# Petrov Types for Riemannian 4-manifolds

Amir Babak Aazami  Clark University Worcester, MA 01610<EMAIL_ADDRESS>
###### Abstract.

On an oriented 4-manifold, we examine the geometry that arises when the
curvature operator of a Riemannian metric $g$ commutes, not with its own Hodge
star operator, but with that of a Lorentzian metric that is a deformation of
$g$. This leads to two complementary notions of “Petrov Type” for $g$, one
directly in terms of $g$’s curvature operator, the other in terms of a variant
of it. Both versions lead to pointwise classifications, and both include among
them Riemannian metrics that deform the Einstein condition in the direction of
a fixed vector field. While the first version is more direct, the second
version has “normal forms,” in that the Petrov Type of $g$ will be determined
by the critical points of a certain quadratic form. We close by generalizing
our construction beyond Lorentzian metrics.

##  1\. Introduction

Of the many attributes of 4-dimensional Riemannian geometry, not least is its
beautiful characterization of Einstein metrics, namely, those metrics $g$
whose Ricci tensor $\text{Ric}_{\scalebox{0.6}{\emph{g}}}$ satisfies

$\displaystyle\text{Ric}_{\scalebox{0.6}{\emph{g}}}=\lambda g$ (1)

for some $\lambda\in\mathbb{R}$. Lurking behind (1) is the Hodge star operator
$*$ of an oriented Riemannian 4-manifold $(M,g)$, which splits the second
exterior product $\Lambda^{2}$ into the direct sum of “self-dual” ($*\xi=\xi$)
and “anti-self-dual” ($*\xi=-\xi$) eigenspaces. As shown in [Ber61] and
[ST69], the Einstein condition (1) arises precisely when $*$ commutes with
$g$’s curvature operator $\hat{R}_{\scalebox{0.6}{\emph{g}}}$. Even more, such
a $g$ will have a “normal form,” in that $\hat{R}_{\scalebox{0.6}{\emph{g}}}$
will be determined by knowledge of just the critical points and values of
$g$’s sectional curvature function, viewed as the quadratic form of
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$. Two important ingredients go into making
this characterization possible: i) Because
$\hat{R}_{\scalebox{0.6}{\emph{g}}},*\colon\Lambda^{2}\longrightarrow\Lambda^{2}$
both derive from the same metric $g$, they are both self-adjoint with respect
to the $g$-induced inner product
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$ on $\Lambda^{2}$; ii)
the positive-definiteness of
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$ then makes their
diagonalization possible (simultaneously so, since they commute), which,
together with the eigenstructure of $*$, ultimately makes $g$’s normal form
possible. A similar construction arises in Lorentzian geometry — but with one
important difference: As shown in [Tho69], although oriented Lorentzian
Einstein 4-manifolds $(M,g_{\scalebox{0.4}{\emph{L}}})$ are characterized by
the same commuting property as above, the critical points of
$g_{\scalebox{0.4}{\emph{L}}}$’s sectional curvature function do not by
themselves determine the curvature operator
$\hat{R}_{\scalebox{0.4}{\emph{L}}}$. Rather, they can only determine
$g_{\scalebox{0.4}{\emph{L}}}$’s “Petrov Type” [Pet69], a classification of
4-dimensional spacetimes prominent within general relativity. The reason is
because $g_{\scalebox{0.4}{\emph{L}}}$ is not positive-definite, so that i)
holds but not ii). In this paper we combine the Riemannian with the
Lorentzian, and further engage with [Tho69, ST69], by studying the geometry
that arises when the curvature operator $\hat{R}_{\scalebox{0.6}{\emph{g}}}$
of a Riemannian metric $g$ commutes with the Hodge star operator
$*_{\scalebox{0.4}{\emph{L}}}$ of a Lorentzian metric
$g_{\scalebox{0.4}{\emph{L}}}$. If $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ does
not commute with $*_{\scalebox{0.4}{\emph{L}}}$, then we may take its
“symmetric” part, $\hat{S}_{\scalebox{0.6}{\emph{g}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\frac{1}{2}(\hat{R}_{\scalebox{0.6}{\emph{g}}}-*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}),$
which will commute by construction (thus $\hat{S}_{\scalebox{0.6}{\emph{g}}}$
plays a role analogous to the Weyl curvature tensor). Although this sort of
inquiry can be pursued fairly generally — after all, one may take the Hodge
star $\tilde{*}$ of any metric $\tilde{g}$ on $M$ — we focus here on
Lorentzian metrics because our $\tilde{g}$’s are obtained by deforming $g$ in
the direction of a fixed $g$-unit length vector field $T$ on $M$,

$\tilde{g}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g+\varepsilon
T^{\flat}\otimes T^{\flat},$

where $T^{\flat}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g(T,\cdot)$
and $\varepsilon$ is any real number not equal to $-1$. The reason for this
choice is because if $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ commutes with such a
$\tilde{*}$, then $g$’s Ricci tensor will be “close” to being Einstein — see
(3) below. However, only the cases $\varepsilon=0$ or $\varepsilon=-2$ prove
interesting — and $\varepsilon=-2$ is a Lorentzian metric. Now, if one’s
interest is global, then the very existence of $T$ requires $M$ to have Euler
characteristic zero. By [Ber61] (see Lemma 1 below), such manifolds have no
non-flat Einstein metrics — making them, for that reason, a natural place in
which to search for “almost-Einstein” metrics (though in Section 6 we will in
fact generalize beyond Lorentzian metrics). Therefore, whether local or
global, we weaken i) above by asking: If $\hat{R}_{\scalebox{0.6}{\emph{g}}}$
is the curvature operator of $g$, and $*_{\scalebox{0.4}{\emph{L}}}$ is the
Hodge star operator of the Lorentzian metric

$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat},$

then what geometry arises for $g$ if

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}?$
(2)

In fact this immediately weakens ii) as well. Indeed, while
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ is self-adjoint with respect to its
$g$-induced inner product
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$ on $\Lambda^{2}$, and
while $*_{\scalebox{0.4}{\emph{L}}}$ is self-adjoint with respect to the
$g_{\scalebox{0.4}{\emph{L}}}$-induced inner product
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ on $\Lambda^{2}$,
neither is self-adjoint with respect to the other’s inner product. We will
address this issue in two ways, each of which will lead to different
classifications of $g$ but to complementary notions of “almost-Einstein”
metric:

  1. 1.

Work as is with $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ over
$(\Lambda^{2},\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}})$. As we
show in Section 4, in any oriented $g$-orthonormal basis of the form
$\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$,
the Ricci tensor of a metric $g$ satisfying (2) will take the form

$\displaystyle\text{Ric}_{\scalebox{0.6}{\emph{g}}}=\begin{bmatrix}\lambda&*&*&*\\\
*&\lambda&0&0\\\ *&0&\lambda&0\\\ *&0&0&\lambda\end{bmatrix},$ (3)

where $\lambda\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{Ric}(T,T)$
is not constant in general. This is one realization for us of an “almost-
Einstein” metric — a deformation of (1) along $T$. The salient point is that
even though (3) is basis-dependent, (2) is not. Can such a $g$ have a normal
form? The answer is “not quite” — because in losing self-adjointness, we also
lose the algebraic Bianchi identity (of $\hat{R}_{\scalebox{0.6}{\emph{g}}}$
relative to $\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$).
_Nevertheless, as we show Theorems 2 and 3, we can associate a “Petrov Type”
to $g$ by classifying the _complex_ eigenstructure of
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_ that arises from the fact that
_$*_{\scalebox{0.4}{\emph{L}}}^{2}=-1$_. We can also establish a relationship
between these eigenvectors and the critical points of
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_ ’s
_$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_ -quadratic form.
If _$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_ does not commute with
_$*_{\scalebox{0.4}{\emph{L}}}$_ , then Theorems 2 and 3 extend to the
symmetric part, _$\hat{S}_{\scalebox{0.6}{\emph{g}}}$_ , of
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ ;_ see Theorem 4._

  2. 2.

For the second approach, we modify $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ into a
new operator,
“${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$,” that _is_
self-adjoint with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ (see Definition 7).
While ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is not
quite the curvature operator $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ (though it
is close), the advantage gained here is that
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ and
$*_{\scalebox{0.4}{\emph{L}}}$ are now both self-adjoint with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$. Replacing (2) with
the condition

$\displaystyle{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}},$
(4)

the methods of [Tho69] now become fully available to us. _As we show in
Theorems 6 and 7, we can also associate a Petrov Type to $g$ by classifying
the complex eigenstructure of
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ — and
these now have normal forms, in that their Petrov Types will be completely
determined by the critical points of
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ ’s
_$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_ -quadratic form_.
Relative to $\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$,
the Ricci tensors of such $g$’s will take the form

$\displaystyle\text{Ric}_{\scalebox{0.6}{\emph{g}}}=\begin{bmatrix}\lambda&0&0&0\\\
0&\lambda+\lambda_{2}&*&*\\\ 0&*&\lambda+\lambda_{3}&*\\\
0&*&*&\lambda+\lambda_{4}\end{bmatrix},$ (5)

with $\lambda\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{Ric}(T,T)$
once again. Note the complementary forms of (3) and (5), how each deforms (1)
along $T$, but, so to speak, in opposite ways. Again, the salient point is
that (4) is basis-independent. _Finally, in Theorem 8 we extend these results
to _${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ , the
symmetric part of_
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$.

In conclusion, the approach taken in this paper is to have the commuting
condition

$\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*=*\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}$

take priority over the Einstein condition
$\text{Ric}_{\scalebox{0.6}{\emph{g}}}=\lambda g$, by viewing the latter as
merely the special (though most important) case where
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ and $*$ arise from the _same metric $g$_.
We further elaborate on this viewpoint in Section 6, by taking the standard
round metric $\mathring{g}$ on $\mathbb{S}^{4}$ in place of
$g_{\scalebox{0.4}{\emph{L}}}$, and examining (2) or (4) with $\mathring{*}$
instead of $*_{\scalebox{0.4}{\emph{L}}}$ (or any Riemannian metric
$\tilde{g}$ on any oriented 4-manifold $M$). In short, although Einstein
metrics may not be abundant in dimension 4, “normal forms” are.

##  2\. The Hodge star operator and normal forms

In this section we provide a brief overview as well as some historical remarks
regarding normal forms and the Hodge star operator in both the Riemannian and
Lorentzian settings. The study of “normal forms” for curvature operators is
motivated by a well known fact from linear algebra: On any finite-dimensional
inner product space $(V,\langle\,,\rangle)$, a self-adjoint linear
transformation $T\colon V\longrightarrow V$ is determined by the critical
points of its associated _quadratic form_ function on the sphere of unit
length vectors:

$\displaystyle v\mapsto\langle Tv,v\rangle\hskip 14.45377pt,\hskip
14.45377pt|v|=1.$ (6)

The connection to Riemannian geometry is due to the fact that the sectional
curvature of a Riemannian manifold $(M,g)$ can also be realized as a quadratic
form. First, fix $p\in M$ and observe that $\Lambda^{2}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\Lambda^{2}(T_{p}M)$
inherits an inner product
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$ from $g$ via

$\displaystyle\langle{v_{1}\wedge w_{1}},{v_{2}\wedge
w_{2}}\rangle_{\\!\scalebox{0.7}{\emph{g}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{det}\begin{bmatrix}g(v_{1},v_{2})&g(v_{1},w_{2})\\\
g(w_{1},v_{2})&g(w_{1},w_{2})\end{bmatrix}\cdot$ (7)

With this in hand, we may express the action of the Riemann curvature 4-tensor
$R$ as a linear map called the _curvature operator_ ,

$\hat{R}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda^{2}\longrightarrow\Lambda^{2},$

whose action $v\wedge w\mapsto\hat{R}_{\scalebox{0.6}{\emph{g}}}(v\wedge w)$
is defined to be the unique 2-vector satisfying

$\displaystyle\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(v\wedge w)},{x\wedge
y}\rangle_{\\!\scalebox{0.7}{\emph{g}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-R(v,w,x,y)\hskip
14.45377pt\text{for all $x,y\in T_{p}M$}.$ (8)

Owing to the symmetry $R_{ijkl}=R_{klij}$,
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ is self-adjoint on the inner product
space $(\Lambda^{2},\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}})$. It
is with respect to $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ that the sectional
curvature of $g$ is a quadratic form. Indeed, for any orthonormal pair $v,w\in
T_{p}M$, the sectional curvature $\text{sec}_{\scalebox{0.6}{\emph{g}}}$ of
the 2-plane $v\wedge w$ is

$\displaystyle\text{sec}_{\scalebox{0.6}{\emph{g}}}(v\wedge w)\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=R(v,w,w,v)=\underbrace{\,\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(v\wedge
w)},{v\wedge w}\rangle_{\\!\scalebox{0.7}{\emph{g}}}\,}_{\text{``$\langle
Tv,v\rangle$"}}.$ (9)

(Our sign convention is $R(a,b,c,d)\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g(\nabla_{\\!a}{\\!\nabla_{\\!b}{\,c}}-\nabla_{\\!b}{\\!\nabla_{\\!a}{c}}-\nabla_{\\![a,b]}{c},d)$.)
Given this analogy with (6), it is thus natural to study
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ by studying the critical point behavior
of $\text{sec}_{\scalebox{0.6}{\emph{g}}}$ — the goal being to classify those
curvature tensors which are determined by knowledge of just the critical point
structure of $\text{sec}_{\scalebox{0.6}{\emph{g}}}$. Such curvature operators
are then said to have a “normal form.” (Observe that this problem is more
difficult than (6) because not all unit 2-vectors in $\Lambda^{2}$ correspond
to 2-planes in $T_{p}M$. Rather, only those unit 2-vectors $\xi$ that are
decomposable do; i.e., those that can be written as $\xi=v\wedge w$ for
$v,w\in T_{p}M$.) As shown in [ST69], if one restricts attention to the class
of Einstein metrics on a smooth 4-manifold $M$, then such metrics are indeed
determined solely by the critical point structure of
$\text{sec}_{\scalebox{0.6}{\emph{g}}}$ — i.e., the class of Einstein
4-manifolds does indeed possess a “normal form.” As mentioned in the
Introduction, what makes this beautiful result possible is that, in dimension
4, Einstein metrics $g$ are exactly those whose curvature operators
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ commute with the Hodge star operator $*$,

