# On singular Yamabe obstructions

Andreas Juhl and Bent Ørsted  Department of Mathematics of Århus University,
Ny Munkegade 118, 8000 Århus, Denmark<EMAIL_ADDRESS>Humboldt-Universität,
Institut für Mathematik, Unter den Linden 6, 10099 Berlin, Germany
<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We discuss the singular Yamabe obstruction ${\mathcal{B}}_{3}$ of a
hypersurface in a four-dimensional general background. We derive various
explicit formula for ${\mathcal{B}}_{3}$ from the original definition. We
relate these formulas to corresponding formulas in the literature. The proofs
are elementary.

###### Key words and phrases:

conformal geometry, hypersurface invariant, Poincare-Einstein metric, singular
Yamabe problem, Yamabe obstruction, Willmore functional

August 27, 2024

###### Contents

  1. 1 Introduction
  2. 2 Notation
  3. 3 Some second-order identities involving the second fundamental form
  4. 4 The singular Yamabe problem and the obstruction
  5. 5 The singular Yamabe obstruction ${\mathcal{B}}_{2}$
  6. 6 The singular Yamabe obstruction ${\mathcal{B}}_{3}$
    1. 6.1 ${\mathcal{B}}_{3}$ in terms of volume coefficients
    2. 6.2 The volume coefficients
    3. 6.3 Evaluation I
    4. 6.4 Evaluation II
    5. 6.5 Proof of the main result. Equivalences
  7. 7 Variational aspects

##  1\. Introduction

On a smooth manifold equipped with a Riemannian metric, the basic objects for
geometry are the canonical Levi-Civita connection and the corresponding
Riemann curvature tensor. Conformal geometry is partly studying
transformations preserving angles, and partly polynomials in the curvature and
its derivatives transforming in simple ways under conformal changes of the
metric, i.e., multiplying it by a positive smooth function. Already the
decomposition of the curvature into the Weyl tensor, the scalar curvature, and
the trace-free Einstein tensor indicates the role of conformal geometry, since
the Weyl tensor transforms just by multiplication by a power of the conformal
factor. H. Weyl himself was introducing the first gauge theory in physics
exactly via the local change of scale given by a conformal factor, and related
it to the Maxwell equations in relativistic field theory. Given a hypersurface
in a Riemannian manifold one may in addition consider invariants coming from
the embedding, i.e., not only by the metric on the hypersurface induced by the
metric of the ambient space, but also from the normal geometry of the
hypersurface. In the classical Gauss theory of surfaces in a three-dimensional
Euclidean space, the product of the principal curvatures is the intrinsic
Gauss/scalar curvature, whereas the arithmetic mean of the principal
curvatures is the extrinsic mean curvature. A celebrated conformal invariant
is the integral of the square of the mean curvature over the surface, the so-
called Willmore energy, also relevant for the physical theory of surfaces. It
is closely related to the conformally invariant integral

$\mathcal{W}_{2}=\int_{M^{2}}|\mathring{L}|^{2}dvol$

of the conformally invariant squared norm of the trace-free second fundamental
form. In general, curvature invariants of a hypersurface consist of intrinsic
invariants, coming from the induced metric, and extrinsic invariants, coming
from the second fundamental form and the ambient metric. Whereas for a given
manifold it is known how to describe the conformally invariant scalar
curvature quantities using the Fefferman-Graham ambient metric [FG12], an
analogous description of scalar conformal invariants of a hypersurface is not
known. Such a classification would also be of interest in physics [S15].

Recent years have seen attempts to embed the theory of the Feffermann-Graham
ambient metric and of the related Poincaré-Einstein metric into a wider
framework. For instance, Albin [A20] extended parts of the theory to Poincaré-
Lovelock metrics including applications to $Q$-curvature. In another
direction, Gover et al. [G10, GW14, GW15, GGHW19] developed a tractor calculus
approach to the problem of constructing higher-order generalizations of the
Willmore functional $\mathcal{W}_{2}$. Here a central role is played by the
singular Yamabe problem which replaces the Einstein condition. In [ACF92], it
was discovered that the obstruction to the smooth solvability of the singular
Yamabe problem of a hypersurface of dimension $n$ is a scalar conformal
invariant ${\mathcal{B}}_{n}$. The observation of Gover et al. that
${\mathcal{B}}_{2}=0$ is the Euler-Lagrange equation of $\mathcal{W}_{2}$ was
the starting point of their theory. More generally, [G17] identified the
equation ${\mathcal{B}}_{n}=0$ as the Euler-Lagrange equations of a
conformally invariant functional which he termed the Yamabe energy. This
energy is an analog of the integrated (critical) renormalized volume
coefficient of a Poincaré-Einstein metric which in turn is related to the
integrated (critical) Branson $Q$-curvature. Notably this connection to
$Q$-curvature also extends to the present setting [GW17], [JO21].

Formulas for the conformally invariant obstruction in terms of classical
curvature data are not known for $n\geq 4$. But for $n=3$, such a formula for
${\mathcal{B}}_{3}$ was derived in [GGHW19] from a general tractor calculus
formula in [GW15].

In the present paper, we shall take a classical perspective and derive
formulas for ${\mathcal{B}}_{3}$ directly from its very definition. This
approach is independent. We only apply standard linear algebra and tensor
calculations. It confirms and partly corrects results in the literature. As
technical tools we also employ some differential identities involving $L$.
Partly these are classical such as those found by J. Simons, and partly these
are less well-known. Finally, we apply classical style arguments to relate
${\mathcal{B}}_{3}$ to the variation of the conformally invariant functional

$\mathcal{W}_{3}=\int_{M}(\operatorname{tr}(\mathring{L}^{3})+(\mathring{L},\overline{\mathcal{W}}))dvol$

which can be viewed as a natural generalizations of the classical Willmore
functional. This fits with Graham’s theorem [G17, Theorem 3.1]. Our arguments
replace a technique introduced and exploited in [GGHW19] for the same purpose.

The formulation of the main result requires some notation. Let $L$ be the
second fundamental form, $\mathring{L}$ its trace-free part and $H$ the mean
curvature. Let $\overline{W}$ be the Weyl tensor of the background metric. We
also define two contractions $\overline{W}_{0}$ and $\overline{\mathcal{W}}$
of $\overline{W}$ on $M$ by inserting a unit normal vector $\partial_{0}$ at
the last and at the first and the last slot, respectively. We let the operator
$\mathcal{D}$ act on trace-free symmetric bilinear forms $b$ on $M$ by
$\mathcal{D}(b)=\delta\delta(b)+({\sf P},b)$. $\mathcal{D}$ maps trace-free
symmetric bilinear forms to $C^{\infty}(M)$. It is conformally invariant in
the sense that $e^{4\varphi}\hat{\mathcal{D}}(b)=\mathcal{D}(b)$ for
$\varphi\in C^{\infty}(M)$. The Levi-Civita connections on $X$ and $M$ are
denoted by $\bar{\nabla}$ and $\nabla$. For more details see Section 2. For
the definition of the obstructions ${\mathcal{B}}_{n}$ we refer to Section 4.

######  Theorem 1.

Let $\iota:M^{3}\hookrightarrow(X^{4},g)$ be a smooth embedding. Then it holds

$\displaystyle 12{\mathcal{B}}_{3}$
$\displaystyle=6\mathcal{D}((\mathring{L}^{2})_{\circ})+2|\mathring{L}|^{4}+2\mathcal{D}(\overline{\mathcal{W}})$
$\displaystyle-2\mathring{L}^{ij}\bar{\nabla}^{0}(\overline{W})_{0ij0}-4\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}-4H(\mathring{L},\overline{\mathcal{W}})+16(\mathring{L}^{2},\overline{\mathcal{W}})+4|\overline{\mathcal{W}}|^{2}+2|\overline{W}_{0}|^{2}.$
(1.1)

For a conformally flat background, Theorem 1 reduces to the identity

$6{\mathcal{B}}_{3}=3\mathcal{D}((\mathring{L}^{2})_{\circ})+|\mathring{L}|^{4}=\Delta(|\mathring{L}|^{2})-|\nabla\mathring{L}|^{2}+3/2|\delta(\mathring{L})|^{2}-2{\sf
J}|\mathring{L}|^{2}+|\mathring{L}|^{4}.$ (1.2)

The second equality follows from Lemma 3.13. We recall the well-known fact
that for an odd-dimensional $M$, there are Poincaré-Einstein metrics $g_{+}$
such that the conformal compactification $g=r^{2}g_{+}$ is smooth [FG12]. For
such a metric, it holds $\mathring{L}=0$, $\overline{\mathcal{W}}=0$ and even
$\overline{W}_{0}=0$ [G10, Proposition 4.3]. In particular, for $n=3$, the
above formula confirms that ${\mathcal{B}}_{3}=0$. If $\mathring{L}=0$, then

$6{\mathcal{B}}_{3}=\mathcal{D}(\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+|\overline{W}_{0}|^{2}.$

Formula (1) confirms the conformal invariance of ${\mathcal{B}}_{3}$ (of
weight $-4$). In fact, the conformal invariance of $\mathcal{D}$ implies that
$\mathcal{D}((\mathring{L}^{2})_{\circ})$ and
$\mathcal{D}(\overline{\mathcal{W}})$ are individually conformally invariant.
Furthermore, the sum of the first three terms in the second line of (1) is
conformally invariant. In fact, it holds

$\mathring{L}^{ij}(\bar{\nabla}^{0}(\overline{W})_{0ij0}+2\nabla^{k}\overline{W}_{kij0}+2H\overline{\mathcal{W}}_{ij})=(\mathring{L},B)+(\mathring{L}^{2},\overline{\mathcal{W}})+\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}$

(see Lemma 6.27) with a conformally invariant tensor $B$ of weight $-1$, i.e.,
$e^{\varphi}\hat{B}=B$, introduced in [GGHW19, Lemma 2.1] and termed the
hypersurface Bach tensor. All remaining terms in (1) are individually
conformally invariant.

In the course of the proof of Theorem 1 we shall derive a number of equivalent
formulas for ${\mathcal{B}}_{3}$ which are of interest in special cases. In
particular, we find that

$12{\mathcal{B}}_{3}=\Delta(|\mathring{L}|^{2})+6(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}+12|dH|^{2}$
(1.3)

for a flat background (Corollary 6.11). In [JO21, Section 13.7], we derived
this formula in a different way as a consequence of a general expression for
singular Yamabe obstructions [JO21, Theorem 6]. One may also derive the
general case of the above formula along this line. In [JO21], we derived (1.2)
by combining the conformal invariance of ${\mathcal{B}}_{3}$ with (1.3) and
Simons identity.

The heat kernel asymptotics of elliptic boundary value problems for Laplace-
type operators are another rich source of polynomials in the covariant
derivatives of the curvature tensor and of the second fundamental form; in
[BG90] explicit formulas are given for the first five (integrated) heat
coefficients. It is a natural question to determine the polynomials of this
nature which exhibit a conformal invariance.

The paper is organized as follows. In Section 3, we derive identities for
$\delta\delta(\mathring{L}^{2})$ and $\Delta(|\mathring{L}|^{2})$ which are
crucial for later calculations and may also be of independent interest. They
are closely related to some identities of J. Simons [S68]. In Section 4 we
define the singular Yamabe problem and the resulting obstructions in general
dimensions. In Section 5 we derive a formula for ${\mathcal{B}}_{2}$ and
connect it with the Willmore equation. Section 6 is devoted to the derivation
of formulas for ${\mathcal{B}}_{3}$ in terms of standard curvature quantities.
The starting point will be a formula in terms of the volume expansion of $g$
in geodesic normal coordinates. Along the way we derive several equivalent
formulas for ${\mathcal{B}}_{3}$ which might be of interest under specific
additional assumptions on the background metric and the embedding. The proof
of the main theorem Theorem 1 is contained in the last subsections. It is here
where we need the material of Section 3. In the final section, we derive the
right-hand side of (1) by variation of the functional $\mathcal{W}_{3}$ under
normal variations of the embedding reproving a result in [GGHW19], [G17].
Finally, we note that the main result is equivalent to [GGHW19, Proposition
1.1] in the arXiv-version, but differs from its printed version - we clarify
that issue in Remark 6.28.

In view of the possible applications to physics, such as string and membrane
theory, and the principles of holography, we have tried to be very explicit
throughout.

##  2\. Notation

All manifolds $X$ are smooth. For a manifold $X$, $C^{\infty}(X)$ and
$\Omega^{p}(X)$ denote the respective spaces of smooth functions and smooth
$p$-forms. Let $\mathfrak{X}(X)$ be the space smooth vector fields on $X$.
Metrics on $X$ usually are denoted by $g$. $dvol_{g}$ is the Riemannian volume
element defined by $g$. The Levi-Civita connection of $g$ is denoted by
$\nabla_{X}^{g}$ or simply $\nabla_{X}$ for $X\in\mathfrak{X}(X)$ if $g$ is
understood. In these terms, the curvature tensor $R$ of the Riemannian
manifold $(X,g)$ is defined by
$R(X,Y)Z=\nabla_{X}\nabla_{Y}(Z)-\nabla_{Y}\nabla_{X}(Z)-\nabla_{[X.Y]}(Z)$
for vector fields $X,Y,Z\in\mathfrak{X}(X)$. The components of $R$ are defined
by $R(\partial_{i},\partial_{j})(\partial_{k})={R_{ijk}}^{l}\partial_{l}$. We
also set $\nabla_{X}(u)=\langle du,X\rangle$ for $X\in\mathfrak{X}(X)$ and
$u\in C^{\infty}(X)$. $\operatorname{Ric}$ and $\operatorname{scal}$ are the
Ricci tensor and the scalar curvature of $g$. On a manifold $(X,g)$ of
dimension $n$, we set $2(n-1){\sf J}=\operatorname{scal}$ and define the
Schouten tensor ${\sf P}$ of $g$ by $(n-2){\sf P}=\operatorname{Ric}-{\sf
J}g$. Let $W$ be the Weyl tensor.

For a metric $g$ on $X$ and $u\in C^{\infty}(X)$, let
$\operatorname{grad}_{g}(u)$ be the gradient of $u$ with respect to $g$, i.e.,
it holds $g(\operatorname{grad}_{g}(u),V)=\langle du,V\rangle$ for all vector
fields $V\in\mathfrak{X}(X)$. $g$ defines pointwise scalar products
$(\cdot,\cdot)$ and norms $|\cdot|$ on $\mathfrak{X}(X)$, on forms
$\Omega^{*}(X)$ and on general tensors. Then
$|\operatorname{grad}(u)|^{2}=|du|^{2}$. In these definitions, we use the
metric as a subscript if needed for clarity. $\delta^{g}$ is the divergence
operator on differential forms or symmetric bilinear forms. On forms it
coincides with the negative adjoint $-d^{*}$ of the exterior differential $d$
with respect to the Hodge scalar product defined by $g$. Let
$\Delta_{g}=\delta^{g}d$ be the non-positive Laplacian on $C^{\infty}(X)$. On
the Euclidean space ${\mathbb{R}}^{n}$, it equals $\sum_{i}\partial_{i}^{2}$.
In addition, $\Delta$ will also denote the Bochner-Laplacian (when acting on
$L$).

A metric $g$ on a manifold $X$ with boundary $M$ induces a metric $h$ on $M$.
In such a setting, the curvature quantities of $g$ and $h$ will be
distinguished by adding a bar to those of $g$. In particular, the covariant
derivative, the curvature tensor and the Weyl tensor of $(X,g)$ are
$\bar{\nabla}$, $\bar{R}$ and $\overline{W}$. Similarly,
$\overline{\operatorname{Ric}}$ and $\overline{\operatorname{scal}}$ are the
Ricci tensor and the scalar curvature of $g$.

A hypersurface usually is given by an embedding $\iota:M\hookrightarrow X$.
Accordingly, tensors on $X$ are pulled back by $\iota^{*}$ to $M$. In
formulas, we often omit this pull back. For a hypersurface
$\iota:M\hookrightarrow X$ with the induced metric $h=\iota^{*}(g)$ on $M$,
the second fundamental form $L$ is defined by $L(X,Y)=-h(\nabla^{g}_{X}(Y),N)$
for vector fields $X,Y\in\mathfrak{X}(M)$ and a unit normal vector field
$\partial_{0}=N$. We set $nH=\operatorname{tr}_{h}(L)$ if $M$ has dimension
$n$. $H$ is the mean curvature of $M$. Let $\mathring{L}=L-Hh$ be the trace-
free part of $L$. We often identify $L$ with the shape operator $S$ defined by
$h(X,S(Y))=L(X,Y)$.

We use metrics as usual to raise and lower indices. In particular, we set
$(L^{2})_{ij}=L_{i}^{k}L_{kj}=h^{lk}L_{il}L_{kj}$ and similarly for higher
powers of $L$. We always sum over repeated indices.

The $1$-form $\overline{\operatorname{Ric}}_{0}\in\Omega^{1}(M)$ is defined by
$\overline{\operatorname{Ric}}_{0}(X)=\overline{\operatorname{Ric}}(X,\partial_{0})$
for $X\in\mathfrak{X}(M)$. Similarly, we write $b_{0}$ for the analogous
$1$-form defined by a bilinear form $b$ and we let $\overline{W}_{0}$ be the
$3$-tensor on $M$ with components by $\overline{W}_{ijk0}$, i.e., we always
insert $\partial_{0}$ into the last slot. Moreover, we set
$\overline{\mathcal{W}}_{ij}=\overline{W}_{0ij0}$.

##  3\. Some second-order identities involving the second fundamental form

In the present section, we derive formulas for
$\delta\delta(\mathring{L}^{2})$ and $\Delta(|\mathring{L}|^{2})$ in terms of
the geometry of the background and the intrinsic geometry of $M$. The second
of these formulas is closely related to a well-known formula of Simons. The
main results are Lemma 3.13 and Lemma 3.14. The latter result will play an
important role in Section 6.

