% The Faraday 2-form in Einstein-Weyl geometry.
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% David M. J. Calderbank
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% Mathematics Institute, University of Warwick. Coventry CV4 7AL.
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% Short title: The Faraday 2-form.
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\begin{document}
\title{The Faraday 2-form in Einstein-Weyl geometry}
\author{David M. J. Calderbank}
\address{Mathematics Institute, University of Warwick\\
Coventry CV4 7AL. England.}
\curraddr{Department of Mathematics and Statistics\\
University of Edinburgh\\ King's Buildings, Mayfield Road\\
Edinburgh EH9 3JZ\\ Scotland.}
\email{davidmjc@maths.ed.ac.uk}
\date{\thismonth}
\maketitle
\section*{Abstract.}\noindent
{\small On a conformal manifold, a compatible torsion free connection $D$ need
not be the Levi-Civita connection of a compatible Riemannian metric.  The
local obstruction is a real $2$-form $F^D$, the Faraday curvature.  It is
shown that, except in four dimensions, $F^D$ necessarily vanishes if it is
divergence free. In four dimensions another differential operator may be
applied to $F^D$ to show that an Einstein-Weyl $4$-manifold with selfdual
Weyl curvature also has selfdual Faraday curvature and so is either Einstein
or locally hypercomplex. More generally, the Bach tensor and the scalar
curvature are shown to control the selfduality of $F^D$. Finally, the
constancy of the sign of the scalar curvature on compact Einstein-Weyl
$4$-manifolds~\cite{PS2} is generalised to higher dimensions. The scalar
curvature need not have constant sign in dimensions two and three.}

\section{Introduction.}\noindent
A \emphdef{Weyl manifold} is a conformal manifold equipped with a compatible
torsion-free connection. Such connections correspond bijectively
to covariant derivatives on a real line bundle, which will be called
\emphdef{Weyl derivatives}. The curvature of a Weyl derivative $D$ is a real
$2$-form $F^D$, whose vanishing implies that the induced torsion-free
connection is locally the Levi-Civita connection of a compatible
Riemannian metric. This is the original gauge theory introduced
by Hermann Weyl~\cite{Weyl} as a geometrisation of classical
electromagnetism: $F^D$ is the \emphdef{Faraday $2$-form} representing the
electromagnetic field.

There has been a resurgence of interest in Weyl geometry recently, not only
because of the original motivation, but also because of a number of more
surprising links with geometry and mathematical physics. The germinal idea to
unify general relativity and electromagnetism led naturally to the notion of
an \emphdef{Einstein-Weyl manifold}, in which the symmetric trace free part
of the Ricci tensor of the Weyl connection is assumed to vanish. However, the
geometry of Einstein-Weyl manifolds initially attracted particular
interest in three dimensions, rather than four, thanks to work of
Cartan~\cite{Cartan}, and Hitchin~\cite{Hitchin3}, who showed that the
Einstein-Weyl condition is the integrability condition for a complex
structure on the space of oriented geodesics of a three dimensional Weyl
manifold. This provides a twistor correspondence between Einstein-Weyl
3-manifolds and complex surfaces containing rational curves with normal
bundle $\mathcal{O}(2)$. Subsequently, Einstein-Weyl manifolds have been much
studied in all dimensions~\cite{CP,DMJC1,ET,Gauduchon4,HP,PS1,PS2,PT,Tod1},
but the three dimensional case retains a special aura, thanks to the above
twistor construction, and the link with selfdual conformal
$4$-manifolds~\cite{JT}.

In four dimensions, locally hypercomplex $4$-manifolds are
Einstein-Weyl~\cite{PS1}, and this class of Einstein-Weyl geometries
has recently been linked to local heterotic geometry~\cite{Tod3}.
One of the main aims of this paper is to prove that every Einstein-Weyl
$4$-manifold with selfdual Weyl tensor is either Einstein or locally
hypercomplex. This local result, like several of the results in this paper,
was previously only known in the compact case.

After setting up a framework for studying Weyl geometry in a gauge
invariant way, I present a very simple result which shows that,
except in four dimensions, the vanishing of the divergence of $F^D$
(with respect to the Weyl connection) is enough to ensure that $F^D$
itself vanishes. A consequence of this is that $F^D$ necessarily
vanishes on scalar flat Einstein-Weyl $n$-manifolds, for $n\neq4$.
In four dimensions, scalar-flatness only implies that $F^D$ is source-free,
i.e., coclosed; in other words its selfdual and antiselfdual parts
$F^D_+$ and $F^D_-$ are both closed. I next show, in section~\ref{Bsect},
that the vanishing of one of $F^D_+$ and $F^D_-$ implies that the Bach tensor
of the underlying conformal structure is identically zero.
The converse also holds, as long as we continue to assume vanishing
scalar curvature. These results are established by giving a formula
for the Bach tensor on an Einstein-Weyl $4$-manifold (Theorem~\ref{Bach})
generalising a formula of~\cite{PS2} to the local setting.

Also in section~\ref{Bsect}, I present some identities relating to the Bach
tensor, which may be used to verify (locally) that it vanishes when the Weyl
tensor is (anti)selfdual. This fact is not used in the proof of
Theorem~\ref{Bach}, nor is it new: see~\cite{Besse,ER} for the statement,
and~\cite{GNotes} for a proof using the normal conformal connection.
However, these calculations do provide motivation for considering the Bach
tensor in a local context, and they are not readily accessible in the
literature, so I hope it will be of some interest to sketch them briefly here.

In~\cite{ET}, Eastwood and Tod have shown that the Faraday $2$-form vanishes
on conformally flat Einstein-Weyl manifolds. In section~\ref{ETsect}, I
reformulate their result and demonstrate that the same ideas may be used to
prove that half conformally flat Einstein-Weyl $4$-manifolds have
(anti)selfdual Faraday curvature, and are therefore Einstein or locally
hypercomplex by~\cite{PS1}. This was previously only known in the compact
case~\cite{Gauduchon4,PS2} and shows that the twistor theory of Pedersen and
Swann~\cite{PS1} deals with all Einstein-Weyl structures on an (anti)selfdual
conformal $4$-manifold.

I end the paper with a discussion of the sign of the scalar curvature
on compact Einstein-Weyl manifolds. This was shown to be constant in
four dimensions by Pedersen and Swann~\cite{PS2}. I show that the same
is true in higher dimensions, but not in dimensions two and three,
contrary to previous assertions~\cite{PS2,PT}. The higher dimensional
result follows by applying the maximum principle to a formula
in~\cite{PS2}, while the low dimensional counterexamples can be found
in~\cite{DMJC1,PT,Tod1}.

\section*{Acknowledgements.}\noindent
Several of the results in this paper arose during the preparation
of a survey paper on Einstein-Weyl geometry in collaboration with Henrik
Pedersen~\cite{CP}, and I gratefully acknowledge many useful and instructive
conversations with him. The approach taken in section~\ref{ETsect} is strongly
influenced by numerous discussions with Tammo Diemer on linear field theories
in conformal geometry, and I am likewise indebted to him.  Finally, I would
like to thank Paul Gauduchon, Henrik Pedersen and Andrew Swann for discussing
with me the sign of the scalar curvature.