$\displaystyle*\colon\Lambda^{2}\longrightarrow\Lambda^{2}\hskip
14.45377pt,\hskip 14.45377pt\text{$g$ Einstein
$\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*=*\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}$},$
(10)

where the Hodge star operator $*$ is defined as sending any 2-plane $v\wedge
w$ to its $\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$-orthogonal
complement. The precise definition is

$\displaystyle\xi\wedge*\eta\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\langle{\xi},{\eta}\rangle_{\\!\scalebox{0.7}{\emph{g}}}\,dV\hskip
14.45377pt,\hskip 14.45377pt\xi,\eta\in\Lambda^{2},$ (11)

where $dV$ is the orientation form in the 1-dimensional space
$\Lambda^{4}(T_{p}M)$. (As we’ll see below, these definitions would not change
if we replaced the Riemannian metric $g$ with a Lorentzian metric
$g_{\scalebox{0.4}{\emph{L}}}$, as we will very soon do.) Thanks to $*$, there
is in fact another way to express the Einstein condition in dimension 4 (see
[ST69]):

$g$ is Einstein $\iff\text{sec}_{\scalebox{0.6}{\emph{g}}}(v\wedge
w)=\text{sec}_{\scalebox{0.6}{\emph{g}}}(*(v\wedge w))$. (12)

I.e., $g$ is Einstein if and only if the sectional curvature of each 2-plane
is equal to that of its orthogonal complement. With these preliminaries out of
the way, let us comment briefly on what else has been done regarding normal
forms in the Riemannian setting. Continuing the study of the pointwise
sectional curvature function $\text{sec}_{\scalebox{0.6}{\emph{g}}}$, Thorpe
analyzed its zero set when $\text{sec}_{\scalebox{0.6}{\emph{g}}}\geq 0$ in
[Tho71], and also gave a condition for a 4-manifold to have
$\text{sec}_{\scalebox{0.6}{\emph{g}}}>0$, in [Tho72]. Dimension 4 is
particularly rich: Aside from Einstein metrics, the class of Kähler
4-manifolds also possesses a normal form, as shown in [Joh80a]. In fact
Johnson also found normal forms for 6-dimensional Kähler manifolds with
positive sectional curvature, in [Joh80b]. As one may imagine, however, in
higher dimensions the behavior of $\text{sec}_{\scalebox{0.6}{\emph{g}}}$
becomes more difficult to analyze in general, as shown in [Zol79]; indeed,
aside from the Kähler case in dimension 6, the author knows of only one other
higher-dimensional result, in [Zol80], who found normal forms for a special
class of curvature tensors in dimension 5. The study of sectional curvature
related to and inspired by these works continues to the present day; e.g., on
the eigenvalues of $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ and their relationship
to the topology of the underlying manifold $M$, or on a variant of
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ known as the _curvature operator of the
second kind_ (see, e.g., the recent works [BM17, PW21, CGT23, Li23]). So, too,
does the quest for normal forms — even in dimension 4. Indeed, [CT16, Remark
2.6] posed the question of whether gradient Ricci 4-solitons, which satisfy

$\text{Ric}_{\scalebox{0.6}{\emph{g}}}+\text{Hess}\,f=\lambda g$

for some smooth function $f$ on $M$, also possess a normal form. Therefore,
the pursuit of normal forms remains to this day an active one. Now we turn to
Lorentzian geometry. First, recall that a _Lorentzian metric_
$g_{\scalebox{0.4}{\emph{L}}}$ on a smooth manifold $M$ is a smooth
nondegenerate metric with signature $(-\\!+\\!+\\!\cdots+)$. The absence of
positive-definiteness implies that all nonzero vectors $X\in TM$ come in three
flavors, which terminology we will use freely in this paper:

$\displaystyle\text{$X$
is}~{}\left\\{\begin{array}[]{ccc}\text{``spacelike"}&\text{if}&\text{$g_{\scalebox{0.4}{\emph{L}}}(X,X)>0$},\\\
\text{``timelike"}&\text{if}&\text{$g_{\scalebox{0.4}{\emph{L}}}(X,X)<0$},\\\
\text{``lightlike"}&\text{if}&\text{$g_{\scalebox{0.4}{\emph{L}}}(X,X)=0$}.\end{array}\right.$
(16)

The Hodge star operator $*_{\scalebox{0.4}{\emph{L}}}$ of a Lorentzian
4-manifold $(M,g_{\scalebox{0.4}{\emph{L}}})$ is defined in perfect analogy
with that of a Riemannian metric $g$ on $M$; however, owing to a certain sign
difference (see (29) below), let us go through the derivation with some care.
To begin with, our $g_{\scalebox{0.4}{\emph{L}}}$ will not be chosen
arbitrarily, but rather “directly” from $g$ itself, as follows: For a suitable
choice of $g$-unit length vector field $T$, let us form the metric

$\displaystyle g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat},$ (17)

where $T^{\flat}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g(T,\cdot)$
is the one-form $g$-metrically equivalent to $T$. Notice that
$g_{\scalebox{0.4}{\emph{L}}}(T,T)=-1$, so that $T$ is unit _timelike_ with
respect to $g_{\scalebox{0.4}{\emph{L}}}$. _Note also that any $g$-orthonormal
basis containing $T$ is automatically a _$g_{\scalebox{0.4}{\emph{L}}}$_
-orthonormal basis, and vice-versa_ — though this is not true for arbitrary
$g$-orthonormal bases. Generally speaking, the more “distinguished” $T$ is —
e.g., if it is closed, $dT^{\flat}=0$, or if it is a Killing vector field,
$\mathfrak{L}_{T}g=0$ — the more similar the properties of $g$ and
$g_{\scalebox{0.4}{\emph{L}}}$ will be; see [Ole14] for a careful treatment.
Having said that, whatever the choice of $T$, if we take an oriented local
$g_{\scalebox{0.4}{\emph{L}}}$-orthonormal frame
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ with $e_{1}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=T$,
then the 2-vectors

$\displaystyle\\{e_{1}\wedge e_{2}\,,\,e_{1}\wedge e_{3}\,,\,e_{1}\wedge
e_{4}\,,\,e_{3}\wedge e_{4}\,,\,e_{4}\wedge e_{2}\,,\,e_{2}\wedge e_{3}\\}$
(18)

will be an orthonormal basis for $\Lambda^{2}$ with respect to the Lorentzian
inner product on $\Lambda^{2}$, which is analogous to (7) and defined by

$\displaystyle\langle{v_{1}\wedge w_{1}},{v_{2}\wedge
w_{2}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{det}\begin{bmatrix}g_{\scalebox{0.4}{\emph{L}}}(v_{1},v_{2})&g_{\scalebox{0.4}{\emph{L}}}(v_{1},w_{2})\\\
g_{\scalebox{0.4}{\emph{L}}}(w_{1},v_{2})&g_{\scalebox{0.4}{\emph{L}}}(w_{1},w_{2})\end{bmatrix}\cdot$
(19)

(Note that the first three basis elements in (18) are all timelike,

$\langle{e_{1}\wedge e_{i}},{e_{1}\wedge
e_{i}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=-1\hskip 14.45377pt,\hskip
14.45377pti=2,3,4,$

so that $\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ has
signature $(-\\!-\\!-\\!+\\!++)$.) Now we define the Hodge star operator
$*_{\scalebox{0.4}{\emph{L}}}$ with respect to $g_{\scalebox{0.4}{\emph{L}}}$,
in perfect analogy with (11):

$\xi\wedge*_{\scalebox{0.4}{\emph{L}}}\eta\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\langle{\xi},{\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,dV\hskip
14.45377pt,\hskip 14.45377pt\xi,\eta\in\Lambda^{2}.$

Bearing in mind that $g_{\scalebox{0.4}{\emph{L}}}(e_{1},e_{1})=-1$, observe
that the action of $*_{\scalebox{0.4}{\emph{L}}}$ on the basis (18) is

$\displaystyle\left\\{\begin{array}[]{lr}*_{\scalebox{0.4}{\emph{L}}}(e_{1}\wedge
e_{2})=-e_{3}\wedge e_{4},\\\ *_{\scalebox{0.4}{\emph{L}}}(e_{1}\wedge
e_{3})=-e_{4}\wedge e_{2},\\\ *_{\scalebox{0.4}{\emph{L}}}(e_{1}\wedge
e_{4})=-e_{2}\wedge e_{3},\end{array}\right.\hskip 14.45377pt,\hskip
14.45377pt\left\\{\begin{array}[]{lr}*_{\scalebox{0.4}{\emph{L}}}(e_{3}\wedge
e_{4})=e_{1}\wedge e_{2},\\\ *_{\scalebox{0.4}{\emph{L}}}(e_{4}\wedge
e_{2})=e_{1}\wedge e_{3},\\\ *_{\scalebox{0.4}{\emph{L}}}(e_{2}\wedge
e_{3})=e_{1}\wedge e_{4},\end{array}\right.$ (26)

or in block matrix form,

$\displaystyle*_{\scalebox{0.4}{\emph{L}}}=\begin{bmatrix}O&I\\\
-I&O\end{bmatrix},$ (27)

where $I$ is the $3\times 3$ identity matrix. By contrast, the Riemannian
Hodge star (11) is

$\displaystyle*=\begin{bmatrix}O&I\\\ I&O\end{bmatrix}$ (28)

with respect to the same basis (18). While $*$ is diagonalizable,
$*_{\scalebox{0.4}{\emph{L}}}$ is not (even though
$*_{\scalebox{0.4}{\emph{L}}}$ is self-adjoint with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$). Furthermore,

$\displaystyle*^{2}=1\hskip 14.45377pt\text{whereas}\hskip
14.45377pt*_{\scalebox{0.4}{\emph{L}}}^{2}=-1.$ (29)

The almost complex structure that the latter defines on each $\Lambda^{2}$
will be crucial to our construction of Riemannian Petrov Type below — just as
it was in the original Lorentzian construction of Petrov Type; see, e.g.,
[Tho69] and [O’N95, Chapter 5]. Let us describe this now. Because
$*_{\scalebox{0.4}{\emph{L}}}^{2}=-1$, one loses the self-dual/anti-self-dual
splitting of $\Lambda^{2}$. Having said that, one does gain a _complex
structure_ , via $i\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=*_{\scalebox{0.4}{\emph{L}}}$.
Then the commuting condition
$\widehat{W}_{\scalebox{0.4}{\emph{L}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ\widehat{W}_{\scalebox{0.4}{\emph{L}}},$
where $\widehat{W}_{\scalebox{0.4}{\emph{L}}}$ is the Weyl curvature operator
of $g_{\scalebox{0.4}{\emph{L}}}$ — which, being trace-free, automatically
commutes with $*_{\scalebox{0.4}{\emph{L}}}$ — turns
$\widehat{W}_{\scalebox{0.4}{\emph{L}}}$ into a _complex-linear_ map on
$\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$. The “Petrov Type of
$g_{\scalebox{0.4}{\emph{L}}}$” is then precisely the complex eigenstructure
of $\widehat{W}_{\scalebox{0.4}{\emph{L}}}$ (see [O’N95, Chapter 5]). This
eigenstructure was shown in [Tho69] to coincide with the _number_ of critical
points of $g_{\scalebox{0.4}{\emph{L}}}$’s sectional curvature function (see
also [Bes07, pp. 98ff.]). For example, the Petrov Type of vacuum black holes
is equivalent to their sectional curvature functions having exactly one
“spacelike” critical point at any point on $M$. Therefore, in having
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ commute with a Lorentzian
$*_{\scalebox{0.4}{\emph{L}}}$, we can associate complex eigenstructures and
thus Petrov Types. (Compare this with [AG97], wherein it was shown that an
oriented Einstein 4-manifold will admit a (locally) Hermitian structure if and
only if its self-dual Weyl tensor ${\widehat{W}}^{+}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\frac{1}{2}(\widehat{W}+\widehat{W}\circ*)$
has at least two of its three eigenvalues equal.) In doing so, the following
is helpful to keep in mind:

######  Definition 1.

Let _$(M,g_{\scalebox{0.4}{\emph{L}}})$_ be an oriented Lorentzian 4-manifold.
At any $p\in M$, let $P$ denote an oriented 2-dimensional subspace of
$T_{p}M$. Then $P$ is _nondegenerate_ if the restriction of
_$g_{\scalebox{0.4}{\emph{L}}}$_ to $P$, _$g_{\scalebox{0.4}{\emph{L}}}|_{P}$_
, is nondegenerate. The _sign of $P$_, denoted
_$\epsilon_{\scalebox{0.4}{\emph{L}}}(P)=\pm 1$_ , is defined to be $-1$ if
_$g_{\scalebox{0.4}{\emph{L}}}|_{P}$_ is Lorentzian and $+1$ if
_$g_{\scalebox{0.4}{\emph{L}}}|_{P}$_ is positive-definite. The 2-plane
_$g_{\scalebox{0.4}{\emph{L}}}$ -orthogonal_ to a nondegenerate $P$, denoted
by _$P^{\perp_{\scalebox{0.35}{\emph{L}}}}$_ , is defined to be __

$P^{\perp_{\scalebox{0.35}{\emph{L}}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=*_{\scalebox{0.4}{\emph{L}}}P.$

Finally, following [Tho69], let $G_{+}(p)\cup
G_{-}(p)\subseteq\Lambda^{2}(T_{p}M)$ denote the 2-Grassmannians of all
decomposable 2-vectors of length $\pm 1$, respectively _;_ i.e., the set of
all oriented, nondegenerate 2-dimensional subspaces of $T_{p}M$. Note that
_$\epsilon_{\scalebox{0.4}{\emph{L}}}(P)=\langle{P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_
for any $P\in G_{\pm}(p)$.

