe that the former is in fact purely imaginary,

ε=ig(TX,Y),\displaystyle\varepsilon=ig({\nabla_{\!T}{X}},{Y}), (24)

and the latter, when TT=0\nabla_{\!T}{T}=0, is given by (23). The following “gauge freedom” simultaneously enjoyed by these two spin coefficients will prove useful in the proof of Theorem 1:

Proposition 1.

Let TT be a unit length Killing vector field with twist function ω2\omega^{2} and {T,𝐦,𝐦¯}\{T,{\boldsymbol{m}},\overline{\boldsymbol{m}}\} a complex frame. Then there exists a smooth real-valued function ϑ\vartheta such that the complex frame {T,𝐦,𝐦¯}\{T,{\boldsymbol{m}}_{*},\overline{\boldsymbol{m}}_{*}\} defined by the rotation

𝒎=eiϑ𝒎,𝒎¯=eiϑ𝒎¯{\boldsymbol{m}}_{*}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=e^{i\vartheta}{\boldsymbol{m}}\hskip 14.45377pt,\hskip 14.45377pt\overline{\boldsymbol{m}}_{*}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=e^{-i\vartheta}\overline{\boldsymbol{m}}

has spin coefficients κ=σ=0,ρ=ρ\kappa_{*}=\sigma_{*}=0,\rho_{*}=\rho,

ε=ρ,Re(β)=0,T(β)=0.\displaystyle\varepsilon_{*}=\rho\hskip 14.45377pt,\hskip 14.45377pt\emph{\text{Re}}(\beta_{*})=0\hskip 14.45377pt,\hskip 14.45377ptT(\beta_{*})=0. (25)
Proof.

By definition,

κ=g(TT,𝒎)=eiϑκ=0;\kappa_{*}=-g({\nabla_{\!T}{T}},{{\boldsymbol{m}}_{*}})=e^{i\vartheta}\kappa=0;

similarly, σ=e2iϑσ=0\sigma_{*}=e^{2i\vartheta}\sigma=0, and ρ=ρ\rho_{*}=\rho (in particular, ω2=ω2\omega_{*}^{2}=\omega^{2}). Next,

ε\displaystyle\varepsilon_{*}\!\! =\displaystyle= g(T𝒎,𝒎¯)\displaystyle\!\!g({\nabla_{\!T}{{\boldsymbol{m}}_{*}}},{\overline{\boldsymbol{m}}_{*}}) (26)
=\displaystyle= eiϑg(T(eiϑ𝒎),𝒎¯)\displaystyle\!\!e^{-i\vartheta}g({\nabla_{\!T}{(e^{i\vartheta}{\boldsymbol{m}})}},{\overline{\boldsymbol{m}}})
=\displaystyle= ε+eiϑT(eiϑ)\displaystyle\!\!\varepsilon+e^{-i\vartheta}T(e^{i\vartheta})
=\displaystyle= ε+iT(ϑ).\displaystyle\!\!\varepsilon+iT(\vartheta).

Similarly,

β\displaystyle\beta_{*}\!\! =\displaystyle= g(𝒎𝒎,𝒎¯)\displaystyle\!\!g({\nabla_{\!{\boldsymbol{m}}_{*}}{{\boldsymbol{m}}_{*}}},{\overline{\boldsymbol{m}}_{*}})
=\displaystyle= g(𝒎(eiϑ𝒎),𝒎¯)\displaystyle\!\!g({\nabla_{\!{\boldsymbol{m}}}{(e^{i\vartheta}{\boldsymbol{m}})}},{\overline{\boldsymbol{m}}})
=\displaystyle= eiϑ(β+i𝒎(ϑ)).\displaystyle\!\!e^{i\vartheta}(\beta+i{\boldsymbol{m}}(\vartheta)).

By (24) and (26), we may choose a locally defined function ϑ\vartheta so that

ε=ρ=ρ.\varepsilon_{*}=\rho_{*}=\rho.

Now, choose any other function ψ\psi satisfying T(ψ)=0T(\psi)=0 and rotate 𝒎,𝒎¯{\boldsymbol{m}}_{*},\overline{\boldsymbol{m}}_{*} by ψ\psi; let {T,𝒎o,𝒎¯o}\{T,{\boldsymbol{m}}_{o},\overline{\boldsymbol{m}}_{o}\} denote the corresponding frame. Then the analogue of (26) for the frame {T,𝒎o,𝒎¯o}\{T,{\boldsymbol{m}}_{o},\overline{\boldsymbol{m}}_{o}\} shows that εo\varepsilon_{o} remains unchanged,

εo=ε=ρ=ρ,\varepsilon_{o}=\varepsilon_{*}=\rho_{*}=\rho,

so that our task would be complete if we can find a ψ\psi satisfying

T(ψ)=0,Re(βo)=0,T(βo)=0.\displaystyle T(\psi)=0\hskip 14.45377pt,\hskip 14.45377pt\text{Re}(\beta_{o})=0\hskip 14.45377pt,\hskip 14.45377ptT(\beta_{o})=0. (27)

To do so, go back to the complex frame {T,𝒎,𝒎¯}\{T,{\boldsymbol{m}}_{*},\overline{\boldsymbol{m}}_{*}\} and observe that when ε=ρ\varepsilon_{*}=\rho_{*}, then

[T,𝒎]=(11)0.\displaystyle[T,{\boldsymbol{m}}_{*}]\overset{\eqref{eqn:LB1}}{=}0. (28)

Let {T,X,Y}\{T,X_{*},Y_{*}\} denote the underlying orthonormal frame corresponding to {T,𝒎,𝒎¯}\{T,{\boldsymbol{m}}_{*},\overline{\boldsymbol{m}}_{*}\}. Since [T,X]=0[T,X_{*}]=0, there exist local coordinates (t,u,v)(t,u,v) and functions p,q,rp,q,r such that

T=t,X=u,Y=pt+qu+rv,T=\partial_{t}\hskip 14.45377pt,\hskip 14.45377ptX_{*}=\partial_{u}\hskip 14.45377pt,\hskip 14.45377ptY_{*}=p\partial_{t}+q\partial_{u}+r\partial_{v},

with p,q,rp,q,r functions of u,vu,v only, since [T,Y]=0[T,Y_{*}]=0, and with rr nowhere vanishing. The coframe metrically equivalent to {T,X,Y}\{T,X_{*},Y_{*}\} is therefore

T=dtprdv,X=duqrdv,Y=1rdv.T^{\flat}=dt-\frac{p}{r}dv\hskip 14.45377pt,\hskip 14.45377ptX_{*}^{\flat}=du-\frac{q}{r}dv\hskip 14.45377pt,\hskip 14.45377ptY_{*}^{\flat}=\frac{1}{r}dv.

