# Sufficient conditions for the existence of limiting Carleman weights

Pablo Angulo-Ardoy  Department of Mathematics, Universidad Autónoma de Madrid
<EMAIL_ADDRESS>,  Daniel Faraco  Department of Mathematics, Universidad
Autónoma de Madrid, and ICMAT CSIC-UAM-UCM-UC3M<EMAIL_ADDRESS>and
Luis Guijarro  Department of Mathematics, Universidad Autónoma de Madrid, and
ICMAT CSIC-UAM-UCM-UC3M<EMAIL_ADDRESS>
###### Abstract.

In [1], we found some necessary conditions for a Riemannian manifold to admit
a local limiting Carleman weight (LCW), based upon the Cotton-York tensor in
dimension $3$ and the Weyl tensor in dimension $4$. In this paper, we find
further necessary conditions for the existence of local LCWs that are often
sufficient. For a manifold of dimension $3$ or $4$, we classify the possible
Cotton-York, or Weyl tensors, and provide a mechanism to find out whether the
manifold admits local LCW for each type of tensor. In particular, we show that
a product of two surfaces admits a LCW if and only if at least one of the two
surfaces is of revolution. This provides an example of a manifold satisfying
the eigenflag condition of [1] but not admitting $LCW$.

The authors were supported by research grants MTM2011-22612, MTM2011-28198,
MTM2014-57769-1-P and MTM2014-57769-3-P from the Ministerio de Ciencia e
Innovación (MCINN), by ICMAT Severo Ochoa projects SEV-2011-0087 and
SEV-2015-0554 (MINECO), and by the ERC 301179

P.ANGULO-ARDOY, D. FARACO AND L. GUIJARRO

##  1\. Introduction

Since the 1987 foundational paper of Sylvester and Uhlmann [14] (for more
recent results see [5, 9, 10]), the only effective strategy to solve the
Calderón inverse problem, has been based on the existence of Complex Geometric
Optic solutions, CGO for short. In the Riemannian setting, it was discovered
in [6] that CGO solutions depend on the existence of so called limiting
Carleman weights. The existence of such functions was shown to be a problem in
conformal geometry:

######  Theorem 1.1 ([6, Theorem 1.2]).

If $(M,g)$ is a open manifold having a limiting Carleman weight, then some
conformal multiple of the metric $g$, called $\tilde{g}\in[g]$, admits a
parallel unit vector field. For simply connected manifolds, the converse is
true.

For further developments see [3, 7, 8, 11, 12, 13]. To avoid the simply
connected hypothesis, we will focus on existence of a _local_ LCW at a point
$p$, which is a LCW defined on some neighborhood of $p$.

Since this condition relates to the conformal class, the paper [1] studied
such condition in terms of the classical tensors of conformal geometry, i.e,
the Weyl and Cotton tensors (see section $2$ in [1] for the basics on
algebraic curvature operators and bivectors over a real vector space $V$; we
will stick to the notation from that paper). In [1] we introduced the
following notion:

######  Definition 1.2 ([1]).

Let $W$ be a _Weyl tensor_ in $S^{2}(\Lambda^{2}V)$. We say that $W$ satisfies
the _eigenflag condition_ if and only if there is a vector $v\in V$ such that
$W(v\wedge v^{\perp})\subset v\wedge v^{\perp}$.

By examining the Weyl tensor of metrics with an $\mathbb{R}$-factor, we
obtained in [1] the following obstruction for the existence of LCW’s.

######  Theorem 1.3 ([1]).

Let $(M,g)$ be a Riemannian manifold of dimension $n\geqslant 4$. Assume that
a metric $\tilde{g}\in[g]$ admits a parallel vector field. Then for any $p\in
M$, $W_{p}$ satisfies the eigenflag condition. In particular, for any $p\in
M$, $W_{p}\in S^{2}(\Lambda^{2}(T_{p}^{\ast}M))$ has at least $n-1$ linearly
independent eigenvectors which are simple.

In dimension 3, the obstruction was described in terms of the Cotton-York
tensor $CY$.

######  Theorem 1.4 ([1]).

Let $n=3$. If a metric $\tilde{g}\in[g]$ admits a parallel vector field, then
for any $p\in M$, there is a tangent vector $v\in T_{p}M$ such that

$CY_{p}(v,v)=CY_{p}(w_{1},w_{2})=0$

for any pair of vectors $w_{1},w_{2}\in v^{\perp}$.

Both theorems only gave necessary conditions for the existence of LCW’s. The
results of the current paper study whether the converse to the above results
hold. Specifically, we provide sufficient conditions for the existence of
limiting Carleman weights, and thus CGO solutions, and develop them in detail
in dimensions three and four.

For 3- and 4-dimensional manifolds, the results in this paper and in [1] can
answer whether the manifold admits a limiting Carleman weight, and identify
them if they exist, except for some corner cases that may require some ad hoc
work. We show in section 7 how to deal with these cases.

The proof of our results combines a more precise analysis of the algebraic
structure of the Weyl and Cotton tensor and an analysis of how distributions
in $TM$ get affected by conformal changes in the metric. With a slight abuse
of notation we define

######  Definition 1.5.

$D\subset TM$ is a conformal factor of a metric $g$ if it is a smooth
distribution of constant rank, such that a conformal multiple of $g$ is a
product metric with $D$ and $D^{\perp}$ tangent to the factors.

We prove that for a concrete distribution $D$, the behaviour of the Lie
derivatives or of the Covariant derivatives characterises conformal factors
(see our Theorem 2.4 ). Thus if for a metric $(M,g)$ there are a finite number
of eigenflag conditions the combination of this result for distributions of
rank $1$ and theorems 1.3 and 1.4 answers the question of the existence of LCW

In this way we can analyse all three dimensional manifolds.

######  Theorem 1.6.

Let $(M,g)$ be a 3-dimensional manifold, and $p\in M$ a given point.

  1. (1)

If $det\operatorname{CY}_{p}\neq 0$, or there is a sequence of points $p_{k}$
converging to $p$ such that $det\operatorname{CY}_{p_{k}}\neq 0$, there are no
local LCWs at $p$

  2. (2)

If there is a neighborhood $U$ of $p$ where $\operatorname{CY}\neq 0$ but
$\det\operatorname{CY}=0$, then $U$ admits a LCW if and only if one of the two
one-dimensional distributions of eigenflag directions for $\operatorname{CY}$
satisfies one of the equivalent conditions in Theorem 2.4

  3. (3)

If there is a neighborhood $U$ of $p$ where $\operatorname{CY}|_{U}\equiv 0$,
then the metric is conformally flat in $U$, and it admits the same LCWs as a
subset of $\mathbb{R}^{3}$.

The four dimensional case is contained in the following Theorem; the
definition of the types of Weyl tensors is deferred to Lemma 1.9.

######  Theorem 1.7.

Let $(M,g)$ be a 4-dimensional manifold and $p\in M$.

  * •

If $W$ is of type $A$ at $p$, or there is a sequence of points with Weyl
tensors of type $A$ converging to $p$, there are not local LCWs at $p$.

  * •

If $W$ is of type B at all points in a neighborhood of $p$, then $M$ admits a
local LCW at $p$ if and only if at least one of the four $1$-dimensional
distributions defined by eigenflags as in Lemma 1.9 satisfies the hypothesis
of Theorem 2.4.

  * •

If $W$ is of type C at all points in a neighborhood of $p$, and the two
complementary distributions of eigenflag directions satisfy the hypothesis of
Theorem 2.4, then $M$ is locally conformal to a product of surfaces, and it
admits a local LCW if and only if at least one of the two integral factors is
a surface of revolution.