($P^{\perp_{\scalebox{0.35}{\emph{L}}}}$ should be distinguished from its
Riemannian version $P^{\perp}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=*P$.
Also, recall that a 2-vector $\xi\in\Lambda^{2}$ is decomposable
$\Leftrightarrow\xi\wedge\xi=0\Leftrightarrow\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=0$.)
We close this section by writing the matrix of
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ with respect to (18). Denoting by
“$R_{ijkl}$” the components of the Riemann curvature 4-tensor of $g$, we have
(recall the minus sign in (8))

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}=-\\!\begin{bmatrix}R_{1212}&R_{1312}&R_{1412}&R_{3412}&R_{4212}&R_{2312}\\\
R_{1213}&R_{1313}&R_{1413}&R_{3413}&R_{4213}&R_{2313}\\\
R_{1214}&R_{1314}&R_{1414}&R_{3414}&R_{4214}&R_{2314}\\\
R_{1234}&R_{1334}&R_{1434}&R_{3434}&R_{4234}&R_{2334}\\\
R_{1242}&R_{1342}&R_{1442}&R_{3442}&R_{4242}&R_{2342}\\\
R_{1223}&R_{1323}&R_{1423}&R_{3423}&R_{4223}&R_{2323}\end{bmatrix},$ (30)

which, owing to the symmetry $R_{ijkl}=R_{klij}$, has the block form

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}=-\\!\begin{bmatrix}A&B\\\
B^{t}&D\end{bmatrix},$ (31)

with $A$ and $D$ symmetric $3\times 3$ matrices and $B^{t}$ the transpose of
the $3\times 3$ matrix $B$, which is not symmetric in general. Now, in order
to motivate our construction in Section 3 below, let us return for a moment to
the fully Riemannian setting and recall the Einstein condition (10),

$\text{Ric}=\lambda
g\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*=*\circ\hat{R}_{\scalebox{0.6}{\emph{g}}},$

which, we now see from (31), is the case if and only if $A=D$ and $B^{t}=B$.
And as we saw in (12),

$\displaystyle\underbrace{\,\text{sec}_{\scalebox{0.6}{\emph{g}}}(P)=\text{sec}_{\scalebox{0.6}{\emph{g}}}(P^{\perp})\,}_{\text{for
any 2-plane
$P$}}\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*=*\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}=-\\!\begin{bmatrix}A&B\\\
B&A\end{bmatrix},$ (32)

where $P^{\perp}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=*P$.
If instead we had required that $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ and $*$
_anti_ -commute, then $\text{sec}_{\scalebox{0.6}{\emph{g}}}(P^{\perp})$ would
have the opposite sign; in fact,

$\displaystyle\underbrace{\,\text{sec}(P)=-\text{sec}(P^{\perp})\,}_{\text{for
any 2-plane
$P$}}\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*=-(*\circ\hat{R}_{\scalebox{0.6}{\emph{g}}})\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}=-\\!\begin{bmatrix}A&B\\\
-B&-A\end{bmatrix}\cdot$ (33)

In the next section, we are going to study geometries that are in some sense
_intermediate between ( 32) and (33)_ — see (36) below.

##  3\. almost-Einstein metrics

We now begin our construction, by first of all defining the class of “almost-
Einstein” metrics whose geometry we wish to investigate:

######  Definition 2 (almost-Einstein metric).

An oriented Riemannian 4-manifold $(M,g)$ is _almost-Einstein_ if there is a
nowhere vanishing vector field $T$ on $M$, together with local ordered frames
$\\{T,X_{1},X_{2},X_{3}\\}$ in a neighborhood of each point, with respect to
which the Ricci tensor Ric of $g$ uniformly takes exactly one of the following
forms _:_

$\displaystyle\underbrace{\,\emph{\text{Ric}}=\begin{bmatrix}\lambda&\psi_{1}&\psi_{2}&\psi_{3}\\\
\psi_{1}&\lambda&0&0\\\ \psi_{2}&0&\lambda&0\\\
\psi_{3}&0&0&\lambda\end{bmatrix}\,}_{\emph{\text{``Type A"}}}\hskip
7.22743pt,\hskip
7.22743pt\underbrace{\,\emph{\text{Ric}}=\begin{bmatrix}\lambda&0&0&0\\\
0&\lambda+\lambda_{2}&\psi_{2}&\psi_{3}\\\
0&\psi_{2}&\lambda+\lambda_{3}&\psi_{4}\\\
0&\psi_{3}&\psi_{4}&\lambda+\lambda_{4}\end{bmatrix}\,}_{\emph{\text{``Type
B"}}},$ (34)

Here $\lambda\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\emph{\text{Ric}}(T,T)$
and $\lambda_{i},\psi_{i}$ are smooth functions.

Einstein metrics are locally, if not always globally, almost-Einstein: With
respect to any local orthonormal frame, the Ricci tensor of an Einstein metric
will be of either Type in (34) (with $\lambda$ a constant and each
$\psi_{i}=\lambda_{i}=0$). Almost-Einstein metrics therefore deform the
Einstein condition “in the direction of a fixed vector field $T$,” and the two
types complement each other. Here is one reason why they are worth
considering:

######  Lemma 1.

Let $g$ be an almost-Einstein metric on a closed 4-manifold. If $g$ is
Einstein, then it is flat.

###### Proof.

If the closed 4-manifold $M$ globally admits a nowhere vanishing vector field
$T$, then it must have Euler characteristic zero: $\chi(M)=0$. But by a
classical result of Berger [Ber61], on such a 4-manifold any Einstein metric
$g$ must be flat; indeed, by the Chern-Gauss-Bonnet formula in dimension 4,

$\chi(M)=\frac{1}{8\pi^{2}}\int_{M}|W|^{2}-\frac{1}{2}|\cancelto{0}{\mathring{\text{Ric}}}|^{2}+\frac{1}{24}|\text{scal}|^{2},$

where $\mathring{\text{Ric}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{Ric}-\frac{\text{scal}}{4}g$
is the trace-free Ricci tensor, which vanishes identically for Einstein
metrics. ∎

Note that almost-Einstein metrics certainly do exist locally:

######  Lemma 2.

Let $\lambda,\psi_{1},\psi_{2},\psi_{3}$ be smooth functions on
$\mathbb{R}^{4}$ such that

$(\psi_{1}^{2}+\psi_{2}^{2}+\psi_{3}^{2})\Big{|}_{\bf
0}\neq\lambda\Big{|}_{\bf 0}\neq 0$

at the origin ${\bf 0}\in\mathbb{R}^{4}$. Then there exists a Type A almost-
Einstein metric in a neighborhood of ${\bf 0}$.

###### Proof.

Type A in (34) has eigenvalues

$\lambda\hskip 7.22743pt,\hskip 7.22743pt\lambda\hskip 7.22743pt,\hskip
7.22743pt\lambda\pm\sqrt{\psi_{1}^{2}+\psi_{2}^{2}+\psi_{3}^{2}}.$

If $(\psi_{1}^{2}+\psi_{2}^{2}+\psi_{3}^{2})\big{|}_{\bf
0}\neq\lambda\big{|}_{\bf 0}\neq 0$, then these eigenvalues are all nonzero at
${\bf 0}$, in which case the 4-tensor defined in the coordinate frame on
$\mathbb{R}^{4}$ by (34) is invertible at ${\bf 0}$. As shown in [DeT81] (see
also [Bes07]), this guarantees the existence of a smooth Riemannian metric $g$
on a neighborhood of ${\bf 0}$ whose Ricci tensor is, with respect to the
coordinate frame $\\{\partial_{1},\partial_{2},\partial_{3},\partial_{4}\\}$,
equal to the Type A Ricci tensor in (34). Setting $T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\partial_{1}$
completes the proof. ∎

We now show that there is an important subclass of almost-Einstein metrics
which has a rich underlying goemetry. Here we will deal with the case of Type
A in (34), postponing Type B to Section 5:

######  Proposition 1.

Let $(M,g)$ be an oriented Riemannian 4-manifold and $T$ a unit length vector
field on $M$. If the curvature operator of $g$ commutes with the Hodge star
operator _$*_{\scalebox{0.4}{\emph{L}}}$_ of the Lorentzian metric
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_ , then $g$ is an almost-Einstein metric of Type A.

###### Proof.

We will express the commuting condition

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}$
(35)

with respect to a local $g$-orthonormal frame of the form $\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$.
The matrices of $*_{\scalebox{0.4}{\emph{L}}}$ and
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ relative to the corresponding basis (18)
for $\Lambda^{2}$ were already given in (27) and (31), respectively. It
follows that (35) holds if and only if $A=D$ and $B^{t}=-B$:

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}=-\\!\begin{bmatrix}A&B\\\
-B&A\end{bmatrix}\cdot$ (36)

(Note how this is intermediate between (32) and (33).) We now determine the
Ricci tensor of such a $g$. For reference, recall from (30) that, with respect
to $\\{T=e_{1},e_{2},e_{3},e_{4}\\}$,

$\hat{R}_{\scalebox{0.6}{\emph{g}}}=-\\!\begin{bmatrix}R_{1212}&R_{1312}&R_{1412}&R_{3412}&R_{4212}&R_{2312}\\\
R_{1213}&R_{1313}&R_{1413}&R_{3413}&R_{4213}&R_{2313}\\\
R_{1214}&R_{1314}&R_{1414}&R_{3414}&R_{4214}&R_{2314}\\\
R_{1234}&R_{1334}&R_{1434}&R_{3434}&R_{4234}&R_{2334}\\\
R_{1242}&R_{1342}&R_{1442}&R_{3442}&R_{4242}&R_{2342}\\\
R_{1223}&R_{1323}&R_{1423}&R_{3423}&R_{4223}&R_{2323}\end{bmatrix}\cdot$

To begin with, $B^{t}=-B$ yields the relations

$R_{1334}=-R_{4212}\hskip 7.22743pt,\hskip 7.22743ptR_{1434}=-R_{2312}\hskip
7.22743pt,\hskip 7.22743ptR_{1442}=-R_{2313}.$

(It also yields the identities

$R(e_{3},e_{4},T,e_{2})=R(e_{4},e_{2},T,e_{3})=R(e_{2},e_{3},T,e_{4})=0,$

though we will not need these in what follows.) As a consequence,

$\displaystyle\left\\{\begin{array}[]{lr}\text{Ric}_{12}=R_{3123}+R_{4124}=-2R_{3123},\\\
\text{Ric}_{13}=R_{2132}+R_{4134}=-2R_{2132},\\\
\text{Ric}_{14}=R_{2142}+R_{3143}=-2R_{2142},\end{array}\right.$

which functions we’ll denote by $\psi_{1},\psi_{2},\psi_{3}$. On the other
hand, the relation $A=D$ yields

$\displaystyle\left\\{\begin{array}[]{lr}\text{Ric}_{23}=R_{1231}+R_{4234}=0,\\\
\text{Ric}_{24}=R_{1241}+R_{3243}=0,\\\
\text{Ric}_{34}=R_{1341}+R_{2342}=0.\end{array}\right.$

Next, the components

$\displaystyle\left\\{\begin{array}[]{lr}\text{Ric}_{11}=R_{2112}+R_{3113}+R_{4114},\\\
\text{Ric}_{22}=R_{1221}+R_{3223}+R_{4224},\\\
\text{Ric}_{33}=R_{1331}+R_{2332}+R_{4334},\\\
\text{Ric}_{44}=R_{1441}+R_{2442}+R_{3443},\end{array}\right.$

yield, together with $A=D$, the following three identities:

$\displaystyle\left\\{\begin{array}[]{lr}\text{(1)}\ \
\text{Ric}_{11}-\text{Ric}_{22}+\text{Ric}_{33}-\text{Ric}_{44}=2R_{1331}-2R_{2442}=0,\\\
\text{(2)}\ \
\text{Ric}_{11}+\text{Ric}_{22}-\text{Ric}_{33}-\text{Ric}_{44}=2R_{1221}-2R_{3443}=0,\\\
\text{(3)}\ \
\text{Ric}_{11}-\text{Ric}_{22}-\text{Ric}_{33}+\text{Ric}_{44}=2R_{4114}-2R_{2332}=0.\\\
\end{array}\right.$

The combinations $(1)+(2)$, $(1)+(3)$, and $(2)+(3)$ then yield, respectively,

$\text{Ric}_{11}=\text{Ric}_{44}\hskip 14.45377pt,\hskip
14.45377pt\text{Ric}_{11}=\text{Ric}_{22}\hskip 14.45377pt,\hskip
14.45377pt\text{Ric}_{11}=\text{Ric}_{33}.$

Setting $\lambda\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{Ric}_{11}$
now puts the Ricci tensor of $g$, expressed with respect to the orthonormal
frame $\\{T,e_{2},e_{3},e_{4}\\}$, precisely in the Type A form of (34). (Note
that because the traceless Ricci tensor does not vanish, one cannot use
Schur’s Lemma to prove that $\lambda$ must be constant, as of course would be
the case were $g$ an Einstein metric.) ∎

Let us pause to give the metrics of Proposition 1 a more suggestive name:

######  Definition 3 ($*_{\scalebox{0.4}{\emph{L}}}$-Einstein metric).

An oriented Riemannian 4-manifold $(M,g)$ is _$*_{\scalebox{0.4}{\emph{L}}}$
-Einstein_ if there exists a unit length vector field $T$ on $M$ such that the
curvature operator of $g$ commutes with the Hodge star operator
_$*_{\scalebox{0.4}{\emph{L}}}$_ of the Lorentzian metric
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_.

As there are always two metrics under consideration here, $g$ and
$g_{\scalebox{0.4}{\emph{L}}}$, the following points are important to keep in
mind:

  1. i.

While $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ is self-adjoint with respect to its
own $g$-induced inner product (7) on $\Lambda^{2}$, it is not self-adjoint
with respect to the $g_{\scalebox{0.4}{\emph{L}}}$-induced inner product on
$\Lambda^{2}$,

$\displaystyle\langle{v_{1}\wedge w_{1}},{v_{2}\wedge
w_{2}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{det}\begin{bmatrix}g_{\scalebox{0.4}{\emph{L}}}(v_{1},v_{2})&g_{\scalebox{0.4}{\emph{L}}}(v_{1},w_{2})\\\
g_{\scalebox{0.4}{\emph{L}}}(w_{1},v_{2})&g_{\scalebox{0.4}{\emph{L}}}(w_{1},w_{2})\end{bmatrix}\cdot$
(41)

Nor is $(\Lambda^{2},\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}})$
positive-definite; in fact its signature is $(-\\!-\\!-\\!+\\!++)$. Later on
in Section 5 below, we will in fact introduce a new “curvature operator” for
$g$ to rectify this.

  2. ii.