\section{Weyl derivatives.}\noindent
In order to carry out computations in a gauge invariant way, it is convenient
to work with densities. If $V$ is a real $n$-dimensional vector space and $w$
any real number, then the oriented one dimensional linear space $L^w=L^w(V)$
carrying the representation $A\mapsto|\det A|^{w/n}$ of $\GL(V)$ will be
called the space of \emphdef{densities of weight $w$} or
\emphdef{$w$-densities}. It can be constructed canonically as the space of
maps $\rho\colon(\Lambda^nV)\setdif0\to\R$ such that
$\rho(\lambda\omega)=|\lambda|^{-w/n}\rho(\omega)$ for all
$\lambda\in\R\mult$ and $\omega\in(\Lambda^nV)\setdif0$.

\begin{defn}
Suppose $\Mn$ is any $n$-manifold. Then the \emphdef{density line bundle}
$L^w=L^w(T\Mn)$ of $\Mn$ is defined to be the bundle whose fibre at
$x\in\Mn$ is $L^w(T_x\Mn)$. Equivalently it is the associated bundle
$\GL(\Mn)\times_{\GL(n)} L^w(n)$ where $\GL(\Mn)$ is the frame bundle of $\Mn$
and $L^w(n)$ is the space of $w$-densities of $\R^n$.
\end{defn}

The density bundles are oriented (hence trivialisable) real line bundles,
but there is no preferred trivialisation. Sections of $L=L^1$ may be
thought of as scalar fields with dimensions of length. This geometric
dimensional analysis may also be applied to tensors:
the tensor bundle $L^w\tens(T\Mn)^j\tens(T\dual\Mn)^k$
(and any subbundle, quotient bundle, element or section) will be said to
have \emphdef{weight} $w+j-k$, or dimensions of [length]$^{w+j-k}$.

Note that sections of $L^{-n}$ may be invariantly integrated, and that
an orientation of $\Mn$ defines an isomorphism between $L^{-n}$ and
$\Lambda^nT\dual\Mn$. Also note that $L^{w_1}\tens L^{w_2}$ is
canonically isomorphic to $L^{w_1+w_2}$ and $L^0$ is the trivial bundle.
When tensoring a vector bundle with some $L^w$, I shall often
omit the tensor product sign. A nonvanishing (usually positive)
section of $L^1$ (or $L^w$ for $w\neq0$) will be called a
\emphdef{length scale} or \emphdef{gauge} (of weight $w$).

\begin{defn} A \emphdef{Weyl derivative} is a covariant derivative $D$
on $L^1$. It induces covariant derivatives on $L^w$ for all $w$.
The curvature of $D$ is a real $2$-form $F^D$ which will be called the
\emphdef{Faraday curvature} or \emphdef{Faraday $2$-form}.
\end{defn}

If $F^D=0$ then $D$ is said to be \emphdef{closed}. It follows that
there are local length scales $\mu$ with $D\mu=0$. If such a length
scale $\mu$ exists globally then $D$ is said to be \emphdef{exact}.
Conversely, a length scale $\mu$ induces an exact Weyl derivative
$D^\mu$ such that $D^\mu\mu=0$. Note that Weyl derivatives form
an affine space modelled on the linear space of $1$-forms,
while closed and exact Weyl derivatives are affine subspaces modelled
on the linear spaces of closed and exact $1$-forms respectively.

A \emphdef{gauge transformation} on $\Mn$ is a positive function $e^f$
which rescales a gauge $\mu\in\Cinf(\Mn,L^w)$ to give $e^{wf}\mu$.
If $D$ is any Weyl derivative, then 
$D=D^\mu+\omega^\mu$ for the $1$-form $\omega^\mu=\mu^{-1}D\mu$,
and consequently, $\omega^{e^f\mu}=\omega^\mu+df$.

The notion of density allows one to define a conformal structure,
not as an equivalence class of inner products, but a genuine inner product.

\begin{defn} A \emphdef{conformal structure} on a manifold $\Mn$ is
a \emphdef{normalised} $L^2$ valued inner product on $T\Mn$. More precisely
it is a section $\conf\in\Cinf(\Mn,L^2S^2T\dual\Mn)$ which is everywhere
positive definite. The conformal inner product of tangent vector fields $X,Y$
will be denoted $\cip{X,Y}$, and is a section of $L^2$.

A conformal structure may also be viewed as a metric on the
\emphdef{weightless tangent bundle} $L^{-1}T\Mn$. The normalisation
condition states that $|\det\conf|=1$, which makes sense since
the densities of $L^{-1}T\Mn$ are canonically trivial.
\end{defn}

Conformal structures and Weyl derivatives fit together very well: for
instance, Riemannian geometry is the geometry of an exact Weyl derivative
on a conformal manifold. The existence and uniqueness of the Levi-Civita
connection is a special case of the following central result.

\begin{fundthm}\label{FTCG}\tcite{Weyl}
On a conformal manifold $\Mn$ there is an affine bijection
between covariant derivatives on $L^1$ and torsion free connections on $T\Mn$
preserving the conformal structure. More explicitly, the covariant derivative
on $T\Mn$ is determined from the one on $L^1$ by the Koszul formula
\begin{align*}
2\cip{D_XY,Z}&=D_X\,\cip{Y,Z}+D_Y\,\cip{X,Z}-D_Z\,\cip{X,Y}\\
&+\cip{[X,Y],Z}-\cip{[X,Z],Y}-\cip{[Y,Z],X},
\end{align*}
where $X,Y,Z$ are vector fields and the conformal inner products
are sections of $L^2$.
\end{fundthm}

A conformal manifold equipped with a Weyl derivative $D$ is called a
\emphdef{Weyl manifold}. Henceforth $D$ will be identified with the
induced connection on $T\Mn$ and all associated bundles.

It is sometimes useful to compare two Weyl derivatives on a conformal
manifold: suppose that $\tilde D=D+\gamma$, where $\gamma$ is a $1$-form,
and suppose that the induced connections on $L^{w-1}T\Mn$ are related
by $\tilde D=D+\Gamma$ for some $\lie{co}(T\Mn)$-valued $1$-form $\Gamma$.
Then $\Gamma$ is given in terms of $\gamma$ by the linearised Koszul formula:
$$\Gamma_X=w\gamma(X)\iden+\gamma\skwend X,\quad \text{where}\quad
(\gamma\skwend X)(Y)=\gamma(Y)X-\cip{X,Y}\gamma.$$ Here free use is being
made of the natural isomorphism $\sharp\colon T\dual\Mn\to L^{-2}T\Mn$ given
by the conformal structure.

A basic formula in Weyl geometry (and indeed Riemannian geometry)
is the following decomposition of the curvature $R^{D,w}$ of $D$
on $L^{w-1}T\Mn$: 
\begin{equation}\label{curv} R^{D,w}_{X,Y}=\W_{X,Y}+wF^D(X,Y)\iden
-r^D(X)\skwend Y+r^D(Y)\skwend X.
\end{equation}
Here $\W_{X,Y}$ is the Weyl curvature (independent of $D$), $F^D$ is
the Faraday curvature (independent of $\conf$), and $r^D$ is a
section of $L^{-2}\End T\Mn$ called the \emphdef{normalised Ricci tensor}.
(Convention: $R^{D,w}_{X,Y}=D^2_{X,Y}-D^2_{Y,X}=[D_X,D_Y]-D_{[X,Y]}$.)