  * •

If $W$ is of type D at all points in a neighborhood of $p$, then it is locally
conformally flat at $p$, and admits the same LCWs as a subset of
$\mathbb{R}^{4}$.

To write a concrete example of a manifold with an eigenflag vector field, but
without any local LCW (showing that the necessary condition from theorem 1.3
is not sufficient) recall that _an ellipsoid_ in $\mathbb{R}^{3}$ is _scalene_
if its three axis have different lengths.

######  Theorem 1.8.

Let $(M,g)=(S_{1},g_{1})\times(S_{2},g_{2})$ where $S_{1}$ and $S_{2}$ are two
scalene ellipsoids with the metric induced by $\mathbb{R}^{3}$. Then any open
subset of $(M,g)$ satisfies the eigenflag condition but does not admit any
LCW.

The paper starts in Section 2 by considering a metric that becomes a product
after a conformal change. The 3-dimensional case is examined in Section 3.
Here Theorem 1.4 says that there are only a finite number of directions than
can be factors of a product after conformal change; we analyze them by means
of Theorem 2.4. This will provide a proof of Theorem 1.6.

Section 4 classifies the possible algebraic Weyl tensors that may arise in a
$4$-dimensional manifold in terms of its eigenflag directions and eigenspaces.
This allows us to identify all the eigenflag directions.

######  Lemma 1.9.

The algebraic Weyl operators $W$ in a vector space of dimension $4$ fall into
one of the following types:

A:

$W$ has no eigenflag directions.

B:

$W$ has at least one eigenflag direction and three different eigenspaces of
dimension $2$. In this case, $W$ has exactly four eigenflag directions.

C:

$W$ has at least one eigenflag direction and two eigenspaces with dimensions
four and two. In this case, the eigenflag directions for $W$ consist of the
union of two orthogonal 2-planes.

D:

$W$ is null. All directions are eigenflag.

We use this information in Section 5; as explained above, Theorem 2.4 suffices
to deal with case B. In Section 6, we give an interesting example where three
of the four eigenflag directions are conformal factors. This is a manifold
with three LCWs with orthogonal level sets, that is not conformal to a product
of surfaces.

We start Section 5 by proving that product metrics of surfaces have Weyl
operators of Types C and D; then we show that if such a product of surfaces
admits a LCW and its Weyl operator is of type C (not the trivial conformally
flat case), we can choose coordinates where the metric assumes a specially
nice form. As a result, we are able to prove

######  Theorem 1.10.

Let $(S_{1},g_{1})$ and $(S_{2},g_{2})$ be open subsets of $\mathbb{R}^{2}$
with Riemannian metrics. Assume that the Weyl operator of the product metric
does not vanish at any point. The following are equivalent:

  * •

$(S_{1},g_{1})$ is locally isometric to a surface of revolution

  * •

$(S_{1},g_{1})$ has a non-trivial Killing vector field

  * •

$(S_{1}\times S_{2},g_{1}\times g_{2})$ admits a LCW that is everywhere
tangent to the first factor

Hence Theorem 1.8 is a corollary of this, as scalene ellipsoids are not
surfaces of revolution and their product satisfy the condition on the Weyl
operator.

We are left with the case where a manifold may have a Weyl tensor of type C,
but not be conformal to a product of surfaces. Therefore in principle there
are many candidates to be one dimensional conformal factors. In section 7, we
show that this is indeed possible, and show to deal with this situation in an
specific example and explain how to proceed in general.

Finally, we would like to point out that, in principle, a similar analysis
might be conducted in higher dimensions though satisfying the eigenflag
condition is more rare (see [1, Theorem 6.1] for a quantitative statement in
this regard), and the analysis is bound to become much more complicated.

##  2\. Criteria for a conformal product

Suppose we are given a Riemannian manifold $(M,g)$ with a vector distribution
$D\subset TM$ such that both $D$ and its orthogonal complement $D^{\perp}$ are
integrable. The main result of this section establishes criteria that will
assure that $M$ is locally conformal to a product metric with $D$ and
$D^{\perp}$ tangent to the factors. In order to state them, we will need some
notation.

The orthogonal splitting $TM=D\oplus D^{\perp}$ induces bundle projections

$P_{D}:TM\to D,\quad P_{D^{\perp}}:TM\to D^{\perp};$

we will alternatively denote by $X^{D}$ and $X^{\perp}$ to the components of a
vector field under the above splitting.

Given a metric tensor $g$ in $M$, we denote by $g^{D}$ and $g^{\perp}$ the
restrictions of $g$ to $D$ and $D^{\perp}$ respectively, that is

$g^{D}(X,Y):=g(P_{D}(X),P_{D}(y)),\quad
g^{\perp}(X,Y):=g(P_{D^{\perp}}(X),P_{D^{\perp}}(y)).$

It is clear that

$g(X,Y)=g^{D}(X,Y)+g^{\perp}(X,Y).$

for any pair of vector fields $X$, $Y$.

######  Lemma 2.1.

Assume $D$, $D^{\perp}$ are integrable distributions of dimensions $d$ and
$n-d$ respectively. Then for any point $p\in M$, there is a coordinate chart
$(U,\phi)$ with $\phi=(x_{1},\dots,x_{n})$ such that

  1. (1)

$\phi(p)=0$, $\phi(U)=(-1,1)^{n}$;

  2. (2)

the integral manifolds for $D$ in $U$ are given by equations

$(x^{d+1},\dots,x^{n})=(a_{d+1},\dots,a_{n});$

  3. (3)

the integral manifolds for $D^{\perp}$ in $U$ are given by equations

$(x^{1},\dots,x^{d})=(a_{1},\dots,a_{d}).$

###### Proof.

Frobenius theorem gives us charts $(U_{1},\phi_{1})$ and $(U_{2},\phi_{2})$,
with $\phi_{1}=(y_{1},\dots,y_{n})$, $\phi_{2}=(z_{1},\dots,z_{n})$,
$\phi_{1}(p)=\phi_{2}(p)=0$, and such that the integral submanifolds for $D$
and $D^{\perp}$ are given respectively by the equations
$(y_{d+1},\dots,y_{n})=(b_{d+1},\dots,b_{n})$ and
$(z_{1},\dots,z_{d})=(c_{d+1},\dots,c_{n})$ for constants $b_{i}$, $c_{j}$.
The map $\phi=(z_{1},\dots,z_{d},y_{d+1},\dots,y_{n})$, defined in a
neighborhood $U$ of $p$, is a local diffeomorphism at $p$ with $\phi(p)=0$,
thus it defines a coordinate chart in some neighborhood of $p$. Taking a
smaller neighborhood if needed, and with the help of a linear change of
coordinates in $\mathbb{R}^{n}$, we can assume that its image is $(-1,1)^{n}$
as required. ∎

It is clear from the Lemma that, if $N$ and $N^{\perp}$ are the integral
manifolds for $D$ and $D^{\perp}$ through $p$ in $U$, then $U$ is
diffeomorphic to $N\times N^{\perp}$. The aim of this section is to find
conditions on a metric $g$ in $U$ such that $(\phi^{-1})^{*}g$ is conformal to
a product metric on $N\times N^{\perp}$. In order to do this, we need to
introduce the following $1$–form $\Phi$ in $M$:

(1)
$\Phi(X):=\operatorname{Tr}_{g}\left(\frac{1}{d}\mathcal{L}_{X^{\perp}}g^{D}+\frac{1}{n-d}\mathcal{L}_{X^{D}}g^{\perp}\right)$

Although the Lie derivative of some tensor $T$,
$X\rightarrow\mathcal{L}_{X}T$, is not tensorial in $X$, we have that

(2)
$\mathcal{L}_{fX^{\perp}}g^{D}(Y,Z)=f\mathcal{L}_{X^{\perp}}g^{D}(Y,Z)-g^{D}(YfX^{\perp},Z)-g^{D}(Y,ZfX^{\perp})=\\\
=f\mathcal{L}_{X^{\perp}}g^{D}(Y,Z)$

and similarly for $\mathcal{L}_{X^{D}}g^{\perp}$, so $\Phi$ is effectively a
1-form.