######  Lemma 3.1.

For $n=3$, it holds

$\delta\delta(\mathring{L}^{2})=2\mathring{L}_{jk}\nabla^{j}\delta(\mathring{L})^{k}+|\nabla\mathring{L}|^{2}+\frac{1}{2}|\delta(\mathring{L})|^{2}-\frac{1}{2}|\overline{W}_{0}|^{2}+\kappa_{1}$
(3.1)

with

$\kappa_{1}\stackrel{{\scriptstyle\text{def}}}{{=}}(\nabla^{i}\nabla^{j}\mathring{L}^{k}_{i}-\nabla^{j}\nabla^{i}\mathring{L}^{k}_{i})\mathring{L}_{kj}.$
(3.2)

###### Proof.

First, we calculate

$\displaystyle\delta\delta(\mathring{L}^{2})$
$\displaystyle=\nabla^{i}\nabla^{j}\mathring{L}^{2}_{ij}=(\nabla^{i}\nabla^{j}\mathring{L}_{i}^{k})\mathring{L}_{kj}+(\nabla^{j}\mathring{L}_{i}^{k})(\nabla^{i}\mathring{L}_{kj})+(\nabla^{i}\mathring{L}_{i}^{k})(\nabla^{j}\mathring{L}_{kj})+\mathring{L}_{i}^{k}(\nabla^{i}\nabla^{j}\mathring{L}_{kj})$
$\displaystyle=(\nabla^{i}\nabla^{j}\mathring{L}^{k}_{i})\mathring{L}_{kj}+(\nabla^{j}\mathring{L}_{i}^{k})(\nabla^{i}\mathring{L}_{kj})+\delta(\mathring{L})^{k}\delta(\mathring{L})_{k}+\mathring{L}_{i}^{k}\nabla^{i}\delta(\mathring{L})_{k}.$

In the first term, we interchange covariant derivatives. This generates the
curvature term

$\kappa_{1}\stackrel{{\scriptstyle\text{def}}}{{=}}(\nabla^{i}\nabla^{j}\mathring{L}_{i}^{k})\mathring{L}_{kj}-(\nabla^{j}\nabla^{i}\mathring{L}_{i}^{k})\mathring{L}_{kj}.$

In the second term, we apply the Codazzi-Mainardi equation:

$\nabla_{j}L_{ik}-\nabla_{i}L_{jk}=\bar{R}_{ijk0}.$

Its trace-free part gives

$\nabla^{j}\mathring{L}^{k}_{i}-\nabla_{i}\mathring{L}^{jk}=-\frac{1}{2}\delta(\mathring{L})^{j}h_{i}^{k}+\frac{1}{2}\delta(\mathring{L})_{i}h^{jk}+{{\overline{W}_{i}}^{jk}}_{0}.$
(3.3)

In particular, we get

$(\nabla^{j}\mathring{L}_{i}^{k}-\nabla_{i}\mathring{L}^{jk})\nabla^{i}\mathring{L}_{jk}=-\frac{1}{2}(\delta(\mathring{L}),\delta(\mathring{L}))+{{\overline{W}_{i}}^{jk}}_{0}\nabla^{i}\mathring{L}_{jk}.$

But

$\overline{W}_{ijk0}\nabla^{i}\mathring{L}^{jk}=\frac{1}{2}\left(\overline{W}_{ijk0}-\overline{W}_{jik0}\right)\nabla^{i}\mathring{L}^{jk}=\frac{1}{2}\overline{W}_{ijk0}\left(\nabla^{i}\mathring{L}^{jk}-\nabla^{j}\mathring{L}^{ki}\right)=-\frac{1}{2}|\overline{W}_{0}|^{2},$

where
$|\overline{W}_{0}|^{2}\stackrel{{\scriptstyle\text{def}}}{{=}}\overline{W}_{ijk0}\overline{W}^{ijk0}$.
Thus

$\displaystyle(\nabla^{j}\mathring{L}_{i}^{k})(\nabla^{i}\mathring{L}_{kj})$
$\displaystyle=(\nabla^{j}\mathring{L}_{i}^{k}-\nabla_{i}\mathring{L}^{jk})\nabla^{i}\mathring{L}_{kj}+(\nabla_{i}\mathring{L}^{jk})(\nabla^{i}\mathring{L}_{kj})$
$\displaystyle=(\nabla_{i}\mathring{L}^{jk})(\nabla^{i}\mathring{L}_{kj})-\frac{1}{2}(\delta(\mathring{L}),\delta(\mathring{L}))-\frac{1}{2}|\overline{W}_{0}|^{2}$
$\displaystyle=(\nabla\mathring{L},\nabla\mathring{L})-\frac{1}{2}(\delta(\mathring{L}),\delta(\mathring{L}))-\frac{1}{2}|\overline{W}_{0}|^{2}.$

These observations show that

$\delta\delta(\mathring{L}^{2})=2\mathring{L}_{jk}\nabla^{j}\delta(\mathring{L})^{k}+|\nabla\mathring{L}|^{2}+\frac{1}{2}|\delta(\mathring{L})|^{2}-\frac{1}{2}|\overline{W}_{0}|^{2}+\kappa_{1}$
(3.4)

with $\kappa_{1}$ as being defined in (3.2). ∎

######  Lemma 3.2.

$\kappa_{1}=3(\mathring{L}^{2},{\sf P})+{\sf J}|\mathring{L}|^{2}.$

###### Proof.

By definition, we have

$\kappa_{1}=(\nabla^{i}\nabla^{j}\mathring{L}^{k}_{i}-\nabla^{j}\nabla^{i}\mathring{L}^{k}_{i})\mathring{L}_{kj}=\mathcal{R}^{ij}(\mathring{L})_{i}^{k}\mathring{L}_{kj},$

where $\mathcal{R}$ denotes the curvature operator of $M$. We also observe
that

$\mathcal{R}^{ij}(\mathring{L})_{i}^{k}\mathring{L}_{kj}=\mathcal{R}^{ij}(L)_{i}^{k}L_{kj}.$

Hence by

$\mathcal{R}_{ij}(L)_{kl}=-L_{l}^{m}R_{ijkm}-L^{m}_{k}R_{ijlm}$

and the decomposition

$R_{ijkl}=-{\sf P}_{ik}h_{jl}+{\sf P}_{jk}h_{il}-{\sf P}_{jl}h_{ik}+{\sf
P}_{il}h_{jk}$ (3.5)

(the Weyl tensor vanishes in dimension $3$) we get (see also Remark 3.11)

$\displaystyle\kappa_{1}$
$\displaystyle=-(L^{m}_{i}{R^{ij}}_{km}+L^{km}{{R^{ij}}_{im}})L_{kj}$ (3.6)
$\displaystyle=L^{m}_{i}({\sf P}^{i}_{k}h^{j}_{m}-{\sf
P}_{k}^{j}h_{m}^{i}+{\sf P}^{j}_{m}h_{k}^{i}-{\sf
P}_{m}^{i}h_{k}^{j})L_{kj}+L^{km}\operatorname{Ric}^{j}_{m}L_{kj}$
$\displaystyle=2(L^{2},{\sf P})-6H(L,{\sf P})+(L^{2},\operatorname{Ric})$
$\displaystyle=3(L^{2},{\sf P})-6H(L,{\sf P})+{\sf J}|L^{2}|$
$\displaystyle=3(\mathring{L}^{2},{\sf P})+{\sf J}|\mathring{L}|^{2}.$

Alternatively, combining the Simons’ identity (3.9) with the Gauss formula for
the curvature endomorphism yields the first identity in (3.6). This completes
the proof. ∎

######  Example 3.3.

For flat backgrounds, it holds

$\kappa_{1}=3H\operatorname{tr}(L^{3})-|L|^{4}=3H\operatorname{tr}(\mathring{L}^{3})+3H^{2}|\mathring{L}|^{2}-|\mathring{L}|^{4}.$

###### Proof.

The Gauss identity gives

${\sf J}=-\frac{1}{4}|\mathring{L}|^{2}+\frac{3}{2}H^{2}$

and the identity

$\mathcal{F}\stackrel{{\scriptstyle\text{def}}}{{=}}\iota^{*}\bar{{\sf
P}}-{\sf
P}+H\mathring{L}+\frac{1}{2}H^{2}h\stackrel{{\scriptstyle!}}{{=}}\mathring{L}^{2}-\frac{1}{4}|\mathring{L}|^{2}h+\overline{\mathcal{W}}$
(3.7)

for the conformally invariant Fialkov tensor $\mathcal{F}$ of weight $0$ [J09,
Lemma 6.23.3] implies

${\sf
P}=-\mathring{L}^{2}+\frac{1}{4}|\mathring{L}|^{2}h+H\mathring{L}+\frac{1}{2}H^{2}h.$

Hence

$3(\mathring{L}^{2},{\sf P})+{\sf
J}|\mathring{L}|^{2}=-3\operatorname{tr}(\mathring{L}^{4})+\frac{1}{2}|\mathring{L}|^{4}+3H\operatorname{tr}(\mathring{L}^{3})+3H^{2}|\mathring{L}|^{2}$

and it suffices to apply the identity
$2\operatorname{tr}(\mathring{L}^{4})=|\mathring{L}|^{4}$ (Corollary 6.8). ∎

Now let

$\kappa_{2}\stackrel{{\scriptstyle\text{def}}}{{=}}(\mathring{L},\Delta(\mathring{L}))-\frac{3}{2}\mathring{L}^{ij}\nabla_{j}\delta(\mathring{L})_{i}.$
(3.8)

######  Lemma 3.4.

$\kappa_{1}-\kappa_{2}=\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}.$

###### Proof.

The trace-free part of the Codazzi-Mainardi equation reads

$\nabla_{i}\mathring{L}_{kj}-\nabla_{k}\mathring{L}_{ij}-\frac{1}{2}\delta(\mathring{L})_{k}h_{ij}+\frac{1}{2}\delta(\mathring{L})_{i}h_{kj}=\overline{W}_{kij0}.$
(3.9)

Hence

$\nabla^{k}\nabla_{i}\mathring{L}_{kj}-\nabla^{k}\nabla_{k}\mathring{L}_{ij}-\frac{1}{2}\nabla^{k}\delta(\mathring{L})_{k}h_{ij}+\frac{1}{2}\nabla^{k}\delta(\mathring{L})_{i}h_{kj}=\nabla^{k}\overline{W}_{kij0}.$

We commute the covariant derivatives in the first term and obtain

$\mathring{L}^{ij}\nabla_{i}\delta(\mathring{L})_{j}+\kappa_{1}-(\mathring{L},\Delta(\mathring{L}))+\frac{1}{2}\mathring{L}^{ij}\nabla_{j}\delta(\mathring{L})_{i}=\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}.$

Therefore, we get

$\frac{3}{2}\mathring{L}^{ij}\nabla_{i}\delta(\mathring{L})_{j}-(\mathring{L},\Delta(\mathring{L}))+\kappa_{1}=\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}.$

In other words, we have

$\kappa_{1}-\kappa_{2}=\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}.$

The proof is complete. ∎

One should compare this result with [GGHW19, (2.12)].

######  Corollary 3.5.

$\kappa_{2}=3(\mathring{L}^{2},{\sf P})+{\sf
J}|\mathring{L}|^{2}-\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}.$ (3.10)

######  Corollary 3.6.

$(\mathring{L},\Delta(\mathring{L}))=3(\mathring{L},\operatorname{Hess}(H))+3(\mathring{L}^{2},{\sf
P})+{\sf J}|\mathring{L}|^{2}+3\mathring{L}^{ij}\nabla_{i}(\bar{{\sf
P}}_{0})_{j}-\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}.$

###### Proof.

We calculate

$\displaystyle(\mathring{L},\Delta(\mathring{L}))$
$\displaystyle=\frac{3}{2}\mathring{L}^{ij}\nabla_{j}\delta(\mathring{L})_{i}+\kappa_{2}\qquad\mbox{(by
\eqref{M2})}$
$\displaystyle=3(\mathring{L},\operatorname{Hess}(H))+3\mathring{L}^{ij}\nabla_{j}\bar{{\sf
P}}_{0i}+\kappa_{2}\qquad\mbox{(by Codazzi-Mainardi)}.$

Now we apply Corollary 3.5. ∎

Alternatively, we outline how Corollary 3.6 can be derived by using a Simons
type formula. First, we prove

######  Lemma 3.7.

In general dimensions, it holds

$\nabla_{k}\nabla_{l}(L)_{ij}=\nabla_{i}\nabla_{j}(L)_{kl}-\nabla_{i}\bar{R}_{kjl0}-\nabla_{k}\bar{R}_{lij0}+{{R_{ki}}^{m}}_{l}L_{mj}+{{R_{ki}}^{m}}_{j}L_{lm}.$

###### Proof.

We start with the Codazzi-Mainardi equation

$\nabla_{i}(L)_{lj}-\nabla_{l}(L)_{ij}=\bar{R}_{lij0}.$

Differentiation gives

$\nabla_{k}\nabla_{i}(L)_{lj}-\nabla_{k}\nabla_{l}(L)_{ij}=\nabla_{k}\bar{R}_{lij0}.$

Now we commute the derivatives in the first term using

$\nabla_{k}\nabla_{i}(L)_{lj}-\nabla_{i}\nabla_{k}(L)_{lj}={{R_{ki}}^{m}}_{l}L_{mj}+{{R_{ki}}^{m}}_{j}L_{lm}.$

Hence

$\nabla_{i}\nabla_{k}(L)_{lj}-\nabla_{k}\nabla_{l}(L)_{ij}=\nabla_{k}\bar{R}_{lij0}-{{R_{ki}}^{m}}_{l}L_{mj}-{{R_{ki}}^{m}}_{j}L_{lm}.$
(3.11)

Similarly, we differentiate the Codazzi-Mainardi equation

$\nabla_{j}(L)_{kl}-\nabla_{k}(L)_{jl}=\bar{R}_{kjl0}$

and obtain

$\nabla_{i}\nabla_{j}(L)_{kl}-\nabla_{i}\nabla_{k}(L)_{jl}=\nabla_{i}\bar{R}_{kjl0}.$
(3.12)

Adding (3.11) and (3.12) proves the assertion. ∎

Lemma 3.7 implies

$\displaystyle(\nabla^{i}\nabla^{j}(L)_{ik}-\nabla^{j}\nabla^{i}(L)_{ik})L^{k}_{j}$
$\displaystyle=(\nabla^{i}\nabla^{j}(L)_{ik}-\nabla^{k}\nabla^{i}(L)_{ij})L^{k}_{j}\quad\mbox{(by
the symmetry of $L$)}$
$\displaystyle=R_{ikmj}L^{mi}L^{kj}+{R_{ikm}}^{i}L^{jm}L^{k}_{j}.$

One can easily check that this identity also holds if $L$ is replaced by
$\mathring{L}$. This reproves (3.6).

Now taking a trace in Lemma 3.7 gives

######  Lemma 3.8.

In general dimensions, it holds

$\Delta(L)_{ij}=n\operatorname{Hess}_{ij}(H)-\nabla^{k}\bar{R}_{kij0}+\nabla_{i}(\overline{\operatorname{Ric}}_{0})_{j}+{{R_{ik}}^{k}}_{m}L_{j}^{m}-R_{kijm}L^{km}.$

In particular, this gives a

###### Second proof of Corollary 3.6..

Lemma 3.8 and $(L,\Delta(L))=(\mathring{L},\Delta(\mathring{L}))+3H\Delta(H)$
imply

$\displaystyle(\mathring{L},\Delta(\mathring{L}))$
$\displaystyle=3(\mathring{L},\operatorname{Hess}(H))+L^{ij}\nabla_{i}(\overline{\operatorname{Ric}}_{0})_{j}-L^{ij}\nabla^{k}\bar{R}_{kij0}+L^{ij}{{R_{ik}}^{k}}_{m}L_{j}^{m}-L^{ij}R_{kijm}L^{km}$
$\displaystyle=3(\mathring{L},\operatorname{Hess}(H))+(L^{2})^{im}{{R_{ik}}^{k}}_{m}-L^{ij}L^{kl}R_{kijl}+L^{ij}\nabla_{i}(\overline{\operatorname{Ric}}_{0})_{j}-L^{ij}\nabla^{k}\bar{R}_{kij0}.$

Now

$(L^{2})^{im}{{R_{ik}}^{k}}_{m}=(L^{2})^{ij}\operatorname{Ric}_{ij}=(L^{2},{\sf
P})+{\sf J}|L|^{2},$

and (3.5) implies

$\displaystyle L^{ij}L^{kl}R_{kijl}$ $\displaystyle=-2(L^{2},{\sf
P})+6H(L,{\sf P}).$

Hence

$\displaystyle(\mathring{L},\Delta(\mathring{L}))=3(\mathring{L},\operatorname{Hess}(H))+3(L^{2},{\sf
P})+{\sf J}|L|^{2}-6H(L,{\sf P})+2L^{ij}\nabla_{i}(\bar{{\sf
P}}_{0})_{j}-L^{ij}\nabla^{k}\bar{R}_{kij0}.$

In order to simplify this formula, we note that

$3(L^{2},{\sf P})+{\sf J}|L|^{2}-6H(L,{\sf P})=3(\mathring{L}^{2},{\sf
P})+{\sf J}|\mathring{L}|^{2}$

and

$\nabla^{k}\bar{R}_{kij0}=\nabla^{k}(\bar{{\sf
P}}_{0})_{k}h_{ij}-\nabla_{j}(\bar{{\sf
P}}_{0})_{i}+\nabla^{k}\overline{W}_{kij0}.$

Hence

$\displaystyle(\mathring{L},\Delta(\mathring{L}))=3(\mathring{L},\operatorname{Hess}(H))+3(\mathring{L}^{2},{\sf
P})+{\sf J}|\mathring{L}|^{2}+3L^{ij}\nabla_{i}(\bar{{\sf
P}}_{0})_{j}-3H\nabla^{k}(\bar{{\sf
P}}_{0})_{k}-L^{ij}\nabla^{k}\overline{W}_{kij0}.$

The last term is unchanged if we replace $L$ by $\mathring{L}$. This completes
the proof. ∎

Combining Lemma 3.7 with some arguments using the Gauss formula relating the
curvature tensors of $X$ and $M$ leads to the following well-known identities
which are due to Simons [S68, SSY75, HP99, V13].

######  Proposition 3.9.

For any hypersurface $M^{n}\hookrightarrow X^{n+1}$ with the second
fundamental form $L$, it holds

$\displaystyle\nabla_{i}\nabla_{j}(L)_{kl}$
$\displaystyle=\nabla_{k}\nabla_{l}(L)_{ij}+L_{ij}L^{2}_{kl}-L_{kl}L^{2}_{ij}+L_{il}L^{2}_{jk}-L_{jk}L^{2}_{il}$
$\displaystyle-
L_{i}^{m}\bar{R}_{jklm}-L_{j}^{m}\bar{R}_{iklm}+L_{k}^{m}\bar{R}_{lijm}+L_{l}^{m}\bar{R}_{kijm}$
$\displaystyle+L_{ij}\bar{R}_{0kl0}-L_{kl}\bar{R}_{0ij0}+\bar{\nabla}_{i}(\bar{R})_{kjl0}+\bar{\nabla}_{k}(\bar{R})_{lij0}.$
(3.13)

Taking a trace gives

$\displaystyle\Delta(L)_{ij}$
$\displaystyle=n\operatorname{Hess}_{ij}(H)+nHL^{2}_{ij}-L_{ij}|L|^{2}$
$\displaystyle+L^{s}_{j}\bar{R}_{ikks}+L_{i}^{s}\bar{R}_{jkks}-2L^{rs}\bar{R}_{rijs}$
$\displaystyle+nH\bar{R}_{0ij0}-L_{ij}\overline{\operatorname{Ric}}_{00}+\bar{\nabla}_{k}(\bar{R})_{ikj0}+\bar{\nabla}_{i}(\bar{R})_{jkk0}.$
(3.14)

For flat backgrounds, Proposition 3.9 specializes to

######  Proposition 3.10.

For any hypersurface $M^{n}\hookrightarrow{\mathbb{R}}^{n+1}$ with the second
fundamental form $L$, it holds

$\nabla_{i}\nabla_{j}(L)_{kl}=\nabla_{k}\nabla_{l}(L)_{ij}+L_{ij}L^{2}_{kl}-L_{kl}L^{2}_{ij}+L_{il}L^{2}_{kj}-L_{kj}L^{2}_{il}.$
(3.15)

Hence

$\Delta(L)=n\operatorname{Hess}(H)+nHL^{2}-L|L|^{2}$ (3.16)

and

$\frac{1}{2}\Delta(|L|^{2})=n(L,\operatorname{Hess}(H))+|\nabla
L|^{2}+nH\operatorname{tr}(L^{3})-|L|^{4}.$ (3.17)

######  Remark 3.11.

The first part of Proposition 3.9 again confirms Lemma 3.2. In fact, by the
symmetry of $L$, we obtain

$(\nabla^{i}\nabla^{j}(L)_{ki}-\nabla^{j}\nabla^{i}(L)_{ki})L^{k}_{j}=(\nabla^{i}\nabla^{j}(L)_{ki}-\nabla^{k}\nabla^{i}(L)_{ji})L^{k}_{j}.$

In this identity one can replace $L$ by $\mathring{L}$. Hence (3.9) implies

$\kappa_{1}=(L^{2})^{kl}\bar{R}_{kiil}-L^{il}L^{jk}\bar{R}_{kilj}.$

By the Gauss identity, we obtain

$\displaystyle\kappa_{1}$
$\displaystyle=3H\operatorname{tr}(L^{3})-|L|^{4}+(L^{2})^{kl}R_{kiil}-L^{il}L^{jk}R_{kilj}$
$\displaystyle+(L^{2})^{kl}(L_{ki}L_{il}-L_{kl}L_{ii})-L^{il}L^{jk}(L_{kl}L_{ij}-L_{kj}L_{il})$
$\displaystyle=(L^{2})^{kl}R_{kiil}-L^{il}L^{jk}R_{kilj}.$

The remaining arguments are as in the proof of Lemma 3.2.