The normalised Ricci tensor may be computed from $R^{D,w}$ by taking a
trace (since $W$ is tracefree)---such a trace does not necessarily
produce a symmetric tensor. An alternative description of $r^D$ is given
by its decomposition under the orthogonal group:
$$r^D=r^D_0+\tfrac1{2n(n-1)}\scal^D\iden-\tfrac12 F^D,$$
where $r^D_0$ is symmetric and tracefree and $\scal^D$ is the
\emphdef{scalar curvature} of $D$ (which is a section of $L^{-2}$).

\medbreak

Two of the results in this paper are based on the idea of twisting
a complex of differential operators by a connection. The most familiar
example of this is the deRham complex. If we have an auxiliary
bundle $E$, then a connection on $E$ induces a sequence of twisted
exterior derivatives on $E$-valued differential forms. This sequence
is no longer a complex in general: the composite of two twisted exterior
derivatives is given by the natural action of the curvature of the
connection.

This idea will be applied to the density bundles and Weyl derivatives,
but using other differential complexes.
The first such result is a very simple one. Instead of considering
the deRham complex of exterior derivatives
$$\Mn\times\R\rto[d]T\dual\Mn\rto[d]\Lambda^2T\dual\Mn\rto\cdots$$
consider the adjoint complex of exterior divergences
$$L^{-n}\lto[\delta]L^{-n}T\Mn\lto[\delta] L^{-n}\Lambda^2T\Mn\lto\cdots$$
Here $\delta=\trace D$, where
$D$ is any torsion free connection and the trace is with respect to
the first entry: thus $\delta$ is minus the formal adjoint of $d$.
Both are complexes of invariant operators. Using the conformal
structure, we obtain a complex of conformally invariant operators:
$$L^{-n}\lto[\delta]L^{2-n}T\dual\Mn\lto[\delta]
L^{4-n}\Lambda^2T\dual\Mn\lto\cdots$$
It is this complex which I shall twist by a Weyl derivative.

\begin{prop}\label{DivF} Let $D$ be a Weyl derivative on a conformal manifold
$\Mn$. Then $(\delta^D)^2F^D=-(n-4)|F^D|^2$. If $n\neq4$ it follows
that $\delta^DF^D=0$ iff $F^D=0$.
\proofof{prop} Since $F^D$ is a $2$-form (of weight $-2$), the exterior
divergence has been twisted by $D$ on $L^{n-4}$. Suppose that
$D=D^0+\omega$ on $L^1$. Then for any $2$-form $\phi$,
\begin{equation}\label{delta}
\delta^{D^0+\omega}\phi=\delta^{D^0}\phi+(n-4)\phi(\omega,.).
\end{equation}
If $D^0$ is closed, it follows that
\begin{align*}
(\delta^D)^2\phi&=\delta^{D^0+\omega}\delta^{D^0+\omega}\phi\\
&=\delta^{D^0}\delta^{D^0}\phi+(n-4)\bigl(\delta^{D^0}(\phi(\omega,.))
+(\delta^{D^0}\phi)(\omega)\bigr)\\
&=(n-4)\trace\phi(D^0\omega,.),
\end{align*}
where $(\delta^{D^0})^2=0$ because $L^{n-4}$ may be locally trivialised
by a $D^0$-parallel section (equivalently, when we twist a complex by a
flat connection, it remains a complex).
Therefore
\begin{align*}
(\delta^D)^2\phi&=(n-4)\sum_{i,j}D^0_{e_i}\omega(e_j)\phi(e_j,e_i)\\
&=(n-4)\sum_{i<j}d\omega(e_i,e_j)\phi(e_j,e_i)
\end{align*}
and $d\omega=F^D$ since $D^0$ is closed. Substituting $\phi=F^D$
proves the proposition.
\end{proof}

As in the Riemannian case, the second Bianchi identity is a useful tool in
Weyl geometry. The differential Bianchi identity for the Weyl derivative $D$
on $L^1$ gives the simple fact that $dF^D=0$, while for the induced covariant
derivative on $L^{w-1}T\Mn$, we have $d^DR^{D,w}=0$, where $R^{D,w}$ is
viewed as an endomorphism valued $2$-form. In particular:
$$(D_XR^{D,0})_{Y,Z}U+(D_ZR^{D,0})_{X,Y}U+(D_YR^{D,0})_{Z,X}U=0.$$
If we take the trace of this identity as a function of $Z$,
and separate out the trace using the conformal structure and~(\ref{curv}),
then we obtain, for $n>2$, the contracted Bianchi identities:
\begin{align}\label{cy}
\trace D\W_{X,Y}U&=(n-3)C^D_{X,Y}U\\
C^D_{X,e_i}e_i&=0.\label{ccy}
\end{align}
Here $C^D_{X,Y}=(d^Dr^D)_{X,Y}=(D_Xr^D)(Y)-(D_Yr^D)(X)$ is the
\emphdef{Cotton-York tensor} of $D$, and there is an implied
summation over the weightless orthonormal basis $e_i$.

\section{Einstein-Weyl geometry.}\noindent
A Weyl manifold of dimension $n>2$ is said to be \emphdef{Einstein-Weyl} iff
$r^D_0=0$. This class of Weyl manifolds has been extensively studied
(see~\cite{CP} and references therein). The Einstein-Weyl manifolds with
exact Weyl derivative are precisely the (homothety classes of) Einstein
metrics, and the closed Einstein-Weyl manifolds are ``locally conformally
Einstein''. There are also many examples with nonzero Faraday
curvature~\cite{Bonneau,JT,Madsen2,PS1,PS2,PT}.

The contracted Bianchi identity gives an important relationship
between the Faraday and scalar curvatures. Writing out~(\ref{ccy})
explicitly gives,
$$\divg^Dr^D_0-\tfrac1{2n}D\scal^D+\tfrac12\delta^DF^D=0,$$
from which we immediately have the following.

\begin{prop}\label{EWBianchi}\tcite{PT,Gauduchon4} Suppose $\Mn$
is Einstein-Weyl of dimension $n>2$. Then $D\scal^D - n\,\delta^D F^D=0$.
\end{prop}
When $n=2$, we \emph{define} Einstein-Weyl manifolds by this
identity~\cite{DMJC1}, generalising the metrics of constant scalar
curvature.