######  Definition 2.2.

A distribution $D$ is said to be _umbilical_ if there exists a vector field
$H\in D^{\perp}$, called the _mean curvature vector field of $D$_, such that
for $X,Y\in D$ and $Z\in D^{\perp}$ it holds that

(3) $\langle\nabla_{X}Y,Z\rangle=\langle X,Y\rangle\langle Z,H\rangle$

The relation of umbilicity to LCWs was already noted in [6].

######  Remark 2.3.

If $D$ is an arbitrary smooth distribution, then we can define the second
fundamental form of the distribution

$II(X,Y)=P_{D^{\perp}}(\nabla_{X}Y)$

where $P_{D^{\perp}}$ is the projection onto $D^{\perp}$ and $X$, $Y$, are
vector fields tangent to $D$. It is easy to check that the distribution is
integrable if and only this form is symmetric, and in particular, an umbilical
distribution is integrable.

######  Theorem 2.4.

Let $(M,g)$ be a Riemannian metric, and $D$, $D^{\perp}$ integrable
distributions as above.

The following are equivalent:

  1. (1)

$g$ is locally conformal to the product of the metric restricted to an
integral leave of $D$ and the metric restricted to an integral leave of
$D^{\perp}$;

  2. (2)

the Lie derivative of $g^{D}$ with respect to any vector field in $D^{\perp}$
is a multiple of $g^{D}$, the Lie derivative of $g^{\perp}$ with respect to
any vector field in $D$ is a multiple of $g^{\perp}$, and the 1-form $\Phi$ is
closed;

  3. (3)

the distributions $D$ and $D^{\perp}$ are umbilic and, if $H_{1}$ and $H_{2}$
are the respective mean curvature vector fields, $H_{1}+H_{2}$ is a _gradient
vector field_ for the metric $g$.

###### Proof.

$\text{\ref{uno}}\implies\text{\ref{dos}}$ Let $\tilde{g}=e^{\alpha}g$ be a
product metric of two metrics on the leaves of $D$ and $D^{\perp}$. Using the
chart $(U,\phi)$, there are symmetric tensors in $U$ such that $\tilde{g}$ is
written as

$\tilde{g}=\left(\begin{array}[]{cc}e^{\alpha}g_{1}&\\\
&e^{\alpha}g_{2}\end{array}\right)$

where the first block does not depend on the coordinates of the second factor
and vice versa. This is equivalent to

$\mathcal{L}_{X^{\perp}}(e^{\alpha}g_{1})=\mathcal{L}_{X^{D}}(e^{\alpha}g_{2})=0,$

and also to

$\displaystyle\mathcal{L}_{X^{\perp}}(g_{1})$ $\displaystyle=$
$\displaystyle-X^{\perp}(\alpha)g_{1}$
$\displaystyle\mathcal{L}_{X^{D}}(g_{2})$ $\displaystyle=$
$\displaystyle-X^{D}(\alpha)g_{2}$

Thus the first part of (2) is immediate, and the second follows because

$\Phi(X)=-X(\alpha)=-d\alpha(X).$

$\text{\ref{dos}}\implies\text{\ref{uno}}$ $D$ is integrable since, for
$X,Y\in D$ and $Z\in D^{\perp}$:

$\displaystyle g^{\perp}([X,Y],Z)$ $\displaystyle=$
$\displaystyle-(\mathcal{L}_{X}g^{\perp})(Y,Z)+X(g^{\perp}(Y,Z))-g^{\perp}(Y,[X,Z])$
$\displaystyle=$ $\displaystyle-cg^{\perp}(Y,Z)$ $\displaystyle=$
$\displaystyle 0$

where we are using that $\mathcal{L}_{X}g^{\perp}=cg^{\perp}$ for some
function $c$. The integrability of $D^{\perp}$ is proved similarly.

$U$ is simply connected, thus $\Phi$ being closed yields some function
$\alpha$ such that $\Phi=-d\alpha$. For an arbitrary vector field $X$, the
hypothesis in part (2) gives some function $c$ such that
$\mathcal{L}_{X^{\perp}}g^{D}=cg^{D}$; taking traces in both sides, and using
that the dimension of $D$ is $d$, we get that

$c=\frac{1}{d}\operatorname{Tr}_{g}\left(\mathcal{L}_{X^{\perp}}g^{D}\right)=\Phi(X^{\perp})=-d\alpha(X^{\perp})=-X^{\perp}(\alpha),$

thus $\mathcal{L}_{X^{\perp}}g^{D}+X^{\perp}(\alpha)=0$, and a simple
computation yields $\mathcal{L}_{X^{\perp}}(e^{\alpha}g^{D})=0$. A similar
procedure gives us $\mathcal{L}_{X^{D}}(e^{\alpha}g^{\perp})=0$, so the metric
$\tilde{g}=e^{\alpha}g$ is a product of two metrics on the leaves of $D$ and
$D^{\perp}$.

$\text{\ref{uno}}\implies\text{\ref{tres}}$ Let $\tilde{g}=e^{\alpha}g$ be a
product of two metrics on the leaves of $D$ and $D^{\perp}$ as before. A
simple formula relates the Levi-Civita connections of $g$ and $\tilde{g}$:

(4) $\nabla_{X}Y=\tilde{\nabla}_{X}Y+g(X,Y)U-g(X,U)Y-g(Y,U)X$

where $U=-\nabla\alpha$. Thus for $X,Y\in D$ and $Z,W\in D^{\perp}$,

$\begin{array}[]{lll}g(\nabla_{X}Y,Z)&=&g(X,Y)g(U,Z)\end{array}$

so the projection of $U$ onto $D^{\perp}$ is the mean curvature of $D$ and
vice versa. It follows that $H_{1}+H_{2}=U$, which is a gradient.

$\text{\ref{tres}}\implies\text{\ref{uno}}$ Suppose
$H_{1}+H_{2}=-\nabla\alpha$, and define $\tilde{g}=e^{-\alpha}g$. Equation (4)
shows that

$\begin{array}[]{ll}g(\tilde{\nabla}_{X}Y,Z)&=\end{array}0$
$\begin{array}[]{ll}g(\tilde{\nabla}_{Z}W,X)&=\end{array}0$

for $X,Y\in D$ and $Z,W\in D^{\perp}$.

It follows from equation (2.2) that umbilic distributions are integrable, so
by Lemma 2.1, we can find a coordinate basis adapted simultaneously to $D$ and
$D^{\perp}$, i.e. $\\{\partial_{1},\ldots,\partial_{d}\\}$ span $D$ and
$\\{\partial_{d+1},\ldots,\partial_{n}\\}$ span $D^{\perp}$. Then for
$i\in\\{1\ldots d\\}$, $j,k\in\\{d+1\ldots n\\}$:

$\displaystyle\partial_{i}\tilde{g}_{jk}$ $\displaystyle=$
$\displaystyle\partial_{i}(\tilde{g}(\partial_{j},\partial_{k}))$
$\displaystyle=$
$\displaystyle\tilde{g}(\tilde{\nabla}_{\partial_{i}}\partial_{j},\partial_{k})+\tilde{g}(\partial_{j},\tilde{\nabla}_{\partial_{i}}\partial_{k})$
$\displaystyle=$
$\displaystyle-\tilde{g}(\tilde{\nabla}_{\partial_{j}}\partial_{i},\partial_{k})-\tilde{g}(\partial_{j},\tilde{\nabla}_{\partial_{k}}\partial_{i})-\tilde{g}([\partial_{i},\partial_{j}],\partial_{k})-\tilde{g}([\partial_{i},\partial_{k}],\partial_{j})$
$\displaystyle=$
$\displaystyle-\partial_{j}(\tilde{g}(\partial_{i},\partial_{k}))+\tilde{g}(\partial_{i},\tilde{\nabla}_{\partial_{j}}\partial_{k})-\partial_{k}(\tilde{g}(\partial_{i},\partial_{j}))+\tilde{g}(\partial_{i},\tilde{\nabla}_{\partial_{k}}\partial_{j})$
$\displaystyle=$ $\displaystyle 0$

so $\tilde{g}$ is a product metric. ∎

Related conditions can be found in the literature. For example, [15]
investigates the case of warped products. However we have preferred to keep
our criteria as simple as possible.

