categorifications even from, for example, a consistent dimer model on a torus.
See [2, §6].

######  Definition 7.10.

We say that a weakly consistent dimer model $Q$ is boundary-finite if
$\widehat{A}_{Q}/\langle e\rangle$ is finite-dimensional. We say that $Q$ is
Noetherian if $\widehat{A}_{Q}$ is Noetherian.

Hence, if $Q$ is a finite weakly consistent dimer model, then in order to
apply Corollary 7.8 we must only check Noetherianness and boundary-finiteness
of $Q$. The authors are not aware of an example of a consistent Noetherian
dimer model on a surface other than a disk, annulus, or torus. Dimer models on
tori and disks have been studied extensively.

  * •

Any consistent dimer model on a torus is Noetherian, but since the boundary
idempotent is zero such a dimer model is never boundary-finite. On the other
hand, as in Remark 7.9 we may use a larger idempotent to apply Theorem 7.7.

  * •

If $Q$ is a consistent dimer model on a disk with no digon-faces, Pressland
showed in [38, Proposition 4.4] that $Q$ is Noetherian and boundary-finite.
Hence, Corollary 7.8 can be applied to any consistent dimer model on a disk.
Moreover, in the same paper it is shown that
$(\textup{GP}(\widehat{B}_{Q}),e\widehat{A}_{Q})$ is a Frobenius 2-Calabi-Yau
realization of the cluster algebra $\mathscr{A}_{Q}$ given by the ice quiver
$Q$.

On the other hand, dimer models on annuli have seen comparatively little
attention. In a future paper, we will study consistent dimer models on annuli.
For now, we give a few examples pertaining to Corollary 7.8. Figure 15 shows
that a dimer model on an annulus may be Noetherian but not boundary-finite, or
boundary-finite but not Noetherian. Example 7.11 shows that there exist
(strongly) consistent dimer models on annuli satisfying both of these
conditions simultaneously.

Figure 15: On the left is a dimer model on an annulus which is Noetherian but
not boundary-finite. On the right is a dimer model on an annulus which is
boundary-finite but not Noetherian.

######  Example 7.11.

Figure 16 and Figure 17 show strongly consistent dimer models satisfying
Noetherianness and boundary finiteness. Both figures suggest general
constructions. One may, as in Figure 16, construct a strongly consistent dimer
model on an annulus such that the subquiver of internal vertices looks like
“some number of layers of an alternating affine type $\widetilde{A}_{2n}$
quiver” where corresponding vertices of adjacent layers are connected by
arrows. Similarly, one may as in Figure 17 construct a strongly consistent
dimer model on an annulus such that the subquiver of internal vertices looks
like some number of layers of any affine type $\widetilde{A}_{n}$ quiver. Note
that in all such examples, no strand starts and ends at the same boundary
component.

Figure 16: Two strongly consistent dimer models on annuli satisfying
Noetherianness and boundary finiteness. Figure 17: A strongly consistent dimer
model on an annulus satisfying Noetherianness and boundary finiteness. The
left and right sides should be identified. The strands going up are shown in
red, while the strands going down are left out for readability.

Corollary 7.8 may be applied to the annulus models of Example 7.11 to show
that the Gorenstein-projective module categories over their boundary algebras
categorify the cluster algebra given by their underlying quivers.

##  8 Extra Results about Equivalence Classes

For this section, we suppose that $Q$ is a weakly consistent dimer model
unless otherwise specified. We use Theorem 5.3 to get some interesting results
about the path equivalence classes in weakly consistent quivers. These results
are not used anywhere else in the paper.

######  Proposition 8.1.

Let $p$ and $q$ be distinct elementary paths in a weakly consistent dimer
model $\widetilde{Q}$ with the same start and end vertices. If $p$ is to the
right of $q$, then either $p$ has a left-morphable arrow or $q$ has a strictly
higher c-value than $p$.

###### Proof.

We may reduce to the case where $p$ and $q$ share no vertices except for the
start and end vertices. Then $q^{-1}p$ defines a disk submodel $Q^{\prime}$ of
$\widetilde{Q}$ in which $p$ consists only of boundary arrows in counter-
clockwise faces and $q$ consists of boundary arrows in clockwise faces. If $p$
has a strictly higher c-value than $q$ in $\widetilde{Q}$, then we are done.
Suppose the c-value of $p$ is less than or equal to that of $q$ in
$\widetilde{Q}$. Then $[p]=[qf^{m}]$ for some $m$ in $\widetilde{Q}$. By
Theorem 5.3, the same is true in $Q^{\prime}$. It follows that $p$ has a left-
morphable arrow in $Q^{\prime}$, and hence in $\widetilde{Q}$. This proves the
desired result. ∎

######  Corollary 8.2.

Let $Q=\widetilde{Q}$ be a simply connected dimer model. Let $v$ and $w$ be
vertices of $Q$. Suppose that there is a leftmost minimal path $p$ from $v$ to
$w$. Then $p$ is to the left of every minimal path and to the right of every
leftmost path.