A consequence of the Einstein-Weyl Bianchi identity is the following
formula, given by Pedersen and Swann in the compact case:
\begin{prop}\label{LapScal}\textup{(cf.~\cite{PS2})} Let $\Mn$ be an
$n$-dimensional Einstein-Weyl manifold. Then
$\Laplace^D\scal^D=-n(n-4)|F^D|^2$.
\proofof{prop} Here $\Laplace^D=\trace D^2$, so this follows by
applying $D$ to the formula in~\ref{EWBianchi} and taking trace,
where Proposition~\ref{DivF} is used to compute
$\trace D(\delta^DF^D)=(\delta^D)^2 F^D$.
\end{proof}
I will make use of this later.

\begin{thm}\label{EWthm} If $(M^n,D)$ is Einstein-Weyl,
the following are equivalent:
\begin{enumerate}
\item\label{A1} Either $D$ is closed or $n=4$, $\Mn$ is noncompact
and $F^D$ is harmonic.
\item\label{A2} $\delta^DF^D=0$.
\item\label{A3} $D\scal^D=0$.
\item\label{A4} Either $D$ is exact or $\scal^D$ is identically zero.
\end{enumerate}
\end{thm}
Before proving this theorem, let me make some remarks. There are
two important consequences to be extracted from these equivalences.
The first of these is the fact that if $D$ is closed then $D$ is
exact or $\scal^D=0$. This is not new: it was established by
Gauduchon~\cite{Gauduchon4} using~\ref{EWBianchi}, and by Higa~\cite{Higa},
using the contracted Bianchi identity on the universal cover (which
is Einstein). The second consequence is that if $\scal^D=0$ then
$D$ is closed, except in the local four dimensional case.
This was previously only known for compact manifolds~\cite{Gauduchon4,PT}.
The new part of the above theorem follows easily from~\ref{DivF}.

\begin{proof} \ref{A2} and~\ref{A3} are equivalent by~\ref{EWBianchi},
and clearly~\ref{A3}$\implies$\ref{A4}$\implies$\ref{A2} or~\ref{A3}.
Also~\ref{A1} immediately implies~\ref{A2}---note that the divergence
on $2$-forms is conformally invariant in four dimensions---so it remains
to prove the converse. This follows from~\ref{DivF} unless $\Mn$ is four
dimensional, in which case $F^D$ is an exact coclosed $2$-form,
and so it is certainly harmonic.
If $\Mn$ is compact, an exact coclosed $2$-form necessarily vanishes:
write $F^D=d\gamma$ and integrate the section $|F^D|^2$ of $L^{-4}$ by parts.
\end{proof}

\section{Four dimensions and the Bach tensor.}\label{Bsect}\noindent
The four dimensional case is singled out by Theorem~\ref{EWthm} for
good reason: it is only in four dimensions that the divergence is
conformally invariant on $2$-forms. Closely related to this is the
fact that the bundle of $2$-forms is not irreducible, but splits into
the selfdual and antiselfdual $2$-forms. Correspondingly, the
Faraday $2$-form decomposes as $F^D=F^D_+ + F^D_-$.

As shown by Pedersen and Swann~\cite{PS1}, hypercomplex manifolds provide
examples of scalar flat four dimensional Einstein-Weyl manifolds with
nonzero Faraday curvature.
In this case, if we fix the orientation by
taking the complex structures to be antiselfdual, then the Faraday
$2$-form and Weyl tensor are both selfdual.
[There are explicit
examples amongst the Pedersen/LeBrun metrics: these are $\Un(2)$-symmetric
selfdual conformal structures, which are both conformally
Einstein (with nonzero scalar curvature)~\cite{HP} and conformally
scalar flat K\"ahler (with antiselfdual complex structure)~\cite{LeBrun}.
They may be constructed from harmonic functions (``monopoles'') on
$S^3$ or $\mathcal H^3$. Those coming from $S^3$ also admit a
hypercomplex structure with nonzero Faraday curvature~\cite{Madsen2}.]

On the other hand, the theory in the previous section only shows
that $F^D_+$ and $F^D_-$ are both closed, or equivalently, $F^D$ is
a Maxwell field (an abelian Yang-Mills field). It is therefore natural
to ask whether the Einstein-Weyl condition implies that $F^D$ is an
\emph{(anti)selfdual} Maxwell field. We might then be able to sharpen
Theorem~\ref{EWthm}.

In this section I will show that this question is closely related to
the Bach tensor, the weight $-4$ symmetric traceless bilinear
form which arises as the gradient of the functional
$$\conf\mapsto\int_\Mn |W|^2.$$ Here $W$ is the Weyl curvature of $\conf$,
so that $|W|^2$ is a section of $L^{-4}$. It is well known~\cite{Besse} that
the signature formula may be used to show that compact conformal
$4$-manifolds with (anti)selfdual Weyl are absolute minima of this
functional. It is also known, but perhaps less widely appreciated, that
there is also a local result: $4$-manifolds with (anti)selfdual Weyl
tensor have vanishing Bach tensor. Since I have not found a convenient
reference for the local theory, I will sketch it here.

\begin{defn} Let $\Mn,\conf,D$ be a Weyl $n$-manifold ($n>2$). Then the
\emphdef{Bach tensor} of $\Mn$ is defined by the formula
$$B^D(X,Y)=D_{e_i}C^D_{e_i,X}Y+r^D(e_i,\W_{e_i,X}Y)$$
where $r^D(X,Y)=\langle r^D(X),Y\rangle$ and there is an implied
summation over the weightless orthonormal basis $e_i$.
\end{defn}
It is not immediately clear that this is symmetric, although it is
clearly tracefree by~(\ref{ccy}). The symmetry turns out to be a consequence
of a contracted ``third Bianchi identity''. The first Bianchi identity
shows that the Weyl curvature has the symmetries of a ``Weyl tensor'',
while the second Bianchi identity shows that it lies in the kernel of a
conformally invariant first order differential operator. In four
dimensions, however, this first order condition is vacuous, and
the Weyl curvature lies in the kernel of a conformally invariant
\emph{second} order differential operator. This second order operator arises
from the third Bianchi identity.

All three Bianchi identities are consequences of the differential
Bianchi identity for the conformal connection~\cite{GNotes}. However,
I shall prove the following directly.

\begin{prop} Let $D$ be any Weyl derivative a conformal manifold. Then
\begin{equation}\label{tbi}
d^DC^D_{X,Y,Z}U+r^D(X,\W_{Y,Z}U)+r^D(Z,\W_{X,Y}U)+r^D(Y,\W_{Z,X}U)=0.
\end{equation}
Here $d^DC^D$ denotes the exterior derivative of $C^D$ as a covector valued
$2$-form.
\proofof{prop} Since $C^D=d^Dr^D$, we readily see that:
\begin{align*}
d^DC^D_{X,Y,Z}=\quad&(D^2_{X,Y}r^D)(Z)+(D^2_{Z,X}r^D)(Y)+(D^2_{Y,Z}r^D)(X)\\
-&(D^2_{Y,X}r^D)(Z)-(D^2_{X,Z}r^D)(Y)-(D^2_{Z,Y}r^D)(X).
\end{align*}
Note that $(D^2_{X,Y}r^D)(Z,U)-(D^2_{Y,X}r^D)(Z,U)
=-r^D(R^D_{X,Y}Z,U)-r^D(Z,R^D_{X,Y}U)$, where $R^D=R^{D,1}$, and
the first term cancels with its cyclic permutations by the first Bianchi
identity. For the second term we compute:
\begin{align*}
r^D(Z,R^D_{X,Y}U)&=r^D(Z,\W_{X,Y}U)
+r^D(Z,(r^D(Y)\skwend X)\,U)-r^D(Z,(r^D(X)\skwend Y)\,U)\\
&\quad+F^D(X,Y)r^D(Z,U)\\
&=\cip{r^D(Z),\W_{X,Y}U}+\cip{r^D(X),r^D(Z)}\cip{Y,U}
-\cip{r^D(Y),r^D(Z)}\cip{X,U}\\
&\quad+r^D(Z,Y)r^D(X,U)-r^D(Z,X)r^D(Y,U)+F^D(X,Y)r^D(Z,U).
\end{align*}
On taking cyclic permutations over $X,Y,Z$, the second and third terms cancel
by symmetry, and the last three terms cancel because the skew symmetric part
of $r^D$ is $-\frac12 F^D$. The first term and its cyclic permutations give
the stated formula.
\end{proof}