##  3\. Conformal factors in dimension three

This section combines the results of the previous section with an examination
of the Cotton-York tensor of a metric $g$ to describe when a 3-dimensional
manifold has a LCW. We refer the reader to [1] for some background on the
Cotton-York tensor. Recall that the space of algebraic Cotton-York tensors at
some given point $p\in M$ coincides with the space of traceless symmetric
operators in $T_{p}M$.

######  Definition 3.1.

An eigenflag direction of a traceless symmetric operator in a three-
dimensional euclidean space $V$ is a one-dimensional vector subspace $L$ such
that for any $v\in L$ and $w_{1},w_{2}\in L^{\perp}$, we have

$CY_{p}(v,v)=CY_{p}(w_{1},w_{2})=0$

Suppose we are given a metric $g$ in $M$; Theorem 1.6 in [1] shows that if a
conformal metric $\tilde{g}$ admits a parallel vector field, the subspace $L$
that generates is an eigenflag direction of the Cotton-York tensor of $g$ at
each point of $M$.

We start by classifying the possible algebraic Cotton-York tensors in terms of
their eigenflag directions.

######  Lemma 3.2.

An algebraic Cotton-York tensor $\operatorname{CY}$ falls into one of the
following categories:

  * •

Every direction in $V$ is eigenflag for $\operatorname{CY}$; this agrees with
the case when $\operatorname{CY}$ is null.

  * •

There are exactly two eigenflag directions for $\operatorname{CY}$; this
agrees with the case when $\operatorname{CY}$ is not null and
$\det(\operatorname{CY})=0$.

  * •

There are not eigenflag directions; this is equivalent to the case
$\det(\operatorname{CY})\neq 0$.

###### Proof.

Lemma 5.1 in [1] shows that $\det(\operatorname{CY})=0$ if and only if
$\operatorname{CY}$ has an eigenflag direction, so assume that $L$ is an
eigenflag direction for $\operatorname{CY}$. If $v$ is a unit vector along
$L$, and $\\{v,w_{1},w_{2}\\}$ is an orthonormal basis of $V$, the matrix of
$\operatorname{CY}$ is given by

$\left(\begin{array}[]{ccc}0&a&b\\\ a&0&0\\\ b&0&0\end{array}\right)$

(see Lemma 1.7 in [1]).

A further rotation with axis $L$ changes the matrix of $\operatorname{CY}$ to
the form

$\left(\begin{array}[]{ccc}0&c&0\\\ c&0&0\\\ 0&0&0\end{array}\right)$

If $\operatorname{CY}$ is not zero, the null directions for this symmetric
operator is the union of the planes $\\{x=0\\}$ and $\\{y=0\\}$. Thus only the
orthogonals to those two planes can be eigenflag directions.

∎

###### Proof of Theorem 1.6.

In the first case, the metric is conformally flat; in the second it follows
from Lemma 3.2 and Theorem 2.4. ∎

##  4\. Classification of algebraic Weyl tensors of $4$-manifolds

We would like to carry out a similar analysis for 4-dimensional manifolds to
the one we did in Section 3; however, the extra dimension brings out a higher
complexity, even at the level of algebraic Weyl operators. So in this section,
we will start by classifying such operators with respect to the size of their
set of eigenflag directions.

###### Proof of Lemma 1.9.

Let $W$ be a Weyl operator with an eigenflag direction $L=\langle v\rangle$.
The operator $W|_{\langle v\wedge v^{\perp}\rangle}$ is symmetric, and
diagonalizes in an orthonormal basis $v\wedge e_{2},v\wedge e_{3},v\wedge
e_{4}$. Define $e_{1}=v$, and $e_{ij}=e_{i}\wedge e_{j}$. It follows from the
properties of the Weyl operator that $e_{23}$, $e_{24}$ and $e_{34}$ are
eigenvectors of $W$ with the same eigenvalues as $e_{14}$, $e_{13}$, $e_{12}$
respectively (see the proof of Theorem 6.1 in [1]). Thus $W$ diagonalizes as

(5) $W=\left(\begin{array}[]{cccccc}\lambda_{12}&&&&&\\\ &\lambda_{13}&&&&\\\
&&\lambda_{14}&&&\\\ &&&\lambda_{12}&&\\\ &&&&\lambda_{13}&\\\
&&&&&\lambda_{14}\end{array}\right)$

Recall that $W$ is traceless, so $\lambda_{12}+\lambda_{13}+\lambda_{14}=0$.
If the three numbers $\lambda_{12},\lambda_{13},\lambda_{14}$ are different,
the operator has exactly three eigenspaces, each of dimension $2$. If two
numbers coincide, there is one eigenspace of dimension $4$ and a second
eigenspace of dimension $2$. Finally, the remaining possibility is that the
Weyl operator vanishes. This will account for the different possibilities in
the statement of the Theorem, once we have related them to the eigenflags. We
will do this case by case.

Three different eigenvalues. Suppose
$\lambda_{12}\neq\lambda_{13}\neq\lambda_{14}$; then the eigenspace for
$\lambda_{12}$ is the set of bivectors of the form $ae_{12}+be_{34}$,
$a,b\in\mathbb{R}$. The Plücker relations imply that such bivector is simple
only when $ab=0$. In other words, the only simple bivectors in the eigenspace
for $\lambda_{12}$ are the multiples of either $e_{12}$ or of $e_{34}$.
Changing $i,j$ we get that every simple eigenvector of $W$ is a multiple of
some $e_{ij}$, and consequently each one of the $e_{i}$’s is an eigenflag
direction.

Suppose $v$ were a vector spanning a different eigenflag direction. Then
$v\wedge v^{\perp}$ would be an eigenspace of $W$ consisting of simple
bivectors, so there should be three orthogonal unit vectors $w_{k}\in
v^{\perp}$ such that $v\wedge w_{k}$ are eigenvectors for $W$. It follows that
$v\wedge w_{1}=e_{ij}$ for some $i,j$, which implies that
$v,w_{1}\in\operatorname{span}(e_{i},e_{j})$, and we can assume without lost
of generality that $i=1,j=2$. Then
$w_{2}\in(v,w_{1})^{\perp}=\operatorname{span}(e_{3},e_{4})$, and $v\wedge
w_{2}$ can only be a coordinate $2$-plane if $v=e_{1}$ or $v=e_{2}$.

Two different eigenvalues. Suppose
$\lambda_{12}=\lambda_{13}\neq\lambda_{14}$. Let $v=ae_{2}+be_{3}$ be any
vector in $\operatorname{span}(e_{2},e_{3})$. Then

$v\wedge e_{1}=-ae_{12}-be_{13},\quad v\wedge e_{4}=ae_{24}+be_{34},$

are eigenvectors of eigenvalue $\lambda_{12}$, while

$v\wedge(-be_{2}+ae_{3})=(ae_{2}+be_{3})\wedge(-be_{2}+ae_{3})=(a^{2}+b^{2})e_{23}$

is an eigenbivector of eigenvalue $\lambda_{14}$. Therefore $v\wedge
v^{\perp}$ is an invariant subspace, and $v$ is an eigenflag direction. A
similar argument applies to any vector in $\operatorname{span}(e_{1},e_{4})$
to show that it is an eigenflag direction, hence we only need to prove that
there are no additional eigenflag directions.