Next, since (X)2+(Y)2(X_{*}^{\flat})^{2}+(Y_{*}^{\flat})^{2} defines a Riemannian metric on the 2-manifold with coordinates {(u,v)}\{(u,v)\}, and since any Riemannian 2-manifold is locally conformally flat (see, e.g., [Che55]), it follows that there exist coordinates (x,y)(x,y) and a smooth function λ(x,y)\lambda(x,y) such that

(X)2+(Y)2=e2λ(dx2+dy2).(X_{*}^{\flat})^{2}+(Y_{*}^{\flat})^{2}=e^{2\lambda}(dx^{2}+dy^{2}).

By a rotation in x,yx,y if necessary, we may further assume that

X=eλdx,Y=eλdy.X_{*}^{\flat}=e^{\lambda}dx\hskip 14.45377pt,\hskip 14.45377ptY_{*}^{\flat}=e^{\lambda}dy.

In the new coordinates (t,x,y)(t,x,y), we thus have that

T=t,X=eλ(x+at),Y=eλ(y+bt),T=\partial_{t}\hskip 14.45377pt,\hskip 14.45377ptX_{*}=e^{-\lambda}(\partial_{x}+a\partial_{t})\hskip 14.45377pt,\hskip 14.45377ptY_{*}=e^{-\lambda}(\partial_{y}+b\partial_{t}),

for some smooth functions a(x,y),b(x,y)a(x,y),b(x,y). With these coordinates in hand, we now return to the task of satisfying Re(βo)=T(βo)=0\text{Re}(\beta_{o})=T(\beta_{o})=0 in (27). For the former, ψ(x,y)\psi(x,y) should satisfy Re(βo)=Re(eiψ(β+i𝒎(ψ)))=0\text{Re}(\beta_{o})=\text{Re}\big{(}e^{i\psi}(\beta_{*}+i{\boldsymbol{m}}_{*}(\psi))\big{)}=0, or

eiψ(β+i𝒎(ψ))+eiψ(β¯i𝒎¯(ψ))=0.\displaystyle e^{i\psi}(\beta_{*}+i{\boldsymbol{m}}_{*}(\psi))+e^{-i\psi}(\bar{\beta}_{*}-i\,\overline{\boldsymbol{m}}_{*}(\psi))=0. (29)

When expanded, and using the fact that

divX=λxeλ,divY=λyeλ,\text{div}\,X_{*}=\frac{\lambda_{x}}{e^{\lambda}}\hskip 14.45377pt,\hskip 14.45377pt\text{div}\,Y_{*}=\frac{\lambda_{y}}{e^{\lambda}},

(29) is a quasilinear first-order PDE in ψ\psi,

(sinψ)ψx(cosψ)ψy=(cosψ)λx+(sinψ)λy,(\sin\psi)\psi_{x}-(\cos\psi)\psi_{y}=(\cos\psi)\lambda_{x}+(\sin\psi)\lambda_{y},

which has a solution by the method of characteristics. For the latter, (15) and (16) together yield

T(βo)𝒎(εo)𝒎o(ρo)=(16)Ric(T,𝒎o)𝒎o(ρo)by(15)T(βo)=0,T(\beta_{o})-\underbrace{\,{\boldsymbol{m}}(\varepsilon_{o})\,}_{{\boldsymbol{m}}_{o}(\rho_{o})}\overset{\eqref{eqn:S5}}{=}-\!\!\underbrace{\,\text{Ric}(T,{\boldsymbol{m}}_{o})\,}_{{\boldsymbol{m}}_{o}(\rho_{o})~\text{by}~\eqref{eqn:S3}}\hskip 14.45377pt\Rightarrow\hskip 14.45377ptT(\beta_{o})=0,

completing the proof. ∎

The following Corollary collects together what we’ve established so far:

Corollary 1.

Let (M,g)(M,g) be a Riemannian 3-manifold and TT a unit length Killing vector field with twist function ω2\omega^{2}. Then there exists an orthonormal frame {T,X,Y}\{T,X,Y\} satisfying

κ=σ=0,ρ=ε=i2ω,β=i2divY,\displaystyle\kappa=\sigma=0\hskip 14.45377pt,\hskip 14.45377pt\rho=\varepsilon=-\frac{i}{2}\omega\hskip 14.45377pt,\hskip 14.45377pt\beta=-\frac{i}{\sqrt{2}}\,\emph{\text{div}}\,Y, (30)

and with T(ω)=T(β)=0T(\omega)=T(\beta)=0. In this frame, (21) takes the form

Y(divY)=(divY)212(S+ω22)\displaystyle Y(\emph{\text{div}}\,Y)=-(\emph{\text{div}}\,Y)^{2}-\frac{1}{2}\Big{(}S+\frac{\omega^{2}}{2}\Big{)}\cdot (31)

Notice that (31) implies that such a frame may not always exist globally; e.g., if MM is compact and SS is nonnegative and positive somewhere, then a standard Riccati argument yields that in such a case the only complete solution to (31) is one where divY=S+ω22=0\text{div}\,Y=S+\frac{\omega^{2}}{2}=0, which is impossible. We now proceed to our local classification.

5. The Local Classification

Theorem 1 follows from one further modification to the orthonormal frame satisfying (30):

Theorem 1.

Let (M,g)(M,g) be a Riemannian 3-manifold that admits a unit length Killing vector field TT. Then there exist local coordinates (t,r,θ)(t,r,\theta) and a smooth function φ(r,θ)\varphi(r,\theta) such that

T=t,g=(T)2+dr2+φ2dθ2,\displaystyle T=\partial_{t}\hskip 14.45377pt,\hskip 14.45377ptg=(T^{\flat})^{2}+dr^{2}+\varphi^{2}d\theta^{2}, (32)

and where the quotient metric dr2+φ2dθ2dr^{2}+\varphi^{2}d\theta^{2} has Gaussian curvature

φrrφ=12(S+Ric(T,T)),\displaystyle-\frac{\varphi_{rr}}{\varphi}=\frac{1}{2}\big{(}S+\emph{\text{Ric}}(T,T)\big{)}, (33)

with SS and Ric the scalar curvature and Ricci tensor of gg, respectively. If (1) is given globally on M=3M=\mathbb{R}^{3} (with r,θr,\theta polar coordinates on 2\mathbb{R}^{2}), and if Ric(T,T)=0,φ(0,θ)=0\emph{\text{Ric}}(T,T)=0,\varphi(0,\theta)=0, φr(0,θ)=1\varphi_{r}(0,\theta)=1, φ(r,θ+π)=φ(r,θ)\varphi(-r,\theta+\pi)=-\varphi(r,\theta), and φ>0\varphi>0 when r>0r>0, then gg is complete if and only if

limrinf|p|rS|p0.\lim_{r\to\infty}\inf_{|p|\geq r}S\big{|}_{p}\leq 0.
Proof.