The symmetry of the Bach tensor follows from the following properties
of the Cotton-York tensor.
\begin{prop} Let $C^D$ be the Cotton-York tensor of a Weyl $n$-manifold.
Then
\begin{align}\label{altfree}
C^D_{X,Y}Z+C^D_{Y,Z}X+C^D_{Z,X}Y&=0\\
D_{e_i}C^D_{X,Y}e_i+r^D(e_i,\W_{X,Y}e_i)&=0.\label{divfree}
\end{align}
In the second formula, only the skew symmetric part of $r^D$ contributes.
\proofof{prop}
The first part follows easily from the fact that the skew symmetric
part of $r^D$, namely $-\frac12 F^D$, is closed. The second formula
generalises the fact that the Cotton-York tensor on a $3$-manifold is
divergence free and follows immediately by taking the trace over
$Z$ and $U$ of the third Bianchi identity~(\ref{tbi}).
\end{proof}
Putting~(\ref{altfree}) and~(\ref{divfree}) together gives
\begin{align*} B^D(X,Y)-B^D(Y,X)&=D_{e_i}C^D_{e_i,X}Y-D_{e_i}C^D_{e_i,Y}X\\
&\quad+r^D(e_i,\W_{e_i,X}Y)-r^D(e_i,\W_{e_i,Y}X)\\
&=-D_{e_i}C^D_{X,Y}e_i-r^D(e_i,\W_{X,Y}e_i)=0.
\end{align*} From this it follows
that an alternative (perhaps better) formula for the Bach tensor is:
$$B^D(X,Y)=\tfrac12\bigl(D_{e_i}C^D_{e_i,X}Y+D_{e_i}C^D_{e_i,Y}X\bigr)
+r^D_0(e_i,\W_{e_i,X}Y).$$

In dimension four, $B=B^D$ is independent of the choice of $D$.
To check this, consider another Weyl derivative $\tilde D=D+\gamma$.
Then direct calculations give the following formulae:
\begin{align}
r^{\tilde D}&=
r^D-(D\gamma-\gamma\tens\gamma+\tfrac12\cip{\gamma,\gamma}\iden)\\
C^{\tilde D}_{X,Y}U&=C^D_{X,Y}U+\gamma(\W_{X,Y}U)\\
B^{\tilde D}(X,Y)&=B^D(X,Y)
+(n-4)\bigl(C^D_{\gamma,X}Y+C^D_{\gamma,Y}X+\gamma(\W_{\gamma,X}Y)\bigr).
\end{align}

For $n>3$, the third Bianchi identity and the definition of the Bach tensor
can be rewritten using the second Bianchi identity~(\ref{cy}):
\begin{align*}
&(d^D\divg^DW)_{X,Y,Z}U
+r^D(Z,\W_{X,Y}U)+r^D(Y,\W_{Z,X}U)+r^D(X,\W_{Y,Z}U)=0\\
&(\delta^D\divg^DW)_XY+(n-3)r^D(e_i,\W_{e_i,X}Y)=(n-3)B^D(X,Y).
\end{align*}
($d^D$ and $\delta^D$ are the exterior derivative and divergence
on covector valued $2$-forms.)

In order to interpret this, we need to choose an orientation
and introduce the conformal Hodge star operator
$*\colon L^{w+k}\Lambda^kT\dual\Mn\to L^{w+n-k}\Lambda^{n-k}T\dual\Mn$.
In four dimensions, if we view $W$ as an endomorphism valued $2$-form,
and its divergence $C^D$ as a covector valued $2$-form, then the
star operator decomposes $W$ and $C^D$ into selfdual and antiselfdual
parts. We then see that $*d^D\divg^DW=\delta^D\divg^D*W$, and
consequently, the Bach tensor may be computed from $W^+$ or $W^-$:
\begin{align*}
B(X,Y)&=2(\delta^D\divg^DW^+)_XY+2r^D(e_i,W^+_{e_i,X}Y)\\
&=2(\delta^D\divg^DW^-)_XY+2r^D(e_i,W^-_{e_i,X}Y).
\end{align*}
Hence we finally have that the Bach tensor vanishes on (anti)selfdual
conformal $4$-manifolds.

These formulae also relate to conformally invariant
second order differential operators acting on weight $-2$ Weyl
tensors in dimension four (cf.~\cite{Branson,ER}).

\medbreak

I will now turn to the impact of the Bach tensor on Einstein-Weyl geometry.
In~\cite{PS2}, Pedersen and Swann gave a formula for the Bach tensor
of a \emph{compact} Einstein-Weyl manifold in terms of the $1$-form
of the Weyl structure in the Gauduchon gauge. They deduced that
on a compact Bach flat ($B=0$) Einstein-Weyl $4$-manifold,
the Faraday $2$-form necessarily vanishes.

The Bach tensor and Faraday $2$-form are intimately linked even locally. 

\begin{thm}\label{Bach} Let $\Mn$ be an Einstein-Weyl $4$-manifold. Then
$$B(X,Y)=\tfrac1{24}\bigl(D^2_{X,Y}\scal^D+D^2_{Y,X}\scal^D\bigl)
-\cip{F^D_+(X),F^D_-(Y)}-\cip{F^D_+(Y),F^D_-(X)}.$$
Consequently the following are equivalent:
\begin{enumerate}
\item\label{B1} $F^D$ is either selfdual or antiselfdual.
\item\label{B2} $B=0$ and either $\scal^D=0$ or $D$ is exact.
\end{enumerate}
\proofof{thm}On an Einstein-Weyl $4$-manifold,
$r^D=\frac1{24}\scal^D-\frac12F^D$ and so the Cotton-York tensor is:
$$C^D_{U,X}Y=\tfrac1{24}\bigl(D_U\scal^D\cip{X,Y}-D_X\scal^D\cip{U,Y}\bigr)
+\tfrac12D_YF^D(U,X)$$
where I have used the fact that $dF^D=0$. Consequently,
using~\ref{EWBianchi} and~\ref{LapScal},
\begin{align*}
B(X,Y)&=\tfrac1{24}\bigl(D^2_{X,Y}\scal^D+D^2_{Y,X}\scal^D\bigr)\\
&\quad+\tfrac14\bigl(F^D(R^D_{X,e_i}e_i,Y)+F^D(R^D_{Y,e_i}e_i,X)
+F^D(e_i,R^D_{X,e_i}Y+R^D_{Y,e_i}X)\bigr).
\end{align*}
After some work, the curvature terms reduce to
$-\cip{F^D(X),F^D(Y)}+\frac12|F^D|^2\cip{X,Y}$. Writing $F^D=F^D_+ + F^D_-$
gives the formula in the theorem, since the symmetric tracefree part of
$\cip{F^D_+(X),F^D_+(Y)}$ is zero, and similarly for $F^D_-$. The equivalence
of~\ref{B1} and~\ref{B2} follows from this formula, together with the fact
that Einstein-Weyl manifolds with (anti)selfdual Faraday satisfy $D\scal^D=0$
(see~\cite{Gauduchon4} or~\ref{EWthm}).  To see that~\ref{B2}
implies~\ref{B1}, observe that $D\scal^D=0$ and $B=0$ together imply that one
of $F^D_+,F^D_-$ must vanish at each point, since
$\Lambda^2_+T\dual\Mn\tens\Lambda^2_-T\dual\Mn\cong L^{-2}S^2_0T\dual\Mn$.
Now, Einstein-Weyl manifolds are real analytic~\cite{PS2,PT} and so (assuming
$\Mn$ is connected), $F^D$ must be everywhere selfdual or antiselfdual (or
both, i.e., zero).
\end{proof}