For this reason, the components of $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ do
not, in general, satisfy the algebraic Bianchi identity with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$, even for
$g$-orthonormal bases of the form $\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$.
Indeed, by (8), for such a basis we have

$\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{i}\wedge
e_{j})=-\\!\sum_{k<l}R_{ijkl}\,e_{k}\wedge e_{l},$

in which case the usual permuted sum

$\hskip
18.06749pt\underbrace{\,\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{i}\wedge
e_{j})},{e_{k}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$-R_{ijkl}\,\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge
e_{l})$}}+\underbrace{\,\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{j}\wedge
e_{k})},{e_{i}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$-R_{jkil}\,\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{i}\wedge
e_{l})$}}+\underbrace{\,\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{k}\wedge
e_{i})},{e_{j}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$-R_{kijl}\,\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{j}\wedge
e_{l})$}}$

need not vanish in general, since the
$\epsilon_{\scalebox{0.4}{\emph{L}}}$-terms need not all have the same sign.
(By contrast, such a sum will always vanish for the Lorentzian curvature
operator $\hat{R}_{\scalebox{0.4}{\emph{L}}}$ because
$\hat{R}_{\scalebox{0.4}{\emph{L}}}$ is defined by the condition
$({\text{Rm}_{\scalebox{0.4}{\emph{L}}}})_{ijkl}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\langle{\hat{R}_{\scalebox{0.4}{\emph{L}}}(e_{i}\wedge
e_{j})},{e_{k}\wedge e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$; i.e.,
without $\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge e_{l})$, precisely to
ensure that $\hat{R}_{\scalebox{0.4}{\emph{L}}}$ will be self-adjoint. Thus

$\hskip
18.06749pt\underbrace{\,\langle{\hat{R}_{\scalebox{0.4}{\emph{L}}}(e_{i}\wedge
e_{j})},{e_{k}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$({\text{Rm}_{\scalebox{0.4}{\emph{L}}}})_{ijkl}$}}+\underbrace{\,\langle{\hat{R}_{\scalebox{0.4}{\emph{L}}}(e_{j}\wedge
e_{k})},{e_{i}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$({\text{Rm}_{\scalebox{0.4}{\emph{L}}}})_{jkil}$}}+\underbrace{\,\langle{\hat{R}_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge
e_{i})},{e_{j}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$({\text{Rm}_{\scalebox{0.4}{\emph{L}}}})_{kilj}$}}$

will always vanish by the (Lorentzian) algebraic Bianchi identity.)

Both of these facts will have important consequences for us. In any case,
since $*_{\scalebox{0.4}{\emph{L}}}^{2}=-1$, we can define complex scalar
multiplication on $\Lambda^{2}$ by

$\displaystyle i\xi\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=*_{\scalebox{0.4}{\emph{L}}}\,\xi.$
(42)

As a result, the real $6$-dimensional vector space $\Lambda^{2}$ becomes a
$3$-dimensional complex vector space, which we’ll henceforth denote by
$\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$. We may take as a basis
for $\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$ the first three
elements in (18), since now

$e_{3}\wedge e_{4}=-i\,e_{1}\wedge e_{2}\hskip 14.45377pt,\hskip
14.45377pte_{4}\wedge e_{2}=-i\,e_{1}\wedge e_{3}\hskip 14.45377pt,\hskip
14.45377pte_{2}\wedge e_{3}=-i\,e_{1}\wedge e_{4}.$

More importantly, for $*_{\scalebox{0.4}{\emph{L}}}$-Einstein metrics, the
condition
$\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}$
now renders $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ into a _complex-linear_ map
on $\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$:

$\hat{R}_{\scalebox{0.6}{\emph{g}}}(i\xi)=\hat{R}_{\scalebox{0.6}{\emph{g}}}(*_{\scalebox{0.4}{\emph{L}}}(\xi))=*_{\scalebox{0.4}{\emph{L}}}(\hat{R}_{\scalebox{0.6}{\emph{g}}}(\xi))=i\hat{R}_{\scalebox{0.6}{\emph{g}}}(\xi).$

Now, given that $*_{\scalebox{0.4}{\emph{L}}}$ is not the Riemannian Hodge
star operator associated with $g$, and therefore not self-adjoint with respect
to (7), we will need to replace $g$’s sectional curvature
$\text{sec}_{\scalebox{0.6}{\emph{g}}}$ with one that is “adapted” to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ (the reason for this
will become clear in Theorem 1 below). In reading the following definition,
recall the terminology and notation of Definition 1:

######  Definition 4 ($L$-sectional curvature).

Let $(M,g)$ be an oriented Riemannian 4-manifold with curvature operator
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_ and $T$ a unit length vector field on
$M$. Consider the Lorentzian metric
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_ and its inner product
_$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_ on $\Lambda^{2}$.
Then the function _$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_
defined on each $G_{+}(p)\cup G_{-}(p)\subseteq T_{p}M$ by

$\displaystyle\text{\emph{$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$}},$
(43)

is called the $L$-_sectional curvature of $g$_.

Observe that $\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$ is not
defined on _degenerate_ 2-planes $P$ (i.e., those 2-dimensional subspaces
$P\subseteq T_{p}M$ on which $g_{\scalebox{0.4}{\emph{L}}}|_{P}$ is
degenerate). Having said that, there are 2-planes $P$ on which
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}=\text{sec}_{\scalebox{0.6}{\emph{g}}}$:

######  Lemma 3.

If $P\in G_{+}(p)\cup G_{-}(p)$ contains $T$ or is orthogonal to $T$, then
_$\text{sec}_{\scalebox{0.6}{\emph{g}}}(P)=\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)$_.

###### Proof.

If $P\in G_{+}(p)\cup G_{-}(p)$ contains $T$, then we may write $P=T\wedge w$
for some $g_{\scalebox{0.4}{\emph{L}}}$-unit $w$ that is
$g_{\scalebox{0.4}{\emph{L}}}$-orthogonal (hence also $g$-orthogonal) to $T$
($w$ must be spacelike; see, e.g., [O’N83, Lemma 26, p. 141]). Now form a
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$-orthonormal basis
(18) for $\Lambda^{2}$ with $T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1}$
and $w\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{2}$,
and note that

$\epsilon_{\scalebox{0.4}{\emph{L}}}(P)=\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{1}\wedge
e_{2})=\langle{e_{1}\wedge e_{2}},{e_{1}\wedge
e_{2}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=-1.$

Since this basis is also
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$-orthonormal by
construction, we have that $\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{1}\wedge
e_{2})=-R_{1212}\,e_{1}\wedge e_{2}+\cdots$. Then

$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)=\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{1}\wedge
e_{2})},{e_{1}\wedge
e_{2}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=-R_{1212}=R_{1221}\overset{\eqref{sec:v0}}{=}\text{sec}_{\scalebox{0.6}{\emph{g}}}(P).$

If $P$ is orthogonal to $T$, then choose a basis (18) with $T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1}$
and $P=e_{2}\wedge e_{3}$, where, in this case,
$\epsilon_{\scalebox{0.4}{\emph{L}}}(P)=\langle{e_{2}\wedge
e_{3}},{e_{2}\wedge e_{3}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=+1$.
Writing $\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{2}\wedge e_{3})=\cdots-
R_{2323}\,e_{2}\wedge e_{3}+\cdots$, we have

$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)=\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{2}\wedge
e_{3})},{e_{2}\wedge
e_{3}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=-R_{2323}=R_{2332}=\text{sec}_{\scalebox{0.6}{\emph{g}}}(P),$

which completes the proof. ∎

Here now is the reason for considering
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$ instead of
$\text{sec}_{\scalebox{0.6}{\emph{g}}}$; note the crucial role played by the
self-adjoint property of $*_{\scalebox{0.4}{\emph{L}}}$ on
$(\Lambda^{2},\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}})$:

######  Theorem 1.

If $(M,g)$ is _$*_{\scalebox{0.4}{\emph{L}}}$ -Einstein_, then

$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})=\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)$

for all $P\in G_{+}(p)\cup G_{-}(p)$.

###### Proof.

Our proof follows that of [Tho69, Theorem, p. 5] (which Theorem is with
respect to the Lorentzian curvature operator, not the Riemannian one that we
are dealing with here; see our remark after the proof). Suppose that $g$ is a
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein metric with respect to some choice of
unit length vector field $T$ on $M$. Noting that
$\epsilon_{\scalebox{0.4}{\emph{L}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})=-\epsilon_{\scalebox{0.4}{\emph{L}}}(P)$
(this follows from the self-adjoint property of
$*_{\scalebox{0.4}{\emph{L}}}$), we have

$\displaystyle\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})\\!\\!$
$\displaystyle=$
$\displaystyle\\!\\!\epsilon_{\scalebox{0.4}{\emph{L}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(*_{\scalebox{0.4}{\emph{L}}}P)},{*_{\scalebox{0.4}{\emph{L}}}P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle\overset{\eqref{eqn:p00}}{=}$
$\displaystyle\\!\\!-\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{*_{\scalebox{0.4}{\emph{L}}}(\hat{R}_{\scalebox{0.6}{\emph{g}}}P)},{*_{\scalebox{0.4}{\emph{L}}}P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle=$
$\displaystyle\\!\\!\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle=$
$\displaystyle\\!\\!\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P),$

where in (3) we used the self-adjoint property of
$*_{\scalebox{0.4}{\emph{L}}}$ once again. ∎

It’s natural to ask whether the converse is true — as it is in the fully
Riemannian [Ber61, ST69] or fully Lorentzian [Tho69] cases. In fact it is not
true here, as we now demonstrate. Suppose that
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})=\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)$
for all nondegenerate 2-planes $P$, and we would like to conclude from this
that
$*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}=\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}$
(equivalently, that
$*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=-\hat{R}_{\scalebox{0.6}{\emph{g}}}$).
Using the self-adjoint property of $*_{\scalebox{0.4}{\emph{L}}}$ once again,
we would have

$\displaystyle\underbrace{\,-\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{*_{\scalebox{0.4}{\emph{L}}}\hat{R}_{\scalebox{0.6}{\emph{g}}}*_{\scalebox{0.4}{\emph{L}}}\\!P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})$,
via
\eqref{eqn:secsec}}}\,\,=\underbrace{\,\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}(P)$}}.$
(46)

This means that the operators
$*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}$
and $-\hat{R}_{\scalebox{0.6}{\emph{g}}}$ have the same
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$-quadratic forms on
all nondegenerate 2-planes $P$. Now, if two curvature operators both satisfy
the algebraic Bianchi identity and have equivalent quadratic forms, then by a
standard argument they must be equal (even if their quadratic forms are equal
only on _nondegenerate_ 2-planes, by a continuity argument they would still be
equal on all 2-planes; see [Tho69]). However, as we remarked above,
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ does not satisfy the algebraic Bianchi
identity with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ (and hence neither
does
$*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}$).

##  4\. The Petrov Type of $g$ via its curvature operator
$\hat{R}_{\scriptscriptstyle g}$

We are now ready to classify $*_{\scalebox{0.4}{\emph{L}}}$-Einstein metrics,
by defining a Riemannian version of “Petrov Type” in terms of
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$.

######  Definition 5 (Petrov Type of $*_{\scalebox{0.4}{\emph{L}}}$-Einstein
metric).

Let $(M,g)$ be _$*_{\scalebox{0.4}{\emph{L}}}$_ -Einstein, with complex-linear
curvature operator
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}\longrightarrow\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$_.
Then $g$ has _Petrov Type I, II, or III at $p\in M$_ if
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_ has, respectively, 3, 2, or 1 linearly
independent complex eigenvectors at $p$.

Note that the Petrov Type of $g$ is a pointwise condition; it needn’t be
uniform across $M$. Three distinct eigenvalues is the generic case; the
remaining cases are “algebraically special.” Jordan normal-form theory now
yields the following matrix characterization of Petrov Type (cf. [Tho69,
Theorem, p. 3]):

######  Theorem 2.

Let $(M,g)$ be _$*_{\scalebox{0.4}{\emph{L}}}$_ -Einstein. At each $p\in M$,
its complex-linear curvature operator
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}\longrightarrow\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$_
will be similar to one of

$\displaystyle\underbrace{\,\begin{bmatrix}\lambda_{1}&0&0\\\
0&\lambda_{2}&0\\\ 0&0&\lambda_{3}\end{bmatrix}\,}_{\emph{\text{Type
I}}}\hskip 14.45377pt,\hskip
14.45377pt\underbrace{\,\begin{bmatrix}\lambda_{1}&0&0\\\ 0&\lambda_{2}&1\\\
0&0&\lambda_{2}\end{bmatrix}\,}_{\emph{\text{Type II}}}\hskip
14.45377pt,\hskip 14.45377pt\underbrace{\,\begin{bmatrix}\lambda_{1}&1&0\\\
0&\lambda_{1}&1\\\ 0&0&\lambda_{1}\end{bmatrix}\,}_{\emph{\text{Type III}}},$
(47)

with $\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$ not necessarily
distinct.