######  Remark 3.12.

For flat backgrounds, it holds
$\kappa_{2}=\kappa_{1}=3H\operatorname{tr}(L^{3})-|L|^{4}$ (Example 3.3) and
the above results yield

$\frac{1}{2}\Delta(|\mathring{L}|^{2})=3(\mathring{L},\operatorname{Hess}(H))+|\nabla\mathring{L}|^{2}+3H\operatorname{tr}(L^{3})-|L|^{4}$

and

$\delta\delta(\mathring{L}^{2})=4(\mathring{L},\operatorname{Hess}(H))+|\nabla\mathring{L}|^{2}+2|dH|^{2}+3H\operatorname{tr}(L^{3})-|L|^{4}.$

As a consequence, we find the difference formula

$\frac{1}{2}\Delta(|\mathring{L}|^{2})-\delta\delta(\mathring{L}^{2})=-(\mathring{L},\operatorname{Hess}(H))-2|dH|^{2}.$
(3.18)

Now combining Lemma 3.1 with (3.8), we obtain

$\displaystyle\delta\delta(\mathring{L}^{2})$
$\displaystyle=\frac{4}{3}(\mathring{L},\Delta(\mathring{L}))+|\nabla\mathring{L}|^{2}+\frac{1}{2}|\delta(\mathring{L})|^{2}-\frac{1}{2}|\overline{W}_{0}|^{2}-\frac{4}{3}\kappa_{2}+\kappa_{1}.$

Hence

$\displaystyle\delta\delta((\mathring{L}^{2})_{\circ})$
$\displaystyle=\delta\delta(\mathring{L}^{2})-\frac{1}{3}\Delta(|\mathring{L}|^{2})$
$\displaystyle=\delta\delta(\mathring{L}^{2})-\frac{2}{3}|\nabla\mathring{L}|^{2}-\frac{2}{3}(\mathring{L},\Delta(\mathring{L}))$
$\displaystyle=\frac{2}{3}(\mathring{L},\Delta(\mathring{L}))+\frac{1}{3}|\nabla\mathring{L}|^{2}+\frac{1}{2}|\delta(\mathring{L})|^{2}-\frac{1}{2}|\overline{W}_{0}|^{2}-\frac{1}{3}\kappa_{1}+\frac{4}{3}(\kappa_{1}-\kappa_{2}).$
(3.19)

Thus, using

$({\sf P},(\mathring{L}^{2})_{\circ})=({\sf
P},\mathring{L}^{2})-\frac{1}{3}{\sf J}|\mathring{L}|^{2},$

Lemma 3.2 and Lemma 3.4, we obtain

######  Lemma 3.13.

$\displaystyle\delta\delta((\mathring{L}^{2})_{\circ})+({\sf
P},(\mathring{L}^{2})_{\circ})$
$\displaystyle=\frac{2}{3}(\mathring{L},\Delta\mathring{L})+\frac{1}{3}|\nabla\mathring{L}|^{2}+\frac{1}{2}|\delta(\mathring{L})|^{2}-\frac{2}{3}{\sf
J}|\mathring{L}|^{2}$
$\displaystyle-\frac{4}{3}\mathring{L}^{ij}\nabla^{k}(\overline{W}_{0})_{kij}-\frac{1}{2}|\overline{W}_{ikj0}|^{2}.$

Lemma 3.13 confirms [GGHW19, Proposition 2.4] up to the sign of the term
$|\overline{W}_{0}|^{2}$.

The following result extends the difference formula (3.18) to general
backgrounds. It will play an important role in Section 6.

######  Lemma 3.14.

It holds

$\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2})=-2(\mathring{L},\operatorname{Hess}(H))-2(\mathring{L},\nabla(\bar{{\sf
P}}_{0}))-|\delta(\mathring{L})|^{2}-2\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}+|\overline{W}_{0}|^{2}.$
(3.20)

###### Proof.

We recall that

$\displaystyle\Delta(|\mathring{L}|^{2})$
$\displaystyle=2(\mathring{L},\Delta(\mathring{L}))+2|\nabla(\mathring{L})|^{2}$
$\displaystyle=6(\mathring{L},\operatorname{Hess}(H))+2\kappa_{1}+6(\mathring{L},\nabla(\bar{{\sf
P}}_{0}))-2\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}+2|\nabla(\mathring{L})|^{2}$

(by Lemma 3.2 and Lemma 3.6) and

$2\delta\delta(\mathring{L}^{2})=8(\mathring{L},\operatorname{Hess}(H))+8(\mathring{L},\nabla(\bar{{\sf
P}}_{0}))+2|\nabla(\mathring{L})|^{2}+|\delta(\mathring{L})|^{2}-|\overline{W}_{0}|^{2}+2\kappa_{1}$

(by (3.4) and $\delta(\mathring{L})=2dH+2\bar{{\sf P}}_{0}$ (Codazzi-
Mainardi)). The difference of both sums equals

$-2(\mathring{L},\operatorname{Hess}(H))-2(\mathring{L},\nabla(\bar{{\sf
P}}_{0}))-|\delta(\mathring{L})|^{2}-2\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}+|\overline{W}_{0}|^{2}.$
(3.21)

The proof is complete. ∎

Note that the left-hand side of (3.20) is a total divergence, i.e., integrates
to $0$ on a closed $M$. The fact that the sum of the first three terms on the
right-hand side of (3.20) is a total divergence follows by partial integration
and the Codazzi-Mainardi equation. In fact, for closed $M$, we calculate

$\displaystyle\int_{M}-2(\mathring{L},\operatorname{Hess}(H))-2(\mathring{L},\nabla(\bar{{\sf
P}}_{0}))-|\delta(\mathring{L})|^{2}dvol_{h}$
$\displaystyle=\int_{M}2(\delta(\mathring{L}),dH)+2(\delta(\mathring{L}),\bar{{\sf
P}}_{0})-|\delta(\mathring{L})|^{2}dvol_{h}=0$

by $2dH+\bar{{\sf P}}_{0}=\delta(\mathring{L})$. The fact that the additional
terms on the right-hand side of (3.20) also form a total divergence can be
seen as follows. Partial integration gives

$\displaystyle-2\int_{M}\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}dvol_{h}=2\int_{M}\nabla^{k}(\mathring{L})^{ij}\overline{W}_{kij0}dvol_{h}.$

By the trace-free part of the Codazzi-Mainardi equation

$\nabla_{k}(\mathring{L})_{ij}-\nabla_{i}(\mathring{L})_{kj}-\frac{1}{2}\delta(\mathring{L})_{i}h_{kj}+\frac{1}{2}\delta(\mathring{L})_{k}h_{ij}=\overline{W}_{ikj0}=-\overline{W}_{kij0}$

and partial integration, this integral equals

$\displaystyle
2\int_{M}\nabla^{i}(\mathring{L})^{kj}\overline{W}_{kij0}dvol_{h}-2\int_{M}\overline{W}^{kij0}\overline{W}_{kij0}dvol_{h}$
$\displaystyle=-2\int_{M}\mathring{L}^{kj}\nabla^{i}\overline{W}_{kij0}dvol_{h}-2\int_{M}|\overline{W}_{kij0}|^{2}dvol_{h}$

Hence

$\int_{M}(-4\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}+2|\overline{W}_{kij0}|^{2})dvol_{h}=0.$

This proves the claim.

##  4\. The singular Yamabe problem and the obstruction

The material in this section rests on [ACF92] and [GW15].

Let $(X^{n+1},g)$ be a compact manifold with boundary $M$ of dimension $n$.
The singular Yamabe problem asks to find a defining function $\sigma$ of $M$
so that

$\operatorname{scal}(\sigma^{-2}g)=-n(n+1).$ (4.1)

The conformal transformation law of scalar curvature shows that

$\operatorname{scal}(\sigma^{-2}g)=-n(n+1)|d\sigma|_{g}^{2}+2n\sigma\Delta_{g}(\sigma)+\sigma^{2}\operatorname{scal}(g).$

Following [GW15], we write this equation in the form

$\operatorname{scal}(\sigma^{-2}g)=-n(n+1)\mathcal{S}(g,\sigma),$

where

$\mathcal{S}(g,\sigma)\stackrel{{\scriptstyle\text{def}}}{{=}}|d\sigma|_{g}^{2}+2\rho\sigma,\quad(n+1)\rho\stackrel{{\scriptstyle\text{def}}}{{=}}-\Delta_{g}(\sigma)-\sigma{\sf
J}\quad\mbox{and}\quad 2n{\sf J}=\operatorname{scal}(g).$

In these terms, $\sigma$ is a solution of (4.1) iff $\mathcal{S}(g,\sigma)=1$.
Although such $\sigma$ exist and are unique, in general, $\sigma$ is not
smooth up to the boundary. The smoothness is obstructed by a locally
determined conformally invariant scalar function on $M$ which is called the
singular Yamabe obstruction.

In order to describe the structure of a solution $\sigma$ of the singular
Yamabe problem more precisely, we use geodesic normal coordinates. Let $r$ be
the distance function of $M$ for the background metric $g$. Then there are
uniquely determined coefficients $\sigma_{(k)}\in C^{\infty}(M)$ for $2\leq
k\leq n+1$ so that the smooth defining function

$\sigma_{F}\stackrel{{\scriptstyle\text{def}}}{{=}}r+\sigma_{(2)}r^{2}+\dots+\sigma_{(n+1)}r^{n+1}$
(4.2)

satisfies

$\mathcal{S}(g,\sigma_{F})=1+Rr^{n+1}$ (4.3)

with a smooth remainder term $R$. The coefficients are recursively determined.
In geodesic normal coordinates, the metric $g$ takes the form $dr^{2}+h_{r}$
with a one-parameter family $h_{r}$ of metrics on $M$. The condition (4.3) is
equivalent to

$|d\sigma_{F}|_{g}^{2}-\frac{2}{n+1}\sigma_{F}\Delta_{g}(\sigma_{F})-\frac{1}{n(n+1)}\sigma_{F}^{2}\operatorname{scal}(g)=1+Rr^{n+1}.$

We write the left-hand side of this equation in the form

$\displaystyle\partial_{r}(\sigma_{F})^{2}+h_{r}^{ij}\partial_{i}(\sigma_{F})\partial_{j}(\sigma_{F})$
$\displaystyle-\frac{2}{n+1}\sigma_{F}\left(\partial_{r}^{2}(\sigma_{F})+\frac{1}{2}\operatorname{tr}(h_{r}^{-1}h_{r}^{\prime})\partial_{r}(\sigma_{F})+\Delta_{h_{r}}(\sigma_{F})\right)-\frac{1}{n(n+1)}\sigma_{F}^{2}\operatorname{scal}(g)$
(4.4)

and expand this sum into a Taylor series in the variable $r$. Then the
vanishing of the coefficient of $r^{k}$ for $k\leq n$ is equivalent to an
identity of the form

$(k-1-n)\sigma_{(k+1)}=LOT,$

where $LOT$ involves only lower-order Taylor coefficients of $\sigma$. The
latter relation also indicates that there is a possible obstruction to the
existence of an improved solution $\sigma_{F}^{\prime}$ which contains a term
$\sigma_{(n+2)}r^{n+2}$ and satisfies
$\mathcal{S}(g,\sigma_{F}^{\prime})=1+Rr^{n+2}$. Following [ACF92], we define
the singular Yamabe obstruction by

${\mathcal{B}}_{n}\stackrel{{\scriptstyle\text{def}}}{{=}}\left(r^{-n-1}(\mathcal{S}(g,\sigma_{F})-1)\right)|_{r=0}.$
(4.5)

Since $\sigma_{F}$ is determined by $g$, ${\mathcal{B}}_{n}$ is a functional
of $g$. It is a key result that ${\mathcal{B}}_{n}$ is a conformal invariant
of $g$ of weight $-(n+1)$. More precisely, we write $\hat{{\mathcal{B}}}_{n}$
for the obstruction defined by $\hat{g}=e^{2\varphi}g$ with $\varphi\in
C^{\infty}(X)$. Then

######  Lemma 4.1.

$e^{(n+1)\iota^{*}(\varphi)}\hat{{\mathcal{B}}}_{n}={\mathcal{B}}_{n}$.

Let us be a bit more precise about the above algorithm. We set
$S_{k}=\sum_{j=1}^{k}\sigma_{(j)}$ so that $S_{n+1}=\sigma_{F}$. Then the
coefficients of $\sigma_{F}$ are recursively determined by the conditions

$\mathcal{S}(S_{k})=1+O(r^{k}).$

More precisely, we recursively find

$\mathcal{S}(S_{k})=1+r^{k-1}(c(n-k+2)\sigma_{(k)}+\cdots)+\cdots$

with $c=2k/(n+1)$. Then the condition $\mathcal{S}(S_{k})-1=O(r^{k})$ with an
unknown coefficient $\sigma_{(k)}$ is satisfied iff the coefficient of
$r^{k-1}$ in this expansion vanishes. This can be solved for $\sigma_{(k)}$ if
$k=2,3,\dots,n+1$. In the case $k=n+1$, we obtain

$\mathcal{S}(S_{n+1})=1+O(r^{n+1})$

and the restriction of the latter remainder is the obstruction
${\mathcal{B}}_{n+1}$.

In the following, we shall need explicit formulas for the coefficients
$\sigma_{(k)}$ for $k\leq 4$.

First, we consider flat backgrounds.

We approximately solve the equation $\mathcal{S}(g,\sigma_{F})=1$ for the flat
metric $g$ by differentiation of the relation

$\displaystyle\partial_{r}(\sigma_{F})^{2}+h_{r}^{ij}\partial_{i}(\sigma_{F})\partial_{j}(\sigma_{F})-\frac{2}{n+1}\sigma_{F}\left(\partial_{r}^{2}(\sigma_{F})+\frac{1}{2}\operatorname{tr}(h_{r}^{-1}h_{r}^{\prime})\partial_{r}(\sigma_{F})+\Delta_{h_{r}}(\sigma_{F})\right)=0$
(4.6)

in the variable $r$. Then, for general $n\geq 3$, we find the solution

$\sigma_{F}=r+\frac{r^{2}}{2}H-\frac{r^{3}}{3(n-1)}|\mathring{L}|^{2}+r^{4}\sigma_{(4)}+\cdots$
(4.7)

with the coefficient

$\sigma_{(4)}=\frac{1}{24(n-2)}\left(6\operatorname{tr}(\mathring{L}^{3})+\frac{7n-11}{n-1}H|\mathring{L}|^{2}+3\Delta(H)\right).$
(4.8)

Note that $\sigma_{(3)}$ is singular for $n=1$ and $\sigma_{(4)}$ is singular
for $n=2$. In particular, we have

$\sigma_{F}=r+\frac{r^{2}}{2}H-\frac{r^{3}}{6}|\mathring{L}|^{2}+\frac{r^{4}}{24}\left(6\operatorname{tr}(\mathring{L}^{3})+5H|\mathring{L}|^{2}+3\Delta(H)\right)+\cdots$

if $n=3$ ([GG19, (2.16)-(2.18]). These results are determined by the
conditions

$\displaystyle\mathcal{S}(S_{2})=1+O(r^{2}),\quad\mathcal{S}(S_{3})=1+O(r^{3}),\quad\mathcal{S}(S_{4})=1+O(r^{4}).$

In particular, the obstructions ${\mathcal{B}}_{2}$ and ${\mathcal{B}}_{3}$
are the restrictions of the remainder terms in the second and the third
expansions. More precisely, for ${\mathcal{B}}_{2}$ ($n=2$), we find

${\mathcal{B}}_{2}=(r^{-3}(\mathcal{S}(S_{3})-1))|_{0}=-\frac{1}{3}(H|\mathring{L}|^{2}+\Delta(H))-\frac{2}{3}\operatorname{tr}(\mathring{L}^{3}).$
(4.9)

Since for $n=2$ the term $\operatorname{tr}(\mathring{L}^{3})$ vanishes, we
get

${\mathcal{B}}_{2}=-\frac{1}{3}(H|\mathring{L}|^{2}+\Delta(H)).$

Similarly, for $n=3$, we get

$\displaystyle{\mathcal{B}}_{3}$
$\displaystyle=(r^{-4}(\mathcal{S}(S_{4})-1))|_{0}$
$\displaystyle=\frac{1}{12}(|\mathring{L}|^{4}-6H\Delta(H)+\Delta(|\mathring{L}|^{2})+6H\operatorname{tr}(\mathring{L}^{3})+3|dH|^{2}-3\Delta^{\prime}(H)).$

where $\Delta_{h_{r}}=\Delta_{h}+r\Delta^{\prime}_{h}+\cdots$. By the
variation formula $\Delta^{\prime}(u)=-2(L,\operatorname{Hess}(u))-3(dH,du)$
(see the proof of Lemma 6.2), this leads to

$12{\mathcal{B}}_{3}=\Delta(|\mathring{L}|^{2})+6(\mathring{L},\operatorname{Hess}(H))+|\mathring{L}|^{4}+6H\operatorname{tr}(\mathring{L}^{3})+12|dH|^{2}.$
(4.10)

For general background, we shall express the coefficients $\sigma_{(k)}$ in
terms of the volume coefficients $v_{k}$ of $h_{r}$, which are defined by the
expansion

$v(r)=\sum_{k\geq 0}r^{k}v_{k}$

of

$v(r)\stackrel{{\scriptstyle\text{def}}}{{=}}dvol(h_{r})/dvol(h)=(\det(h_{r})/\det(h))^{\frac{1}{2}}.$

This is convenient since the identity

$\frac{v^{\prime}(r)}{v(r)}=\frac{1}{2}\operatorname{tr}(h_{r}^{-1}h_{r}^{\prime})$
(4.11)

provides natural formulas for the expansion of the coefficient
$\operatorname{tr}(h_{r}^{-1}h_{r}^{\prime})$ in (4).

Note that for the background ${\mathbb{R}}^{n+1}$ it holds

$v(r)=\det(\operatorname{Id}+rL)$ (4.12)

(see [G04, Section 3.4]). Hence $v_{n+1}=0$ in this case. For a general
backgrounds, the volume coefficient $v_{n+1}$ does not vanish, however.

The calculation of the remainder term in the expansion of $\mathcal{S}(S_{k})$
requires $k$ normal derivatives of the equation (4.6). Since the expansion of
$S_{k}$ has a vanishing zeroth-order term, this amounts to take $k-1$
derivatives of the trace term. In turn, this involves volume coefficients of
the metric $h_{r}$ up to order $k$. In general, we find

${\mathcal{B}}_{n}\stackrel{{\scriptstyle\text{def}}}{{=}}(r^{-(n+1)}(\mathcal{S}(\sigma_{F})-1))|_{0}=(\cdots)-2v_{n+1}.$

In particular, ${\mathcal{B}}_{2}$ involves the coefficient $v_{3}$ and
${\mathcal{B}}_{3}$ involves $v_{4}$.