A general bivector in the $\lambda_{12}$-eigenspace has the form
$ae_{12}+be_{34}+ce_{13}+de_{24}$, and it is simple when

$ab-cd=0$

This equation defines a $3$-dimensional quadric in a $4$ dimensional space,
and does not contain any linear space of dimension $3$. If $v$ is an eigenflag
direction, $v\wedge v^{\perp}$ is an invariant subspace, and the restriction
of $W$ to $v\wedge v^{\perp}$ diagonalizes in subspaces of the eigenspaces for
$W$. Thus, $v\wedge v^{\perp}$ must intersect the eigenspace associated to
$\lambda_{14}$. The intersection is spanned by a bivector $v\wedge w$, for
some $w\in v^{\perp}$, but the only simple eigenbivectors in the eigenspace
associated to $\lambda_{14}$ are $e_{14}$ and $e_{23}$. If $v\wedge w=e_{14}$,
this implies that $v\in\operatorname{span}(e_{1},e_{4})$, while $v\wedge
w=e_{23}$ implies $v\in\operatorname{span}(e_{2},e_{3})$. ∎

##  5\. Conformal factors in dimension four

A 4-dimensional manifold may be conformal to a product in two different ways:
a product $\mathbb{R}\times N$, with $N$ a 3-dimensional manifold, or as a
product of two surfaces. Since we are going to use the Weyl tensor to
distinguish between the two cases, we will start by computing the Weyl
operator of a product of two surfaces.

######  Lemma 5.1.

Let $M_{1}$ and $M_{2}$ be two Riemannian surfaces.

  1. (1)

The Weyl operator of $M_{1}\times M_{2}$ has type $C$ or $D$ at any point.

  2. (2)

It has type $D$ at every point if and only if both surfaces have constant
curvature $s_{1}$ and $s_{2}$, and $s_{1}+s_{2}=0$.

  3. (3)

If it has type $C$ at a point $p$, then the two planes of eigenflag directions
of $W_{p}$ of Lemma 1.9 are the distributions tangent to the factors $M_{1}$
and $M_{2}$ at $p$.

###### Proof.

Taking isothermal coordinates $(t,x)$ on $M_{1}$ and $(y,z)$ on $M_{2}$, the
product metric is written as

$g=\left(\begin{array}[]{cccc}f(t,x)&&&\\\ &f(t,x)&&\\\ &&h(y,z)&\\\
&&&h(y,z)\end{array}\right).$

Denote the normalized coordinate fields as

$\hat{\partial_{t}}=\frac{1}{\sqrt{f}}\partial_{t},\quad\hat{\partial_{x}}=\frac{1}{\sqrt{f}}\partial_{x},\quad\hat{\partial_{y}}=\frac{1}{\sqrt{h}}\partial_{y},\quad\hat{\partial_{z}}=\frac{1}{\sqrt{h}}\partial_{z}.$

In the basis of $\Lambda^{2}T_{p}M$

$\hat{\partial}_{t}\wedge\hat{\partial}_{x},\quad\hat{\partial}_{t}\wedge\hat{\partial}_{y},\quad\hat{\partial}_{t}\wedge\hat{\partial}_{z},\quad\hat{\partial}_{y}\wedge\hat{\partial}_{z},\quad\hat{\partial}_{x}\wedge\hat{\partial}_{z},\quad\hat{\partial}_{x}\wedge\hat{\partial}_{y},$

the Weyl operator is diagonal, with

(6) $\begin{array}[]{c}\lambda\mathop{=}\limits^{\operatorname{def}}\langle
W(\hat{\partial}_{t}\wedge\hat{\partial}_{x}),\hat{\partial}_{t}\wedge\hat{\partial}_{x}\rangle=\langle
W(\hat{\partial}_{y}\wedge\hat{\partial}_{z}),\hat{\partial}_{y}\wedge\hat{\partial}_{z}\rangle=\\\
\\\
\dfrac{h_{y}^{2}+h_{z}^{2}-hh_{yy}-hh_{zz}}{6h^{3}}+\dfrac{f_{t}^{2}+f_{x}^{2}-ff_{tt}-ff_{xx}}{6f^{3}}\end{array}$

and

(7) $\begin{array}[]{c}\langle
W(\hat{\partial}_{t}\wedge\hat{\partial}_{y}),\hat{\partial}_{t}\wedge\hat{\partial}_{y}\rangle=\langle
W(\hat{\partial}_{t}\wedge\hat{\partial}_{z}),\hat{\partial}_{t}\wedge\hat{\partial}_{z}\rangle=\\\
\\\ \langle
W(\hat{\partial}_{x}\wedge\hat{\partial}_{y}),\hat{\partial}_{x}\wedge\hat{\partial}_{y}\rangle=\langle
W(\hat{\partial}_{x}\wedge\hat{\partial}_{z}),\hat{\partial}_{x}\wedge\hat{\partial}_{z}\rangle=\\\
\\\
-\dfrac{h_{y}^{2}+h_{z}^{2}-hh_{yy}-hh_{zz}}{12h^{3}}-\dfrac{f_{t}^{2}+f_{x}^{2}-ff_{tt}-ff_{xx}}{12f^{3}}=-\lambda/2\end{array}$

Thus $W$ falls into type $C$ or $D$ depending on whether $\lambda$ is
different or equal to zero.

We recognize

$\lambda(t,x,y,z)=\frac{-1}{3}\left(s_{1}(t,x)+s_{2}(y,z)\right),$

where $s_{i}$ is the Gaussian curvature of $M_{i}$. Thus if $\lambda$ vanishes
identically, $s_{1}$ and $s_{2}$ are global constants whose sum is $0$.

∎

We are now ready to distinguish those metric products of surfaces that admit,
simultaneously, a LCW.

######  Lemma 5.2.

Let $(M,g)$ be a product of surfaces that admits a limiting Carleman weight
$\varphi$. Let $p\in M$ with $W_{p}\neq 0$; then there are coordinates
$(t,x,y,z)$ around $p$ in which $g$ is written as

(8) $g=\left(\begin{array}[]{cccc}e^{J(x)}&&&\\\ &e^{J(x)}&&\\\
&&e^{K(y,z)}&\\\ &&&e^{K(y,z)}\end{array}\right)$

for some functions $J=J(x)$, $K=K(y,z)$.

###### Proof.

Since $W_{p}\neq 0$, Lemma 5.1 implies that $W$ is of type $C$ in some
neighborhood of $p$, and the set of eigenflag directions for the Weyl tensor
of $g$ and $\tilde{g}$ is the union of the two planes
$\langle\partial_{t},\partial_{x}\rangle$ and
$\langle\partial_{y},\partial_{z}\rangle$.

Choose some coordinate system $(t,x,y,z)$ adapted to the product structure
$M_{1}\times M_{2}$, as in the proof of Lemma 5.1.

Let $\tilde{g}$ be the rescaled metric

$\tilde{g}=|\nabla\varphi|^{2}g.$

Lemmas 3.10 and 3.11 in [12] give that the vector field
$A=\tilde{\nabla}\varphi$ is a parallel vector field for $\tilde{g}$. Theorem
1.3 in [1] shows that $\langle A\rangle$ must be an eigenflag direction.