###### Proof.

If $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ has three linearly independent
eigenvectors (Type I) at $p\in M$, then it is diagonalizable, with eigenvalues
$\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$ not necessarily distinct;
this is the leftmost matrix. If $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ has two
linearly independent eigenvectors (Type II), then it is not diagonalizable and
splits into two cases: 1) Two of its eigenvalues are equal, or 2) all three
are equal. Since in Type II the total geometric multiplicity is 2, either case
leads to two Jordan blocks and thus to the middle matrix above (if all three
eigenvalues are equal, then $\lambda_{1}=\lambda_{2}$). The final case, with
only one linearly independent eigenvector (Type III), has geometric
multiplicity 1, hence one Jordan block. ∎

Now we would like to relate a $*_{\scalebox{0.4}{\emph{L}}}$-Einstein metric’s
Petrov Type to its $L$-sectional curvature
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$. To do that, we will need
to characterize the critical points of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$. In doing so, recall once
again that the curvature operator $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ of $g$
is not self-adjoint with respect to the Lorentzian inner product
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ on $\Lambda^{2}$.
What this means, in practice, is the following: In an arbitrary
$g_{\scalebox{0.4}{\emph{L}}}$-orthonormal frame
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ with timelike direction $e_{1}$ (we don’t
assume that $T=e_{1}$), the components

$\displaystyle K_{ijkl}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{i}\wedge
e_{j})},{e_{k}\wedge e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}},$ (48)

while they clearly satisfy $K_{jikl}=-K_{ijkl}$ and $K_{ijlk}=-K_{ijkl}$, are
however not pairwise symmetric in general: $K_{ijkl}\neq K_{klij}$. If
$e_{1}=T$, however, then the following can be said:

######  Lemma 4.

With respect to any oriented _$g_{\scalebox{0.4}{\emph{L}}}$_ -orthonormal
basis of the form $\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$,
the components (48) satisfy

$\displaystyle\text{\emph{$K_{ijkl}=R_{ijkl}\,\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge
e_{l})$}}.$ (49)

Hence in such a basis,

$\displaystyle\text{\emph{$K_{ijkl}=K_{klij}\iff\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{i}\wedge
e_{j})=\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge e_{l})$}}.$ (50)

###### Proof.

The key is that such a basis is also $g$-orthonormal, hence by (8),

$K_{ijkl}=-\langle{\\!\\!\\!\\!\\!\\!\underbrace{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{i}\wedge
e_{j})}_{\text{$-R_{ijkl}\,e_{k}\wedge
e_{l}+\cdots$}}\\!\\!\\!\\!\\!\\!},{e_{k}\wedge
e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=R_{ijkl}\,\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge
e_{l}).$

Since $R_{ijkl}=R_{klij}$, it’s clear that $K_{ijkl}=K_{klij}$ if and only if
$\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{i}\wedge
e_{j})=\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge e_{l})$. ∎

Now we are in a position to characterize the critical points of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$:

######  Proposition 2.

Let $(M,g)$ be an oriented Riemannian 4-manifold and $T$ a unit length vector
field on $M$. Let _$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_ be the
$L$-sectional curvature of $g$ with respect to
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_. Then at any $p\in M$, $P\in G_{+}(p)$ is a critical point of
_$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_ if and only if with
respect to any oriented orthonormal frame $\\{e_{1},e_{2},e_{3},e_{4}\\}$ of
$T_{p}M$ with timelike direction $e_{1}$ and with $P=e_{3}\wedge e_{4}$,

$\displaystyle K_{34ij}=-K_{ij34}\hskip 14.45377pt,\hskip
14.45377pt\text{$(i,j)\neq(3,4)$ or $(1,2)$},$ (51)

with $K_{ijkl}$ given by (48). Similarly, for $P\in G_{-}(p)$ with
$P=e_{1}\wedge e_{2}$,

$\displaystyle K_{12ij}=-K_{ij12}\hskip 14.45377pt,\hskip
14.45377pt\text{$(i,j)\neq(3,4)$ or $(1,2)$}.$ (52)

If $e_{1}$ can be chosen to equal $T$, then $P=e_{3}\wedge e_{4}$ is a
critical point of _$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_ if and
only if $R_{3442}=R_{3423}=0$, and $P=e_{1}\wedge e_{2}$ is a critical point
of _$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_ if and only if
$R_{1213}=R_{1214}=0$.

###### Proof.

We employ the same coordinate chart $\phi$ as in the proof of [Tho69, Lemma,
p. 5], though with minor index changes. Given a critical 2-plane $P\in
G_{+}(p)$, we may express it as $P=e_{3}\wedge e_{4}$ for two spacelike
$g_{\scalebox{0.4}{\emph{L}}}$-orthonormal vectors $e_{3},e_{4}$, and then
extend these to a $g_{\scalebox{0.4}{\emph{L}}}$-orthonormal basis
$\\{e_{1},e_{2},e_{3},e_{4}\\}$ for $T_{p}M$ with $e_{1}$ timelike (we do not
assume that $e_{1}=T$, or even that $g$ is
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein). Now define a coordinate chart $\phi$
from $\mathbb{R}^{4}$ into $G_{+}(p)$ by

$\phi(x_{1},x_{2},x_{3},x_{4})\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\frac{(e_{3}+x_{3}e_{1}+x_{4}e_{2})\wedge(e_{4}+x_{1}e_{1}+x_{2}e_{2})}{\sqrt{(1-x_{3}^{2}+x_{4}^{2})(1-x_{1}^{2}+x_{2}^{2})-(x_{2}x_{4}-x_{1}x_{3})^{2}}}\cdot$

(Note that the 2-Grassmannians $G_{\pm}(p)\subseteq\Lambda^{2}$ are each
4-dimensional. Note also that $\phi({\bf 0})=P$, that $\phi$ is a
diffeomorphism on a neighborhood of ${\bf 0}\in\mathbb{R}^{4}$, and that
$\langle{\phi(x)},{\phi(x)}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=+1$ for
every $x=(x_{1},x_{2},x_{3},x_{4})$ in its domain.) Then $P$ is a critical
point of $\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$ if and only if

$\frac{\partial(\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}\circ\phi)}{\partial
x_{i}}\bigg{|}_{\bf 0}=0\hskip 14.45377pt,\hskip 14.45377pti=1,2,3,4.$

As we now show, each partial derivative will yield one of the cases in (51).
To compute the case $i=1$, let us set $x_{2}=x_{3}=x_{4}=0$ from the outset,
which simplifies the computation:

$\displaystyle\underbrace{\,\frac{\partial}{\partial x_{1}}\bigg{|}_{\bf
0}\\!\\!(\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}\circ\phi){\scriptstyle\Big{|}_{\text{$(x_{1},0,0,0)$}}}\,}_{0}\\!\\!\\!\\!\\!\\!\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\frac{\partial}{\partial
x_{1}}\bigg{|}_{\bf
0}\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}\bigg{(}\frac{e_{3}\wedge(e_{4}+x_{1}e_{1})}{\sqrt{1-x_{1}^{2}}}\bigg{)}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\frac{\partial}{\partial
x_{1}}\bigg{|}_{\bf
0}\frac{\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{4})+x_{1}\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{1})},{e_{3}\wedge(e_{4}+x_{1}e_{1})}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}}{1-x_{1}^{2}}$
$\displaystyle\overset{\eqref{eqn:K}}{=}$
$\displaystyle\\!\\!\\!\\!-\frac{\partial}{\partial x_{1}}\bigg{|}_{\bf
0}\frac{K_{3434}+x_{1}(K_{3431}+K_{3134})+x_{1}^{2}K_{3131}}{1-x_{1}^{2}}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!-(K_{3431}+K_{3134}),$

so that $K_{3431}=-K_{3134}$. For $i=2$,

$\displaystyle\underbrace{\,\frac{\partial}{\partial x_{2}}\bigg{|}_{\bf
0}\\!\\!(\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}\circ\phi){\scriptstyle\Big{|}_{\text{$(0,x_{2},0,0)$}}}\,}_{0}\\!\\!\\!\\!\\!\\!\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\frac{\partial}{\partial
x_{2}}\bigg{|}_{\bf
0}\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}\bigg{(}\frac{e_{3}\wedge(e_{4}+x_{2}e_{2})}{\sqrt{1+x_{2}^{2}}}\bigg{)}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\frac{\partial}{\partial
x_{2}}\bigg{|}_{\bf
0}\frac{\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{4})+x_{2}\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{2})},{e_{3}\wedge(e_{4}+x_{2}e_{2})}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}}{1+x_{2}^{2}}$
$\displaystyle\overset{\eqref{eqn:K}}{=}$
$\displaystyle\\!\\!\\!\\!-\frac{\partial}{\partial x_{2}}\bigg{|}_{\bf
0}\frac{K_{3434}+x_{2}(K_{3432}+K_{3234})+x_{2}^{2}K_{3232}}{1+x_{2}^{2}}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!-(K_{3432}+K_{3234}).$

Likewise for the cases $i=3,4$, which yield $K_{3414}=-K_{1434}$ and
$K_{3424}=-K_{2434}$, respectively. Upon suitably modifying $\phi$, similar
computations for $P=e_{1}\wedge e_{2}\in G_{-}(p)$ yield the four cases in
(52). Finally, if a critical 2-plane $P\in G_{+}(p)$ can be expressed as
$P=e_{3}\wedge e_{4}$ with respect to an oriented
$g_{\scalebox{0.4}{\emph{L}}}$-orthonormal basis
$\\{T=e_{1},e_{2},e_{3},e_{4}\\}$ — i.e., if $P$ is orthogonal to $T$ — then
by Lemma 4

$K_{4234}\overset{\eqref{eqn:the_same}}{=}K_{3442}\overset{\eqref{eqn:KK+}}{=}-K_{4234}\hskip
14.45377pt\Rightarrow\hskip
14.45377ptR_{4234}\overset{\eqref{eqn:the_same0}}{=}0.$

Likewise, $R_{3423}=0$. Or if a critical 2-plane $P\in G_{-}(p)$ can be
expressed as $P=e_{1}\wedge e_{2}$ with respect to
$\\{T=e_{1},e_{2},e_{3},e_{4}\\}$ — i.e., if $P$ contains $T$ — then

$K_{1213}\overset{\eqref{eqn:the_same}}{=}K_{1312}\overset{\eqref{eqn:KK-}}{=}-K_{1213}\hskip
14.45377pt\Rightarrow\hskip
14.45377ptR_{1213}\overset{\eqref{eqn:the_same0}}{=}0.$

Likewise, $R_{1214}=0$. This completes the proof. ∎

One consequence of Proposition 2 is that if $P=e_{3}\wedge e_{4}\in G_{+}(p)$
is a critical point of $\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$
that is orthogonal to $e_{1}=T$, then, because $\\{e_{1},e_{2},e_{3},e_{4}\\}$
is both $g$\- and $g_{\scalebox{0.4}{\emph{L}}}$-orthonormal,

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}P=-R_{3434}\,P-R_{3412}\,P^{\perp_{\scalebox{0.35}{\emph{L}}}}-R_{3413}\,e_{1}\wedge
e_{3}-R_{3414}\,e_{1}\wedge e_{4},$ (53)

where we’ve used the facts that $R_{3442}=R_{3423}=0$ and
$P^{\perp_{\scalebox{0.35}{\emph{L}}}}=*_{\scalebox{0.4}{\emph{L}}}P=e_{1}\wedge
e_{2}$. Likewise, if $P=e_{1}\wedge e_{2}\in G_{-}(p)$ is a critical point of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$ such that $e_{1}=T$, then

$\hat{R}_{\scalebox{0.6}{\emph{g}}}P=-R_{1212}\,P+R_{1234}\,P^{\perp_{\scalebox{0.35}{\emph{L}}}}-R_{1242}\,e_{4}\wedge
e_{2}-R_{1223}\,e_{2}\wedge e_{3},$

where this time $P^{\perp_{\scalebox{0.35}{\emph{L}}}}=-e_{3}\wedge e_{4}$. We
now present a sort of converse to this fact in the event that $g$ is
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein, a result that establishes a
relationship between the Petrov Type of $g$ and the critical points of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$:

######  Theorem 3.

Let $(M,g)$ be a _$*_{\scalebox{0.4}{\emph{L}}}$_ -Einstein manifold. If $P\in
G_{+}(p)$ is an eigenvector of
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}\longrightarrow\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$_
that is orthogonal to $T$, then $P$ is a spacelike critical point of
_$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_. Thus the number of such
eigenvectors is bounded above by the number of spacelike critical points of
_$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$_. The same is true if
$P\in G_{-}(p)$ is an eigenvector of _$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_
that contains $T$, with spacelike replaced by timelike.

###### Proof.

Suppose $P\in G_{+}(p)$ and write $P=e_{3}\wedge e_{4}$, where $e_{3},e_{4}$
are part of an oriented $g_{\scalebox{0.4}{\emph{L}}}$-orthonormal basis
$\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$;
since $P$ is assumed orthogonal to $T$, such a basis exists. By (26),
$iP=*_{\scalebox{0.4}{\emph{L}}}P=e_{1}\wedge e_{2}\in G_{-}(p)$. If $P$ is an
eigenvector of $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ with eigenvalue
$a+ib\in\mathbb{C}$, then

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{4})=(a+ib)(e_{3}\wedge e_{4})=b(e_{1}\wedge e_{2})+a(e_{3}\wedge e_{4}).$
(54)

Recalling (48), this yields

$\displaystyle K_{3412}=b\hskip 7.22743pt,\hskip 7.22743ptK_{3434}=-a\hskip
7.22743pt,\hskip 7.22743ptK_{3413}=K_{3414}=K_{3442}=K_{3423}=0.$ (55)

To show that $P$ is a critical point of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$, we must satisfy (51) in
Proposition 2, which, given what we already have in (55), means showing that

$K_{1334}=K_{1434}=K_{4234}=K_{2334}=0.$

But these all follow directly from Lemma 4. We therefore conclude that any
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$-eigenvector $P\in G_{+}(p)$ that is
orthogonal to $T$ is necessarily a critical point of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$. The case when $P\in
G_{-}(p)$ contains $T$ follows similarly. ∎

Note that, by (53), the converse of Theorem 3 is not true in general, unless
$R_{3413}=R_{3414}=0$ (indeed, (49) and (51) yield only $R_{3413}=R_{1334}$
and $R_{3414}=R_{1434}$). Once again, the impediment is that
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ is not self-adjoint with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$. One way to remedy
this — and thus to gain a more direct relationship between Petrov Type and the
critical points of $\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{R}$}}}$ — is to
modify the Riemannian curvature operator
$\hat{R}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda^{2}\longrightarrow\Lambda^{2}$
so as to _make_ it self-adjoint with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$. This will result in
a different notion of Petrov Type. We will present precisely such a
modification in Section 5. But before doing so, let us first move beyond
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein metrics and extend Definition 5 to
_any_ oriented Riemannian 4-manifold admitting a nowhere vanishing vector
field (hence $\chi(M)=0$ if $M$ is compact). This is easy to accomplish: Given
any choice of Lorentzian Hodge star operator $*_{\scalebox{0.4}{\emph{L}}}$,
the curvature operator $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ decomposes into
the sum