Now, the above algorithm shows that, for general backgrounds and in general
dimensions, the coefficients $\sigma_{(k)}$ are given by the formulas

$\displaystyle\sigma_{(2)}$ $\displaystyle=\frac{1}{2n}v_{1},$
$\displaystyle\sigma_{(3)}$
$\displaystyle=\frac{2}{3(n-1)}v_{2}-\frac{1}{3n}v_{1}^{2}+\frac{1}{3(n-1)}\bar{{\sf
J}},$ $\displaystyle\sigma_{(4)}$
$\displaystyle=\frac{3}{4(n-2)}v_{3}-\frac{9n^{2}-20n+7}{12n(n-1)(n-2)}v_{1}v_{2}+\frac{6n^{2}-11n+1}{24n^{2}(n-2)}v_{1}^{3}$
$\displaystyle+\frac{2n-1}{6n(n-1)(n-2)}v_{1}\bar{{\sf
J}}+\frac{1}{4(n-2)}\bar{{\sf
J}}^{\prime}+\frac{1}{4(n-2)}\Delta(\sigma_{(2)}).$ (4.13)

The observation that $\sigma_{(4)}$ has a (formal) pole at $n=2$ reflects the
fact that there is no approximate solution $\sigma_{F}$ up to order $r^{4}$ in
that dimension. Similarly, $\sigma_{(5)}$ has a pole at $n=3$ \- we shall not
display an explicit formula for $\sigma_{(5)}$, however. The obstruction to
the existence of a smooth solution in $n=3$ is defined in terms of
$\mathcal{S}(S_{4})$.

In the flat case, the identity (4.12) implies that the volume coefficients
$v_{k}$ are given by the elementary symmetric polynomials $\sigma_{k}(L)$ in
the eigenvalues of the shape operator defined by $L$. Hence Newton’s formulas
show that

$\displaystyle v_{1}$ $\displaystyle=nH,$ $\displaystyle v_{2}$
$\displaystyle=\frac{1}{2}(nH)^{2}-\frac{1}{2}|L|^{2},$ $\displaystyle v_{3}$
$\displaystyle=\frac{1}{6}(nH)^{3}-\frac{1}{2}nH|L|^{2}+\frac{1}{3}\operatorname{tr}(L^{3}).$
(4.14)

A combination of these formulas with (4) reproduces the expressions in (4.7),
(4.8).

##  5\. The singular Yamabe obstruction ${\mathcal{B}}_{2}$

In this section, we derive an explicit formula for the obstruction
${\mathcal{B}}_{2}$ from its definition in Section 4. This reproves a result
in [ACF92]. We also briefly recall the relation to the conformal Willmore
functional $\mathcal{W}_{2}$.

Let $n=2$. The formula for ${\mathcal{B}}_{2}$ in terms of volume coefficients
reads

${\mathcal{B}}_{2}\stackrel{{\scriptstyle\text{def}}}{{=}}(r^{-3}(\mathcal{S}(S_{3})-1))|_{0}=-2v_{3}-\frac{1}{12}v_{1}^{3}+\frac{1}{3}v_{1}v_{2}-\frac{2}{3}\Delta(\sigma_{(2)})-\frac{2}{3}v_{1}\bar{{\sf
J}}-\frac{2}{3}\bar{{\sf J}}^{\prime}.$ (5.1)

We recall that the term $v_{3}$ vanishes in $n=2$ in the flat case but not in
the curved case. In the flat case, this formula reduces to (4.9). The proof
easily follows using $v_{1}=2H$, $v_{2}=H^{2}-|\mathring{L}|^{2}/2$ (see (4))
and $\sigma_{(2)}=H/2$. In the general case, the formulas for the volume
coefficients in Lemma 6.4 imply

$\displaystyle{\mathcal{B}}_{2}$
$\displaystyle=\left(\frac{1}{3}\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}-\frac{1}{6}\overline{\operatorname{scal}}^{\prime}\right)-\frac{1}{3}H\overline{\operatorname{scal}}+H\overline{\operatorname{Ric}}_{00}-\frac{2}{3}(\mathring{L},\bar{{\mathcal{G}}})-\frac{1}{3}\Delta(H)-\frac{1}{3}H|\mathring{L}|^{2}$

using $\operatorname{tr}(\mathring{L}^{3})=0$ in dimension $n=2$. Now the
second Bianchi identity implies

$\frac{1}{3}\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}-\frac{1}{6}\overline{\operatorname{scal}}^{\prime}=-\frac{1}{3}\delta(\overline{{\sf
P}}_{0})+\frac{1}{3}(\mathring{L},\bar{{\sf
P}})-H\overline{\operatorname{Ric}}_{00}+\frac{1}{3}H\overline{\operatorname{scal}}.$

(see [JO21, (13.6.5)]). Hence using
$(\mathring{L},\bar{{\mathcal{G}}})=(\mathring{L},\bar{{\sf P}})$ we find

${\mathcal{B}}_{2}=-\frac{1}{3}(\Delta(H)+H|\mathring{L}|^{2}+\delta(\overline{{\sf
P}}_{0})+(\mathring{L},\overline{{\sf P}})).$ (5.2)

By Codazzi-Mainardi
$dH=\delta(\mathring{L})-\overline{\operatorname{Ric}}_{0}$, this formula is
equivalent to

${\mathcal{B}}_{2}=-\frac{1}{3}(\delta\delta(\mathring{L})+H|\mathring{L}|^{2}+(\mathring{L},\bar{{\sf
P}})).$

The latter formula for the obstruction was first derived in [ACF92, Theorem
1.3].

######  Remark 5.1.

The coefficient $\sigma_{(4)}$ has a simple (formal) pole in $n=2$. Moreover,
(4) implies

$\operatorname{Res}_{n=2}(\sigma_{(4)})=\frac{3}{4}v_{3}+\frac{1}{32}v_{1}^{3}-\frac{1}{8}v_{1}v_{2}+\frac{1}{4}v_{1}\bar{{\sf
J}}+\frac{1}{4}\bar{{\sf J}}^{\prime}+\frac{1}{4}\Delta(\sigma_{(2)}).$

This formula shows the residue formula

$\operatorname{Res}_{n=2}(\sigma_{(4)})=-\frac{3}{8}{\mathcal{B}}_{2}$

being a special case of [JO21, Lemma 16.3.9]).

Let $K$ be the Gauss curvature of a surface
$M\hookrightarrow{\mathbb{R}}^{3}$. By $2(H^{2}-K^{2})=|\mathring{L}|^{2}$,
the equation $\Delta(H)+H|\mathring{L}|^{2}=0$ is equivalent to

$\Delta(H)+2H(H^{2}-K)=0$

This equation is well-known as the Willmore equation for a surface $M$. It is
the Euler-Lagrange equation of the Willmore functional

$\mathcal{W}_{2}=\int_{M}|\mathring{L}|^{2}dvol_{h}$

for variations of the embedding of $M$ [W93, Section 7.4]. In other words,
${\mathcal{B}}_{2}$ provides the Euler-Lagrange equation of $\mathcal{W}_{2}$.
This fact extends to the curved case (for details see [JO21, Section 13.9]).

##  6\. The singular Yamabe obstruction ${\mathcal{B}}_{3}$

In this section, we determine explicit formulas for the obstruction
${\mathcal{B}}_{3}$. We shall start by expressing the definition (4.5) in
terms of volume coefficients of the background metric and two normal
derivatives of the scalar curvature. We simplify that result by repeated
applications of the second Bianchi identity. A sequence of further
transformations finally leads to Theorem 1.

###  6.1. ${\mathcal{B}}_{3}$ in terms of volume coefficients

For $n=3$, the formulas in (4) read

$\displaystyle\sigma_{(2)}=\frac{1}{6}v_{1},\quad\sigma_{(3)}=\frac{1}{9}(3v_{2}-v_{1}^{2})+\frac{1}{6}\bar{{\sf
J}}$

and

$\sigma_{(4)}=\frac{1}{108}(81v_{3}-42v_{1}v_{2}+11v_{1}^{3})+\frac{5}{36}v_{1}\bar{{\sf
J}}+\frac{1}{4}\bar{{\sf J}}^{\prime}+\frac{1}{4}\Delta(\sigma_{(2)}).$

These quantities define $S_{4}$. We also recall the expansion
$\Delta_{h_{r}}=\Delta_{h}+r\Delta_{h}^{\prime}+\cdots$. In these terms, we
obtain

######  Lemma 6.1.

It holds

$\displaystyle{\mathcal{B}}_{3}\stackrel{{\scriptstyle\text{def}}}{{=}}(r^{-4}(\mathcal{S}(S_{4})-1))|_{0}$
$\displaystyle=-2v_{4}+\frac{1}{2}v_{1}v_{3}+\frac{1}{3}v_{2}^{2}-\frac{7}{18}v_{1}^{2}v_{2}+\frac{2}{27}v_{1}^{4}$
$\displaystyle-\frac{1}{3}\bar{{\sf J}}v_{2}-\frac{5}{12}\bar{{\sf
J}}^{\prime}v_{1}-\frac{1}{4}\bar{{\sf J}}^{\prime\prime}$
$\displaystyle-\frac{1}{2}\Delta(\sigma_{(3)})-\frac{1}{3}v_{1}\Delta(\sigma_{(2)})-\frac{1}{2}\Delta^{\prime}(\sigma_{(2)})+|d\sigma_{(2)}|^{2}.$
(6.1)

This result follows by direct evaluation of the definition of
${\mathcal{B}}_{3}$. We omit the details.

In the remaining part of this section, we evaluate this formula.

First of all, we calculate the last line in (6.1).

######  Lemma 6.2.

It holds

$\displaystyle-\frac{1}{2}\Delta(\sigma_{(3)})-\frac{1}{3}v_{1}\Delta(\sigma_{(2)})-\frac{1}{2}\Delta^{\prime}(\sigma_{(2)})+|d\sigma_{(2)}|^{2}$
$\displaystyle=\frac{1}{12}\Delta(|\mathring{L}|^{2})+\frac{1}{2}(\mathring{L},\operatorname{Hess}(H))+\frac{1}{6}\Delta(\bar{{\sf
P}}_{00})+\frac{1}{2}(dH,\overline{\operatorname{Ric}}_{0})+|dH|^{2}.$

###### Proof.

We recall that $v_{1}=3H$. By

$\displaystyle\sigma_{(2)}=\frac{1}{2}H\quad\mbox{and}\quad\sigma_{(3)}=\frac{1}{6}(-|\mathring{L}|^{2}-2\bar{{\sf
P}}_{00}),$

we obtain

$\displaystyle-\frac{1}{2}\Delta(\sigma_{(3)})-\frac{1}{3}v_{1}\Delta(\sigma_{(2)})-\frac{1}{2}\Delta^{\prime}(\sigma_{(2)})+|d\sigma_{(2)}|^{2}$
$\displaystyle=\frac{1}{12}\Delta(|\mathring{L}|^{2})+\frac{1}{6}\Delta(\bar{{\sf
P}}_{00})-\frac{1}{2}H\Delta
H-\frac{1}{4}\Delta^{\prime}(H)+\frac{1}{4}|dH|^{2}.$

Now the variation formula [B87, Proposition 1.184]

$\displaystyle(d/dt)|_{0}(\Delta_{g+th}(u))=-(\nabla^{g}(du),h)_{g}-(\delta_{g}(h),du)_{g}+\frac{1}{2}(d(\operatorname{tr}_{g}(h)),du)_{g}$
(6.2)

for the Laplacian implies (for $h=2L$ and $g=h$)

$\Delta^{\prime}(u)=-2(L,\operatorname{Hess}(u))-2(\delta(L),du)+3(dH,du).$

By Codazzi-Mainardi it holds $\delta(L)=3dH+2\bar{{\sf
P}}_{0}=3dH+\overline{\operatorname{Ric}}_{0}$. Hence

$\Delta^{\prime}(u)=-2(L,\operatorname{Hess}(u))-3(dH,du)-2(\overline{\operatorname{Ric}}_{0},du).$

These results imply the assertion. ∎

###  6.2. The volume coefficients

The volume coefficients $v_{j}$ can be expressed in terms of the Taylor
coefficients of $h_{r}$. These relations follow by Taylor expansion of the
identity (4.11) in the variable $r$ and solving the resulting relations for
$v_{j}$. We find

$\displaystyle 2v_{1}$ $\displaystyle=\operatorname{tr}(h_{(1)}),$
$\displaystyle 8v_{2}$
$\displaystyle=\operatorname{tr}(h_{(1)})^{2}+4\operatorname{tr}(h_{(2)})-2\operatorname{tr}(h_{(1)}^{2}),$
$\displaystyle 48v_{3}$
$\displaystyle=\operatorname{tr}(h_{(1)})^{3}+12\operatorname{tr}(h_{(1)})\operatorname{tr}(h_{(2)})+24\operatorname{tr}(h_{(3)})-6\operatorname{tr}(h_{(1)})\operatorname{tr}(h_{(1)}^{2})$
$\displaystyle-24\operatorname{tr}(h_{(1)}h_{(2)})+8\operatorname{tr}(h_{(1)}^{3})$

and

$\displaystyle 384v_{4}$
$\displaystyle=\operatorname{tr}(h_{(1)})^{4}+24\operatorname{tr}(h_{(1)})^{2}\operatorname{tr}(h_{(2)})+48\operatorname{tr}(h_{(2)})^{2}+96\operatorname{tr}(h_{(1)})\operatorname{tr}(h_{(3)})+192\operatorname{tr}(h_{(4)})$
$\displaystyle-12\operatorname{tr}(h_{(1)})^{2}\operatorname{tr}(h_{(1)}^{2})-48\operatorname{tr}(h_{(2)})\operatorname{tr}(h_{(1)}^{2})+12\operatorname{tr}(h_{(1)}^{2})^{2}-96\operatorname{tr}(h_{(1)})\operatorname{tr}(h_{(1)}h_{(2)})$
$\displaystyle-192\operatorname{tr}(h_{(1)}h_{(3)})-96\operatorname{tr}(h_{(2)}^{2})+32\operatorname{tr}(h_{(1)})\operatorname{tr}(h_{(1)}^{3})+192\operatorname{tr}(h_{(1)}^{2}h_{(2)})-48\operatorname{tr}(h_{(1)}^{4}.$

These formulas are valid in general dimension. In order to evaluate them, we
apply the following results for the coefficients $h_{(k)}$ for $k\leq 3$
[GG19], [JO21, Proposition 13.2.1].

######  Lemma 6.3.

In general dimensions, it holds

$h_{(1)}=2L,\quad h_{(2)}=L^{2}-\bar{{\mathcal{G}}}\quad\mbox{and}\quad
3(h_{(3)})_{ij}=-\bar{\nabla}_{0}(\bar{R})_{0ij0}-2L_{i}^{k}\bar{{\mathcal{G}}}_{jk}-2L_{j}^{k}\bar{{\mathcal{G}}}_{ik},$

where
$\bar{{\mathcal{G}}}_{ij}\stackrel{{\scriptstyle\text{def}}}{{=}}\bar{R}_{0ij0}$.

As consequences, we find explicit formulas for the volume coefficients $v_{k}$
for $k\leq 3$.

######  Lemma 6.4.

In general dimensions, it holds

$\displaystyle v_{1}$ $\displaystyle=nH$ $\displaystyle 2v_{2}$
$\displaystyle=-\overline{\operatorname{Ric}}_{00}-|\mathring{L}|^{2}+n(n\\!-\\!1)H^{2}=\overline{\operatorname{Ric}}_{00}+\operatorname{scal}-\overline{\operatorname{scal}}$
$\displaystyle 6v_{3}$
$\displaystyle=-\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}+2(\mathring{L},\bar{{\mathcal{G}}})-(3n\\!-\\!2)H\overline{\operatorname{Ric}}_{00}+2\operatorname{tr}(\mathring{L}^{3})-3(n\\!-\\!2)H|\mathring{L}|^{2}+n(n\\!-\\!1)(n\\!-\\!2)H^{3}.$

These formulas coincide with the corresponding terms in the expansion of the
volume form in [AGV, Theorem 3.4]. Note that this is obvious for $v_{1}$ and
$v_{2}$ but requires to apply the Gauss identities

$\displaystyle\overline{\operatorname{scal}}-\operatorname{scal}$
$\displaystyle=2\overline{\operatorname{Ric}}_{00}+|L|^{2}-n^{2}H^{2},$
$\displaystyle\overline{\operatorname{Ric}}-\operatorname{Ric}$
$\displaystyle=\bar{{\mathcal{G}}}-nHL-L^{2}$

for $v_{3}$. Equivalent formulas can be found in [GG19, Section 2].

The coefficient $v_{4}$ is more involved. It depends on $h_{(k)}$ for $k\leq
3$ and $\operatorname{tr}(h_{(4)})$. We shall not discuss an explicit formula
for $h_{(4)}$. For our purpose, it will be enough to prove the following
formula for the quantity $\operatorname{tr}(h_{(4)})$.