Without loss of generality, we will assume that $A$ is in
$\langle\partial_{t},\partial_{x}\rangle$ at every point, hence there are
functions $\alpha,\beta$ such that

$A=\alpha\partial_{t}+\beta\partial_{x}.$

It follows from the definition of parallel vector field that $\Delta$, the
distribution orthogonal to $A$, is integrable (see the proof of lemma 3.12 in
page 57 of [12]). It is clear that

$\Delta=\left<\,-\beta\partial_{t}+\alpha\partial_{x},\partial_{y},\partial_{z}\,\right>.$

Denote $B:=-\beta\partial_{t}+\alpha\partial_{x}$; the integrability condition
implies that

$\left[B,\partial_{y}\right]=(\partial_{y}\beta)\partial_{t}-(\partial_{y}\alpha)\partial_{x}\in\Delta$

and therefore $\left[B,\partial_{y}\right]$ must be in the $B$ direction, i.e,
$\left[B,\partial_{y}\right]=\lambda_{1}B$ for some function $\lambda_{1}$;
thus

$\partial_{y}\beta=-\lambda_{1}\cdot\beta,\quad\partial_{y}\alpha=-\lambda_{1}\cdot\alpha$

Looking at $\left[B,\partial_{z}\right]$, we get

$\partial_{z}\beta=-\lambda_{2}\cdot\beta,\quad\partial_{z}\alpha=-\lambda_{2}\cdot\alpha$

Next, observe that any integral submanifold of $\Delta$ is foliated by
surfaces tangent to $\langle\partial_{y},\partial_{z}\rangle$; since $A$ is a
parallel vector field, its integral curves are given by geodesics. Therefore,
if we consider the maps

$\phi_{(t_{0},x_{0})}(u,v,w)=\exp_{(t_{0},x_{0},v,w)}\left(uA\right),$

we see that for each $(t_{0},x_{0})$, the vector fields $\partial_{u}$,
$\partial_{v}$, $\partial_{w}$ are mapped by the differential of
$\phi_{(t_{0},x_{0})}$ to $A$, $\partial_{y}$ and $\partial_{z}$, and
therefore

$\left[A,\partial_{y}\right]=\left[A,\partial_{z}\right]=0.$

On the other hand, writing $A=\alpha\partial_{t}+\beta\partial_{x}$, we would
get

$\left[A,\partial_{y}\right]=-(\partial_{y}\alpha)\partial_{t}-(\partial_{y}\beta)\partial_{x},\quad\left[A,\partial_{z}\right]=-(\partial_{z}\alpha)\partial_{t}-(\partial_{z}\beta)\partial_{x}.$

The consequence of this is that $\lambda_{1}=\lambda_{2}=0$, and every Lie
bracket between elements in the basis $\\{A,B,\partial_{y},\partial_{z}\\}$
vanishes, and hence they form a coordinate basis for some set of coordinates
about $p$. We will keep on denoting them $(t,x,y,z)$, although only the last
two would coincide with the former. The first two, $(t,x)$, will still,
however, parametrize the first factor of the product structure of $M$. In this
chart, $\tilde{g}$ is written as

(9) $\tilde{g}=\left(\begin{array}[]{cccc}1&&&\\\ &1&&\\\ &&b(t,x,y,z)&\\\
&&&b(t,x,y,z)\end{array}\right)$

where $b$ is actually $h(y,z)|\nabla\varphi|^{2}(t,x,y,z)$.

Because $A$ is parallel, it follows from Koszul formula

(10)
$2g(\nabla_{X}Y,Z)=\partial_{X}(g(Y,Z))+\partial_{Y}(g(X,Z))-\partial_{Z}(g(X,Y))\\\
+g([X,Y],Z)-g([X,Z],Y)-g([Y,Z],X),$

that $b$ does not depend on $t$ (taking $X=A$, $Y=Z=\partial_{y}$).

The metric $\tilde{g}$ is conformal to a product where the factors are tangent
to $\langle\partial_{t},\partial_{x}\rangle>=\langle A,B\rangle$ and
$\langle\partial_{y},\partial_{z}\rangle$. Let
$\hat{\partial}_{y}=\frac{1}{\sqrt{b}}\partial_{y}$ be a unit vector in the
direction of $\partial_{y}$. We compute the Weyl operator in the basis

$A\wedge B,\quad A\wedge\hat{\partial}_{y},\quad
A\wedge\hat{\partial}_{z},\quad\hat{\partial}_{y}\wedge\hat{\partial}_{z},\quad
B\wedge\hat{\partial}_{z},\quad B\wedge\hat{\partial}_{y}.$

The $6\times 6$ matrix of $W$ in this basis has the block structure

$W=\left(\begin{array}[]{cc}W_{1}&0\\\ 0&W_{1}\end{array}\right)$

where the $3\times 3$ block $W_{1}$ is

$\begin{pmatrix}-\Psi&\frac{bb_{xy}-b_{x}b_{y}}{4\,b^{5/2}}&\frac{bb_{xz}-b_{x}b_{z}}{4\,b^{5/2}}\\\
\frac{bb_{xy}-b_{x}b_{y}}{4\,b^{5/2}}&\Psi/2&0\\\
\frac{bb_{xz}-b_{x}b_{z}}{4\,b^{5/2}}&0&\Psi/2\end{pmatrix}$

where
$\Psi=\frac{bb_{x}^{2}-b^{2}b_{xx}-b_{y}^{2}+bb_{yy}-b_{z}^{2}+bb_{zz}}{6\,b^{3}}$.

As mentioned in the proof of Lemma 5.1, if the metric is a product of surfaces
tangent to $\langle\partial_{1},\partial_{2}\rangle$ and
$\langle\partial_{3},\partial_{4}\rangle$, the off-diagonal elements in the
corresponding basis of $\Lambda^{2}(T_{p}M)$ must vanish at every $p$. Thus:

$\displaystyle bb_{xy}$ $\displaystyle=$ $\displaystyle b_{x}b_{y}$
$\displaystyle bb_{xz}$ $\displaystyle=$ $\displaystyle b_{x}b_{z}$

Integrating the first equation we get:

$\displaystyle\frac{(b_{x})_{y}}{b_{x}}$ $\displaystyle=$
$\displaystyle\frac{b_{y}}{b}$ $\displaystyle\log(b_{x})$ $\displaystyle=$
$\displaystyle\log(b)+C(x,z)$ $\displaystyle\frac{b_{x}}{b}$ $\displaystyle=$
$\displaystyle e^{C(x,z)}$ $\displaystyle\log(b)$ $\displaystyle=$
$\displaystyle D(x,z)+E(y,z)$

and similarly, integrating the second, we get $\log(b)=F(x,y)+G(y,z)$. Define
$r=\log(b)$, and observe that $r_{xy}=r_{xz}=0$, which in turn yields
$r_{x}=H(x)$, or $r(x,y,z)=-J(x)+K(y,z)$.

Thus, in the basis $\\{A,B,\partial_{y},\partial_{z}\\}$, the metric
$\tilde{g}$ is written as

$\tilde{g}=\left(\begin{array}[]{cccc}1&&&\\\ &1&&\\\
&&e^{-J(x)}e^{K(y,z)}&\\\ &&&e^{-J(x)}e^{K(y,z)}\end{array}\right)$

Undoing the change from $g$ to $\tilde{g}$ shows that the matrix for $g$ is as
claimed in the statement of the Theorem. ∎

We can now proceed with the proof of Theorem 1.10

###### Proof of Theorem 1.10.

Assume $S_{1}\times S_{2}$ has a LCW tangent to the first factor; Lemma 5.2
shows that there are coordinates in $S_{1}$ such that $g_{1}$ can be written
in the form:

(11) $g=\left(\begin{array}[]{cc}e^{J(x)}&\\\ &e^{J(x)}\end{array}\right)$

Clearly such a metric has $\partial_{t}$ as a Killing field.