Let (M,g)(M,g) be a Riemannian 3-manifold and TT a unit length Killing vector field with twist function ω2\omega^{2}. By Corollary 1, there exist a local orthonormal frame {T,X,Y}\{T,X,Y\} satisfying (30) and coordinates (t,x,y)(t,x,y) in which T=/tT=\partial/\partial t. Let {T,X,Y}\{T^{\flat},X^{\flat},Y^{\flat}\} denote the dual coframe. We now modify the coordinates (t,x,y)(t,x,y) while keeping T=/tT=\partial/\partial t unchanged. The key is that (11) and (12) satisfy

[T,X]=[T,Y]=0,[X,Y]=ωT+(divY)X,[T,X]=[T,Y]=0\hskip 14.45377pt,\hskip 14.45377pt[X,Y]=\omega T+(\text{div}\,Y)X,

from which it follows that YY^{\flat} is closed, dY=0dY^{\flat}=0; hence

Y=drY^{\flat}=dr

for some smooth function r(x,y)r(x,y). Similarly,

dX=(divY)YXX=φdθdX^{\flat}=(\text{div}\,Y)Y^{\flat}\wedge X^{\flat}\hskip 14.45377pt\Rightarrow\hskip 14.45377ptX^{\flat}=\varphi d\theta

for some smooth functions φ(x,y)>0\varphi(x,y)>0 and θ(x,y)\theta(x,y), with the former satisfying

Y(φ)=(divY)φ\displaystyle Y(\varphi)=(\text{div}\,Y)\varphi (34)

(recall that T(β)=0T(\beta)=0). Since X(r)=Y(θ)=0X(r)=Y(\theta)=0, we can define new coordinates (t,r,θ)(t,r,\theta), in terms of which the frame {T,X,Y}\{T,X,Y\} takes the form

T=t,X=ht+1φθ,Y=kt+r,\displaystyle T=\partial_{t}\hskip 14.45377pt,\hskip 14.45377ptX=h\partial_{t}+\frac{1}{\varphi}\partial_{\theta}\hskip 14.45377pt,\hskip 14.45377ptY=k\partial_{t}+\partial_{r}, (35)

for some smooth functions h,kh,k; furthermore, φt=ht=kt=0\varphi_{t}=h_{t}=k_{t}=0 (recall that [T,X]=[T,Y]=0[T,X]=[T,Y]=0), so that φ,h,k\varphi,h,k are all functions of r,θr,\theta only. Thus

g=(T)2+(X)2+(Y)2=(T)2+dr2+φ2dθ2,g=(T^{\flat})^{2}+(X^{\flat})^{2}+(Y^{\flat})^{2}=(T^{\flat})^{2}+dr^{2}+\varphi^{2}d\theta^{2},

confirming (32). Now, the quotient metric dr2+φ2dθ2dr^{2}+\varphi^{2}d\theta^{2} has scalar curvature 2φrr/φ-2\varphi_{rr}/\varphi, hence Gaussian curvature φrr/φ-\varphi_{rr}/\varphi. To relate this to the curvature of (M,g)(M,g), we take a YY-derivative of (34), make use of (31), and note that t(divY)=0\partial_{t}(\text{div}\,Y)=0 by (25), to obtain

φrr=Y(divY)φ+(divY)2φφrrφ=(31)12(S+ω22)\varphi_{rr}=Y(\text{div}\,Y)\varphi+(\text{div}\,Y)^{2}\varphi\hskip 14.45377pt\Rightarrow\hskip 14.45377pt-\frac{\varphi_{rr}}{\varphi}\overset{\eqref{eqn:Ric2}}{=}\frac{1}{2}\Big{(}S+\frac{\omega^{2}}{2}\Big{)}\cdot

Since Ric(T,T)=ω22\text{Ric}(T,T)=\frac{\omega^{2}}{2} by Lemma 3, this confirms (33). There remains, finally, the question of completeness; thus, suppose that on 3={(t,r,θ)}\mathbb{R}^{3}=\{(t,r,\theta)\} with r,θr,\theta polar coordinates on 2\mathbb{R}^{2}, and with metric gg given by (32), we have the globally defined vector fields appearing in (35). If Ric(T,T)=ω22=0\text{Ric}(T,T)=\frac{\omega^{2}}{2}=0, so that g(T,[X,Y])=0g(T,[X,Y])=0, then T=tT=\partial_{t} is parallel in (3,g)(\mathbb{R}^{3},g) and thus T=dtT^{\flat}=dt, from which it follows from (35) that h=k=0h=k=0, hence that {t,r,θ}\{\partial_{t},\partial_{r},\partial_{\theta}\} is orthogonal. Now, let γ(s)=(t(s),r(s),θ(s))\gamma(s)=(t(s),r(s),\theta(s)) be a geodesic in (3,g)(\mathbb{R}^{3},g); since g(T,γ)=g(t,γ)g(T,\gamma^{\prime})=g(\partial_{t},\gamma^{\prime}) is a constant, which constant we denote by cc, the tangent vector γ(s)\gamma^{\prime}(s) takes the form

γ(s)=ct|γ(s)+r˙(s)r|γ(s)+θ˙(s)θ|γ(s).\gamma^{\prime}(s)=c\partial_{t}|_{\gamma(s)}+\dot{r}(s)\partial_{r}|_{\gamma(s)}+\dot{\theta}(s)\partial_{\theta}|_{\gamma(s)}.

Since gg splits as the product of dt2dt^{2} and dr2+φ2dθ2dr^{2}+\varphi^{2}d\theta^{2}, γ(s)\gamma(s) is a geodesic if and only if (r(s),θ(s))(r(s),\theta(s)) is a geodesic in (2,dr2+φ2dθ2)(\mathbb{R}^{2},dr^{2}+\varphi^{2}d\theta^{2}). We now use [KW74]: when r,θr,\theta are polar coordinates, with φ(0,θ)=0,φr(0,θ)=1\varphi(0,\theta)=0,\varphi_{r}(0,\theta)=1, φ(r,θ+π)=φ(r,θ)\varphi(-r,\theta+\pi)=-\varphi(r,\theta), and φ>0\varphi>0 when r>0r>0, then the metric is complete if and only if

limrinf|p|r(φrrφ)|p0.\lim_{r\to\infty}\inf_{|p|\geq r}\Big{(}\!\!-\!\frac{\varphi_{rr}}{\varphi}\Big{)}\Big{|}_{p}\leq 0.

Since φrrφ=12(S+Ric(T,T))=12S-\frac{\varphi_{rr}}{\varphi}=\frac{1}{2}\big{(}S+\text{Ric}(T,T)\big{)}=\frac{1}{2}S, the proof is complete. ∎

A final remark regarding Theorem 1: bear in mind that, since in general

T=dtφhdθkdr,T^{\flat}=dt-\varphi hd\theta-kdr,

the coordinates (t,r,θ)(t,r,\theta) above are not “semigeodesic” (see, e.g., [Lee18]); indeed, the metric components gijg_{ij} in the coordinate basis {t,r,θ}\{\partial_{t},\partial_{r},\partial_{\theta}\} are given by

(gij)=(1kφhk1+k2φhkφhφhkφ2(1+h2))\displaystyle(g_{ij})=\begin{pmatrix}1&-k&-\varphi h\\ -k&1+k^{2}&\varphi hk\\ -\varphi h&\varphi hk&\varphi^{2}(1+h^{2})\end{pmatrix}\cdot (36)

We now proceed to the Lorentzian setting.