######  Lemma 6.5.

In general dimensions, it holds

$12\operatorname{tr}(h_{(4)})=-\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}-6L^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}-4(L^{2},\bar{{\mathcal{G}}})+4(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}}).$
(6.3)

###### Proof.

A calculation of Christoffel symbols shows that

$\bar{R}_{0jk0}=\frac{1}{4}g^{ab}g_{aj}^{\prime}g_{bk}^{\prime}-\frac{1}{2}g_{jk}^{\prime\prime}$
(6.4)

[JO21, (13.2.5)]. The assertion then follows by evaluating the second-order
derivative in $r$ of this equation followed by contraction with $h^{jk}$. Here
are the details. Differentiating (6.4) twice in $r$ at $r=0$ yields

$\displaystyle\partial_{r}^{2}(\bar{R}_{0jk0})$
$\displaystyle=\frac{1}{4}(g^{ab})^{\prime\prime}g_{aj}^{\prime}g_{bk}^{\prime}+\frac{1}{4}g^{ab}(g_{aj})^{\prime\prime\prime}g_{bk}^{\prime}+\frac{1}{4}g^{ab}g_{aj}^{\prime}(g_{bk})^{\prime\prime\prime}$
$\displaystyle+\frac{1}{2}(g^{ab})^{\prime}(g_{aj})^{\prime\prime}g_{bk}^{\prime}+\frac{1}{2}(g^{ab})^{\prime}(g_{aj})^{\prime}(g_{bk})^{\prime\prime}+\frac{1}{2}g^{ab}(g_{aj})^{\prime\prime}(g_{bk})^{\prime\prime}-\frac{1}{2}g_{jk}^{\prime\prime\prime\prime}$
$\displaystyle=2(3L^{2}+\bar{{\mathcal{G}}})^{ab}L_{aj}L_{bk}+3h^{ab}(h_{(3)})_{aj}L_{bk}+3h^{ab}L_{aj}(h_{(3)})_{bk}$
$\displaystyle-4L^{ab}(L^{2}-\bar{{\mathcal{G}}})_{aj}L_{bk}-4L^{ab}L_{aj}(L^{2}-\bar{{\mathcal{G}}})_{bk}+2h^{ab}(L^{2}-\bar{{\mathcal{G}}})_{aj}(L^{2}-\bar{{\mathcal{G}}})_{bk}-12(h_{(4)})_{jk}$
$\displaystyle=2(3L^{2}+\bar{{\mathcal{G}}})^{ab}L_{aj}L_{bk}+h^{ab}(-\bar{\nabla}_{0}(\bar{R})_{0aj0}-2(L\bar{{\mathcal{G}}})_{aj}-2(\bar{{\mathcal{G}}}L)_{aj})L_{bk}$
$\displaystyle+h^{ab}L_{aj}(-\bar{\nabla}_{0}(\bar{R})_{0bk0}-2(L\bar{{\mathcal{G}}})_{bk}-2(\bar{{\mathcal{G}}}L)_{bk})$
$\displaystyle-4L^{ab}(L^{2}-\bar{{\mathcal{G}}})_{aj}L_{bk}-4L^{ab}L_{aj}(L^{2}-\bar{{\mathcal{G}}})_{bk}+2h^{ab}(L^{2}-\bar{{\mathcal{G}}})_{aj}(L^{2}-\bar{{\mathcal{G}}})_{bk}-12(h_{(4)})_{jk}$

using
$(h_{r}^{-1})_{ij}=h^{ij}-2L^{ij}r+(3(L^{2})^{ij}+\bar{{\mathcal{G}}}^{ij})r^{2}+\cdots$.
Hence

$h^{jk}\partial_{r}^{2}(\bar{R}_{0jk0})$

equals

$-2L^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}-2(L^{2},\bar{{\mathcal{G}}})+2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}})-12\operatorname{tr}(h_{(4)}).$
(6.5)

On the other hand, we calculate

$\displaystyle h^{jk}\partial_{r}^{2}(\bar{R}_{0jk0})$
$\displaystyle=\partial_{r}^{2}((h_{r}^{-1})_{jk}\bar{R}_{0jk0})-2(h_{r}^{-1})^{\prime}_{jk}\partial_{r}(\bar{R}_{0jk0})-(h_{r}^{-1})^{\prime\prime}_{jk}\bar{R}_{0jk0}$
$\displaystyle=\partial_{r}^{2}(\overline{\operatorname{Ric}}_{00})+4L^{jk}\partial_{r}(\bar{R}_{0jk0})-2(3L^{2}+\bar{{\mathcal{G}}},{\mathcal{G}}).$

Therefore, the relations

$\displaystyle\partial_{r}^{2}(\overline{\operatorname{Ric}}_{00})$
$\displaystyle=\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00},$
$\displaystyle\partial_{r}(\bar{R}_{0jk0})$
$\displaystyle=\bar{\nabla}_{0}(\bar{R})_{0jk0}+(L\bar{{\mathcal{G}}}+\bar{{\mathcal{G}}}L)_{jk}$

imply

$h^{jk}\partial_{r}^{2}(\bar{R}_{0jk0})=\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}+4L^{jk}\bar{\nabla}_{0}(\bar{R})_{0jk0}+8(L^{2},\bar{{\mathcal{G}}})-6(L^{2},\bar{{\mathcal{G}}})-2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}}).$
(6.6)

Now combining (6.5) and (6.6) proves the assertion. ∎

######  Example 6.6.

Let $n\geq 3$ be general. Assume that $g=r^{2}g_{+}$ for a Poincaré-Einstein
metric $g_{+}=r^{-2}(dr^{2}+(h-{\sf P}r^{2}+{\sf P}^{2}r^{4}/4))$ with
conformally flat conformal infinity $h$. ${\sf P}$ is the Schouten tensor of
$h$. $r$ is the distance in the metric $g$ from the hypersurface $r=0$. The
formula for $g$ shows that $\operatorname{tr}(h_{(4)})=1/4|{\sf P}|^{2}$.
Comparing the coefficients of $r$ and $r^{2}$ in the expansions of $h_{r}$,
shows that $L=0$ and $\bar{{\mathcal{G}}}={\sf P}$. Hence the above formula
reduces to

$12\operatorname{tr}(h_{(4)})=-\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}+4({\sf
P},{\sf P})=-\partial_{r}^{2}(\overline{\operatorname{Ric}}_{00})+4({\sf
P},{\sf P}).$

But [J09, Lemma 6.11.2] shows that $\bar{{\sf
P}}=-1/(2r)\partial_{r}({h}_{r})$. Hence $\bar{{\sf P}}={\sf P}-1/2r^{2}{\sf
P}^{2}$. It follows that $\partial_{r}^{2}(\bar{{\sf P}})=-{\sf P}^{2}$.
Therefore, $\partial_{r}^{2}(\bar{{\sf P}}_{00})=0$ and we conclude that
$\partial_{r}^{2}(\overline{\operatorname{Ric}}_{00})=\partial_{r}^{2}(\bar{{\sf
J}})\stackrel{{\scriptstyle!}}{{=}}|{\sf P}|^{2}$ (for $r=0$) using [J09,
Lemma 6.11.1]. Hence the right-hand side gives $3|{\sf P}|^{2}$, i.e., we
reproduced the result $4\operatorname{tr}(h_{4)})=|{\sf P}|^{2}$. For general
$h$, the Poincareé-Einstein metric also involves the Bach tensor. But since
the Bach tensor is trace-free, we still have
$\operatorname{tr}(h_{(4)})=1/4|{\sf P}|^{2}$ and we get the same conclusion.

The above results imply the following formula for $v_{4}$.

######  Lemma 6.7.

It holds

$\displaystyle 24v_{4}$
$\displaystyle=-\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}+2L^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}-4nH\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle+3(\overline{\operatorname{Ric}}_{00})^{2}-2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}})+8nH(L,\bar{{\mathcal{G}}})-8(L^{2},\bar{{\mathcal{G}}})+6|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-6n(n\\!-\\!1)H^{2}\overline{\operatorname{Ric}}_{00}$
$\displaystyle+24\sigma_{4}(L)$ (6.7)

or, equivalently,

$\displaystyle 24v_{4}$
$\displaystyle=-\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}+2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}-(4n\\!-\\!2)H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle+3(\overline{\operatorname{Ric}}_{00})^{2}-2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}})+8(n\\!-\\!2)H(\mathring{L},\bar{{\mathcal{G}}})-8(\mathring{L}^{2},\bar{{\mathcal{G}}})$
$\displaystyle-2(n\\!-\\!1)(3n\\!-\\!4)H^{2}\overline{\operatorname{Ric}}_{00}+6|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}$
$\displaystyle+24\sigma_{4}(L).$

In particular, for $n=3$, we find

$\displaystyle 24v_{4}$
$\displaystyle=-\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}+2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}-10H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle+3(\overline{\operatorname{Ric}}_{00})^{2}-2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}})+8H(\mathring{L},\bar{{\mathcal{G}}})-8(\mathring{L}^{2},\bar{{\mathcal{G}}})-20H^{2}\overline{\operatorname{Ric}}_{00}+6|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}$
(6.8)

using $\sigma_{4}(L)=0$.

###### Proof.

This is a direct calculation. We omit the details. ∎

The three terms in the first line of (6.7) coincide with the corresponding
terms in the formula for $v_{4}$ in [AGV, Theorem 3.4]. However, the remaining
terms in both formulas are expressed in different ways.

Note that Newton’s identity

$24\sigma_{4}(L)=\operatorname{tr}(L)^{4}-6\operatorname{tr}(L)^{2}|L|^{2}+3|L|^{4}+8\operatorname{tr}(L)\operatorname{tr}(L^{3})-6\operatorname{tr}(L^{4})$

gives

$24\sigma_{4}(L)=n(n-1)(n-2)(n-3)H^{4}-6(n-2)(n-3)H|\mathring{L}|^{2}+8(n-3)H\operatorname{tr}(\mathring{L}^{3})+3(|\mathring{L}|^{4}-2\operatorname{tr}(\mathring{L}^{4})).$

######  Corollary 6.8.

Let $n=3$. Then $|\mathring{L}|^{4}=2\operatorname{tr}(\mathring{L}^{4})$.

This result generalizes the fact that $\operatorname{tr}(\mathring{L}^{3})=0$
for $n=2$.

######  Example 6.9.

Assume that $g=r^{2}g_{+}$ for a Poincaré-Einstein metric $g_{+}$ with
conformal infinity $h$ (as in Example 6.6). By $L=0$, the formula (6.7) reads

$\displaystyle 24v_{4}$
$\displaystyle=-\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}+3(\overline{\operatorname{Ric}}_{00})^{2}-2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}}).$

As above, we obtain

$24v_{4}=-|{\sf P}|^{2}+3{\sf J}^{2}-2|{\sf P}|^{2}$

using the fact that $\bar{{\mathcal{G}}}={\sf P}$ implies
$\overline{\operatorname{Ric}}_{00}={\sf J}$. This yields the well-known
formula

$v_{4}=\frac{1}{8}({\sf J}^{2}-|{\sf P}|^{2}).$

###  6.3. Evaluation I

Now we combine the formula (6.1) for ${\mathcal{B}}_{3}$ with the results in
Section 6.2. A calculation yields the following result.

######  Lemma 6.10.

$12{\mathcal{B}}_{3}$ equals the sum of

$\displaystyle\left(\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}-\frac{1}{2}\overline{\operatorname{scal}}^{\prime\prime}\right)+5H\left(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}-\frac{1}{2}\overline{\operatorname{scal}}^{\prime}\right)+2H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle+2|\bar{{\mathcal{G}}}|^{2}-2(\overline{\operatorname{Ric}}_{00})^{2}+2\overline{\operatorname{Ric}}_{00}\bar{{\sf
J}}+8H^{2}\overline{\operatorname{Ric}}_{00}-12H^{2}\bar{{\sf
J}}+6(dH,\overline{\operatorname{Ric}}_{0})+2\Delta(\overline{{\sf P}}_{00}),$
(6.9)
$-2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}-2H(\mathring{L},\bar{{\mathcal{G}}})+8(\mathring{L}^{2},\bar{{\mathcal{G}}})-4|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}+2|\mathring{L}|^{2}\bar{{\sf
J}}$ (6.10)

and

$\Delta(|\mathring{L}|^{2})+6(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}+12|dH|^{2}.$
(6.11)

Since the terms in (6.10) and (6.10) vanish for the flat metric, we
immediately reproduced formula (4.10). For later reference, we formulate that
result as

######  Corollary 6.11.

For a hypersurface $M$ in the flat background ${\mathbb{R}}^{4}$, the
obstruction ${\mathcal{B}}_{3}$ is given by

$\Delta(|\mathring{L}|^{2})+6(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}+12|dH|^{2}.$
(6.12)

We continue with the discussion of the curved case.

Next, we simplify the sum (6.10) using the second Bianchi identity. This step
is analogous to the usage of the second Bianchi identity in Section 5.

Let
$\bar{G}\stackrel{{\scriptstyle\text{def}}}{{=}}\overline{\operatorname{Ric}}-\frac{1}{2}\overline{\operatorname{scal}}g$
be the Einstein tensor of $g$. The second Bianchi identity implies
$2\delta^{g}(\overline{\operatorname{Ric}})=d\overline{\operatorname{scal}}$.
Hence

$\displaystyle\bar{\nabla}_{0}(\overline{\operatorname{Ric}})(\partial_{0},\partial_{0})$
$\displaystyle=\delta^{g}(\overline{\operatorname{Ric}})(\partial_{0})-g^{ij}\bar{\nabla}_{\partial_{i}}(\overline{\operatorname{Ric}})(\partial_{j},\partial_{0})$
$\displaystyle=\frac{1}{2}\langle
d\overline{\operatorname{scal}},\partial_{0}\rangle-g^{ij}\partial_{i}(\overline{\operatorname{Ric}}(\partial_{j},\partial_{0}))+g^{ij}\overline{\operatorname{Ric}}(\bar{\nabla}_{\partial_{i}}(\partial_{j}),\partial_{0})+g^{ij}\overline{\operatorname{Ric}}(\partial_{j},\bar{\nabla}_{\partial_{i}}(\partial_{0}))$
$\displaystyle=\frac{1}{2}\langle
d\overline{\operatorname{scal}},\partial_{0}\rangle-
h_{r}^{ij}\partial_{i}(\overline{\operatorname{Ric}}(\partial_{j},\partial_{0}))+h_{r}^{ij}\overline{\operatorname{Ric}}(\nabla^{h_{r}}_{\partial_{i}}(\partial_{j})-(L_{r})_{ij}\partial_{0},\partial_{0})+h_{r}^{ij}\overline{\operatorname{Ric}}(\partial_{j},\bar{\nabla}_{\partial_{i}}(\partial_{0}))$
$\displaystyle=\frac{1}{2}\langle
d\overline{\operatorname{scal}},\partial_{0}\rangle-\delta^{h_{r}}(\overline{\operatorname{Ric}}_{0})-nH_{r}\overline{\operatorname{Ric}}_{00}+h_{r}^{ij}\overline{\operatorname{Ric}}(\partial_{j},\bar{\nabla}_{\partial_{i}}(\partial_{0}))$

on any level surface of $r$. Here $\delta^{h_{r}}$ denotes the divergence
operator for the induced metric on the level surfaces of $r$. Similarly,
$L_{r}$ and $H_{r}$ are the second fundamental form and the mean curvature of
these level surfaces. Therefore, using
$\bar{\nabla}_{\partial_{i}}(\partial_{0})=(L_{r})_{ia}h_{r}^{ak}\partial_{k}$,
we obtain

$\displaystyle\bar{\nabla}_{0}(\bar{G})_{00}$
$\displaystyle=-\delta^{h_{r}}(\overline{\operatorname{Ric}}_{0})-nH_{r}\overline{\operatorname{Ric}}_{00}+h_{r}^{ij}h_{r}^{ak}(L_{r})_{ia}\overline{\operatorname{Ric}}_{jk},$

i.e., we have proved the relation

$\bar{\nabla}_{0}(\bar{G})_{00}=-\delta^{h_{r}}(\overline{\operatorname{Ric}}_{0})-nH_{r}\overline{\operatorname{Ric}}_{00}+(L_{r},\overline{\operatorname{Ric}})_{h_{r}}$
(6.13)

on any level surface of $r$. Differentiating the identity (6.13) (for $n=3$)
with respect to $r$ at $r=0$ gives a formula for the term

$\bar{\nabla}_{0}^{2}(\overline{\operatorname{Ric}})_{00}-\frac{1}{2}\overline{\operatorname{scal}}^{\prime\prime}=\bar{\nabla}_{0}^{2}(\bar{G})_{00}=\partial_{r}(\bar{\nabla}_{0}(\bar{G})_{00})$

for $r=0$. For that purpose, we use the variation formulas

$\displaystyle 3H^{\prime}$
$\displaystyle=-|L|^{2}-\overline{\operatorname{Ric}}_{00},$ $\displaystyle
L^{\prime}$ $\displaystyle=L^{2}-\bar{{\mathcal{G}}}$ (6.14)

for the variation of these quantities under the normal exponential map. Here
we denote the derivative in $r$ by a prime. We recall that in normal geodesic
coordinates the metric $g$ takes the form $dr^{2}+h_{r}$ with
$h_{r}=h+2rL+\cdots$. Moreover, let
${\delta}^{\prime}\stackrel{{\scriptstyle\text{def}}}{{=}}(d/dr)|_{0}(\delta^{h_{r}})$.
Then

$\displaystyle{\delta}^{\prime}(\omega)=-2(L,\nabla(\omega))_{h}-2(\delta(L),\omega)_{h}+3(dH,\omega)_{h}$
(6.15)

for $\omega\in\Omega^{1}(M^{3})$ [B87, (1.185]. Note that the latter identity
fits with the variation formula (6.2). Now differentiating (6.13) (for $n=3$)
implies

$\displaystyle\bar{\nabla}_{0}^{2}(\bar{G})_{00}$
$\displaystyle=-{\delta}^{\prime}(\overline{\operatorname{Ric}}_{0})-\delta(\partial_{r}(\overline{\operatorname{Ric}}_{0}))-3{H}^{\prime}\overline{\operatorname{Ric}}_{00}-3H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle+({L}^{\prime},\overline{\operatorname{Ric}})+L^{ij}\partial_{r}(\overline{\operatorname{Ric}}_{ij})-4(L^{2},\overline{\operatorname{Ric}}),$

where $L^{\prime}=(d/dr)|_{0}(L_{r})$. Note that the last term comes from the
differentiation of $h_{r}$. Hence

$\displaystyle\bar{\nabla}_{0}^{2}(\bar{G})_{00}$
$\displaystyle=2(L,\nabla(\overline{\operatorname{Ric}}_{0}))+2(\delta(L),\overline{\operatorname{Ric}}_{0})-3(dH,\overline{\operatorname{Ric}}_{0})$
$\displaystyle-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})-\delta((L\overline{\operatorname{Ric}})_{0})+|L|^{2}\overline{\operatorname{Ric}}_{00}+(\overline{\operatorname{Ric}}_{00})^{2}-3H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle-3(L^{2},\overline{\operatorname{Ric}})-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})+(L,\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))+2(L^{2},\overline{\operatorname{Ric}})$

at $r=0$. Here we used the relations

$\displaystyle\bar{\nabla}_{0}(\overline{\operatorname{Ric}}_{0})=\partial_{r}(\overline{\operatorname{Ric}}_{00})-(L\overline{\operatorname{Ric}})_{0}\quad\mbox{and}\quad\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{ij}=\partial_{r}(\overline{\operatorname{Ric}}_{ij})-(L\overline{\operatorname{Ric}}+\overline{\operatorname{Ric}}L)_{ij}.$

Now, separating the trace-free part of $L$ in some terms, we obtain

$\displaystyle\bar{\nabla}_{0}^{2}(\bar{G})_{00}$
$\displaystyle=2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+2H\delta(\overline{\operatorname{Ric}}_{0})+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-(dH,\overline{\operatorname{Ric}}_{0})$
$\displaystyle-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})-\delta(H\overline{\operatorname{Ric}}_{0})$
$\displaystyle+|L|^{2}\overline{\operatorname{Ric}}_{00}+(\overline{\operatorname{Ric}}_{00})^{2}-3H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}$
$\displaystyle-(L^{2},\overline{\operatorname{Ric}})-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})+(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))+H\overline{\operatorname{scal}}^{\prime}-H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}.$

This leads to the following result.