Assume now $(S_{1},g_{1})$ has a Killing field. Since in dimension two, any
Killing field can be completed to a coordinate chart, we can assume that there
are coordinates where $g_{1}$ is written as before. It is then clear that
after a change of coordinates of the form $(t,x)\rightarrow(t,\lambda(x))$,
with $\lambda^{\prime}(x)=e^{J(x)/2}$, the matrix of $g$ becomes

$g=\left(\begin{array}[]{cc}e^{-J(x)}&\\\ &1\end{array}\right)$

which is a piece of a surface of revolution.

Finally, if $(S_{1},g_{1})$ is a surface of revolution, take a chart as above;
undoing the change of coordinates $(t,x)\rightarrow(t,\lambda(x))$ get a chart
in the product where the metric $g_{1}\times g_{2}$ takes the expression
appearing in Lemma 5.2. Multiplying by the conformal factor $e^{-J(x)}$ shows
that there is a LCW along $\partial_{t}$.

∎

###### Proof of Theorem 1.7.

The main result of [1] shows that if $W_{p}$ has type A, there cannot be LCWs
around $p$. Clearly, the same thing happens if $p$ is in the closure of the
set of points with Type A Weyl operators.

If $p$ has a neighborhood where every point has a Type C Weyl operator, Lemma
1.9 shows that there are two orthogonal distributions $D$ and $D^{\perp}$; the
hypothesis of the Theorem assure that there is a conformal change such that a
neighborhood of $p$ is isometric to a product of surfaces; since one of them
has a Killing field, there is a LCW.

The case of a Type B neighborhood is similarly done. Finally, if $W$ vanishes
in a neighborhood of $p$, the manifold would be conformally flat around $p$. ∎

##  6\. A metric with Weyl tensor of constant type B, three of the four
eigenflag directions are conformal factors

Lemma 1.9 has an interesting consequence: if a manifold admits a LCW, then it
has at least four vector fields of eigenflag directions. In general, only one
of them will really correspond to a conformal factor. Theorem 6.1 below shows
a less usual case of a manifold with Weyl tensor of constant type B that is
conformal to a product along three of the four eigenflag directions.

######  Theorem 6.1.

Take coordinates $(t,x,y,z)$ on an open set

$U\subset\\{(t,x,y,z)\in\mathbb{R}^{4}:x>0\\}.$

Define the metric

(12) $g=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\ 0&0&x&0\\\
0&0&0&x^{2}\end{array}\right)$

on $U$. Then:

  * •

The Weyl tensor of $g$ has type B at every point.

  * •

The eigenflag directions of the Weyl tensor are spanned by the coordinate
vector fields.

  * •

The functions $t,y,z$ are LCWs.

  * •

The function $x$ is not a LCW.

###### Proof.

The non-zero Christoffel symbols for this metric are

$\Gamma^{x}_{yy}=-1/2,\quad\Gamma^{x}_{zz}=-x,\quad\Gamma^{y}_{xy}=\Gamma^{y}_{yx}=\frac{1}{2x},\quad\Gamma^{z}_{xz}=\Gamma^{z}_{zx}=\frac{1}{x},$

the $(4,0)$ Riemann curvature tensor is

$\mathrm{Riem}\left(g\right)=\frac{1}{x}\left(\mathrm{d}x\wedge\mathrm{d}y\right)\otimes\left(\mathrm{d}x\wedge\mathrm{d}y\right)+2x\left(\mathrm{d}y\wedge\mathrm{d}z\right)\otimes\left(\mathrm{d}y\wedge\mathrm{d}z\right),$

the Ricci tensor is

$\mathrm{Ric}\left(g\right)=\frac{1}{4\,x^{2}}\mathrm{d}x\otimes\mathrm{d}x-\frac{1}{4\,x}\mathrm{d}y\otimes\mathrm{d}y-\frac{1}{2}\mathrm{d}z\otimes\mathrm{d}z,$

and the scalar curvature is $-\frac{1}{2x^{2}}$.

We define the normalized vector fields
$\hat{\partial}_{y}=\frac{\partial_{y}}{\sqrt{x}}$ and
$\hat{\partial}_{z}=\frac{\partial_{z}}{x}$.

In the orthonormal basis of $\Lambda^{2}T_{p}U$

$\partial_{t}\wedge\partial_{x},\quad\partial_{t}\wedge\hat{\partial}_{y},\quad\partial_{t}\wedge\hat{\partial}_{z},\quad\hat{\partial}_{y}\wedge\hat{\partial}_{z},\quad\partial_{x}\wedge\hat{\partial}_{z},\quad\partial_{x}\wedge\hat{\partial}_{y}$

the Weyl tensor of $g$ is diagonal, with a $3\times 3$ block repeated twice

$W=\left(\begin{array}[]{rr}W_{1}&\\\ &W_{1}\end{array}\right),\quad
W_{1}=\left(\begin{array}[]{ccc}-\frac{5}{96\,x^{2}}&0&0\\\
0&\frac{1}{96\,x^{2}}&0\\\ 0&0&\frac{1}{24\,x^{2}}\end{array}\right)$

The proof of Lemma 1.9 shows that the four eigenflag directions are spanned by
$\partial_{t}$, $\partial_{x}$, $\partial_{y}$ and $\partial_{z}$.

It is obvious that $\partial_{t}$ is a unit parallel vector field, while
$\partial_{y}$ and $\partial_{z}$ are unit parallel vector fields for the
conformal metrics $\frac{1}{x}g$ and $\frac{1}{x^{2}}g$, respectively. Thus,
according to Theorem 1.1, the coordinate functions $t$, $y$ and $z$ are LCWs.

In order to check if $\langle\partial_{x}\rangle$ is a conformal factor, we
want to apply our Theorem 2.4. The second fundamental form of
$\\{\partial_{x}\\}^{\perp}$ in the orthonormal basis
$\\{\partial_{t},\hat{\partial}_{y},\hat{\partial}_{z}\\}$ is:

$\left(\begin{array}[]{ccc}0&0&0\\\ 0&-\frac{1}{2\,x}&0\\\
0&0&-\frac{1}{x}\end{array}\right)$

which is not a multiple of the identity. This means that
$\\{\partial_{x}\\}^{\perp}$ is not umbilical, and we deduce that
$\langle\partial_{x}\rangle$ is not a conformal factor. ∎

##  7\. A metric not conformal to a product, but with Weyl tensor of constant
type C

In this example we show a explicit metric whose Weyl tensor has type C at all
points in an open set, but which is not conformal to a product. After that, we
show how to find out all its LCWs.

######  Theorem 7.1.

Let $U$ be any open subset of $\\{(t,x,y,z)\in\mathbb{R}^{4}:x>0\\}$. The
metric

(13) $g=\left(\begin{array}[]{cccc}1&0&0&0\\\ 0&1&0&0\\\ 0&0&x^{3}&0\\\
0&0&0&1/x\end{array}\right)$

on $U$ has Weyl tensor of type C at every point, but is not conformal to a
product of surfaces. Moreover there are only three LCW which are the
coordinate functions $t,y,z$.