6. The Lorentzian setting

Before proceeding to a proof of the Lorentzian analogue of Theorem 1, we first collect a few facts from Lorentzian geometry; in what follows we adopt the metric index (++)(-\!+\!+). First, a vector field TT on a Lorentzian manifold (M,gL)(M,g_{\scriptscriptstyle L}) is timelike if gL(T,T)<0g_{\scriptscriptstyle L}(T,T)<0. Second, if a timelike TT has unit length, gL(T,T)=1g_{\scriptscriptstyle L}(T,T)=-1, then

gR=gL+2(TL)2\displaystyle g_{\scriptscriptstyle R}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=g_{\scriptscriptstyle L}+2(T^{\flat_{\scriptscriptstyle L}})^{2} (37)

defines a Riemannian metric on MM (here TL=gL(T,)T^{\flat_{\scriptscriptstyle L}}=g_{\scriptscriptstyle L}(T,\cdot)). Third, the following properties hold between gRg_{\scriptscriptstyle R} and gLg_{\scriptscriptstyle L}:

  1. 1.

    TT is a unit length Killing vector field with respect to gRg_{\scriptscriptstyle R} if and only if TT is a unit timelike Killing vector field with respect to gLg_{\scriptscriptstyle L} (see, e.g., [Ole14]).

  2. 2.

    If TT is a gRg_{\scriptscriptstyle R}-unit length Killing vector field, then

    RicR(T,T)=RicL(T,T)\text{Ric}_{\scriptscriptstyle R}(T,T)=\text{Ric}_{\scriptscriptstyle L}(T,T)

    (consult [Ole14]; this follows because XRT=XLT\nabla^{\scriptscriptstyle R}_{\!X}T=-\nabla^{\scriptscriptstyle L}_{\!X}T for any unit length XX that is gRg_{\scriptscriptstyle R}- or gLg_{\scriptscriptstyle L}-orthogonal to TT, where R\nabla^{\scriptscriptstyle R} and L\nabla^{\scriptscriptstyle L} are, respectively, the Levi-Civita connections of gRg_{\scriptscriptstyle R} and gLg_{\scriptscriptstyle L}), while their scalar curvatures SRS_{\scriptscriptstyle R} and SLS_{\scriptscriptstyle L} satisfy

    SL=SR+2RicR(T,T).S_{\scriptscriptstyle L}=S_{\scriptscriptstyle R}+2\text{Ric}_{\scriptscriptstyle R}(T,T).

    In particular, SR+RicR(T,T)=SLRicL(T,T)S_{\scriptscriptstyle R}+\text{Ric}_{\scriptscriptstyle R}(T,T)=S_{\scriptscriptstyle L}-\text{Ric}_{\scriptscriptstyle L}(T,T).

  3. 3.

    If TT is gRg_{\scriptscriptstyle R}-unit length Killing vector field, then gLg_{\scriptscriptstyle L} is complete if and only if gRg_{\scriptscriptstyle R} is complete (see [RS94]).

With these facts established, the Lorentzian analogue of Theorem 1 now follows easily:

Corollary 2.

Let (M,gL)(M,g_{\scriptscriptstyle L}) be a Lorentzian 3-manifold that admits a unit timelike Killing vector field TT. Then there exists local coordinates (t,r,θ)(t,r,\theta) and a smooth function φ(r,θ)\varphi(r,\theta) such that

T=t,gL=(T)2+dr2+φ2dθ2,\displaystyle T=\partial_{t}\hskip 14.45377pt,\hskip 14.45377ptg_{\scriptscriptstyle L}=-(T^{\flat})^{2}+dr^{2}+\varphi^{2}d\theta^{2}, (38)

and where the quotient metric dr2+φ2dθ2dr^{2}+\varphi^{2}d\theta^{2} has Gaussian curvature

φrrφ=12(SLRicL(T,T)),-\frac{\varphi_{rr}}{\varphi}=\frac{1}{2}\big{(}S_{\scriptscriptstyle L}-\emph{\text{Ric}}_{\scriptscriptstyle L}(T,T)\big{)},

with SLS_{\scriptscriptstyle L} and RicL\emph{\text{Ric}}_{\scriptscriptstyle L} the scalar curvature and Ricci tensor of gLg_{\scriptscriptstyle L}, respectively. Furthermore, gLg_{\scriptscriptstyle L} is complete if and only if gRg_{\scriptscriptstyle R} is complete, where gRg_{\scriptscriptstyle R} is the corresponding Riemannian metric given by (37).

Proof.

By our remarks above, TT is a unit length Killing vector field with respect to the Riemannian metric gRg_{\scriptscriptstyle R}, with SR+RicR(T,T)=SLRicL(T,T)S_{\scriptscriptstyle R}+\text{Ric}_{\scriptscriptstyle R}(T,T)=S_{\scriptscriptstyle L}-\text{Ric}_{\scriptscriptstyle L}(T,T); Corollary 2 therefore follows immediately from Theorem 1. ∎

7. The compact case

We now prove a global obstruction result in the compact setting. Thus, let (M,g)(M,g) be a compact Riemannian 3-manifold equipped with a globally defined unit length Killing vector field. With respect to a local orthonormal frame {T,X,Y}\{T,X,Y\}, we have, by Lemma 3, that

Ric(X,X)=Ric(Y,Y)viaRic(𝒎,𝒎)= 0,Ric(T,T)=ω22.\underbrace{\,\text{Ric}(X,X)=\text{Ric}(Y,Y)\,}_{\text{via}~\text{Ric}({\boldsymbol{m}},{\boldsymbol{m}})\,=\,0}\hskip 14.45377pt,\hskip 14.45377pt\text{Ric}(T,T)=\frac{\omega^{2}}{2}.