In the compact case, we only need \emph{one} of the conditions $B=0$ or
$\scal^D=0$ to ensure that $F^D=0$. It is natural to ask
whether something similar holds in general. However, Bonneau~\cite{Bonneau}
has recently given examples of $\Un(2)$-symmetric Einstein-Weyl
structures with vanishing scalar curvature, for which the Faraday
$2$-form is \emph{not} (anti)selfdual. On the other hand, Bach flat
Einstein-Weyl structures have not been studied locally. Are they
necessarily scalar flat or Einstein? Equivalently, must $F^D$ be
(anti)selfdual?

\section{The Weyl tensor and the Faraday $2$-form.}\label{ETsect}\noindent
Theorem~\ref{Bach} implies in particular that on a
scalar flat Einstein-Weyl manifold with selfdual Weyl tensor, the
Faraday $2$-form is either selfdual or antiselfdual, since such
manifolds are Bach flat. Now selfduality of the Weyl tensor is a
much stronger assumption than Bach flatness, and so one might hope
to improve this result in two ways: firstly, by removing the assumption
of scalar flatness; and secondly, by showing that the Faraday $2$-form
is in fact selfdual. In this section I will prove such a result:

\begin{thm}\label{mine} Let $\Mn$ be an Einstein-Weyl $4$-manifold with
selfdual Weyl tensor. Then the Faraday $2$-form is also selfdual.
\end{thm}
This is similar to the following theorem of Eastwood and Tod:
\begin{thm}\label{theirs}\tcite{ET} Let $\Mn$ be an Einstein-Weyl
$n$-manifold $(n\geq4)$ with vanishing Weyl tensor. Then the Faraday
$2$-form also vanishes.
\end{thm}
I want to explain how Eastwood and Tod's result fits into the framework of
twisting complexes of differential operators, and show that in four
dimensions, one can split the argument into its selfdual and antiselfdual
parts. Therefore, as well as proving~\ref{mine}, I will provide another proof
of~\ref{theirs}.

\medbreak

The reader may have noticed that the underlying ideas behind many of the
results in this paper are representation theoretic in nature; that is, they
involve the representation theory of the orthogonal group and the theory of
conformally invariant differential operators.  I would now like to make these
ideas more explicit, since the proofs of~\ref{mine} and~\ref{theirs} are quite
straightforward from this point of view, whereas the detailed calculations are
not so simple.

Irreducible representations of $\SO(n)$ may be described by $l$-tuples of
integers, where $n=2l$ or $2l+1$, and I will use these $l$-tuples to describe
the associated vector bundles on a Riemannian manifold $\Mn$~\cite{Fegan,FH}.
For example $(1,0,...)$ is the (co)tangent bundle, $(1,1,...)$ is the bundle
of $2$-forms and $(2,0,...)$ is the bundle of symmetric traceless
$2$-tensors. Here the dots indicate that the remaining integers are all
zero. The integers are nonincreasing, and are usually all positive, in which
case the corresponding tensors have rank equal to the sum of the integers. In
four dimensions (and any even dimension), the last integer may be negative:
for instance, the selfdual and antiselfdual $2$-forms are denoted by $(1,+1)$
and $(1,-1)$.

If $\Mn$ is conformal rather than Riemannian, the weight must also be
specified. I will do this by prefixing the $l$-tuple by the weight, separated
by a vertical bar~$|$. Thus $(-2|1,+1)$ denotes the bundle of selfdual
$2$-forms with their usual weight of $-2$. This notation appears ad hoc, but
in fact these $(l{+}1)$-tuples describe the irreducible representations of the
parabolic subgroup of $\SO(n+1,1)$ associated with conformal
geometry~\cite{BE}: this consists of those elements of $\SO(n+1,1)$ which
stabilise a chosen null line in the null cone $S^n$ of $R^{n+1,1}$ (the model
geometry on which all conformal manifolds are based).

On a conformally flat manifold there is a complex
$$(1|1,1,...)\rto[\Twist] (0|2,1,...)\rto[\CLanc](-1|2,2,...)$$ of
conformally invariant first order differential operators. The three bundles
here are the 2-forms of weight $1$, the weightless Cotton-York tensors and the
Weyl tensors of weight $-1$.  The first operator is a kind of twistor
operator, and the second is the (conformal) Lanczos potential operator. This
complex is the beginning of the Bernstein-Gelfand-Gelfand resolution of the
kernel of $\Twist$~\cite{BE}, which (locally, or on $S^n$) is the
representation $(1,1,1,0,...)$ of $\SO(n+1,1)$.

Twisting this complex by a Weyl derivative gives:
$$(-2|1,1,...)\rto[\Twist^D] (-3|2,1,...) 
\rto[\CLanc^D](-4|2,2,...).$$
This will no longer be a complex in general: $\CLanc^D\circ\Twist^D$ will
be given (up to a nonzero constant) by tensoring a $2$-form with $F^D$ and
projecting onto $(-4|2,2,...)$. Now suppose that $D$ is Einstein-Weyl,
and apply these operators to $F^D$ itself.

Since $D$ is Einstein-Weyl, $\Twist^DF^D$ is the Cotton-York tensor of $D$,
which vanishes, since $W=0$. Therefore $0=\CLanc^D\bigl(\Twist^D
F^D\bigr)=\mathrmsl{pr}\,(F^D \tens F^D)$, where \textsl{pr} denotes
projection onto $(-4|2,2,...)$ (up to a nonzero constant).  It easily follows
that $F^D$ is zero.

The same argument applied to half of the sequence
proves the four dimensional result. If $W^-$ vanishes, then the sequence
$$(1|1,-1)\rto[\Twist](0|2,-1)\rto[\CLanc](-1|2,-2)$$
is a complex~\cite{Besse,Hitchin2}. Twisting this with $D$ and applying
it to $F^D_-$ shows that $F^D_-=0$.

\medbreak

The reader may prefer to regard the above as a sketch, rather than a
full proof. The detailed proof is more complicated, because the operators
and bundles involved are complicated. In four dimensions,
it is perhaps simpler to use spinor methods, since the irreducible
representations of $\SO(4)$ are very easily described in terms of
spinors. Nevertheless, in order to see the four dimensional result
in the higher dimensional context, I shall proceed with brutal
tensor methods.