######  Lemma 6.12.

It holds

$\displaystyle\bar{\nabla}_{0}^{2}(\bar{G})_{00}=$
$\displaystyle-4H\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{00}+H\overline{\operatorname{scal}}^{\prime}$
$\displaystyle+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})+(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))$
$\displaystyle+H\delta(\overline{\operatorname{Ric}}_{0})-2(dH,\overline{\operatorname{Ric}}_{0})+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})$
$\displaystyle+|L|^{2}\overline{\operatorname{Ric}}_{00}-(L^{2},\overline{\operatorname{Ric}})+(\overline{\operatorname{Ric}}_{00})^{2}-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}}).$

Lemma 6.12 enables us to replace second-order normal derivatives in Lemma 6.10
by first-order normal and tangential derivatives. More precisely, it follows
that (6.10) equals

$\displaystyle
3H\bar{\nabla}_{0}(\bar{G})_{00}+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})+(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))$
$\displaystyle+H\delta(\overline{\operatorname{Ric}}_{0})+4(dH,\overline{\operatorname{Ric}}_{0})+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})$
$\displaystyle+|L|^{2}\overline{\operatorname{Ric}}_{00}-(L^{2},\overline{\operatorname{Ric}})+(\overline{\operatorname{Ric}}_{00})^{2}-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})$
$\displaystyle+2|\bar{{\mathcal{G}}}|^{2}-2(\overline{\operatorname{Ric}}_{00})^{2}+2\overline{\operatorname{Ric}}_{00}\bar{{\sf
J}}+2\Delta(\overline{{\sf
P}}_{00})+8H^{2}\overline{\operatorname{Ric}}_{00}-12H^{2}\bar{{\sf J}}.$

Now a second application of the second Bianchi identity (6.13) enables us to
replace the first-order normal derivative of the Einstein tensor in this
formula by tangential derivatives. Hence (6.10) equals the sum

$\displaystyle-3H\delta(\overline{\operatorname{Ric}}_{0})-9H^{2}\overline{\operatorname{Ric}}_{00}+3H(L,\overline{\operatorname{Ric}})$
$\displaystyle+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0})))-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})+(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))$
$\displaystyle+H\delta(\overline{\operatorname{Ric}}_{0})+4(dH,\overline{\operatorname{Ric}}_{0})+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})$
$\displaystyle+|L|^{2}\overline{\operatorname{Ric}}_{00}-(L^{2},\overline{\operatorname{Ric}})+(\overline{\operatorname{Ric}}_{00})^{2}-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})$
$\displaystyle+2|\bar{{\mathcal{G}}}|^{2}-2(\overline{\operatorname{Ric}}_{00})^{2}+2\overline{\operatorname{Ric}}_{00}\bar{{\sf
J}}+2\Delta(\overline{{\sf
P}}_{00})+8H^{2}\overline{\operatorname{Ric}}_{00}-12H^{2}\bar{{\sf J}}.$

A slight reordering and simplification of this sum shows that the sum (6.10)
equals

$\displaystyle-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})-2H\delta(\overline{\operatorname{Ric}}_{0})+4(dH,\overline{\operatorname{Ric}}_{0})$
$\displaystyle+(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})$
$\displaystyle+|L|^{2}\overline{\operatorname{Ric}}_{00}-(L^{2},\overline{\operatorname{Ric}})+3H(L,\overline{\operatorname{Ric}})$
$\displaystyle-(\overline{\operatorname{Ric}}_{00})^{2}-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})+2|\bar{{\mathcal{G}}}|^{2}+2\overline{\operatorname{Ric}}_{00}\bar{{\sf
J}}+2\Delta(\overline{{\sf
P}}_{00})-H^{2}\overline{\operatorname{Ric}}_{00}-12H^{2}\bar{{\sf J}}.$
(6.16)

We continue by further simplifying the sum (6.3). First of all, we observe

######  Lemma 6.13.

Let $n=3$. Then

$-(\overline{\operatorname{Ric}}_{00})^{2}-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})+2|\bar{{\mathcal{G}}}|^{2}+2\overline{\operatorname{Ric}}_{00}\bar{{\sf
J}}=2(\bar{{\sf P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}.$

###### Proof.

We recall that $\bar{{\mathcal{G}}}_{ij}=\bar{{\sf P}}_{ij}+\bar{{\sf
P}}_{00}h_{ij}+\overline{\mathcal{W}}_{ij}$. Therefore, we get
$\overline{\operatorname{Ric}}_{ij}=2\bar{{\sf P}}_{ij}+\bar{{\sf
J}}h_{ij}=2\bar{{\mathcal{G}}}_{ij}-2\overline{\mathcal{W}}_{ij}-(2\bar{{\sf
P}}_{00}-\bar{{\sf J}})h_{ij}$. Thus
$(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})=2(\bar{{\mathcal{G}}},\bar{{\mathcal{G}}})-2(\bar{{\mathcal{G}}},\overline{\mathcal{W}})-(2\bar{{\sf
P}}_{00}-\bar{{\sf J}})\overline{\operatorname{Ric}}_{00}.$ This relation
implies

$2|\bar{{\mathcal{G}}}|^{2}-(\bar{{\mathcal{G}}},\overline{\operatorname{Ric}})=(2\bar{{\sf
P}}_{00}-\bar{{\sf
J}})\overline{\operatorname{Ric}}_{00}+2(\bar{{\mathcal{G}}},\overline{\mathcal{W}})=(\overline{\operatorname{Ric}}_{00})^{2}-2\overline{\operatorname{Ric}}_{00}\bar{{\sf
J}}+2(\bar{{\sf P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}.$

The proof is complete. ∎

Next, we have the following identities.

######  Lemma 6.14.

Let $n=3$. Then

$|L|^{2}\overline{\operatorname{Ric}}_{00}-(L^{2},\overline{\operatorname{Ric}})=|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-(\mathring{L}^{2},\overline{\operatorname{Ric}})-2H(\mathring{L},\overline{\operatorname{Ric}})+4H^{2}\overline{\operatorname{Ric}}_{00}-6H^{2}\bar{{\sf
J}}$

and

$(L,\overline{\operatorname{Ric}})=(\mathring{L},\overline{\operatorname{Ric}})+6H\bar{{\sf
J}}-H\overline{\operatorname{Ric}}_{00}.$

###### Proof.

The assertions follow by direct calculation. ∎

By Lemma 6.13 and Lemma 6.14, the last two lines of (6.3) simplify to

$\displaystyle|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-(\mathring{L}^{2},\overline{\operatorname{Ric}})+H(\mathring{L},\overline{\operatorname{Ric}})+2\Delta(\overline{{\sf
P}}_{00})+2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}.$

Therefore, (6.10) equals

$\displaystyle-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})-2H\delta(\overline{\operatorname{Ric}}_{0})+4(dH,\overline{\operatorname{Ric}}_{0})$
$\displaystyle+(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})$
$\displaystyle+|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-(\mathring{L}^{2},\overline{\operatorname{Ric}})+H(\mathring{L},\overline{\operatorname{Ric}})+2\Delta(\overline{{\sf
P}}_{00})+2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}.$

Hence Lemma 6.10 implies

######  Lemma 6.15.

$12{\mathcal{B}}_{3}$ equals the sum of

$-\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})-2\delta(H\overline{\operatorname{Ric}}_{0})+6(dH,\overline{\operatorname{Ric}}_{0})+2\Delta(\overline{{\sf
P}}_{00}),$ (6.17)
$\displaystyle(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))-2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-\delta((\mathring{L}\overline{\operatorname{Ric}})_{0})$
$\displaystyle-3|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-(\mathring{L}^{2},\overline{\operatorname{Ric}})+H(\mathring{L},\overline{\operatorname{Ric}}),$
(6.18)
$-2H(\mathring{L},\bar{{\mathcal{G}}})+8(\mathring{L}^{2},\bar{{\mathcal{G}}})+2|\mathring{L}|^{2}\bar{{\sf
J}},$ (6.19) $2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}$ (6.20)

and the flat terms

$\displaystyle
6(\mathring{L},\operatorname{Hess}(H))+\Delta(|\mathring{L}|^{2})+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}+12|dH|^{2}.$
(6.21)

Note that the first term in (6.17) contains a normal derivative of
$\overline{\operatorname{Ric}}$. Likewise the first two terms in (6.15)
contain normal derivatives of the curvature of $g$. All other terms in
(6.17)–(6.21) live on $M$. The mixed terms in (6.15) and (6.19) involve the
curvature of $g$ and $L$. Finally, the terms in (6.20) involve the Weyl
tensor, and the terms in (6.21) are completely determined by $L$.

######  Example 6.16.

If the background metric $g$ is Einstein, i.e., if
$\overline{\operatorname{Ric}}=\lambda g$, then

$\displaystyle 12{\mathcal{B}}_{3}$
$\displaystyle=-2\mathring{L}^{ij}\bar{\nabla}_{0}(\widebar{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})+8(\mathring{L}^{2},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}$
$\displaystyle+6(\mathring{L},\operatorname{Hess}(H))+\Delta(|\mathring{L}|^{2})+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}+12|dH|^{2}.$

Of course, if in addition $\overline{W}=0$, then this formula reduces to
Corollary 6.11.

###### Proof.

The assumption implies $\bar{{\sf J}}=\frac{2}{3}\lambda$, $\bar{{\sf
P}}=\frac{1}{6}\lambda g$, $\bar{{\mathcal{G}}}=\frac{1}{3}\lambda
h+\overline{\mathcal{W}}$ and
$|\bar{{\mathcal{G}}}|^{2}=\frac{1}{3}\lambda^{2}+|\overline{\mathcal{W}}|^{2}$.
Furthermore, the terms in (6.17) and the terms in the first line of (6.15)
except the second one vanish. The remaining terms in (6.15)–(6.20) read

$-3\lambda|\mathring{L}|^{2}-\lambda|\mathring{L}|^{2}-2H(\mathring{L},\overline{\mathcal{W}})+\frac{8}{3}\lambda|\mathring{L}|^{2}+8(\mathring{L}^{2},\overline{\mathcal{W}})+\frac{4}{3}\lambda|\mathring{L}|^{2}+2|\overline{\mathcal{W}}|^{2}.$

Simplification proves the claim. ∎

###  6.4. Evaluation II

We further simplify Lemma 6.15. The following result shows that, up to some
contributions of the Weyl tensor, the first two terms in (6.15) cancel and
that the last term of (6.15) cancels against the first term of (6.19).

######  Lemma 6.17.

Let $n=3$. Then it holds

$\displaystyle(\mathring{L},\bar{\nabla}_{0}(\overline{\operatorname{Ric}}))+2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}$
$\displaystyle=2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0},$
$\displaystyle(\mathring{L},\overline{\operatorname{Ric}})+2(\mathring{L},\overline{\mathcal{W}})$
$\displaystyle=2(\mathring{L},\bar{{\mathcal{G}}}).$

###### Proof.

By the Kulkarni-Nomizu decomposition $R=-{\sf P}\owedge g+W$, we have

$\bar{{\mathcal{G}}}_{ij}=\bar{R}_{0ij0}=\bar{{\sf P}}_{ij}+\bar{{\sf
P}}_{00}(h_{r})_{ij}+\overline{W}_{0ij0}.$

Hence using

$\bar{\nabla}_{0}(\bar{R})_{0ij0}=\partial_{0}(\bar{R}_{0ij0})-\bar{R}(\partial_{0},\bar{\nabla}_{0}(\partial_{i}),\partial_{j},\partial_{0})-\bar{R}(\partial_{0},\partial_{i},\bar{\nabla}_{0}(\partial_{j}),\partial_{0})$

and $\bar{\nabla}_{0}(\partial_{i})=L_{i}^{k}\partial_{k}$ we find

$\displaystyle\bar{\nabla}_{0}(\bar{R})_{0ij0}-\bar{\nabla}_{0}(\overline{W})_{0ij0}$
$\displaystyle=\partial_{0}(\bar{{\sf P}}_{ij})+\partial_{0}(\bar{{\sf
P}}_{00})h_{ij}+\bar{{\sf P}}_{00}h_{ij}^{\prime}$ $\displaystyle-\bar{{\sf
P}}(\bar{\nabla}_{0}(\partial_{i}),\partial_{j})-\bar{{\sf
P}}(\partial_{i},\bar{\nabla}_{0}(\partial_{j}))-\bar{{\sf
P}}_{00}h(\bar{\nabla}_{0}(\partial_{i}),\partial_{j})-\bar{{\sf
P}}_{00}h(\partial_{i},\bar{\nabla}_{0}(\partial_{j}))$
$\displaystyle=\bar{\nabla}_{0}(\bar{{\sf P}})_{ij}+\partial_{0}(\bar{{\sf
P}}_{00})h_{ij}+2\bar{{\sf P}}_{00}L_{ij}-2\bar{{\sf P}}_{00}L_{ij}$
$\displaystyle=\bar{\nabla}_{0}(\bar{{\sf P}})_{ij}+\partial_{0}(\bar{{\sf
P}}_{00})h_{ij}$

for $r=0$. Therefore,

$2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{R})_{0ij0}=2\mathring{L}^{ij}\bar{\nabla}_{0}(\bar{{\sf
P}})_{ij}+2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}=\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{ij}+2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}.$

This proves the first identity. The second identity follows from the
decomposition $\bar{{\mathcal{G}}}=\bar{{\sf P}}+\bar{{\sf
P}}_{00}h+\overline{\mathcal{W}}$. ∎

Next, we evaluate the first term of (6.17).

######  Lemma 6.18.

In general dimensions, it holds

$\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})=\frac{1}{2}\Delta(\overline{\operatorname{scal}})-n\delta(H\overline{\operatorname{Ric}}_{0})-\delta((L\overline{\operatorname{Ric}})_{0})-\delta\delta(\overline{\operatorname{Ric}}).$
(6.22)

###### Proof.

The result follows from the second Bianchi identity. Combining the identity

$\bar{\nabla}_{0}(\overline{\operatorname{Ric}})(\partial_{0},\partial_{a})=\delta^{g}(\overline{\operatorname{Ric}})(\partial_{a})-h^{ij}\bar{\nabla}_{\partial_{i}}(\overline{\operatorname{Ric}})(\partial_{j},\partial_{a})$

on $M$ with $2\delta^{g}(\operatorname{Ric}^{g})=d\operatorname{scal}^{g}$, we
obtain

$\displaystyle\bar{\nabla}_{0}(\overline{\operatorname{Ric}})(\partial_{0},\partial_{a})$
$\displaystyle=\frac{1}{2}\langle
d\overline{\operatorname{scal}},\partial_{a}\rangle-h^{ij}\partial_{i}(\overline{\operatorname{Ric}}(\partial_{j},\partial_{a}))+h^{ij}\overline{\operatorname{Ric}}(\bar{\nabla}_{\partial_{i}}(\partial_{j}),\partial_{a})+h^{ij}\overline{\operatorname{Ric}}(\partial_{j},\bar{\nabla}_{\partial_{i}}(\partial_{a}))$
$\displaystyle=\frac{1}{2}\langle
d\overline{\operatorname{scal}},\partial_{a}\rangle-h^{ij}\partial_{i}(\overline{\operatorname{Ric}}(\partial_{j},\partial_{a}))$
$\displaystyle+h^{ij}\overline{\operatorname{Ric}}(\nabla_{\partial_{i}}(\partial_{j})-L_{ij}\partial_{0},\partial_{a})+h^{ij}\overline{\operatorname{Ric}}(\partial_{j},\nabla_{\partial_{i}}(\partial_{a})-L_{ia}\partial_{0})$
$\displaystyle=\frac{1}{2}\langle
d\overline{\operatorname{scal}},\partial_{a}\rangle-\delta^{h}(\overline{\operatorname{Ric}})(\partial_{a})-nH\overline{\operatorname{Ric}}(\partial_{0},\partial_{a})-h^{ij}L_{ia}\overline{\operatorname{Ric}}(\partial_{j},\partial_{0}).$

Now we apply $\delta=\delta^{h}$ to this identity of $1$-forms on $M$. We
obtain

$\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})=\frac{1}{2}\Delta(\overline{\operatorname{scal}})-\delta\delta(\overline{\operatorname{Ric}})-n\delta(H\overline{\operatorname{Ric}}_{0})-\delta((L\overline{\operatorname{Ric}})_{0}).$

The proof is complete. ∎

In order to apply Lemma 6.18, we combine it with the following formula for the
last term on the right-hand side of (6.22).

######  Lemma 6.19.

Let $n=3$. Then

$\delta\delta(\overline{\operatorname{Ric}})=2\Delta({\sf J})+\Delta(\bar{{\sf
J}})-\Delta(H^{2})-2\delta\delta(H\mathring{L})+2\delta\delta(\mathring{L}^{2})-\frac{1}{2}\Delta(|\mathring{L}|^{2})+2\delta\delta(\overline{\mathcal{W}}).$

###### Proof.

First, we note that

$\delta\delta(\overline{\operatorname{Ric}})=2\delta\delta(\bar{{\sf
P}})+\delta\delta(\bar{{\sf J}}h)=2\delta\delta(\bar{{\sf
P}})+\Delta(\bar{{\sf J}}).$

Now we utilize the identity (3.7). It follows that

$2\delta\delta(\bar{{\sf P}})=2\delta\delta({\sf
P})-2\delta\delta(H\mathring{L})-\Delta(H^{2})+2\delta\delta(\mathring{L}^{2})-\frac{1}{2}\Delta(|\mathring{L}|^{2})+2\delta\delta(\overline{\mathcal{W}}).$

Combining these results with $\delta({\sf P})=d{\sf J}$ proves the assertion.
∎

Now, combining Lemma 6.18 and Lemma 6.19 with the Gauss identity

$\bar{{\sf J}}-{\sf J}=\bar{{\sf P}}_{00}+1/4|\mathring{L}|^{2}-3/2H^{2}$

gives

$\displaystyle\delta(\bar{\nabla}_{0}(\overline{\operatorname{Ric}})_{0})$
$\displaystyle=2\Delta(\bar{{\sf J}}-{\sf
J})-3\delta(H\overline{\operatorname{Ric}}_{0})-\delta((L\overline{\operatorname{Ric}})_{0})$
$\displaystyle+\Delta(H^{2})+2\delta\delta(H\mathring{L})-2\delta\delta(\mathring{L}^{2})+\frac{1}{2}\Delta(|\mathring{L}|^{2})-2\delta\delta(\overline{\mathcal{W}})$
$\displaystyle=2\Delta(\bar{{\sf
P}}_{00})+\Delta(|\mathring{L}|^{2})-2\Delta(H^{2})-3\delta(H\overline{\operatorname{Ric}}_{0})-\delta((L\overline{\operatorname{Ric}})_{0})$
$\displaystyle+2\delta\delta(H\mathring{L})-2\delta\delta(\mathring{L}^{2})-2\delta\delta(\overline{\mathcal{W}}).$
(6.23)

This result shows that (6.17) equals

$\displaystyle-\Delta(|\mathring{L}|^{2})-2\delta\delta(H\mathring{L})+2\delta\delta(\mathring{L}^{2})+2\Delta(H^{2})$
$\displaystyle+\delta(H\overline{\operatorname{Ric}}_{0})+\delta((L\overline{\operatorname{Ric}})_{0})+6(dH,\overline{\operatorname{Ric}}_{0})+2\delta\delta(\overline{\mathcal{W}}).$
(6.24)

######  Remark 6.20.

The identity (6.4) shows that

$\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2})+2\delta\delta(H\mathring{L})-2\Delta(H^{2})=0$

for a flat background. This relation also is a consequence of the difference
formula (3.18) and the identity

$\delta\delta(H\mathring{L})=(\mathring{L},\operatorname{Hess}(H))+4|dH|^{2}+2H\Delta(H)$
(6.25)

which is a consequence of Lemma 6.21 and the Codazzi-Mainardi relation
$\delta(\mathring{L})=2dH$ for a flat background. More generally, if $g$ is
Einstein, i.e., if $\overline{\operatorname{Ric}}=\lambda g$, then
$\overline{\operatorname{Ric}}_{0}=0$ and the identity (6.4) implies

$\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2})+2\delta\delta(H\mathring{L})-2\Delta(H^{2})-2\delta\delta(\overline{\mathcal{W}})=0.$

By combination with the Codazzi-Mainardi relation $\delta(\mathring{L})=2dH$,
Lemma 3.14 and Lemma 6.21, we conclude the interesting identity

$-2\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}+|\overline{W}_{0}|^{2}-2\delta\delta(\overline{\mathcal{W}})=0.$

######  Lemma 6.21.

In general dimensions, it holds

$\delta\delta(H\mathring{L})=(\mathring{L},\operatorname{Hess}(H))+2(dH,\delta(\mathring{L}))+H\delta\delta(\mathring{L}).$

###### Proof.

The identity is obvious. ∎

Now we combine formula (6.4) with (6.15)–(6.21). Note that the term
$\delta((L\overline{\operatorname{Ric}})_{0})$ in (6.4) sums up with the fifth
term in (6.15) to $\delta(H\overline{\operatorname{Ric}}_{0})$ and that the
term $-\Delta(|\mathring{L}|^{2})$ in (6.4) cancels with the term
$\Delta(|\mathring{L}|^{2})$ in (6.21). We also use Lemma 6.21.