In fact, because

S2ω24=(22)Ric(𝒎,𝒎¯)=12(Ric(X,X)+Ric(Y,Y)),\frac{S}{2}-\frac{\omega^{2}}{4}\overset{\eqref{eqn:S}}{=}\text{Ric}({\boldsymbol{m}},\overline{\boldsymbol{m}})=\frac{1}{2}\big{(}\text{Ric}(X,X)+\text{Ric}(Y,Y)\big{)},

it follows that Ric(X,X)=Ric(Y,Y)=S2ω24\text{Ric}(X,X)=\text{Ric}(Y,Y)=\frac{S}{2}-\frac{\omega^{2}}{4}. Finally, by (15) we get Ric(T,𝒎)=𝒎(ρ)\text{Ric}(T,{\boldsymbol{m}})={\boldsymbol{m}}(\rho), whose real and imaginary parts yield

Ric(T,X)=Y(ω)2,Ric(T,Y)=Y(ω)2\displaystyle\text{Ric}(T,X)=-\frac{Y(\omega)}{2}\hskip 14.45377pt,\hskip 14.45377pt\text{Ric}(T,Y)=\frac{Y(\omega)}{2}\cdot (39)

Thus the Ricci operator Ric:TMTM\text{Ric}\colon TM\longrightarrow TM, defined by

vRic(v)=R(v,T)T+R(v,X)X+R(v,Y)Y,\displaystyle v\mapsto\text{Ric}(v)=R(v,T)T+R(v,X)X+R(v,Y)Y, (40)

has, with respect to the frame {T,X,Y}\{T,X,Y\}, the matrix

Ric=12(ω2Y(ω)X(ω)Y(ω)Sω220X(ω)0Sω22)\displaystyle\text{Ric}=\frac{1}{2}\begin{pmatrix}\omega^{2}&-Y(\omega)&X(\omega)\\ -Y(\omega)&S-\frac{\omega^{2}}{2}&0\\ X(\omega)&0&S-\frac{\omega^{2}}{2}\end{pmatrix}\cdot (41)

The characteristic polynomial of Ric is

(S2ω24λ)(λ2(S2+ω24)λ+ω22(S2ω24)14|ω|g2),\left(\frac{S}{2}-\frac{\omega^{2}}{4}-\lambda\right)\left(\lambda^{2}-\left(\frac{S}{2}+\frac{\omega^{2}}{4}\right)\lambda+\frac{\omega^{2}}{2}\left(\frac{S}{2}-\frac{\omega^{2}}{4}\right)-\frac{1}{4}|\nabla\omega|_{g}^{2}\right),

where ω=X(ω)X+Y(ω)Y\nabla\omega=X(\omega)X+Y(\omega)Y is the gradient of ω\omega; the eigenvalues of Ric are then easily found to be

λ1=S4+ω28+Δ2,λ2=S4+ω28Δ2,λ3=S2ω24,\displaystyle\lambda_{1}=\frac{S}{4}+\frac{\omega^{2}}{8}+\frac{\sqrt{\Delta}}{2}\hskip 14.45377pt,\hskip 14.45377pt\lambda_{2}=\frac{S}{4}+\frac{\omega^{2}}{8}-\frac{\sqrt{\Delta}}{2}\hskip 14.45377pt,\hskip 14.45377pt\lambda_{3}=\frac{S}{2}-\frac{\omega^{2}}{4}, (42)

where Δ=14(S32ω2)2+|ω|g2\Delta\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\frac{1}{4}(S-\frac{3}{2}\omega^{2})^{2}+|\nabla\omega|_{g}^{2}. Note that when the twist is constant, (41) reduces to

Ric=(ω22000S2ω24000S2ω24)\displaystyle\text{Ric}=\begin{pmatrix}\frac{\omega^{2}}{2}&0&0\\ 0&\frac{S}{2}-\frac{\omega^{2}}{4}&0\\ 0&0&\frac{S}{2}-\frac{\omega^{2}}{4}\end{pmatrix}\cdot (43)

As mentioned in the Introduction, the canonical such example is (𝕊3,g̊)(\mathbb{S}^{3},\mathring{g}) with radius RR and Hopf Killing vector field TT:

Ric(T,T)=2R2=ω22,S=6R2S2ω24=ω22\displaystyle\text{Ric}(T,T)=\frac{2}{R^{2}}=\frac{\omega^{2}}{2}\hskip 14.45377pt,\hskip 14.45377ptS=\frac{6}{R^{2}}\hskip 14.45377pt\Rightarrow\hskip 14.45377pt\frac{S}{2}-\frac{\omega^{2}}{4}=\frac{\omega^{2}}{2}\cdot (44)

In any case, owing to Hamilton’s well known result regarding the positivity of the Ricci operator in dimension 3 [Ham82], we have the following global obstruction:

Theorem 2.

Let (M,g)(M,g) be a compact Riemannian 3-manifold and TT a globally defined, unit length Killing vector field. If Ric(T,T)0\emph{\text{Ric}}(T,T)\neq 0 and

S>2|Ric(T)|g2Ric(T,T)Ric(T,T),\displaystyle S>2\frac{|\emph{\text{Ric}}(T)|_{g}^{2}}{\emph{\text{Ric}}(T,T)}-\emph{\text{Ric}}(T,T), (45)

then MM admits a metric of constant positive sectional curvature.

Proof.

Observe that the eigenvalues of Ric in (42) are all positive when

S>|ω|g2ω2+ω22\displaystyle S>\frac{|\nabla\omega|_{g}^{2}}{\omega^{2}}+\frac{\omega^{2}}{2}\cdot (46)

Because

Ric(T)\displaystyle\text{Ric}(T)\!\! =\displaystyle= R(T,T)T+R(T,X)X+R(T,Y)Y\displaystyle\!\!R(T,T)T+R(T,X)X+R(T,Y)Y (47)
=(39)\displaystyle\overset{\eqref{eqn:RR1}}{=} ω22TY(ω)2X+X(ω)2Y,\displaystyle\!\!\frac{\omega^{2}}{2}T-\frac{Y(\omega)}{2}X+\frac{X(\omega)}{2}Y,

it follows that

|Ric(T)|g2=14(ω4+|ω|g2).|\text{Ric}(T)|_{g}^{2}=\frac{1}{4}(\omega^{4}+|\nabla\omega|_{g}^{2}).

Since Ric(T,T)=ω22\text{Ric}(T,T)=\frac{\omega^{2}}{2}, (45) implies (46). By [Ham82], positive Ricci operator implies that MM admits a metric of constant positive sectional curvature. ∎

Note that the positive sectional curvature condition in [Ham82] requires that no eigenvalue of Ric should be larger than the sum of the other two eigenvalues. This requires SS twice as large: S>2|ω|g2ω2+ω2S>2\frac{|\nabla\omega|_{g}^{2}}{\omega^{2}}+\omega^{2}.

8. Criterion for Conformal Flatness

A metric on a 3-manifold is locally conformally flat if and only if its Cotton-York tensor vanishes; since this 2-tensor is symmetric and trace-free, this gives five conditions. The Cotton-York tensor is calculated in Appendix A, where it is written in matrix form as (c1c2c3)\begin{pmatrix}c_{1}&c_{2}&c_{3}\end{pmatrix} with respect to a local orthonormal frame {T,X,Y}\{T,X,Y\} satisfying (35) in the coordinates (t,r,θ)(t,r,\theta); see (57)-(65) below. In what follows, the entry in the ithi^{\text{th}} column and jthj^{\text{th}} row is denoted by cijc_{ij}. With that said, we now proceed to the proof of Theorem 3:

Theorem 3.