First of all, here are the operators $\Twist^D$ and $\CLanc^D$:
\begin{align*}
(\Twist^D\phi)_{X,Y}U&=D_U\phi(X,Y)
-\tfrac13(d^D\phi)(U,X,Y)\\
&\qquad\qquad-\tfrac1{n-1}\bigl(\cip{U,X}\delta^D\phi(Y)
-\cip{U,Y}\delta^D\phi(X)\bigr)\\[.2cm]
\cip{(\CLanc^D\LP)_{X,Y}U,V}&=\tfrac12\bigl(D_U\LP_{X,Y}V-D_V\LP_{X,Y}U
+D_X\LP_{U,V}Y-D_Y\LP_{U,V}X\bigr)\\
&\hbox{\hskip2cm\llap{$\displaystyle-\frac1{n-2}\Bigl(\:\:$}}
\cip{U,X}\Sdivg^D\LP(V,Y)-\cip{U,Y}\Sdivg^D\LP(V,X)\\
&\hbox{\hskip2cm\llap{$-\,$}}
\cip{V,X}\Sdivg^D\LP(U,Y)+\cip{V,Y}\Sdivg^D\LP(U,X)
\:\:\Bigr),
\end{align*} where
$\Sdivg^D\LP(U,X)=\frac12\bigl(D_{e_i}\LP_{e_i,U}X+D_{e_i}\LP_{e_i,X}U\bigr)$.
In these formulae, $\phi$ and $\LP$ can have any weight $w$, but $\Twist^D$
is only conformally invariant (independent of $D$) when $w=1$,
whereas $\CLanc^D$ is only conformally invariant when $w=0$.

To check the conformal invariance, and to find out what happens when $\phi$
and $\LP$ have the ``wrong'' weight, suppose that $\tilde D=D+\gamma$ on
$L^1$. Then, if $\phi$ has weight $w$,
$\tilde D_U\phi(X,Y)=D_U\phi(X,Y)+w\gamma(U)\phi(X,Y)
-\phi\bigl((\gamma\skwend U)X,Y\bigr)-\phi\bigl(X,(\gamma\skwend U)Y\bigr).$

\noindent
If we now use this in the formula for $\Twist^{\tilde D}$ we obtain
\begin{multline*}
(\Twist^{\tilde D}\phi)_{X,Y}U-(\Twist^D\phi)_{X,Y}U\\
=(w-1)\Bigl(\tfrac23\gamma(U)\phi(X,Y)
-\tfrac13\bigl(\gamma(X)\phi(Y,U)-\gamma(Y)\phi(X,U)\bigr)\\
\quad-\tfrac1{n-1}\bigl(\cip{U,X}\phi(\gamma,Y)
-\cip{U,Y}\phi(\gamma,X)\bigr)\Bigr).
\end{multline*}
A similar formula holds for $\CLanc^D$. Such formulae verify special
cases of the general theory of conformally invariant first order
differential operators due to Fegan~\cite{Fegan}. Equation~(\ref{delta})
is another instance of this.

Since the principal symbols of $\CLanc^D$ and $\Twist^D$ are
canonically defined in terms of the conformal structure, the composite
$\CLanc^D\circ\Twist^D\,\phi$ is obtained by taking the second
derivative $D^2\phi$ and applying an equivariant bundle homomorphism
$T\dual\Mn\tens T\dual\Mn\tens(w|1,1,...)\to(w-2|2,2,...)$.
Now the representation $(w-2|2,2,...)$ does not appear
in the decomposition of the tensor product of $(w|1,1,...)$ with
$S^2T\dual\Mn=(-2|0,...)+(-2|2,0,...)$, and so
$\CLanc^D\circ\Twist^D$ is a zero order, given by the action of the curvature
of $D$. More precisely, $\CLanc^D\circ\Twist^D\,\phi$ is obtained
by applying the composite
$$\Lambda^2T\dual\Mn\tens \lie{co}(T\Mn)\tens(w|1,1,...)
\to\Lambda^2T\dual\Mn\tens(w|1,1,...)\to(w-2|2,2,...)$$
to $R^D\tens\phi$ where the first map is the natural action of
$\lie{co}(T\Mn)$, and the second is projection.
Again since $(w-2|2,2,...)$ is not a subrepresentation of
$S^2T\dual\Mn\tens(w|1,1,...)$, the symmetric part of the Ricci tensor of
$D$ does not contribute. Therefore, on a conformally flat manifold
$\CLanc^{D^0}\circ\Twist^{D^0}=0$ for any closed Weyl derivative $D^0$,
and in particular, the conformally invariant composite
$\CLanc\circ\Twist$ is zero on $(1|1,1,...)$.
In four dimensions $\Twist^D$ and $\CLanc^D$ split into separate
operators on the selfdual and antiselfdual parts of the corresponding
tensors. The composite $\CLanc^{D^0}\circ\Twist^{D^0}$ will be zero
on $(w|1,-1)$ provided $W^-=0$. The representation theoretic reasons behind
this are well known: on antiselfdual 2-forms, $\Twist$ is the operator
$\overline D_2$ in~\cite{Besse}, section 13C.

Therefore we only need to compute the contribution of Faraday curvature
to the composite $\CLanc^D\circ\Twist^D$ on $(w|1,1,...)$.
This can be done by writing $D=D^0+\omega$ with $D^0$ exact, then
applying $\CLanc^{D^0}$ to $\Twist^D\phi$ and looking for the
$D^0\omega$ terms. The $D^0\omega$ terms in $(D^0_V\Twist^D\phi)_{X,Y}U$
are
\begin{align*}
(w-1)\Bigl(\tfrac23D^0_V\omega(U)\phi(X,Y)
&-\tfrac13\bigl(D^0_V\omega(X)\phi(Y,U)-D^0_V\omega(Y)\phi(X,U)\bigr)\\
&-\tfrac1{n-1}\bigl(\cip{U,X}\phi(D^0_V\omega,Y)
-\cip{U,Y}\phi(D^0_V\omega,X)\bigr)\Bigr).
\end{align*}
On applying the symmetries defining $\CLanc^{D^0}$, the result has
to be a constant multiple of the projection of $d\omega\tens\phi$
onto $(w-2|2,2,...)$. Indeed the only reason we need to do the
calculation is to check that for $w\neq1$, the constant multiple is
nonzero. This turns out to be the case: the constant is $w-1$.

It remains to apply $\CLanc^D\circ\Twist^D$ to $F^D$ itself.
The result is the projection of $-3F^D\tens F^D$ onto $(-4|2,2,...)$.
As shown by Eastwood and Tod~\cite{ET}, this
vanishes iff $F^D=0$. In four dimensions,
only $F^D_-\tens F^D_-$ contributes to the projection onto $(-4|2,-2)$.
It is immediate that this projection vanishes iff $F^D_-$
itself vanishes, because $(-4|2,-2)=S^2_0(-2|1,-1)$.

Hence the argument outlined earlier goes through: the formula for $\Twist^D$,
together with the Einstein-Weyl condition and Proposition~\ref{EWBianchi},
shows that $\Twist^DF^D$ is the Cotton-York tensor of $D$. By the contracted
Bianchi identity~(\ref{cy}), this vanishes if $W$ vanishes, and similarly, in
four dimensions, it is selfdual if $W$ is selfdual. Consequently,
$\CLanc^D(\Twist^DF^D)$ vanishes (or is selfdual), which in turn implies that
$F^D$ vanishes (or is selfdual).  The proofs of Theorems~\ref{mine}
and~\ref{theirs} are therefore complete.