###### Proof.

The non-zero Christoffel symbols for this metric are

$\Gamma^{x}_{yy}=-\frac{3}{2}x^{2},\quad\Gamma^{x}_{zz}=\frac{1}{2\,x^{2}},\quad\Gamma^{y}_{xy}=\Gamma^{y}_{yx}=\frac{3}{2\,x},\quad\Gamma^{z}_{xz}=\Gamma^{z}_{zx}=-\frac{1}{2\,x},$

the $(4,0)$-Riemann curvature tensor is

(14)
$\mathrm{Riem}\left(g\right)=-3x\left(\mathrm{d}x\wedge\mathrm{d}y\right)\otimes\left(\mathrm{d}x\wedge\mathrm{d}y\right)\\\
-\frac{3}{x^{3}}\left(\mathrm{d}x\wedge\mathrm{d}z\right)\otimes\left(\mathrm{d}x\wedge\mathrm{d}z\right)-3\left(\mathrm{d}y\wedge\mathrm{d}z\right)\otimes\left(\mathrm{d}y\wedge\mathrm{d}z\right),$

the Ricci tensor is

$\mathrm{Ric}\left(g\right)=-\frac{3}{2\,x^{2}}\mathrm{d}x\otimes\mathrm{d}x,$

and the scalar curvature is $-\frac{3}{2x^{2}}$.

We define the normalized vector fields
$\hat{\partial}_{y}=x^{-3/2}\partial_{y}$ and
$\hat{\partial}_{z}=\sqrt{x}\partial_{z}$.

In the orthonormal basis of $\Lambda^{2}T_{p}U$

$\partial_{t}\wedge\partial_{x},\quad\partial_{t}\wedge\hat{\partial}_{y},\quad\partial_{t}\wedge\hat{\partial}_{z},\quad\hat{\partial}_{y}\wedge\hat{\partial}_{z},\quad\partial_{x}\wedge\hat{\partial}_{z},\quad\partial_{x}\wedge\hat{\partial}_{y}$

the Weyl tensor of $g$ is diagonal, with a $3\times 3$ block repeated twice

$W=\left(\begin{array}[]{rr}W_{1}&\\\ &W_{1}\end{array}\right)\quad
W_{1}=\left(\begin{array}[]{rrr}\frac{1}{8\,x^{2}}&&\\\
&-\frac{1}{16\,x^{2}}&\\\ &&-\frac{1}{16\,x^{2}}\end{array}\right)$

The proof of 1.9 shows that the two $2$-planes distributions of eigenflag
vectors are spanned by $\\{\partial_{t},\partial_{x}\\}$ and
$\\{\partial_{y},\partial_{z}\\}$. We deduce from Lemma 5.1 that our manifold
can only be conformal to a product of two dimensional manifolds tangent to the
planes $\langle\partial_{t},\partial_{x}\rangle$ and
$\langle\partial_{y},\partial_{z}\rangle$.

However, Theorem 2.4 implies that if our metric were conformal to a product
metric, the distribution $\langle\partial_{y},\partial_{z}\rangle$ would be
umbilical. But in that case, it follows from Definition 2.2 that the following
two numbers should be the same

(15)
$\begin{array}[]{rcl}g(\nabla_{\hat{\partial}_{y}}\hat{\partial}_{y},\partial_{x})&=&-\frac{3}{2x},\\\
g(\nabla_{\hat{\partial}_{z}}\hat{\partial}_{z},\partial_{x})&=&\frac{1}{2x},\end{array}$

and we conclude that the metric is not conformal to a product metric.

Thus, the metric in the above Theorem is not covered by our Theorem 1.7.

It is obvious that $\partial_{t}$ is a parallel vector field, while
$\partial_{y}$ and $\partial_{z}$ are unit parallel vector fields for the
conformal metrics $\frac{1}{x^{3}}g$ and $xg$, respectively. Thus, according
to Theorem 1.1, the coordinate functions $t$, $y$ and $z$ are LCWs.

In order to check if $\langle\partial_{x}\rangle$ is a conformal factor, we
want to apply our Theorem 2.4. The second fundamental form of
$\\{\partial_{x}\\}^{\perp}$ in the orthonormal basis
$\\{\partial_{t},\hat{\partial}_{y},\hat{\partial}_{z}\\}$ is:

$\left(\begin{array}[]{rrr}0&0&0\\\ 0&-\frac{3}{2\,x}&0\\\
0&0&\frac{1}{2\,x}\end{array}\right)$

which is not a multiple of the identity. This means that
$\\{\partial_{x}\\}^{\perp}$ is not umbilical, and we deduce that
$\langle\partial_{x}\rangle$ is not a conformal factor.

The analysis of the Weyl tensor shows that any possible conformal factor must
be contained in either $\langle\partial_{t},\partial_{x}\rangle$ or
$\langle\partial_{y},\partial_{z}\rangle$. Let us push Theorem 2.4 a little
bit further to find all of them.

Let $X$ be a unit vector field in $\langle\partial_{t},\partial_{x}\rangle$.
$X$ can be written in the form

$X=\cos(\alpha)\partial_{t}+\sin(\alpha)\partial_{x}$

for a real valued function $\alpha:M\rightarrow\mathbb{R}$.

If $X$ spans a conformal factor for some $\alpha$, then $\langle
X\rangle^{\perp}$ is an umbilical distribution.
$X\in\langle\partial_{t},\partial_{x}\rangle$ implies
$\langle\partial_{y},\partial_{z}\rangle\subset\langle X\rangle^{\perp}$, so
in particular the form

$Z\rightarrow g(\nabla_{Z}Z,X)$

for $Z\in\langle\partial_{y},\partial_{z}\rangle$ must be a multiple of the
identity. This tensor can be written as a linear combination

$g(\nabla_{Z}Z,X)=\cos(\alpha)g(\nabla_{Z}Z,\partial_{t})+\sin(\alpha)g(\nabla_{Z}Z,\partial_{x})$

but we notice that the first summand is zero

$Z\rightarrow g(\nabla_{Z}Z,\partial_{t})=0$

and we saw in (15) that the matrix of $Z\rightarrow
g(\nabla_{Z}Z,\partial_{x})$ in the orthonormal basis
$\\{\hat{\partial}_{y},\hat{\partial}_{z}\\}$ is not a multiple of the
identity. Thus, the only combination of them that produces a multiple of the
identity is $\cos(\alpha)=1,\sin(\alpha)=0$.

The same trick will not help us decide whether there are conformal factors of
dimension $1$ contained in $\langle\partial_{y},\partial_{z}\rangle$. Instead,
we define

$Z=\cos(\alpha)\hat{\partial}_{y}+\sin(\alpha)\hat{\partial}_{z}$

and compute the second fundamental form of $Z^{\perp}$ in the basis

$\partial_{t},\quad\partial_{x},\quad-\sin(\alpha)\hat{\partial}_{y}+\cos(\alpha)\hat{\partial}_{z}.$

We remark that for some choices of $\alpha$ the distribution $Z^{\perp}$ is
not integrable, which is why the matrix

$II=\begin{pmatrix}0&0&0\\\ 0&0&-2\cos(\alpha)\sin(\alpha)\\\
-\partial_{t}\alpha&-\partial_{x}\alpha&\frac{x^{2}\cos(\alpha)\partial_{z}\alpha-\sin(\alpha)\partial_{y}\alpha}{x^{3/2}}\end{pmatrix}$

is not always symmetric. If $Z^{\perp}$ is umbilical for some choice of
$\alpha$, then the above matrix must be a multiple of the identity, and hence
it would vanish identically. In particular, $-2\cos(\alpha)\sin(\alpha)$ is
zero, and since $\alpha$ is continuous, we only have two choices:
$Z=\hat{\partial}_{y}$ and $Z=\hat{\partial}_{z}$. ∎

######  Remark 7.2.

The above example, shows how even if the analysis of the Weyl tensor yields an
infinite number of candidates to be one dimensional conformal factors, a use
of Theorem 2.4 allows to rule out the fake ones. Notice that in this example,
it happened that for the false candidates, $Z^{\perp}$ was not umbilical. In
the rare event (we do not know of any example) that for false candidates
$Z^{\perp}$ was umbilical, the conditions on $H_{1}+H_{2}$ being a gradient
field would rule out those eigenflags directions not arising from LCW.

## References

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