Let (M,g)(M,g) be a Riemannian 3-manifold that admits a unit length Killing vector field TT. If gg is locally conformally flat, then

4|Ric(T)|g2=3Ric(T,T)22BRic(T,T)+C\displaystyle 4|\emph{\text{Ric}}(T)|_{g}^{2}=3\emph{\text{Ric}}(T,T)^{2}-2B\emph{\text{Ric}}(T,T)+C (48)

for some constants B,CB,C, where Ric(T)\emph{\text{Ric}}(T) is the Ricci operator. Conversely, given (48), there exist coordinates (r,θ)(r,\theta) on the quotient metric in (1) with respect to which gg is conformally flat when

ωθ=0,ωr2+14(ω2+2B)2=C+B2,φ=h(θ)ωr,\omega_{\theta}=0\hskip 14.45377pt,\hskip 14.45377pt\omega_{r}^{2}+\frac{1}{4}(\omega^{2}+2B)^{2}=C+B^{2}\hskip 14.45377pt,\hskip 14.45377pt\varphi=h(\theta)\omega_{r},

where ω2=2Ric(T,T)\omega^{2}=2\emph{\text{Ric}}(T,T), φ\varphi is as in Theorem 1, and h(θ)h(\theta) is a smooth function. If Ric(T,T)\emph{\text{Ric}}(T,T) is constant, then gg is locally conformally flat if and only if S=3Ric(T,T)S=3\emph{\text{Ric}}(T,T).

Proof.

We start by setting the entry c32c_{32} equal to zero,

12Y(X(ω))=12r(1φωθ)=0\frac{1}{2}Y(X(\omega))=\frac{1}{2}\frac{\partial}{\partial r}\left(\frac{1}{\varphi}\frac{\partial\omega}{\partial\theta}\right)=0

(recall from Lemma 3 that tω=T(ω)=0\partial_{t}\omega=T(\omega)=0), which implies that

ωθ=A(θ)φ\omega_{\theta}=A(\theta)\varphi (49)

for some function A(θ)A(\theta). Next, c21=c31=0c_{21}=c_{31}=0 together yield

S=52ω2+B1=5Ric(T,T)+B1S=\frac{5}{2}\omega^{2}+B_{1}=5\text{Ric}(T,T)+B_{1} (50)

for some constant B1B_{1}. It follows that 12(S+Ric(T,T))=φrrφ=B12+32ω2\frac{1}{2}\big{(}S+\text{Ric}(T,T)\big{)}=-\frac{\varphi_{rr}}{\varphi}=\frac{B_{1}}{2}+\frac{3}{2}\omega^{2}. The remaining two conditions are c22=c33=0c_{22}=c_{33}=0. Substituting (49) and (50) into c33=0c_{33}=0, and recalling (34), gives

ω(ω2+B1)=2ωrφr+A(θ)φ,\omega(\omega^{2}+B_{1})=-2\frac{\omega_{r}\varphi_{r}+A^{\prime}(\theta)}{\varphi},

which, after rearranging, becomes

φω(ω2+B1)+2ωrφr=2A(θ).\varphi\,\omega(\omega^{2}+B_{1})+2\omega_{r}\varphi_{r}=-2A^{\prime}(\theta). (51)

Finally, from c22=0c_{22}=0 we get

ω(ω2+B1)=2ωrr,\omega(\omega^{2}+B_{1})=-2\omega_{rr},

and after multiplying through by ωr-\omega_{r} and integrating yields

f(θ)=ω4+2B1ω2+4ωr2\displaystyle f(\theta)=\omega^{4}+2B_{1}\omega^{2}+4\omega_{r}^{2} (52)

for some function f(θ)f(\theta). To relate f(θ)f(\theta) and A(θ)A(\theta), take a θ\theta-derivative of ff,

f(θ)\displaystyle f^{\prime}(\theta) =4ω3ωθ+4B1ωωθ+4ωrωrθ,\displaystyle=4\omega^{3}\omega_{\theta}+4B_{1}\omega\omega_{\theta}+4\omega_{r}\omega_{r\theta},
=(49)4ω3A(θ)φ+4B1ωA(θ)φ+8ωrA(θ)φr,\displaystyle\overset{\eqref{omeq}}{=}4\omega^{3}A(\theta)\varphi+4B_{1}\omega A(\theta)\varphi+8\omega_{r}A(\theta)\varphi_{r},
=4A(θ)(φω(ω2+B1)+2ωrφr),\displaystyle=4A(\theta)\big{(}\varphi\omega(\omega^{2}+B_{1})+2\omega_{r}\varphi_{r}\big{)},
=(51)8A(θ)A(θ),\displaystyle\overset{\eqref{y33}}{=}-8A(\theta)A^{\prime}(\theta),

and integrate, to obtain

f(θ)=4A2(θ)+4Cf(\theta)=-4A^{2}(\theta)+4C

for some constant CC. Inserting this back into (52) gives

4ωr2=4A2(θ)+4C+B12(ω2+B1)2.4\omega_{r}^{2}=-4A^{2}(\theta)+4C+B_{1}^{2}-(\omega^{2}+B_{1})^{2}.

Substituting (49) for A(θ)A(\theta), dividing through by 44, and setting B=B1/2B\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=B_{1}/2, yields

ωr2+ωθ2φ2=Cω44Bω2.\displaystyle\omega_{r}^{2}+\frac{\omega_{\theta}^{2}}{\varphi^{2}}=C-\frac{\omega^{4}}{4}-B\omega^{2}. (53)

The left-hand side can be further simplified; indeed, since Ric(T,T)=ω22\text{Ric}(T,T)=\frac{\omega^{2}}{2},

Ric(T)\displaystyle\text{Ric}(T)\!\! =(35)\displaystyle\overset{\eqref{eqn:coord1}}{=} ω22Tωr2X+ωθ2φY\displaystyle\!\!\frac{\omega^{2}}{2}T-\frac{\omega_{r}}{2}X+\frac{\omega_{\theta}}{2\varphi}Y

(recall (47)), so that

|Ric(T)|g2=ω44+ωr24+ωθ24φ2|\text{Ric}(T)|_{g}^{2}=\frac{\omega^{4}}{4}+\frac{\omega_{r}^{2}}{4}+\frac{\omega_{\theta}^{2}}{4\varphi^{2}}\cdot

Substituting this into (53) yields

4|Ric(T)|g2=3Ric(T,T)22BRic(T,T)+C,4|\text{Ric}(T)|_{g}^{2}=3\text{Ric}(T,T)^{2}-2B\text{Ric}(T,T)+C,

which is precisely (48). Conversely, suppose that (48) holds; then (53) holds and we see that |dω|g2=Cω44Bω2=|ω|g2|d\omega|_{g}^{2}=C-\frac{\omega^{4}}{4}-B\omega^{2}=|\nabla\omega|_{g}^{2}. Next, observe that the vector field

X~=2Ric(T)ω2T=X(ω)YY(ω)X\widetilde{X}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=2\text{Ric}(T)-\omega^{2}T=X(\omega)Y-Y(\omega)X

is divergence-free and satisfies both |X~|g=|ω|g|\widetilde{X}|_{g}=|\nabla\omega|_{g} and g(X~,ω)=0g(\widetilde{X},\nabla\omega)=0, in which case its normalization will also be divergence-free:

div(X~|X~|g)=g(|X~|g,X~)|X~|g2=0.\text{div}\bigg{(}\frac{\widetilde{X}}{|\widetilde{X}|_{g}}\bigg{)}=-\frac{g(\nabla|\widetilde{X}|_{g},\widetilde{X})}{|\widetilde{X}|_{g}^{2}}=0.