$\displaystyle\hat{R}_{\scalebox{0.6}{\emph{g}}}=\underbrace{\,\frac{1}{2}(\hat{R}_{\scalebox{0.6}{\emph{g}}}-*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})\,}_{\text{``$\hat{S}_{\scalebox{0.6}{\emph{g}}}\,$"}}+\underbrace{\,\frac{1}{2}(\hat{R}_{\scalebox{0.6}{\emph{g}}}+*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})\,}_{\text{``$\hat{A}_{\scalebox{0.6}{\emph{g}}}$"}},$
(56)

where the “symmetric” and “anti-symmetric” operators
$\hat{S}_{\scalebox{0.6}{\emph{g}}}$ and $\hat{A}_{\scalebox{0.6}{\emph{g}}}$
satisfy

$\hat{S}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ\hat{S}_{\scalebox{0.6}{\emph{g}}}\hskip
14.45377pt,\hskip
14.45377pt\hat{A}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=-*_{\scalebox{0.4}{\emph{L}}}\circ\hat{A}_{\scalebox{0.6}{\emph{g}}}.$

In particular, $\hat{S}_{\scalebox{0.6}{\emph{g}}}$ is
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein, hence a complex-linear map on
$\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$. Obviously,
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ is
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein
$\iff\hat{A}_{\scalebox{0.6}{\emph{g}}}=0\iff\hat{R}_{\scalebox{0.6}{\emph{g}}}=\hat{S}_{\scalebox{0.6}{\emph{g}}}$.
(The symmetric operator $\hat{S}_{\scalebox{0.6}{\emph{g}}}$ is the analogue
of the Weyl curvature 4-tensor $\widehat{W}$. Indeed, in the purely Riemannian
or purely Lorentzian settings, it is given by
$\hat{S}_{\scalebox{0.6}{\emph{g}}}=\widehat{W}+\frac{\text{scal}_{\scalebox{0.5}{\emph{g}}}}{12}I$,
where $\text{scal}_{\scalebox{0.5}{\emph{g}}}$ is the scalar curvature of
$g$.) It is now an easy matter to extend our notion of Petrov Type to any
oriented Riemannian 4-manifold $(M,g)$ that admits a nowhere vanishing vector
field:

######  Definition 6 (Petrov Type, general version).

Let $(M,g)$ be an oriented Riemannian 4-manifold with curvature operator
_$\hat{R}_{\scalebox{0.6}{\emph{g}}}$_ , let $T$ be a unit length vector field
on $M$, and let _$*_{\scalebox{0.4}{\emph{L}}}$_ be the Hodge star operator of
the Lorentzian metric _$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_. Then at any $p\in M$, $g$ has _Petrov Type I, II, or III_ if the
operator _$\hat{S}_{\scalebox{0.6}{\emph{g}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\frac{1}{2}(\hat{R}_{\scalebox{0.6}{\emph{g}}}-*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})$_
, viewed as a complex-linear map
_$\hat{S}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}\longrightarrow\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$_
, has 3, 2, or 1 linearly independent eigenvectors at $p$, respectively.

######  Theorem 4.

Defining the “$L$-sectional curvature of
_$\hat{S}_{\scalebox{0.6}{\emph{g}}}$_ ,” in analogy with (43), to be

$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{S}$}}}(P)\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{\hat{S}_{\scalebox{0.6}{\emph{g}}}P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}},$

it follows that the corresponding results in Theorems 1, 2, and 3 remain true
with respect to _$\hat{S}_{\scalebox{0.6}{\emph{g}}}$_ and
_$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{S}$}}}$_.

###### Proof.

The analogues of Theorems 1 and 2 follow exactly as before. For the analogue
of Theorem 3, first observe that the components $S_{ijkl}$ satisfy

$\displaystyle S_{ijkl}\\!\\!$ $\displaystyle\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=$
$\displaystyle\\!\\!-\langle{\hat{S}_{\scalebox{0.6}{\emph{g}}}(e_{i}\wedge
e_{j})},{e_{k}\wedge e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle=$
$\displaystyle\\!\\!-\frac{1}{2}\langle{(\hat{R}_{\scalebox{0.6}{\emph{g}}}-*_{\scalebox{0.4}{\emph{L}}}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})(e_{i}\wedge
e_{j})},{e_{k}\wedge e_{l}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle\overset{\eqref{eqn:K}}{=}$
$\displaystyle\\!\\!\frac{1}{2}\Big{(}K_{ijkl}+\underbrace{\,\langle{(\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})(e_{i}\wedge
e_{j})},{*_{\scalebox{0.4}{\emph{L}}}(e_{k}\wedge
e_{l})}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\,}_{\text{``$-\epsilon_{ij}\epsilon_{kl}K_{*_{\scalebox{0.4}{\emph{L}}}(ij)*_{\scalebox{0.4}{\emph{L}}}(kl)}$"}}\Big{)},$

where $\epsilon_{ab}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\epsilon_{\scalebox{0.4}{\emph{L}}}(e_{a}\wedge
e_{b})$. Thus

$\displaystyle
S_{ijkl}=\frac{1}{2}(K_{ijkl}-\epsilon_{ij}\epsilon_{kl}K_{*_{\scalebox{0.4}{\emph{L}}}(ij)*_{\scalebox{0.4}{\emph{L}}}(kl)}).$
(57)

Now suppose that $P\in G_{+}(p)$ is an eigenvector of
$\hat{S}_{\scalebox{0.6}{\emph{g}}}$ that is
$g_{\scalebox{0.4}{\emph{L}}}$-orthogonal to $T$, and once again write
$T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1}$
and $P=e_{3}\wedge e_{4}$ as before. If $P$ has eigenvalue
$a+ib\in\mathbb{C}$, then

$\hat{S}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge e_{4})=(a+ib)(e_{3}\wedge
e_{4})=b(e_{1}\wedge e_{2})+a(e_{3}\wedge e_{4}),$

so that

$S_{3412}=b\hskip 7.22743pt,\hskip 7.22743ptS_{3434}=-a\hskip 7.22743pt,\hskip
7.22743ptS_{3413}=S_{3414}=S_{3442}=S_{3423}=0.$

Since Proposition 2 holds with “$S_{ijkl}$” in place of “$K_{ijkl}$,” to show
that $P$ is a critical point of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{S}$}}}$ means showing that

$S_{1334}=S_{1434}=S_{4234}=S_{2334}=0.$

Consider $S_{1334}$; via (57) and Lemma 4,

$S_{3413}=\frac{1}{2}(\,\underbrace{K_{3413}\,}_{\text{$-R_{3413}$}}+\underbrace{K_{1242}}_{\text{$R_{1242}$}}\,)=0\hskip
14.45377pt\Rightarrow\hskip 14.45377ptR_{3413}=R_{1242}.$

Thus

$S_{1334}=\frac{1}{2}(\,\underbrace{K_{1334}\,}_{\text{$R_{1334}$}}+\underbrace{K_{4212}}_{\text{$-R_{4212}$}}\,)=\frac{1}{2}(R_{1334}-R_{4212})=0.$

Similarly, $S_{1434}=0$. For $S_{4234}$,

$S_{3442}=\frac{1}{2}(\,\underbrace{K_{3442}\,}_{\text{$R_{3442}$}}-\underbrace{K_{1213}}_{\text{$-R_{1213}$}}\,)=0\hskip
14.45377pt\Rightarrow\hskip 14.45377ptR_{3442}=-R_{1213}.$

Thus

$S_{4234}=\frac{1}{2}(\,\underbrace{K_{4234}\,}_{\text{$R_{4234}$}}-\underbrace{K_{1312}}_{\text{$-R_{1312}$}}\,)=\frac{1}{2}(R_{4234}+R_{1312})=0.$

Similarly, $S_{2334}=0$. We conclude that any
$\hat{S}_{\scalebox{0.6}{\emph{g}}}$-eigenvector $P\in G_{+}(p)$ that is
orthogonal to $T$ is necessarily a critical point of
$\text{sec}_{\scalebox{0.4}{\emph{$L$-$\hat{S}$}}}$. The case when $P\in
G_{-}(p)$ contains $T$ once again follows similarly. ∎

Finally, do note that $\hat{S}_{\scalebox{0.6}{\emph{g}}}$ is not self-adjoint
with respect to $\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$ _or_
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$.

##  5\. The Petrov Type of $g$ via a modified curvature operator

We now consider the second approach mentioned in the Introduction, namely,
that of modifying $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ so as to make it self-
adjoint with respect to the inner product space
$(\Lambda^{2},\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}})$. This
is easily accomplished:

######  Definition 7.

Let $(M,g)$ be an oriented Riemannian 4-manifold, $T$ a unit length vector
field on $M$, and _$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_. Define an endomorphism
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ on
$\Lambda^{2}$ by setting
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(v\wedge w)$_
to be the unique 2-vector satisfying

$\displaystyle\text{\emph{$\langle{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(v\wedge
w)},{x\wedge y}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-R(v,w,x,y)\hskip
14.45377pt\text{for all $x,y\in T_{p}M$},$ (58)

where $R$ is the Riemann curvature 4-tensor of $g$.

Note first of all that
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is well
defined: Because $R$ is the Riemann curvature 4-tensor of $g$, its symmetries
allow it to be regarded as a symmetric bilinear form
$\Lambda^{2}\times\Lambda^{2}\longrightarrow\mathbb{R}$; one may then realize
it as a linear map
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda^{2}\longrightarrow\Lambda^{2}$
by taking any nondegenerate inner product $\langle{\,},{}\rangle$ on
$\Lambda^{2}$ and defining
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ by

$\text{$\langle{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(v\wedge
w)},{x\wedge y}\rangle$}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-R(v,w,x,y)\hskip
14.45377pt\text{for all $v,w,x,y\in T_{p}M$}.$

In (58), we are simply doing this via
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$, not
$\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$. (See, e.g., [O’N95, p.
300].) While ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$
is not quite equal to $g$’s curvature operator
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ (compare (30) with (70)), it does satisfy
two key properties that $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ does not:

######  Lemma 5.

_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ is self-
adjoint, and satisfies the algebraic Bianchi identity, with respect to
_$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_.

###### Proof.

This is immediate from (58). ∎

We now replace $\hat{R}_{\scalebox{0.6}{\emph{g}}}$ with our new self-adjoint
operator ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$.
For starters, here are the analogues of Definitions 3, 4, and 5:

######  Definition 8
(${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein
metric).

An oriented Riemannian 4-manifold $(M,g)$ is
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ -Einstein_
if there exists a unit length vector field $T$ on $M$ such that the operator
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ defined by
(58) commutes with the Hodge star operator _$*_{\scalebox{0.4}{\emph{L}}}$_ of
the Lorentzian metric _$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_.

######  Definition 9
(${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-sectional
curvature).

Let $(M,g)$ be an oriented Riemannian 4-manifold and $T$ a unit length vector
field on $M$. Consider the Lorentzian metric
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_ and its inner product
_$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_ on $\Lambda^{2}$.
Then the function
_$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_
, defined on each $G_{+}(p)\cup G_{-}(p)\subseteq T_{p}M$ by

$\displaystyle\text{\emph{$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$}}(P)\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{\emph{$\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$}},$
(59)

is called the
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ -sectional
curvature of $g$_.

######  Definition 10
(${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov
Type).

An _${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_
-Einstein metric has
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$__-Petrov
Type I, II,_ or _III_ at $p\in M$ if the complex-linear map
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}\longrightarrow\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$_
has 3, 2, or 1 linearly independent complex eigenvectors at $p$, respectively.

To begin with, note how the analogue of Theorem 1 is stronger for
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein
metrics:

######  Theorem 5.

An oriented Riemannian 4-manifold $(M,g)$ is
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Einstein
if and only if
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})=\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}(P)$
for every nondegenerate 2-plane $P$.

###### Proof.

The proof goes through exactly as it did in the proof of Theorem 1, but this
time the converse also holds, since
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ does indeed
satisfy the algebraic Bianchi identity with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$. Because of this, we
can conclude that if
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}(P^{\perp_{\scalebox{0.35}{\emph{L}}}})=\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}(P)$,
then
$*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=-{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$
(see the remark immediately below (46)). ∎

The analogue of Theorem 2 is also stronger with respect to
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$:

######  Theorem 6.

Let $(M,g)$ be an
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Einstein
manifold. At each $p\in M$, the complex-linear operator
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\colon\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}\longrightarrow\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$_
will be similar to one of

$\displaystyle\underbrace{\,\begin{bmatrix}\lambda_{1}&0&0\\\
0&\lambda_{2}&0\\\ 0&0&\lambda_{3}\end{bmatrix}\,}_{\emph{\text{Type
I}}}\hskip 14.45377pt,\hskip
14.45377pt\underbrace{\,\begin{bmatrix}\lambda_{1}&0&0\\\ 0&\lambda_{2}&1\\\
0&0&\lambda_{2}\end{bmatrix}\,}_{\emph{\text{Type II}}}\hskip
14.45377pt,\hskip 14.45377pt\underbrace{\,\begin{bmatrix}\lambda_{1}&1&0\\\
0&\lambda_{1}&1\\\ 0&0&\lambda_{1}\end{bmatrix}\,}_{\emph{\text{Type III}}},$
(60)

with $\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$ not necessarily
distinct. Furthermore,

  1. 1.

For _${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_
-Petrov Type I,
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ has a
basis of orthogonal spacelike 2-planes.

  2. 2.

For _${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_
-Petrov Type II,
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ has one
spacelike and one lightlike 2-plane as eigenvectors, orthogonal to each other.

  3. 3.

For _${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_
-Petrov Type III, the one linearly independent eigenvector of
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ is
necessarily a lightlike 2-plane.