By Lemmas 6.17–6.21, the formula in Lemma 6.15 turns into the sum of

$2\Delta(H^{2})+2\delta(H\overline{\operatorname{Ric}}_{0})\stackrel{{\scriptstyle!}}{{=}}2\delta(H\delta(\mathring{L}))=2H\delta\delta(\mathring{L})+2(dH,\delta(\mathring{L}))$
(6.26)

(by Codazzi-Mainardi
$\overline{\operatorname{Ric}}_{0}=\delta(\mathring{L})-2dH$),

$6(dH,\overline{\operatorname{Ric}}_{0}),$
$\displaystyle-2\delta\delta(H\mathring{L})+2\delta\delta(\mathring{L}^{2})+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-3|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-(\mathring{L}^{2},\overline{\operatorname{Ric}})$
$\displaystyle\stackrel{{\scriptstyle!}}{{=}}-2(\mathring{L},\operatorname{Hess}(H))-4(dH,\delta(\mathring{L}))+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-2H\delta\delta(\mathring{L})+2\delta\delta(\mathring{L}^{2})$
$\displaystyle+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))-3|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}-(\mathring{L}^{2},\overline{\operatorname{Ric}})$
(6.27)

(by Lemma 6.21),

$\displaystyle
8(\mathring{L}^{2},\bar{{\mathcal{G}}})+2|\mathring{L}|^{2}\bar{{\sf J}}$
$\displaystyle=8(\mathring{L}^{2},\bar{{\sf P}})+8|\mathring{L}|^{2}\bar{{\sf
P}}_{00}+2|\mathring{L}|^{2}\bar{{\sf
J}}+8(\mathring{L}^{2},\overline{\mathcal{W}})$ (by
$\bar{{\mathcal{G}}}=\bar{{\sf P}}+\bar{{\sf
P}}_{00}h+\overline{\mathcal{W}}$)
$\displaystyle=4(\mathring{L}^{2},\overline{\operatorname{Ric}})-2|\mathring{L}|^{2}\bar{{\sf
J}}+8|\mathring{L}|^{2}\bar{{\sf
P}}_{00}+8(\mathring{L}^{2},\overline{\mathcal{W}})$
$\displaystyle\stackrel{{\scriptstyle!}}{{=}}4(\mathring{L}^{2},\overline{\operatorname{Ric}})-6|\mathring{L}|^{2}\bar{{\sf
J}}+4|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}+8(\mathring{L}^{2},\overline{\mathcal{W}}),$
$2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+2\delta\delta(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})$
(6.28)

and

$\displaystyle
6(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}+12|dH|^{2}.$

Note that the contributions $2H\delta\delta(\mathring{L})$ in (6.26) and (6.4)
cancel.

By Codazzi-Mainardi, we find

$(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})-(\delta(\mathring{L}),dH)=|\delta(\mathring{L})|^{2}-3(\delta(\mathring{L}),dH).$

Hence

$2(dH,\delta(\mathring{L}))-4(dH,\delta(\mathring{L}))+6(dH,\overline{\operatorname{Ric}}_{0})+2(\delta(\mathring{L}),\overline{\operatorname{Ric}}_{0})\stackrel{{\scriptstyle!}}{{=}}2|\delta(\mathring{L})|^{2}-12|dH|^{2}.$

Thus, we have proved

######  Proposition 6.22.

$12{\mathcal{B}}_{3}$ equals the sum of

$\displaystyle 2\delta\delta(\mathring{L}^{2})+2|\delta(\mathring{L})|^{2},$
(6.29)
$2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}+3(\mathring{L}^{2},\overline{\operatorname{Ric}})-6|\mathring{L}|^{2}\bar{{\sf
J}},$ (6.30)

the Weyl-curvature terms

$2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+2\delta\delta(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})+8(\mathring{L}^{2},\overline{\mathcal{W}})$
(6.31)

and

$4(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}.$
(6.32)

Note that there is no Laplace term in that formula.

###  6.5. Proof of the main result. Equivalences

Proposition 6.22 has the disadvantage that the conformal invariance of
${\mathcal{B}}_{3}$ is not obvious. Therefore, it is natural to reformulate
the results in a way which makes the conformal invariance transparent. For
this purpose, we relate Proposition 6.22 to the formula

$\displaystyle 12{\mathcal{B}}_{3}$
$\displaystyle=6\mathcal{D}((\mathring{L}^{2})_{\circ})+2|\mathring{L}|^{4}+\star$
$\displaystyle=6\delta\delta(\mathring{L}^{2})-2\Delta(|\mathring{L}|^{2})+6(\mathring{L}^{2},{\sf
P})-2|\mathring{L}|^{2}{\sf J}+2|\mathring{L}|^{4}+\star$ (6.33)

in [GGHW19, Proposition 1.1], where111Here we refer to arXiv:1508.01838v1.
This result differs from the version in the published paper.

$\star\stackrel{{\scriptstyle\text{def}}}{{=}}2\mathcal{D}(\overline{\mathcal{W}})+4|\overline{\mathcal{W}}|^{2}+2|\overline{W}_{0}|^{2}-2(\mathring{L},B)+14(\mathring{L}^{2},\overline{\mathcal{W}})-2\mathring{L}^{ab}\mathring{L}^{cd}\overline{W}_{cabd}.$
(6.34)

Here $B$ is a certain conformally invariant symmetric bilinear form of weight
$-1$ which will be defined in (6.42). Here we took into account that in
[GGHW19] the signs of the components of the curvature tensor and the Weyl
tensor are opposite to ours. All terms in (6.34) are conformally invariant.
The difference of both formulas is

$\displaystyle
2(\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2}))+2|\delta(\mathring{L})|^{2}$
$\displaystyle+4(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})-|\mathring{L}|^{4}$
$\displaystyle+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}+3(\mathring{L}^{2},\overline{\operatorname{Ric}})-6|\mathring{L}|^{2}\bar{{\sf
J}}-6(\mathring{L}^{2},{\sf P})+2|\mathring{L}|^{2}{\sf J}$
$\displaystyle+2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+2\delta\delta(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})+8(\mathring{L}^{2},\overline{\mathcal{W}})-\star.$
(6.35)

######  Lemma 6.23.

The sum (6.5) vanishes.

In other words, Proposition 6.22 is equivalent to [GGHW19, Proposition 1.1].
The proof of this result will also establish the equivalence to Theorem 1.

######  Remark 6.24.

Lemma 6.23 holds for a flat background metric. In this case $\star=0$. In
fact, the identity (3.7) implies

$6(\mathring{L}^{2},{\sf P})-2{\sf
J}|\mathring{L}|^{2}=6H\operatorname{tr}(\mathring{L}^{3})-|\mathring{L}|^{4}.$
(6.36)

By $\delta(\mathring{L})=2dH$ (Codazzi-Mainardi), the sum (6.5) equals

$\displaystyle
2(\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2}))+8|dH|^{2}$
$\displaystyle+4(\mathring{L},\operatorname{Hess}(H))+6H\operatorname{tr}(\mathring{L}^{3})-|\mathring{L}|^{4}-6H\operatorname{tr}(\mathring{L}^{3})+|\mathring{L}|^{4}$
$\displaystyle=2(\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2}))+8|dH|^{2}+4(\mathring{L},\operatorname{Hess}(H)).$

The identity (3.18) shows that this sum vanishes.

A key role in the argument in Remark 6.24 is played by the formula (3.18) for
the divergence term
$\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2})$. Lemma 3.14
extends this result to general backgrounds. We also need the following curved
analog of (6.36).

######  Lemma 6.25.

If $n=3$, then it holds

$6(\mathring{L}^{2},{\sf P})-2|\mathring{L}|^{2}{\sf
J}=6H\operatorname{tr}(\mathring{L}^{3})-|\mathring{L}|^{4}+6(\mathring{L}^{2},\bar{{\sf
P}})-2|\mathring{L}|^{2}\bar{{\sf J}}+2|\mathring{L}|^{2}\bar{{\sf
P}}_{00}-6(\mathring{L}^{2},\overline{\mathcal{W}}).$

###### Proof.

The identity (3.7) yields

$\iota^{*}\bar{{\sf P}}-{\sf
P}=\mathring{L}^{2}-\frac{1}{4}|\mathring{L}|^{2}h-H\mathring{L}-\frac{1}{2}H^{2}h+\overline{\mathcal{W}}.$

Taking the trace yields the Gauss identity

$\bar{{\sf J}}-\bar{{\sf P}}_{00}-{\sf
J}=|\mathring{L}|^{2}-\frac{3}{4}|\mathring{L}|^{2}-\frac{3}{2}H^{2}=\frac{1}{4}|\mathring{L}|^{2}-\frac{3}{2}H^{2}.$

These relations imply the assertion. ∎

Now, by Lemma 6.25, (6.5) simplifies to

$\displaystyle
2(\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2}))+2|\delta(\mathring{L})|^{2}+4(\mathring{L},\operatorname{Hess}(H))$
$\displaystyle+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))+|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}+3(\mathring{L}^{2},\overline{\operatorname{Ric}})-4|\mathring{L}|^{2}\bar{{\sf
J}}-6(\mathring{L}^{2},\bar{{\sf P}})-2|\mathring{L}|^{2}\bar{{\sf P}}_{00}$
$\displaystyle+2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+2\delta\delta(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})+14(\mathring{L}^{2},\overline{\mathcal{W}})-\star$
(6.37)

But

$|\mathring{L}|^{2}\overline{\operatorname{Ric}}_{00}+3(\mathring{L}^{2},\overline{\operatorname{Ric}})-4|\mathring{L}|^{2}\bar{{\sf
J}}-6(\mathring{L}^{2},\bar{{\sf P}})-2|\mathring{L}|^{2}\bar{{\sf
P}}_{00}=0,$

i.e., the second last line of (6.5) reduces to
$2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))$. Therefore, (6.5)
further simplifies to

$\displaystyle
2(\Delta(|\mathring{L}|^{2})-2\delta\delta(\mathring{L}^{2}))+2|\delta(\mathring{L})|^{2}+4(\mathring{L},\operatorname{Hess}(H))+2(\mathring{L},\nabla(\overline{\operatorname{Ric}}_{0}))$
$\displaystyle+2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+2\delta\delta(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})+14(\mathring{L}^{2},\overline{\mathcal{W}})-\star.$

Now, by Lemma 3.14, this sum equals

$\displaystyle
4\mathring{L}^{ij}\nabla^{k}\overline{W}_{ikj0}+2|\overline{W}_{0}|^{2}$
$\displaystyle+2(\bar{{\sf
P}},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}+2\delta\delta(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-2H(\mathring{L},\overline{\mathcal{W}})+14(\mathring{L}^{2},\overline{\mathcal{W}})-\star.$
(6.38)

Now we apply the identity

$(\bar{{\sf P}},\overline{\mathcal{W}})=({\sf
P},\overline{\mathcal{W}})+(\mathring{L}^{2},\overline{\mathcal{W}})-H(\mathring{L},\overline{\mathcal{W}})+|\overline{\mathcal{W}}|^{2}$

(see (3.7)). Hence the sum (6.5) equals

$\displaystyle 2\delta\delta(\overline{\mathcal{W}})+2({\sf
P},\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}-4H(\mathring{L},\overline{\mathcal{W}})$
$\displaystyle+16(\mathring{L}^{2},\overline{\mathcal{W}})+4|\overline{\mathcal{W}}|^{2}+2|\overline{W}_{0}|^{2}+4\mathring{L}^{ij}\nabla^{k}\overline{W}_{ikj0}-\star.$

Therefore, Lemma 6.23 holds true iff

$\displaystyle\star$
$\displaystyle=2\mathcal{D}(\overline{\mathcal{W}})-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}+4\mathring{L}^{ij}\nabla^{k}\overline{W}_{ikj0}$
$\displaystyle-4H(\mathring{L},\overline{\mathcal{W}})+16(\mathring{L}^{2},\overline{\mathcal{W}})+4|\overline{\mathcal{W}}|^{2}+2|\overline{W}_{0}|^{2}.$
(6.39)

Equivalently, Lemma 6.23 holds true iff

$\displaystyle-2(\mathring{L},B)+14(\mathring{L}^{2},\overline{\mathcal{W}})-2\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}$
$\displaystyle=-2\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}+4\mathring{L}^{ij}\nabla^{k}\overline{W}_{ikj0}-4H(\mathring{L},\overline{\mathcal{W}})+16(\mathring{L}^{2},\overline{\mathcal{W}})$
(6.40)

It remains to prove (6.5). As a preparation, we observe

######  Lemma 6.26.

Let $n=3$. Then

$\mathring{L}^{ij}\bar{\nabla}^{k}\overline{W}_{ikj0}=\mathring{L}^{ij}\nabla^{k}\overline{W}_{ikj0}+(\mathring{L}^{2},\overline{\mathcal{W}})-3H(\mathring{L},\overline{\mathcal{W}})+\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}.$

###### Proof.

By
$\bar{\nabla}_{i}(\partial_{j})=\nabla_{i}(\partial_{j})-L_{ij}\partial_{0}$
and $\bar{\nabla}_{k}(\partial_{0})=L_{k}^{m}\partial_{m}$, we find

$\displaystyle\bar{\nabla}^{k}\overline{W}_{ikj0}$
$\displaystyle=\nabla^{k}\overline{W}_{ikj0}-L^{kl}\overline{W}_{ikjl}+L^{k}_{i}\overline{W}_{0kj0}+3H\overline{W}_{i0j0}+L^{k}_{j}\overline{W}_{ik00}$
$\displaystyle=\nabla^{k}\overline{W}_{ikj0}+L^{kl}\overline{W}_{kijl}+L^{k}_{i}\overline{W}_{0kj0}-3H\overline{W}_{0ij0}$
$\displaystyle=\nabla^{k}\overline{W}_{ikj0}+\mathring{L}^{kl}\overline{W}_{kijl}-H\overline{W}_{0ij0}+\mathring{L}_{i}^{k}\overline{W}_{0kj0}+H\overline{W}_{0ij0}-3H\overline{W}_{0ij0}$
$\displaystyle=\nabla^{k}\overline{W}_{ikj0}+\mathring{L}^{kl}\overline{W}_{kijl}+\mathring{L}_{i}^{k}\overline{W}_{0kj0}-3H\overline{W}_{0ij0}.$

The assertion follows by contraction with $\mathring{L}^{ij}$. ∎

######  Lemma 6.27.

It holds

$(\mathring{L},B)=\mathring{L}^{ij}\bar{\nabla}^{0}(\widebar{W})_{0ij0}+2H(\mathring{L},\overline{\mathcal{W}})-2\mathring{L}^{ij}\nabla^{k}\overline{W}_{jki0}-(\mathring{L}^{2},\overline{\mathcal{W}})-\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}.$
(6.41)

###### Proof.

We first restate the definition of $B$ in our conventions:222We recall that
our signs of the components of $\overline{W}$ are opposite.

$\displaystyle B_{ij}$
$\displaystyle=\bar{C}_{0(ij)}-H\overline{\mathcal{W}}_{ij}+\nabla^{k}\overline{W}_{0(ij)k}$
(6.42)
$\displaystyle=\bar{\nabla}^{k}(\overline{W})_{0(ij)k}-H\overline{\mathcal{W}}_{ij}+\nabla^{k}\overline{W}_{0(ij)k}.$

Here

$(n-3)C_{ijk}\stackrel{{\scriptstyle\text{def}}}{{=}}\nabla^{l}(W)_{ijkl}$

defines the Cotton tensor $C$ on a manifold of dimension $n$. It satisfies the
conformal transformation law

$\hat{C}_{ijk}=C_{ijk}+W_{ijk\operatorname{grad}(\varphi)}.$

We emphasize that, in the definition (6.42) of $B$, the index $k$ in the first
term runs over the tangential and the normal vectors. $(ij)$ denotes
symmetrization. We first verify that the symmetric tensor $B$ satisfies the
conformal transformtion law $e^{\varphi}\hat{B}=B$. For this purpose, we first
observe that

$e^{\varphi}\hat{\bar{C}}_{\hat{0}ij}=\hat{\bar{C}}_{0ij}=\bar{C}_{0ij}+\overline{W}_{0ij\operatorname{grad}(\varphi)}=\bar{C}_{0ij}+\overline{W}_{0ij\operatorname{grad}^{t}(\varphi)}+\overline{\mathcal{W}}_{ij}\partial_{0}(\varphi),$

where $\hat{0}=\hat{\partial}_{0}=e^{-\varphi}\partial_{0}$ and
$\operatorname{grad}^{t}(\varphi)$ is the tangential component of the
gradient. We also recall that

$e^{-2\varphi}\hat{\widebar{W}}_{ijkl}=\widebar{W}_{ijkl},\quad\hat{\bar{\overline{\mathcal{W}}}}_{ij}=\bar{\overline{\mathcal{W}}}_{ij}\quad\mbox{and}\quad
e^{\varphi}\hat{H}=H+\partial_{0}(\varphi).$

We calculate

$\displaystyle e^{2\varphi}\hat{\nabla}^{k}\hat{\widebar{W}}_{\hat{0}ijk}$
$\displaystyle=\partial^{k}(e^{\varphi}\widebar{W}_{0ijk})$
$\displaystyle-e^{\varphi}\widebar{W}(\partial_{0},\hat{\nabla}^{k}(\partial_{i}),\partial_{j},\partial_{k})-e^{\varphi}\widebar{W}(\partial_{0},\partial_{i},\hat{\nabla}^{k}(\partial_{j}),\partial_{k})-e^{\varphi}\widebar{W}(\partial_{0},\partial_{i},\partial_{j},\hat{\nabla}^{k}(\partial_{k})).$

Now the general transformation law

$\hat{\nabla}_{i}(\partial_{j})=\nabla_{i}(\partial_{j})+\partial_{i}(\varphi)\partial_{j}+\partial_{j}(\varphi)\partial_{i}-g_{ij}\operatorname{grad}(\varphi)$

implies that in general dimensions ($\dim(M)=n$)333That identity corrects
[GGHW19, (2.10)].