Then, setting Y~=ω|ω|g\widetilde{Y}\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\frac{\nabla\omega}{|\nabla\omega|_{g}} gives an orthonormal frame {T,X~,Y~}\{T,\widetilde{X},\widetilde{Y}\} satisfying (25). Working in this frame, set X=X~X\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\widetilde{X}, Y=Y~Y\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\widetilde{Y} and adjust the coordinates r,θr,\theta accordingly. Then in these new coordinates ωθ=0\omega_{\theta}=0 and (53) becomes the following ODE:

ωr2+14(ω2+2B)2=C+B2.\omega_{r}^{2}+\frac{1}{4}(\omega^{2}+2B)^{2}=C+B^{2}.

This has the form of a conservation of energy equation with positive potential. The potential is a single well when B0B\geq 0 and a double well when B<0B<0. Thus there will be periodic solutions for generic constants BB, CC and initial value ω|r=0=ω0\omega|_{r=0}=\omega_{0} satisfying C+B214(ω02+2B)2C+B^{2}\geq\frac{1}{4}(\omega_{0}^{2}+2B)^{2}. This is not enough to guarantee conformal flatness, as (50) and (51) must also be satisfied. In light of (33), we now show that these require that φ=h(θ)ωr\varphi=h(\theta)\omega_{r} for some function h(θ)h(\theta). Indeed, taking an rr derivative of (53) yields

2ωrrωr=ω(ω2+B1)ωr.2\omega_{rr}\omega_{r}=-\omega(\omega^{2}+B_{1})\omega_{r}.

Since (49) implies that AA is zero, the above implies that (51) can be written as

2ωrrφ+2ωrφr=0.-2\omega_{rr}\varphi+2\omega_{r}\varphi_{r}=0.

This requires ω\omega constant or φ=h(θ)ωr\varphi=h(\theta)\omega_{r} for some function h(θ)h(\theta). Taking two rr derivatives yields

φrr\displaystyle\varphi_{rr}\!\! =\displaystyle= h(θ)ωrrr\displaystyle\!\!h(\theta)\omega_{rrr}
=\displaystyle= h(θ)(32ω2B12)ωr\displaystyle\!\!h(\theta)\left(-\frac{3}{2}\omega^{2}-\frac{B_{1}}{2}\right)\omega_{r}

Now using (33), this gives (50), showing that for this choice of ϕ\phi, the metric gg is conformally flat. Finally, as (66) in Appendix A shows, if Ric(T,T)\text{Ric}(T,T) is constant, then the Cotton-York tensor vanishes if and only if the scalar curvature satisfies S=3Ric(T,T)S=3\text{Ric}(T,T). ∎

Appendix A Derivation of the Cotton-York Tensor

In order to compute the Cotton tensor, we will use the Cartan formalism (see, e.g., [Pet16, pp. 111-12]). First recall the Schouten tensor PP:

P=RicS4g.P\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\mathrm{Ric}-\frac{S}{4}g.

Using the fact that PΩ1(TM)P\in\Omega^{1}(TM), the Cotton tensor is given by

Cot(U,V)=dP(U,V)=(UP)(V)(VP)(U).\mathrm{Cot}(U,V)\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=d^{\nabla}P(U,V)=(\nabla_{U}P)(V)-(\nabla_{V}P)(U).

We now work in a frame {T,X,Y}\{T,X,Y\} satisfying (30); with respect to it, the Levi-Civita connection form 𝝎\boldsymbol{\omega} (not to be confused with the twist function ω\omega) is

𝝎=(0ω2Yω2Xω2Y0ω2T+div(Y)Xω2Xω2Tdiv(Y)X0)\boldsymbol{\omega}=\begin{pmatrix}0&-\frac{\omega}{2}Y^{\flat}&\frac{\omega}{2}X^{\flat}\\ \frac{\omega}{2}Y^{\flat}&0&\frac{\omega}{2}T^{\flat}+\mathrm{div}(Y)X^{\flat}\\ -\frac{\omega}{2}X^{\flat}&-\frac{\omega}{2}T^{\flat}-\mathrm{div}(Y)X^{\flat}&0\\ \end{pmatrix}\cdot

Together with the Ricci tensor (41), the Schouten tensor is thus

P=12(ω2S2Y(ω)X(ω)Y(ω)S2ω220X(ω)0S2ω22)P=\frac{1}{2}\begin{pmatrix}\omega^{2}-\frac{S}{2}&-Y(\omega)&X(\omega)\\ -Y(\omega)&\frac{S}{2}-\frac{\omega^{2}}{2}&0\\ X(\omega)&0&\frac{S}{2}-\frac{\omega^{2}}{2}\end{pmatrix}\cdot

The frame formula for the exterior covariant derivative of PP is just

dP=dP+[𝝎,P].d^{\nabla}P=dP+[\boldsymbol{\omega},P].

Using this, and recalling that SS and ω\omega have zero derivative in the direction of TT, we have

Cot(X,Y)\displaystyle\mathrm{Cot}(X,Y) =(X(P)+[𝝎(X),P])(001)(Y(P)+[𝝎(Y),P])(010)\displaystyle=(X(P)+[\boldsymbol{\omega}(X),P])\left(\begin{array}[]{c}0\\ 0\\ 1\end{array}\right)-(Y(P)+[\boldsymbol{\omega}(Y),P])\left(\begin{array}[]{c}0\\ 1\\ 0\end{array}\right)
=(34ω3+12Sω+12div(Y)Y(ω)+12[X(X(ω))+Y(Y(ω))]14Y(S)+54ωY(ω)14X(S)54ωX(ω)),\displaystyle=\left(\begin{array}[]{c}-\frac{3}{4}\omega^{3}+\frac{1}{2}S\omega+\frac{1}{2}\mathrm{div}(Y)Y(\omega)+\frac{1}{2}[X(X(\omega))+Y(Y(\omega))]\\ -\frac{1}{4}Y(S)+\frac{5}{4}\omega Y(\omega)\\ \frac{1}{4}X(S)-\frac{5}{4}\omega X(\omega)\end{array}\right),
Cot(Y,T)\displaystyle\mathrm{Cot}(Y,T) =(Y(P)+[𝝎(Y),P])(100)[𝝎(T),P](001)\displaystyle=(Y(P)+[\boldsymbol{\omega}(Y),P])\left(\begin{array}[]{c}1\\ 0\\ 0\end{array}\right)-[\boldsymbol{\omega}(T),P]\left(\begin{array}[]{c}0\\ 0\\ 1\end{array}\right)
=(54ωY(ω)14Y(S)38ω314Sω12Y(Y(ω))12Y(X(ω))),\displaystyle=\left(\begin{array}[]{c}\frac{5}{4}\omega Y(\omega)-\frac{1}{4}Y(S)\\ \frac{3}{8}\omega^{3}-\frac{1}{4}S\omega-\frac{1}{2}Y(Y(\omega))\\ \frac{1}{2}Y(X(\omega))\end{array}\right),
Cot(T,X)\displaystyle\mathrm{Cot}(T,X) =[𝝎(T),P](010)(X(P)+[𝝎(X),P])(100)\displaystyle=[\boldsymbol{\omega}(T),P]\left(\begin{array}[]{c}0\\ 1\\ 0\end{array}\right)-(X(P)+[\boldsymbol{\omega}(X),P])\left(\begin{array}[]{c}1\\ 0\\ 0\end{array}\right)
=(14X(S)54ωX(ω)12X(Y(ω))12X(ω)div(Y)38ω314Sω12Y(ω)div(Y)12X(X(ω)))\displaystyle=\left(\begin{array}[]{c}\frac{1}{4}X(S)-\frac{5}{4}\omega X(\omega)\\ \frac{1}{2}X(Y(\omega))-\frac{1}{2}X(\omega)\mathrm{div}(Y)\\ \frac{3}{8}\omega^{3}-\frac{1}{4}S\omega-\frac{1}{2}Y(\omega)\mathrm{div}(Y)-\frac{1}{2}X(X(\omega))\end{array}\right)\cdot