###### Proof.

As for the forms of these matrices, the proof is identical to that of Theorem
2. But because our operator
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is now self-
adjoint with respect to
$\langle{\,},{}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$, we can describe its
eigenvectors, too. Our proof follows that of [Tho69, Theorem, p. 3], though we
will write it out in detail here since our operator
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is not the
Lorentzian curvature operator $\hat{R}_{\scalebox{0.4}{\emph{L}}}$ to be found
therein. The first step in the proof is to use the following complex scalar
product on $\Lambda^{2}$ defined in [Tho69, Eqn. (3)]:

${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi,\eta)\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\langle{\xi},{\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}-i\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}.$

This is indeed a nondegenerate scalar product on
$\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$ (for a proof, see [O’N95,
Lemma 5.2.5]; note also that ${\bf g}_{\scalebox{0.4}{\emph{L}}}$ is not
Hermitian). Moreover,
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is self-
adjoint with respect to ${\bf g}_{\scalebox{0.4}{\emph{L}}}$ whenever $g$ is
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein:

$\displaystyle{\bf
g}_{\scalebox{0.4}{\emph{L}}}({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\xi,\eta)\\!\\!$
$\displaystyle=$
$\displaystyle\\!\\!\langle{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\xi},{\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}-i\langle{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle=$
$\displaystyle\\!\\!\langle{\xi},{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}-i\langle{\xi},{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(*_{\scalebox{0.4}{\emph{L}}}\,\eta)}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle=$
$\displaystyle\\!\\!\langle{\xi},{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\eta}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}-i\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\eta)}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$
$\displaystyle=$ $\displaystyle\\!\\!{\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi,{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\eta).$

Finally, note that

$\displaystyle{\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi,\xi)=\pm
1\iff\langle{\xi},{\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=\pm
1~{}~{}\text{and}~{}~{}\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=0\iff\xi\in
G_{\pm}(p).$ (61)

(Recall that
$\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=0$
is equivalent to $\xi\wedge\xi=0$, and the latter guarantees decomposability.)
With ${\bf g}_{\scalebox{0.4}{\emph{L}}}$ in hand, we now commence with the
proof, starting with the case of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
I at $p$. Thus, let
$\\{\xi_{1},\xi_{2},\xi_{3}\\}\subseteq\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$
be a basis of eigenvectors for
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$, with
corresponding eigenvalues $\lambda_{1},\lambda_{2},\lambda_{3}\in\mathbb{C}$
not necessarily distinct. If $\lambda_{i}\neq\lambda_{j}$, then because
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is self-
adjoint, we must have ${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{i},\xi_{j})=0$,
by the standard argument:

$\displaystyle\lambda_{i}\,{\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi_{i},\xi_{j})={\bf
g}_{\scalebox{0.4}{\emph{L}}}({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\xi_{i},\xi_{j})={\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi_{i},{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\xi_{j})=\lambda_{j}\,{\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi_{i},\xi_{j}).$ (62)

It follows that if $\lambda_{1},\lambda_{2},\lambda_{3}$ are distinct and some
$\xi_{j}$ satisfies ${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{j},\xi_{j})=0$,
then that $\xi_{j}$ will be ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-orthogonal to
_all_ 2-vectors in $\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$, which
is impossible as ${\bf g}_{\scalebox{0.4}{\emph{L}}}$ is nondegenerate. Thus
if $\lambda_{1},\lambda_{2},\lambda_{3}$ are distinct, then each ${\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi_{j},\xi_{j})\neq 0$, in which case, since

$\displaystyle{\bf g}_{\scalebox{0.4}{\emph{L}}}((a+ib)\xi,(a+ib)\xi)\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!(a^{2}-b^{2}+i2ab){\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi,\xi)$ $\displaystyle=$
$\displaystyle\\!\\!\Big{(}(a^{2}-b^{2})\langle{\xi},{\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}+2ab\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\Big{)}+$
$\displaystyle\hskip
14.45377pti\Big{(}\\!\\!-\\!(a^{2}-b^{2})\langle{\xi},{*_{\scalebox{0.4}{\emph{L}}}\,\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}+2ab\langle{\xi},{\xi}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}\Big{)},$

we may, for each eigenvector $\xi_{1},\xi_{2},\xi_{3}$, choose
$\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{C}$ such that each ${\bf
g}_{\scalebox{0.4}{\emph{L}}}(\alpha_{j}\xi_{j},\alpha_{j}\xi_{j})=+1$. Such
complex scalar multiplication will neither change their status as eigenvectors
of ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ nor
change their eigenvalues, since

${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}((a_{j}+ib_{j})\xi_{j})={\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}((a_{j}I+b\,*_{\scalebox{0.4}{\emph{L}}})\xi_{j})=(a_{j}+ib_{j}){\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\xi_{j}=\lambda_{j}((a_{j}+ib_{j})\xi_{j}).$

By (61), we are thus ensured of a basis of orthogonal spacelike 2-planes that
are eigenvectors of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$. Now suppose
that two eigenvalues are equal, say $\lambda_{1}=\lambda_{2}\neq\lambda_{3}$.
Then ${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{3},\xi_{3})\neq 0$, by the same
reasoning as before. In fact the two-dimensional $\lambda_{1}$-eigenspace must
be ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-nondegenerate, too, hence must contain
a ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-orthogonal basis. Thus if
$\lambda_{1}=\lambda_{2}\neq\lambda_{3}$, then we have a ${\bf
g}_{\scalebox{0.4}{\emph{L}}}$-orthogonal basis
$\\{\xi_{1},\xi_{2},\xi_{3}\\}$ of eigenvectors of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$, no element
of which is ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-lightlike; hence, as before,
each can be scaled to yield a spacelike eigenvector. Finally, consider the
case of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
I but with $\lambda_{1}=\lambda_{2}=\lambda_{3}$. In this case, every 2-vector
is an eigenvector of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$, hence we
may simply choose any three orthogonal spacelike 2-planes in $G_{+}(p)$ as a
basis. This concludes the case of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
I. If $g$ has
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
II at $p$, then whether or not $\lambda_{1}$ and $\lambda_{2}$ are distinct,
there is always a 2-vector
$\eta\in\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$ satisfying
$({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-\lambda
I)\eta=\xi_{1}$ for one of the eigenvectors, say $\xi_{1}$ (the existence of
$\eta$ follows from standard Jordan-normal theory). Then

$\displaystyle{\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{1},\xi_{1})={\bf
g}_{\scalebox{0.4}{\emph{L}}}(({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-\lambda
I)\eta,\xi_{1})={\bf
g}_{\scalebox{0.4}{\emph{L}}}(\eta,({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-\lambda
I)\xi_{1})=0,$ (64)

so that $\xi_{1}$ must be ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-lightlike. But
this is the case if and only if

$\langle{\xi_{1}},{\xi_{1}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=0~{}~{}\text{and}~{}~{}\langle{\xi_{1}},{*_{\scalebox{0.4}{\emph{L}}}\,\xi_{1}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=0,$

which in turn is the case if and only if $\xi_{1}$ is a lightlike 2-plane;
i.e., $\xi_{1}=v\wedge w$ with
$\langle{\xi_{1}},{\xi_{1}}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}=0$.
Consider now the other eigenvector, $\xi_{2}$. The first thing to observe is
that $\xi_{2}$ must be ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-orthogonal to
$\xi_{1}$. Indeed, if $\lambda_{1}\neq\lambda_{2}$, then this follows directly
from (62); if $\lambda_{1}=\lambda_{2}$, then

${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{1},\xi_{2})={\bf
g}_{\scalebox{0.4}{\emph{L}}}(({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-\lambda_{1}I)\eta,\xi_{2})={\bf
g}_{\scalebox{0.4}{\emph{L}}}(\eta,({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-\lambda_{1}I)\xi_{2})=0.$

Now suppose that ${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{2},\xi_{2})=0$ also,
so that both $\xi_{1}$ and $\xi_{2}$ are orthogonal lightlike 2-planes.
Because ${\bf g}_{\scalebox{0.4}{\emph{L}}}$ is nondegenerate and
$\\{\xi_{1},\xi_{2},\eta\\}$ is a basis, each ${\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi_{j},\eta)\neq 0$. And yet, because any
2-vector $\beta\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\alpha_{1}\xi_{1}+\alpha_{2}\xi_{2}$
will be ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-lightlike and orthogonal to
$\xi_{1},\xi_{2}$ (by (5), ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-lightlike
2-vectors are invariant under scalar multiplication), we may choose
$\alpha_{1},\alpha_{2}\in\mathbb{C}$ such that ${\bf
g}_{\scalebox{0.4}{\emph{L}}}(\beta,\eta)=0$, contradicting the nondegeneracy
of ${\bf g}_{\scalebox{0.4}{\emph{L}}}$. Therefore we must have ${\bf
g}_{\scalebox{0.4}{\emph{L}}}(\xi_{2},\xi_{2})\neq 0$, in which case we may
scale it to satisfy ${\bf g}_{\scalebox{0.4}{\emph{L}}}(\xi_{2},\xi_{2})=+1$.
This concludes the case of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
II. Finally, if $g$ has
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
III at $p$, let $\xi\in\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$
denote ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$’s one
linearly independent eigenvector, with eigenvalue $\lambda_{1}$. Once again,
there is a 2-vector $\eta$ satisfying
$({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-\lambda_{1}I)\eta=\xi$,
so that, via (64), $\xi$ is necessarily a lightlike 2-plane once again. ∎

Yet another advantage of using
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ in place of
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ is that Proposition 2 can also be
significantly strengthened, this time yielding a _direct_ link between the
critical points of
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$
and the eigenvectors of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$:

######  Proposition 3.

Let $(M,g)$ be an oriented Riemannian 4-manifold and $T$ a unit length vector
field on $M$. Let
_$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_
be the _${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_
-sectional curvature of $g$ defined with respect to
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_. Then at any $p\in M$, a 2-plane $P\in G_{\pm}(p)$ is a critical
point of
_$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_
if and only if

$\displaystyle\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}P=aP+b\text{\emph{$P^{\perp_{\scalebox{0.35}{\emph{L}}}}$}}$
(65)

for some $a,b\in\mathbb{R}$.

###### Proof.

This proof follows as in [Tho69, Lemma, p. 5], except that we are working with
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$, not
$\hat{R}_{\scalebox{0.4}{\emph{L}}}$. For $P=e_{3}\wedge e_{4}\in G_{+}(p)$,
we have, once again,

$\displaystyle\underbrace{\,\frac{\partial}{\partial x_{1}}\bigg{|}_{\bf
0}\\!\\!(\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}\circ\phi){\scriptstyle\Big{|}_{\text{$(x_{1},0,0,0)$}}}\,}_{0}\\!\\!\\!\\!\\!\\!\\!\\!$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\frac{\partial}{\partial
x_{1}}\bigg{|}_{\bf
0}\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}\bigg{(}\frac{e_{3}\wedge(e_{4}+x_{1}e_{1})}{\sqrt{1-x_{1}^{2}}}\bigg{)}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!\frac{\partial}{\partial
x_{1}}\bigg{|}_{\bf
0}\frac{\langle{{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{4})+x_{1}{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{1})},{e_{3}\wedge(e_{4}+x_{1}e_{1})}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}}{1-x_{1}^{2}}$
$\displaystyle\overset{\eqref{eqn:K}}{=}$
$\displaystyle\\!\\!\\!\\!-\frac{\partial}{\partial x_{1}}\bigg{|}_{\bf
0}\frac{R_{3434}+x_{1}(R_{3431}+R_{3134})+x_{1}^{2}R_{3131}}{1-x_{1}^{2}}$
$\displaystyle=$ $\displaystyle\\!\\!\\!\\!-(R_{3431}+R_{3134}),$

hence $R_{3431}=0.$ Similarly, the cases $i=2,3,4$ yield, respectively,
$R_{3432}=R_{3414}=R_{3424}=0,$ so that

$\displaystyle{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}P\\!\\!$
$\displaystyle=$
$\displaystyle\\!\\!{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(e_{3}\wedge
e_{4})$ $\displaystyle\overset{\eqref{def:newR1}}{=}$
$\displaystyle\\!\\!-R_{3412}\,e_{1}\wedge e_{2}-R_{3434}\,e_{3}\wedge e_{4}$
$\displaystyle=$
$\displaystyle\\!\\!-R_{3434}P-R_{3412}P^{\perp_{\scalebox{0.35}{\emph{L}}}},$

where we recall that
$P^{\perp_{\scalebox{0.35}{\emph{L}}}}=*_{\scalebox{0.4}{\emph{L}}}\,P=e_{1}\wedge
e_{2}$. This is precisely in the form (65) with $a=-R_{3434}$ and
$b=-R_{3412}$. A similar analysis for $P=e_{1}\wedge e_{2}\in G_{-}(p)$ yields
$R_{1213}=R_{1214}=R_{1242}=R_{1223}=0,$ and thus

$\displaystyle{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}P\\!\\!$
$\displaystyle=$
$\displaystyle\\!\\!{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}(e_{1}\wedge
e_{2})$ $\displaystyle\overset{\eqref{def:newR1}}{=}$
$\displaystyle\\!\\!-R_{1212}\,e_{1}\wedge e_{2}-R_{1234}\,e_{3}\wedge e_{4}$
$\displaystyle=$
$\displaystyle\\!\\!-R_{1212}P+R_{1234}P^{\perp_{\scalebox{0.35}{\emph{L}}}},$

where, this time,
$P^{\perp_{\scalebox{0.35}{\emph{L}}}}=*_{\scalebox{0.4}{\emph{L}}}P=-e_{3}\wedge
e_{4}$. This, too, is in the form (65) with $a=-R_{1212}$ and $b=R_{1234}$. ∎

Thanks to this, the
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
of an
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein
metric $g$ is not just bounded by, but this time is completely determined by
the critical point structure of
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$;
i.e.,
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein
Riemannian 4-manifolds possess “normal forms” — at least up to their Petrov
Types:

######  Theorem 7.

Let $(M,g)$ be an
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Einstein
manifold. At any $p\in M$, the
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Petrov
Type of $g$ is determined by the number $n$ of spacelike critical points of
_$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_.
In particular, $g$ has

  1. 1.

_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Petrov
Type I
_$\\!\\!\iff\\!\\!\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_
has $n=3$ or $n=\infty$ spacelike critical points.

  2. 2.

_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Petrov
Type II
_$\\!\\!\iff\\!\\!\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_
has $n=1$ spacelike critical point.

  3. 3.

_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Petrov
Type III
_$\\!\\!\iff\\!\\!\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$_
has $n=0$ spacelike critical points.