$e^{\varphi}\hat{\nabla}^{k}\hat{\widebar{W}}_{\hat{0}ijk}=\nabla^{k}\widebar{W}_{0ijk}+(n-4)\widebar{W}_{0ij\operatorname{grad}^{t}(\varphi)}-\widebar{W}_{0\operatorname{grad}^{t}(\varphi)ij}.$

Hence for $n=3$ we find

$e^{\varphi}\hat{\nabla}^{k}\hat{\widebar{W}}_{\hat{0}(ij)k}=\nabla^{k}\widebar{W}_{0(ij)k}-\widebar{W}_{0(ij)\operatorname{grad}^{t}(\varphi)}.$

Therefore,

$e^{\varphi}\hat{B}_{ij}=B_{ij}+\overline{W}_{0(ij)\operatorname{grad}^{t}(\varphi)}+\overline{\mathcal{W}}_{ij}\partial_{0}(\varphi)-\overline{\mathcal{W}}_{ij}\partial_{0}(\varphi)-\widebar{W}_{0(ij)\operatorname{grad}^{t}(\varphi)}=B_{ij}.$

Now the definition of $B$ gives

$\displaystyle(\mathring{L},B)$
$\displaystyle\stackrel{{\scriptstyle\text{def}}}{{=}}\mathring{L}^{ij}\bar{\nabla}^{k}(\overline{W})_{0ijk}-H(\mathring{L},\overline{\mathcal{W}})+\mathring{L}^{ij}\nabla^{k}\overline{W}_{0ijk}$
$\displaystyle=-\mathring{L}^{ij}\bar{\nabla}^{k}(\overline{W})^{t}_{jki0}+\mathring{L}^{ij}\bar{\nabla}^{0}(\overline{W})_{0ij0}-H(\mathring{L},\overline{\mathcal{W}})-\mathring{L}^{ij}\nabla^{k}\overline{W}_{jki0},$

where the superscript $t$ in the first sum indicates that indices are only
tangential. Now Lemma 6.26 yields

$\displaystyle(\mathring{L},B)$
$\displaystyle=\mathring{L}^{ij}\bar{\nabla}^{0}(\overline{W})_{0ji0}+2H(\mathring{L},\overline{\mathcal{W}})-2\mathring{L}^{ij}\nabla^{k}\overline{W}_{jki0}-(\mathring{L}^{2},\overline{\mathcal{W}})-\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}.$

The proof of (6.41) is complete. ∎

Lemma 6.27 shows that

$-2(\mathring{L},B)=-2\mathring{L}^{ij}\bar{\nabla}^{0}(\widebar{W})_{0ij0}-4H(\mathring{L},\overline{\mathcal{W}})+4\mathring{L}^{ij}\nabla^{k}\overline{W}_{jki0}+2(\mathring{L}^{2},\overline{\mathcal{W}})+2\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}.$

Hence

$-2(\mathring{L},B)+14(\mathring{L}^{2},\overline{\mathcal{W}})-2\mathring{L}^{ij}\mathring{L}^{kl}\overline{W}_{kijl}=16(\mathring{L}^{2},\overline{\mathcal{W}})-2\bar{\nabla}^{0}(\widebar{W})_{0ij0}-4H\overline{\mathcal{W}}_{ij}+4\mathring{L}^{ij}\nabla^{k}\overline{W}_{jki0}$

(note the cancellation!). This proves (6.5) and hence Lemma 6.23.

Now in order to finish the proof of Theorem 1, it suffices to combine (6.5)
with (6.5).

######  Remark 6.28.

The formula for ${\mathcal{B}}_{3}$ in the published version of [GGHW19] reads

$12{\mathcal{B}}_{3}=4\mathcal{D}((\mathring{L}^{2})_{\circ})+2\mathcal{D}(\mathring{\mathcal{F}})-2(\mathring{L},B)+|\mathring{L}|^{4}+4(\mathring{\mathcal{F}},\mathcal{F})+2(\mathring{\mathcal{F}},\mathring{L}^{2})+2|\overline{W}_{0}|^{2}.$

By $\mathring{\mathcal{F}}=(\mathring{L}^{2})_{\circ}+\overline{\mathcal{W}}$,
this formula is equivalent to

$\displaystyle 12{\mathcal{B}}_{3}$
$\displaystyle=6\mathcal{D}((\mathring{L}^{2})_{\circ})+2\mathcal{D}(\overline{\mathcal{W}})-2(\mathring{L},B)+|\mathring{L}|^{4}+2|\overline{W}_{0}|^{2}$
$\displaystyle+2((\mathring{L}^{2})_{\circ}+\overline{\mathcal{W}},\mathring{L}^{2})+4((\mathring{L}^{2})_{\circ}+\overline{\mathcal{W}},\mathring{L}^{2}-\frac{1}{4}|\mathring{L}|^{2}h+\overline{\mathcal{W}}).$

The second line simplifies to

$|\mathring{L}|^{4}+10(\mathring{L}^{2},\overline{\mathcal{W}})+4|\overline{\mathcal{W}}|^{2}.$

If follows that the resulting formula for $12{\mathcal{B}}_{3}$ differs from
(6.5), (6.34).

Finally, we note that for a conformally flat background Theorem 1 states that

$6{\mathcal{B}}_{3}=3\mathcal{D}((\mathring{L}^{2})_{\circ})+|\mathring{L}|^{4}=3\delta\delta((\mathring{L}^{2})_{\circ})+3((\mathring{L}^{2})_{\circ},{\sf
P})+|\mathring{L}|^{4}.$

Lemma 3.13 implies that this formula is equivalent to

$6{\mathcal{B}}_{3}=\Delta(|\mathring{L}|^{2})-|\nabla\mathring{L}|^{2}+3/2|\delta(\mathring{L})|^{2}-2{\sf
J}|\mathring{L}|^{2}+|\mathring{L}|^{4}$ (6.43)

(as also stated in [GW15, Proposition 2.10]).

##  7\. Variational aspects

Let $\iota:M^{3}\hookrightarrow X^{4}$ be an embedding. In this section, we
prove that the conformally invariant equation ${\mathcal{B}}_{3}=0$ is the
Euler-Lagrange equation of the conformally invariant functional

$\mathcal{W}_{3}(\iota)\stackrel{{\scriptstyle\text{def}}}{{=}}\int_{\iota(M)}(\operatorname{tr}(\mathring{L}^{3})+(\mathring{L},\overline{\mathcal{W}}))dvol$
(7.1)

under normal variations of the embedding $\iota$. Let $u\in C^{\infty}(M)$ and
$\partial_{0}$ be a unit normal field of $M$. We set
$\iota_{t}(m)=\exp(tu(m)\partial_{0})$, where $\exp$ is the exponential map.
Then $\iota_{0}=\iota$ and $\iota_{t}$ is a variation of $M$ with variation
field $u\partial_{0}$. Let $\mathcal{W}_{3}(\iota_{t})$ be the analogous
functional for $\iota_{t}$ and define

${\textit{var}}(\mathcal{W}_{3})[u]\stackrel{{\scriptstyle\text{def}}}{{=}}(d/dt)|_{0}(\mathcal{W}_{3}(\iota_{t})).$

######  Theorem 2.

It holds

$-{\textit{var}}(\mathcal{W}_{3})[u]=6\int_{M}u{\mathcal{B}}_{3}dvol.$

This result reproves [GGHW19, Proposition 1.2]. Our arguments are classical
and differ substantially from those in the reference (see the comments after
the proof).

###### Proof.

We first note that the variation of
$\int_{M}\operatorname{tr}(\mathring{L}^{3})dvol$ has been determined in
[JO21, Lemma 13.9.1] for conformally flat backgrounds. The given arguments
easily extend to the general case and yield

$\displaystyle-{\textit{var}}\left(\int_{M}\operatorname{tr}(\mathring{L}^{3})dvol_{h}\right)[u]$
$\displaystyle=3\int_{M}u\left(\delta\delta((\mathring{L}^{2})_{\circ})+({\sf
P},(\mathring{L}^{2})_{\circ})+2(\mathring{L}^{2},\overline{\mathcal{W}})+\frac{1}{3}|\mathring{L}|^{4}\right)dvol_{h}$
$\displaystyle=\int_{M}u\left(3\mathcal{D}((\mathring{L}^{2})_{\circ})+6(\mathring{L}^{2},\overline{\mathcal{W}})+|\mathring{L}|^{4}\right)dvol_{h}.$

In the second part of the proof, we determine the variation of
$\int_{M}(\mathring{L},\overline{\mathcal{W}})dvol_{h}$. We write the
integrand as $h^{ai}h^{bj}\mathring{L}_{ab}\overline{\mathcal{W}}_{ij}$ and
apply the well-known variation formulas [A94, Theorem 3-15], [HP99, Theorem
3.2]

$\displaystyle{\textit{var}}(h)[u]$ $\displaystyle=2uL,$
$\displaystyle{\textit{var}}(L)[u]$
$\displaystyle=-\operatorname{Hess}(u)+uL^{2}-u\bar{{\mathcal{G}}},$
$\displaystyle 3{\textit{var}}(H)[u]$
$\displaystyle=-\Delta(u)-u|L|^{2}-u\overline{\operatorname{Ric}}_{00}$

and

${\textit{var}}(dvol_{h})[u]=3uHdvol_{h}.$

It follows that the variation is given by the integral of the sum of

$-4u(\mathring{L}^{2},\overline{\mathcal{W}})-4uH(\mathring{L},\overline{\mathcal{W}})\qquad\mbox{(by
variation of the metric)},$ (7.2)
$\displaystyle({\textit{var}}(\mathring{L})[u],\overline{\mathcal{W}})$
$\displaystyle=\overline{\mathcal{W}}^{ij}({\textit{var}}(L)_{ij}-H{\textit{var}}(h)_{ij})$
$\displaystyle=\overline{\mathcal{W}}^{ij}(-\operatorname{Hess}_{ij}(u)+u(L^{2})_{ij}-u\bar{{\mathcal{G}}}_{ij})-2uH\overline{\mathcal{W}}^{ij}L_{ij}$
$\displaystyle=-(\operatorname{Hess}(u),\overline{\mathcal{W}})+u(\mathring{L}^{2},\overline{\mathcal{W}})-u(\bar{{\mathcal{G}}},\overline{\mathcal{W}}),$
(7.3) $\displaystyle(\mathring{L},{\textit{var}}(\overline{\mathcal{W}})[u])$
$\displaystyle=u\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}+u\mathring{L}^{ij}(L_{i}^{k}\overline{W}_{0kj0}+L^{k}_{j}\overline{W}_{0ik0})$
$\displaystyle=u\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}+2u(\mathring{L}^{2},\overline{\mathcal{W}})+2uH(\mathring{L},\overline{\mathcal{W}})$
(7.4)

and

$\displaystyle-2\mathring{L}^{ij}\overline{W}_{\operatorname{grad}(u)ij0}\qquad\mbox{(by
variation of the normal vector)},$ $\displaystyle
3uH(\mathring{L},\overline{\mathcal{W}})\qquad\mbox{(by variation of the
volume form)}.$ (7.5)

Now using partial integration we obtain

$\displaystyle{\textit{var}}\left(\int_{M}(\mathring{L},\overline{\mathcal{W}})dvol_{h}\right)[u]$
$\displaystyle=\int_{M}u\left[-\delta\delta(\overline{\mathcal{W}})-(\mathring{L}^{2},\overline{\mathcal{W}})+H(\mathring{L},\overline{\mathcal{W}})-(\bar{{\mathcal{G}}},\overline{\mathcal{W}})+\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}\right]dvol_{h}$
$\displaystyle-2\int_{M}\mathring{L}^{ij}\overline{W}_{\operatorname{grad}(u)ij0}dvol_{h}.$

Since

$(\bar{{\mathcal{G}}},\overline{\mathcal{W}})=(\bar{{\sf
P}},\overline{\mathcal{W}})+(\overline{\mathcal{W}},\overline{\mathcal{W}})=({\sf
P},\overline{\mathcal{W}})-H(\mathring{L},\overline{\mathcal{W}})+(\mathring{L}^{2},\overline{\mathcal{W}})+2|\overline{\mathcal{W}}|^{2}$

by (3.7), we get

$\displaystyle{\textit{var}}\left(\int_{M}(\mathring{L},\overline{\mathcal{W}})dvol_{h}\right)[u]$
$\displaystyle=\int_{M}u\left[-\mathcal{D}(\overline{\mathcal{W}})+2H(\mathring{L},\overline{\mathcal{W}})-2(\mathring{L}^{2},\overline{\mathcal{W}})-2|\overline{\mathcal{W}}|^{2}+\mathring{L}^{ij}\bar{\nabla}_{0}(\overline{W})_{0ij0}\right]dvol_{h}$
$\displaystyle-2\int_{M}(du,\mathring{L}^{ij}\overline{W}_{\cdot
ij0})dvol_{h}.$

By partial integration, we find

$\displaystyle 2\int_{M}(du,\mathring{L}^{ij}\overline{W}_{\cdot
ij0})dvol_{h}$
$\displaystyle=-2\int_{M}u\delta(\mathring{L}^{ij}\overline{W}_{\cdot
ij0})dvol_{h}=-2\int_{M}u\nabla^{k}(\mathring{L}^{ij}\overline{W}_{kij0})dvol_{h}$
$\displaystyle=-2\int_{M}u\mathring{L}^{ij}\nabla^{k}\overline{W}_{kij0}dvol_{h}+\int_{M}u|\overline{W}_{0}|^{2}dvol_{h}$

using the trace-free Codazzi-Mainardi equation (3.9) (similarly as on page 3).
Summarizing these results, proves the claim. ∎

Graham’s theorem [G17, Theorem 3.1] and [JO21, (3.8)] imply that the variation
of $\int v_{3}dvol$ equals $4\int u{\mathcal{B}}_{3}dvol$. Here the singular
Yamabe renormalized volume coefficient $v_{3}$ (as defined in [G17]) satisfies

$12\int_{M}v_{3}dvol=-\int_{M}\mathbf{Q}_{3}dvol=-8\int_{M}(\operatorname{tr}(\mathring{L}^{3})+(\mathring{L},\overline{\mathcal{W}}))dvol=-8\mathcal{W}_{3},$

where ${\bf{Q}}_{3}$ is the extrinsic $Q$-curvature (see [JO21, Example
13.10.2 and (13.10.7)]). In other words, the variation of $\mathcal{W}_{3}$
equals $-6\int u{\mathcal{B}}_{3}dvol$. This shows that Theorem 2 fits with
Graham’s theorem. On the other hand, [GGHW19, Proposition 1.2] states that the
variation of $\mathcal{W}_{3}$ equals $6\int u{\mathcal{B}}_{3}dvol$. The
discrepancy of the sign is due to the altered definition of variations. Note
that the notion of variation exploited in [GGHW19] leads to the result
${\textit{var}}(L)[u]=\operatorname{Hess}(u)$ (see [GGHW19, (3.8)]). This
formula differs from the usual formula used above.

## References

  * [AGV] E. Abbena, A. Gray and L. Vanhecke, Steiner’s formula for the volume of a parallel hypersurface in a Riemannian manifold, Annali Sc. Norm. Sup. Pisa, 8, (3), 473-493 (1981).
  * [A20] P. Albin, Poincaré-Lovelock metrics on conformally compact manifolds, Advances in Math. 367 (2020).
  * [ACF92] L. Andersson, P. Chruściel and H. Friedrich, On the regularity of solutions to the Yamabe equation and the existence of smooth hyperboloidal initial data for Einstein’s field equations, Commun. Math. Phys. 149, 587–612 (1992).
  * [A94] B. Andrews, Contraction of convex hypersurfaces in Riemannian spaces, J. Diff. Geom. 39 (1994), 407–431.
  * [B87] A. Besse, Einstein manifolds, Ergebnisse der Mathematik und ihrer Grenzgebiete, 10, Springer-Verlag, (1987).
  * [BG90] T. Branson and P. Gilkey, The asymptotics of the Laplacian on a manifold with boundary, Comm. Partial Differential Equations, 15, 2 (1990), 245–272.
  * [FG12] C. Fefferman and C. R. Graham, The Ambient Metric. Annals of Math. Studies 178, Princeton University Press, 2012. arXiv:0710.0919
  * [GGHW19] M. Glaros, R. Gover, M. Halbasch and A. Waldron, Variational calculus for hypersurface functionals: singular Yamabe problem Willmore energies, J. Geom. Phys. 138, (2019), 168–193. arXiv:1508.01838v1
  * [G10] R. Gover, Almost Einstein and Poincaré-Einstein manifolds in Riemannian signature, Journal of Geometry and Physics 60, 2 (2010), 182–204. arXiv:0803.3510v1
  * [GW14] R. Gover, A. Waldron, Generalising the Willmore equation: submanifold conformal invariants from a boundary Yamabe problem. arXiv:1407.6742v1
  * [GW15] R. Gover, A. Waldron, Conformal hypersurface geometry via a boundary Loewner-Nirenberg-Yamabe problem, Comm. in Analysis and Geometry (to appear). arXiv:150602723v3
  * [GW17] R. Gover, A. Waldron, Renormalized volume, Comm. in Math. Physics 354, 3, (2017), 1205–1244. arXiv:1603.07367
  * [G17] C. R. Graham, Volume renormalization for singular Yamabe metrics, Proc. Amer. Math. Soc. 145 (2017), 1781–1792. arXiv:1606.00069
  * [GG19] C. R. Graham and M. Gursky, Chern-Gauss-Bonnet formula for singular Yamabe metrics in dimension four. arXiv:1902.01562
  * [G04] A. Gray, Tubes, Progress in Math. 221, Birkhäuser, 2004.
  * [HP99] G. Huisken and A. Polden, Geometric Evolution Equations for Hypersurfaces, in S. Hildebrand, M. Struwe (eds) Calculus of Variations and Geometric Evolution Problems, Lecture Notes in Mathematics 1713, 1999, 45–84.
  * [J09] A. Juhl, Families of conformally covariant differential operators, $Q$-curvature and holography, Birkhäuser, Progress in Mathematics 275, 2009.
  * [J13] A. Juhl, Explicit formulas for GJMS-operators and $Q$-curvatures, Geom. Funct. Anal. 23, 4, (2013), 278–1370. arXiv:1108.0273
  * [JO20] A. Juhl and B. Ørsted, Shift operators, residue families and degenerate Laplacians, Pacific J. Math. 308, 1, (2020), 103–160. arXiv:1806:02556
  * [JO21] A. Juhl and B. Ørsted, Residue families, singular Yamabe problems and extrinsic conformal Laplacians. arXiv:2101.09027v1
  * [SSY75] R. Schoen, L. Simon and S.-T. Yau, Curvature estimates for minimal hypersurfaces, Acta Math. 134, (1975), 276–288.
  * [S68] J. Simons, Minimal Varieties in Riemannian Manifolds, Annals of Math., 88, 1 (1968), 62–105.
  * [S15] S. N. Soloduchin, Boundary terms of conformal anomaly, Physics Letters B 752 (2016), 131—134.
  * [V13] Y. Vyatkin, Manufacturing conformal invariants of hypersurfaces, PhD thesis, University of Auckland, 2013.
  * [W93] T. J. Willmore, Riemannian Geometry, Oxford Science Publications, 1993.

## Index

  * $\bar{\nabla}$ Levi-Civita connection §2
  * $\bar{{\mathcal{G}}}$ Lemma 6.3
  * $\bar{G}$ Einstein tensor §6.3
  * $\bar{{\sf J}}$ §2
  * $\overline{\operatorname{Ric}}_{0}$ §2
  * $\overline{\operatorname{scal}}$ scalar curvature §2
  * $\overline{\mathcal{W}}$ §2
  * $\overline{W}$ Weyl tensor §2
  * $\overline{W}_{0}$ §2
  * $\partial_{0}$ unit normal vector §2
  * $\nabla$ Levi-Civita connection §2
  * $B$ hypersurface Bach tensor §6.5
  * ${\mathcal{B}}_{n}$ singular Yamabe obstruction §4
  * $C$ Cotton tensor §6.5
  * $\mathcal{D}$ §1
  * $dvol_{g}$ volume element of $g$ §2
  * $\operatorname{grad}(u)$ gradient field of $u$ §2
  * $H$ mean curvature §2
  * ${\sf J}$ §2
  * $L$ second fundamental form §2
  * $\mathring{L}$ trace-free part of $L$ §2
  * $L^{2}$, $L^{3}$, $L^{4}$ §2
  * ${\sf P}$ Schouten tensor §2
  * $R$ curvature tensor §2
  * $\operatorname{Ric}$ Ricci tensor §2
  * $\mathcal{S}(g,\sigma)$ §4
  * $\operatorname{scal}$ scalar curvature §2
  * $S_{k}$ §4
  * var variation §7
  * $v_{k}$ volume coefficients §4
  * $W$ Weyl tensor §2
  * $\mathcal{W}_{2}$ conformal Willmore functional §1, §5
  * $\mathcal{W}_{3}$ higher Willmore functional §1, §7
  * $\delta$ divergence operator §2
  * $\Delta$ Laplacian §2
  * $\Delta^{\prime}$ §6.1
  * $\Omega^{p}$ space of $p$-forms §2