The Cotton-York tensor is the Hodge-star of the Cotton tensor:

CY\displaystyle CY =Cot.\displaystyle\mathrel{\hbox to0.0pt{\raisebox{1.29167pt}{$\cdot$}\hss}\raisebox{-1.29167pt}{$\cdot$}}=\star\mathrm{Cot}.

Written as CY=(c1c2c3)CY=\left(\begin{array}[]{ccc}c_{1}&c_{2}&c_{3}\end{array}\right) with columns c1,c2,c3c_{1},c_{2},c_{3}, it is given by

c1\displaystyle c_{1} =(34ω3+12Sω+12div(Y)Y(ω)+12[X(X(ω))+Y(Y(ω))]14Y(S)+54ωY(ω)14X(S)54ωX(ω)),\displaystyle=\left(\begin{array}[]{c}-\frac{3}{4}\omega^{3}+\frac{1}{2}S\omega+\frac{1}{2}\mathrm{div}(Y)Y(\omega)+\frac{1}{2}[X(X(\omega))+Y(Y(\omega))]\\ -\frac{1}{4}Y(S)+\frac{5}{4}\omega Y(\omega)\\ \frac{1}{4}X(S)-\frac{5}{4}\omega X(\omega)\end{array}\right), (57)
c2\displaystyle c_{2} =(54ωY(ω)14Y(S)38ω314Sω12Y(Y(ω))12Y(X(ω))),\displaystyle=\left(\begin{array}[]{c}\frac{5}{4}\omega Y(\omega)-\frac{1}{4}Y(S)\\ \frac{3}{8}\omega^{3}-\frac{1}{4}S\omega-\frac{1}{2}Y(Y(\omega))\\ \frac{1}{2}Y(X(\omega))\end{array}\right), (61)
c3\displaystyle c_{3} =(14X(S)54ωX(ω)12Y(X(ω))38ω314Sω12Y(ω)div(Y)12X(X(ω)))\displaystyle=\left(\begin{array}[]{c}\frac{1}{4}X(S)-\frac{5}{4}\omega X(\omega)\\ \frac{1}{2}Y(X(\omega))\\ \frac{3}{8}\omega^{3}-\frac{1}{4}S\omega-\frac{1}{2}Y(\omega)\mathrm{div}(Y)-\frac{1}{2}X(X(\omega))\end{array}\right)\cdot (65)

Observe that if Ric(T,T)=ω2/2\text{Ric}(T,T)=\omega^{2}/2 is constant, then the Cotton-York tensor simplifies to

c1\displaystyle c_{1} =(34ω3+12Sω14Y(S)14X(S)),\displaystyle=\left(\begin{array}[]{c}-\frac{3}{4}\omega^{3}+\frac{1}{2}S\omega\\ -\frac{1}{4}Y(S)\\ \frac{1}{4}X(S)\end{array}\right),
c2\displaystyle c_{2} =(14Y(S)38ω314Sω0),\displaystyle=\left(\begin{array}[]{c}-\frac{1}{4}Y(S)\\ \frac{3}{8}\omega^{3}-\frac{1}{4}S\omega\\ 0\end{array}\right),
c3\displaystyle c_{3} =(14X(S)038ω314Sω),\displaystyle=\left(\begin{array}[]{c}\frac{1}{4}X(S)\\ 0\\ \frac{3}{8}\omega^{3}-\frac{1}{4}S\omega\end{array}\right),

in which case CY=0CY=0 if and only if

S=3Ric(T,T).\displaystyle S=3\text{Ric}(T,T). (66)

Of related interest is the case when CYCY equals the traceless Ricci tensor,

CY=Ric13Sg,\displaystyle CY=\text{Ric}-\frac{1}{3}Sg, (67)

specifically the case when the scalar curvature SS is constant; see, e.g., [NTC15], where this equality is related to so-called topological massive gravity in dimension 3, and where S=6ΛS=6\Lambda with Λ\Lambda the cosmological constant. We mention here in passing that in the presence of a unit length Killing vector field TT, the condition (67) with SS constant implies that Ric(T,T)=ω22\text{Ric}(T,T)=\frac{\omega^{2}}{2} is also constant. Indeed,

54ωY(ω)\displaystyle\frac{5}{4}\omega Y(\omega)\!\!\! =(57)\displaystyle\overset{\eqref{eqn:Y1}}{=} c12=(67)Ric(T,X)=(15)Y(ω)2,\displaystyle\!\!\!c_{12}\overset{\eqref{eqn:TMG}}{=}\text{Ric}(T,X)\overset{\eqref{eqn:S3}}{=}-\frac{Y(\omega)}{2},
54ωX(ω)\displaystyle-\frac{5}{4}\omega X(\omega)\!\!\! =(57)\displaystyle\overset{\eqref{eqn:Y1}}{=} c13=(67)Ric(T,Y)=(15)X(ω)2,\displaystyle\!\!\!c_{13}\overset{\eqref{eqn:TMG}}{=}\text{Ric}(T,Y)\overset{\eqref{eqn:S3}}{=}\frac{X(\omega)}{2},
12Y(X(ω))\displaystyle\frac{1}{2}Y(X(\omega))\!\!\! =(65)\displaystyle\overset{\eqref{eqn:Y3}}{=} c32=(67)Ric(X,Y)=(14)0,\displaystyle\!\!\!c_{32}\overset{\eqref{eqn:TMG}}{=}\text{Ric}(X,Y)\overset{\eqref{eqn:S2}}{=}0,

together imply that ω\omega is constant, as can be easily verified.

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