\medbreak

In~\cite{PS1}, Pedersen and Swann studied the twistor theory of
Einstein-Weyl structures on a selfdual conformal $4$-manifold.
They found that they could only give a twistor interpretation
of the Einstein-Weyl condition when the Faraday $2$-form was
selfdual. Theorem~\ref{mine} shows that their twistor construction
in fact covers \emph{all} Einstein-Weyl structures on a selfdual
conformal $4$-manifold.

Finally, combining Proposition 5.4 of~\cite{PS1} with Theorem~\ref{mine},
we obtain:
\begin{thm} Let $D$ be a Weyl derivative on a selfdual conformal
$4$-manifold. Then the following are equivalent:
\begin{enumerate}
\item $D$ is Einstein-Weyl
\item Either $D$ is Einstein, or $D$ is the Obata connection of
an antiselfdual locally hypercomplex structure on $\Mn$.
\end{enumerate}
The Faraday $2$-form $F^D$ is then selfdual, and either $F^D=0$
or $\scal^D=0$.
\end{thm}
The hypercomplex Einstein-Weyl manifolds are also related to local heterotic
geometries~\cite{Tod3}. These are Riemannian geometries with torsion, which
arise as target spaces for $(4,0)$-supersymmetric nonlinear $\sigma$-models.

\section{The scalar curvature of a compact Einstein-Weyl manifold.}\noindent
There has been some discussion in the Einstein-Weyl literature on the sign of
the scalar curvature. Einstein manifolds have constant scalar curvature, and
the sign (positive, zero or negative) has implications for the global
topology. On a Weyl manifold, the scalar curvature is a section of $L^{-2}$,
and so the only invariant notion of ``constant scalar curvature'' is the
condition $D\scal^D=0$, which implies $D$ is exact or $\scal^D=0$.
Nevertheless, it \emph{is} meaningful to speak of the sign of $\scal^D$ at
each point, since $L^{-2}$ is oriented. Consequently it is natural to ask
whether the scalar curvature has constant sign on a compact Einstein-Weyl
$n$-manifold. My aim in this section is to prove that this is true when
$n\geq4$ and point out that there are counterexamples when $n=2,3$.  This is
contrary to some assertions in the literature: the scalar curvature was
generally believed to have constant sign in dimensions three and four, but
not in higher dimensions~\cite{PS2,PT}.

One motivation for including this material here is to clarify (I hope)
the situation. Another reason is that the Faraday $2$-form enters
in a fundamental way through~\ref{LapScal}:
$$\Laplace^D\scal^D=-n(n-4)|F^D|^2.$$

The main result of this section will follow by applying the strong
(Hopf) maximum principle to this Laplacian. In order to do this,
we need to turn $\scal^D$ into a (weightless) function, by
choosing a length scale $\mu$. If $D=D^\mu+\omega^\mu$ then on $L^{-2}$,
$$\Laplace^D=\Laplace^\mu+(n-6)D^\mu_{\flat\omega^\mu}
-2\trace D^\mu\omega^\mu-2(n-4)\cip{\omega^\mu,\omega^\mu}.$$

The gauge $\mu$ is said to be a \emphdef{Gauduchon gauge} iff $\trace
D^\mu\omega^\mu=0$. Such a gauge exists uniquely up to homothety if $\Mn$ is
compact (\cite{Gauduchon1}---see also~\cite{CP,Tod1}), and in any such gauge,
$$\Laplace^\mu\scal^D+(n-6)D^\mu_{\flat\omega^\mu}\scal^D
-2(n-4)\cip{\omega^\mu,\omega^\mu}\scal^D=-n(n-4)|F^D|^2.$$
We can now multiply by $\mu^2$ to make $\scal^D$ into a function.

It immediately follows that if $n>4$, $\mu^2\scal^D$ cannot have a negative
local minimum unless $\omega^\mu=0$: at a local minimum
$\Laplace^\mu\scal^D\geq0$ and $D^\mu_{\flat\omega^\mu}\scal^D=0$, from which
it follows that $2\cip{\omega^\mu,\omega^\mu}\scal^D\geq n|F^D|^2\geq0$.
(This can also be established using a formula for $\Laplace^\mu\omega^\mu$
given in~\cite{Gauduchon4}.)

The Hopf maximum principle (see~\cite{GilbT}) provides
the following refinement of this result.
\begin{itemize}
\item If $n\geq4$, $\mu^2\scal^D$ cannot have a nonpositive minimum
unless it is constant.
\item If $n\leq4$, $\mu^2\scal^D$ cannot have a nonpositive maximum
unless it is constant.
\end{itemize}

Therefore when $n=2,3$ we cannot rule out the possibility that
$\mu^2\scal^D$ has negative minima and positive maxima. On the other hand,
for $n\geq4$ we have that in any Gauduchon gauge, either $\mu^2\scal^D$
is constant, or its local minima are all positive
(and in four dimensions the local maxima are also all positive).

In the compact case, the existence of the Gauduchon gauge, together
with the existence of maxima and minima, implies that $\scal^D$ has
constant sign on Einstein-Weyl $n$-manifolds when $n\geq4$.

This result is contrary to Corollary 4.4 in~\cite{PS2}, where the
authors claim that some of their examples have scalar curvature with
non-constant sign. The error seems to involve inconsistent choices
of the branch of an inverse trigonometric function in the evaluation
of the scalar curvature at $\gamma^{-1}(c_1/2)$.

\medbreak

I finish this section by remarking that the above dichotomy between
low dimensions and higher dimensions is genuine in the
sense that in two and three dimensions there \emph{do} exist compact
Einstein-Weyl manifolds with scalar curvature of nonconstant sign. This is
also contrary to some remarks in the literature
(e.g.~\cite{PT} p.~107,~\cite{PS2} p.~99). I will show here that
there are Einstein-Weyl structures with scalar curvature of nonconstant
sign on $S^3$. Examples also exist on $S^2$ and $S^1\times S^1$:
see~\cite{DMJC1}.

In~\cite{PT}, Pedersen and Tod found Einstein-Weyl
structures on $S^3$ parameterised by $a,b,c\in\R$ with $a\neq0$. In his
classification of possible compact Einstein-Weyl geometries~\cite{Tod1},
Tod reobtained these solutions in the following form.
\begin{align*}
 g&=P(v)^{-1}dv^2+P(v)dy^2+v^2(dt+Cv^{-2}dy)^2\\
\omega&=2\lambda v^2(dt+Cv^{-2}dy),\\
D&=D^g+\omega\\
\tag*{\text{where}}  P(v)&=-\lambda^2v^4+Av^2+B-C^2v^{-2}
\end{align*}
and $\lambda,A,B,C$ are arbitrary constants, related to the parameters
$a,b,c$ in~\cite{PT} by:
$$A=-a^2+b^2+c^2,\quad \lambda^2B=a^2b^2+a^2c^2-b^2c^2,
\quad \lambda^4C^2=a^2b^2c^2.$$
Tod's change of coordinates is well defined provided
$b\neq c$ and $b\neq -c$. The scalar curvature is
$\scal^D=-6A+3|\omega|^2=6(a^2-b^2-c^2+2\lambda^2v^2)$.
Also, the range of $\lambda^2v^2$ when $P(v)\geq0$ is the interval $[b^2,c^2]$.
Therefore, for $|b^2-c^2|>a^2$, the scalar curvature has nonconstant sign.

\def\section{\@startsection{section}{1}%
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\bibliographystyle{amsplain}
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