###### Proof.

This follows directly from Theorem 6, and is the same method of proof as in
[Tho69, Theorem, p. 5], although the latter is with respect to the Lorentzian
curvature operator $\hat{R}_{\scalebox{0.4}{\emph{L}}}$, not
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$. Thanks to
Proposition 3, any $P\in G_{\pm}(p)$ will be a critical point of
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$
if and only if it is an eigenvector of the complex-linear map
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$:

${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}P=aP+bP^{\perp_{\scalebox{0.35}{\emph{L}}}}=aP+b*_{\scalebox{0.4}{\emph{L}}}\\!P=(a+ib)P.$

Because of this equivalence, the spacelike critical points of
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$
are in one-to-one correspondence with the timelike critical points of
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$,
since $P\in G_{+}(p)\iff iP=*_{\scalebox{0.4}{\emph{L}}}P\in G_{-}(p)$, and
since $P$ is an eigenvector of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\iff iP$ is an
eigenvector of
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$. Thus if
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$
has, say, no spacelike critical points (hence no timelike critical points,
either), then
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ can only
have ${\bf g}_{\scalebox{0.4}{\emph{L}}}$-lightlike eigenvectors. By Theorem
6, only
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
III satisfies this criterion. Likewise, if
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$
has precisely one spacelike critical point (hence precisely one timelike
critical point), then
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ can only
have one eigenvector that is not ${\bf
g}_{\scalebox{0.4}{\emph{L}}}$-lightlike. By Theorem 6, only
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
II satisfies this criterion. Finally, if
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$
has precisely three spacelike critical points, then
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ must have
that many spacelike eigenvectors. By Theorem 6, only
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
I satisfies this criterion. And if, in
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Petrov Type
I, an eigenvalue is repeated, then this yields infinitely many spacelike
eigenvectors, hence infinitely many spacelike (and timelike) critical points
of
$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}$.
∎

Let us also record the analogues of Definition 6 and Theorem 4. If $g$ is not
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein,
then as in (56) we may decompose
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ as

$\displaystyle{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}=\underbrace{\,\frac{1}{2}({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}-*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})\,}_{\text{``${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\,$"}}+\underbrace{\,\frac{1}{2}({\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}+*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}})\,}_{\text{``${\hat{A}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$"}},$
(66)

where the “symmetric” and “anti-symmetric” operators
${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ and
${\hat{A}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ satisfy

${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\hskip
14.45377pt,\hskip
14.45377pt{\hat{A}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=-*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{A}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}.$

Then ${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ will be
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein,
hence a complex-linear map on
$\Lambda_{\scalebox{0.5}{\emph{$\mathbb{C}$}}}^{2}$, and
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ is
$*_{\scalebox{0.4}{\emph{L}}}$-Einstein
$\iff{\hat{A}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}=0\iff{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}={\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$.
Knowing that ${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$
is self-adjoint because both
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ and
$*_{\scalebox{0.4}{\emph{L}}}$ are so, we can therefore repeat our analysis
with ${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$ in
place of ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$; in
particular, Theorem 6, Proposition 3, and Theorem 7 all have
${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-analogues.
Let us collect this information here:

######  Theorem 8.

Every oriented Riemannian 4-manifold admitting a unit length vector field $T$
can be classified by its
_${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Petrov
Type (66), defined via the Hodge star operator
_$*_{\scalebox{0.4}{\emph{L}}}$_ of the Lorentzian metric
_$g_{\scalebox{0.4}{\emph{L}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=g-2T^{\flat}\otimes
T^{\flat}$_. The
_${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Petrov
Type of $g$ is determined pointwise by the spacelike critical points of its
_${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -sectional
curvature function
_$\text{sec}_{\scalebox{0.45}{\emph{\emph{${\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$}}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\epsilon_{\scalebox{0.4}{\emph{L}}}(P)\langle{{\hat{S}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}P},{P}\rangle_{\\!g_{\scalebox{0.3}{\emph{L}}}}$_.

Finally, let us verify that the Ricci tensor of an
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein
metric is that of Type B in (34):

######  Proposition 4.

The Ricci tensor _$\text{Ric}_{\scalebox{0.6}{\emph{g}}}$_ of an
_${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$_ -Einstein
metric $g$, relative to any $g$-orthonormal frame with $T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1}$,
takes the form

$\displaystyle\text{\emph{$\text{Ric}_{\scalebox{0.6}{\emph{g}}}$}}=\begin{bmatrix}\lambda&0&0&0\\\
0&\lambda+\lambda_{2}&\psi_{2}&\psi_{3}\\\
0&\psi_{2}&\lambda+\lambda_{3}&\psi_{4}\\\
0&\psi_{3}&\psi_{4}&\lambda+\lambda_{4}\end{bmatrix},$ (67)

with _$\lambda\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{Ric}_{\scalebox{0.6}{\emph{g}}}(T,T)$_
,

$\displaystyle\lambda_{2}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-2\\!\sum_{i=3,4}\\!\\!R_{TiiT}\hskip
7.22743pt,\hskip 7.22743pt\lambda_{3}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-2\\!\sum_{i=2,4}\\!\\!R_{TiiT}\hskip
7.22743pt,\hskip 7.22743pt\lambda_{4}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-2\\!\sum_{i=2,3}\\!\\!R_{TiiT},$
(68)

and

$\displaystyle\psi_{2}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=2R_{T23T}\hskip
7.22743pt,\hskip 7.22743pt\psi_{3}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=2R_{T24T}\hskip
7.22743pt,\hskip 7.22743pt\psi_{4}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=2R_{T34T}.$
(69)

###### Proof.

Relative to an oriented $g_{\scalebox{0.4}{\emph{L}}}$-orthonormal basis
$\\{T\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=e_{1},e_{2},e_{3},e_{4}\\}$,
it is easily verified that if
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}\circ*_{\scalebox{0.4}{\emph{L}}}=*_{\scalebox{0.4}{\emph{L}}}\circ{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$,
then ${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$, whose
matrix is

$\displaystyle{\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}=\begin{bmatrix}R_{1212}&R_{1312}&R_{1412}&R_{3412}&R_{4212}&R_{2312}\\\
R_{1213}&R_{1313}&R_{1413}&R_{3413}&R_{4213}&R_{2313}\\\
R_{1214}&R_{1314}&R_{1414}&R_{3414}&R_{4214}&R_{2314}\\\
-R_{1234}&-R_{1334}&-R_{1434}&-R_{3434}&-R_{4234}&-R_{2334}\\\
-R_{1242}&-R_{1342}&-R_{1442}&-R_{3442}&-R_{4242}&-R_{2342}\\\
-R_{1223}&-R_{1323}&-R_{1423}&-R_{3423}&-R_{4223}&-R_{2323}\end{bmatrix},$
(70)

(compare this with (30)), will take the block form

${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}=\begin{bmatrix}A&B\\\
-B&A\end{bmatrix}\cdot$

The $B$ blocks yield the relations

$R_{23T3}=R_{T442}\hskip 14.45377pt,\hskip 14.45377ptR_{23T2}=R_{T434}\hskip
14.45377pt,\hskip 14.45377ptR_{42T2}=R_{T334},$

which imply that the following components of the Ricci tensor — expressed in
the $g$-orthonormal basis $\\{T,e_{2},e_{3},e_{4}\\}$ — now vanish:

$\displaystyle\left\\{\begin{array}[]{lr}\text{Ric}_{T2}=R_{3T23}+R_{4T24}=0,\\\
\text{Ric}_{T3}=R_{2T32}+R_{4T34}=0,\\\
\text{Ric}_{T4}=R_{2T42}+R_{3T43}=0.\end{array}\right.$

On the other hand, the $A$ blocks yield

$R_{T3T2}=-R_{4234}\hskip 7.22743pt,\hskip 7.22743ptR_{T4T2}=-R_{2334}\hskip
7.22743pt,\hskip 7.22743ptR_{T4T3}=-R_{2342},$

which now imply that

$\displaystyle\left\\{\begin{array}[]{lr}\text{Ric}_{23}=R_{T23T}+R_{4234}=2R_{T23T},\\\
\text{Ric}_{24}=R_{T24T}+R_{3243}=2R_{T24T},\\\
\text{Ric}_{34}=R_{T34T}+R_{2342}=2R_{T34T}.\end{array}\right.$

These are precisely $\psi_{1},\psi_{2},\psi_{3}$ in (69). Finally, the
components

$\displaystyle\left\\{\begin{array}[]{lr}\text{Ric}_{TT}=R_{2TT2}+R_{3TT3}+R_{4TT4},\\\
\text{Ric}_{22}=R_{T22T}+R_{3223}+R_{4224},\\\
\text{Ric}_{33}=R_{T33T}+R_{2332}+R_{4334},\\\
\text{Ric}_{44}=R_{T44T}+R_{2442}+R_{3443},\end{array}\right.$

yield, together with the $A$ blocks once again, the following three
identities:

$\displaystyle\left\\{\begin{array}[]{lr}\text{(1)}\ \
\text{Ric}_{TT}-\text{Ric}_{22}+\text{Ric}_{33}-\text{Ric}_{44}=2R_{T33T}-2R_{2442}=4R_{T33T},\\\
\text{(2)}\ \
\text{Ric}_{TT}+\text{Ric}_{22}-\text{Ric}_{33}-\text{Ric}_{44}=2R_{T22T}-2R_{3443}=4R_{T22T},\\\
\text{(3)}\ \
\text{Ric}_{TT}-\text{Ric}_{22}-\text{Ric}_{33}+\text{Ric}_{44}=2R_{T44T}-2R_{2332}=4R_{T44T}.\\\
\end{array}\right.$

The combinations $(1)+(2)$, $(1)+(3)$, and $(2)+(3)$ are now easily seen to
yield precisely the equations in (68), with $\lambda\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\text{Ric}_{11}=\text{Ric}(T,T)$.
Thus with respect to any oriented orthonormal frame with $e_{1}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=T$,
the Ricci tensor of an
${\hat{R}^{\scalebox{0.4}{\emph{L}}}}_{\scalebox{0.6}{\emph{g}}}$-Einstein
metric $g$ takes the form (67). ∎

##  6\. Generalizing beyond Lorentzian metrics

We conclude by returning to an observation we made in the Introduction,
namely, that our inquiry can be pursued fairly generally for any pair of
Riemannian metrics $g,\tilde{g}$ on any oriented 4-manifold $M$, regardless of
$M$’s Euler characteristic. This seems particularly fruitful when one of the
metrics is “canonical” in some sense. For example, let $M=\mathbb{S}^{4}$ and
take $\tilde{g}$ to be the standard round metric $\mathring{g}$ of constant
positive sectional curvature; let $\mathring{*}$ and
$\langle\,,\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}}$ denote its Hodge star
operator and corresponding (positive-definite) inner product on
$\Lambda^{2}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\Lambda^{2}(T\mathbb{S}^{4})$,
respectively. For any other Riemannian metric $g$ on $\mathbb{S}^{4}$,
consider the $\mathring{*}$-symmetric part of its curvature operator
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$,
$\hat{S}_{\scalebox{0.6}{\emph{g}}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=\frac{1}{2}(\hat{R}_{\scalebox{0.6}{\emph{g}}}-\mathring{*}\circ\hat{R}_{\scalebox{0.6}{\emph{g}}}\circ\mathring{*}),$
and view it as an operator on the inner product space
$(\Lambda^{2},\langle\,,\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}})$ (as
opposed to
$(\Lambda^{2},\langle{\,},{}\rangle_{\\!\scalebox{0.7}{\emph{g}}}$)). Now we
ask: _What will be the geometry of those Riemannian metrics $g$ on
$\mathbb{S}^{4}$ for whom_
$\hat{S}_{\scalebox{0.6}{\emph{g}}}=\hat{R}_{\scalebox{0.6}{\emph{g}}}$? Or,
in order to classify metrics as in [ST69], we may instead elect to modify
$\hat{R}_{\scalebox{0.6}{\emph{g}}}$ so as to make it self-adjoint with
respect to the inner product
$\langle\,,\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}}$, by defining an
operator ${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}\colon\Lambda^{2}\longrightarrow\Lambda^{2}$
at each $p\in\mathbb{S}^{4}$ by

$\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\scalebox{0.6}{\hskip
0.72229pt$\mathring{g}$}}(v\wedge w),x\wedge
y\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}}\mathrel{\hbox
to0.0pt{\raisebox{1.07639pt}{$\cdot$}\hss}\raisebox{-1.07639pt}{$\cdot$}}=-R(v,w,x,y)\hskip
14.45377pt\text{for all $v,w,x,y\in T_{p}\mathbb{S}^{4}$},$

where $R$ is the Riemann curvature 4-tensor of $g$. Since
${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$ is self-adjoint and satisfies the
algebraic Bianchi identity with respect to
$\langle\,,\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}}$, _Theorem 2.1 in[ST69]
would apply here to yield normal forms_: Relative to an oriented
$\mathring{g}$-orthonormal basis, $\mathring{*}$ would take the form (28), and
the commuting condition ${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}\circ\mathring{*}=\mathring{*}\circ{\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$ would then put
${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$ precisely in the “Einstein” block
form (32) (though $g$ itself would not be Einstein). One may then proceed with
exactly the same proof as in Theorem 2.1 of [ST69], to conclude that, if

$\displaystyle{\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}\circ\mathring{*}=\mathring{*}\circ{\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}},$ (75)

then at each $p\in\mathbb{S}^{4}$ there exists a $\mathring{g}$-orthonormal
basis such that each pair of vectors in this basis spans a critical plane of
${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$ (or
${\hat{S}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$, if (75) doesn’t hold). The matrix
of ${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$ relative to this basis would then
take the form

${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}=-\\!\begin{bmatrix}A&B\\\
B&A\end{bmatrix},~{}\text{where}~{}A=\begin{bmatrix}a_{1}&&\\\ &a_{2}&\\\
&&a_{3}\end{bmatrix}\text{and}~{}B=\begin{bmatrix}b_{1}&&\\\ &b_{2}&\\\
&&b_{3}\end{bmatrix},$

with each pair $a_{i},b_{i}$ arising from a critical plane of
${\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}$; i.e., a critical point of the
quadratic form $P\mapsto\langle{\hat{R}_{\scalebox{0.6}{\emph{g}}}}^{\hskip
0.72229pt\scalebox{0.6}{$\mathring{g}$}}P,P\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}},$
where $P$ is decomposable and satisfies $\langle
P,P\rangle_{\scalebox{0.6}{\\!$\mathring{g}$}}=1$. _Finally, observe that
nothing would change if we replaced $(\mathbb{S}^{4},\mathring{g})$ with any
oriented Riemannian 4-manifold $(M,\tilde{g})$. _

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