###### Proof.

If $p^{\prime}$ is any other minimal path from $v$ to $w$, Proposition 8.1
shows that $p$ is to the left of $p^{\prime}$. If $q$ is any other leftmost
path from $v$ to $w$, Proposition 8.1 shows that $q$ is to the left of $p$. ∎

If $Q$ is not simply connected, we may still pass to the universal cover and
apply Corollary 8.2 to gain insight into the equivalence classes of paths of
$Q$.

######  Corollary 8.3.

Suppose $Q$ is a weakly consistent dimer model. If $p$ and $q$ are minimal
homotopic leftmost paths with the same start and end vertices, then $p=q$.

###### Proof.

Suppose $p$ and $q$ are homotopic minimal leftmost paths with the same start
and end vertices. Lift them to minimal leftmost paths $\tilde{p}$ and
$\tilde{q}$ of $\widetilde{Q}$. Since $p$ and $q$ are homotopic, these paths
have the same start and end vertices. It suffices to show that
$\tilde{p}=\tilde{q}$. Suppose this is not the case. By taking subpaths, we
may reduce to the case where $\tilde{p}$ and $\tilde{q}$ share only their
start and end vertices. Say $\tilde{p}$ is to the right of $\tilde{q}$. Since
both paths have the same c-value, Proposition 8.1 shows that $\tilde{p}$ has a
left-morphable arrow, a contradiction. ∎

Corollary 8.3 indicates that there is at most one minimal leftmost path
between any two vertices of $Q$. In general, leftmost paths need not exist.
Figure 21 gives an example of a finite weakly consistent dimer model on an
annulus where a minimal path has an infinite equivalence class. Let $\alpha$
be one of the arrows of the digon-face of $Q$. Then an arbitrarily long series
of left-morphs at the other arrow of the digon-face may be applied to
$\alpha$. In other words, there is no leftmost path from the outer boundary
ring to the inner boundary ring. The dimer model on the right of Figure 15 has
no digons and displays the same behavior: a minimal path from a vertex of the
inner boundary to a vertex of the outer boundary may be left-morphed or right-
morphed indefinitely, depending on the path.

On the other hand, by assuming nondegeneracy we guarantee the existence of
leftmost and rightmost paths between any given vertices.

######  Theorem 8.4.

Let $Q$ be a strongly consistent dimer model. Let $v$ and $w$ be distinct
vertices of $Q$ and let $C$ be a homotopy class of paths from $v$ to $w$. Then
there is a unique minimal leftmost path $p$ from $v$ to $w$ in $C$.

###### Proof.

We show the existence of a leftmost path from $v$ to $w$. Uniqueness follows
from Corollary 8.3. Lift $v$ and $w$ to vertices $\tilde{v}$ and $\tilde{w}$
of the universal cover model $\widetilde{Q}$ such that a path from $\tilde{v}$
to $\tilde{w}$ descends to a path of $Q$ in homotopy class $C$. We consider
paths on $\widetilde{Q}$ to have the grading of their corresponding paths on
$Q$. By path-consistency, all minimal paths from $\tilde{v}$ to $\tilde{w}$
are equivalent and hence have the same grading $k$. Then if
$\tilde{q}=\alpha_{m}\dots\alpha_{1}$ is a minimal path from $\tilde{v}$ to
$\tilde{w}$, then since the degree of each arrow is positive we must have
$m\leq k$. This shows that the length of a minimal path from $\tilde{v}$ to
$\tilde{w}$ is bounded by $k$. This condition guarantees that there are a
finite number of minimal paths from $\tilde{v}$ to $\tilde{w}$.

Now consider again the minimal path $\tilde{q}$ from $\tilde{v}$ to
$\tilde{w}$. Let $\tilde{r}$ be any path from $\tilde{w}$ to $\tilde{v}$. If
$\tilde{q}$ is not leftmost, then left-morph it at some arrow $\alpha_{1}$ to
get some $\tilde{q}_{1}$. Then Lemma 2.21 gives that
$\textup{Wind}(\tilde{r}\tilde{q}_{1},F)<\textup{Wind}(\tilde{r}\tilde{q},F)$
if $F$ is one of the faces containing $\alpha_{1}$, and
$\textup{Wind}(\tilde{r}\tilde{q}_{1},F)=\textup{Wind}(\tilde{r}\tilde{q},F)$
otherwise. Continue to left-morph to get some sequence of paths
$\tilde{q},\tilde{q}_{1},\tilde{q}_{2},\dots$. By the above inequalities, for
any $j$ the inequality
$\textup{Wind}(\tilde{r}\tilde{q}_{j+1},F)\leq\textup{Wind}(\tilde{r}\tilde{q}_{j},F)$
holds for all faces $F$, and is strict for some choice of $F$. Repeating this
argument shows that $\tilde{q}_{i}\neq\tilde{q}_{j}$ for $i\neq j$. Since
there are a finite number of minimal paths from $\tilde{v}$ to $\tilde{w}$ and
the sequence never repeats itself, the sequence must terminate with some
leftmost path $\tilde{p}:=\tilde{q}_{l}$ from $\tilde{v}$ to $\tilde{w}$. This
descends to a leftmost path $p$ from $v$ to $w$ in homotopy class $C$. ∎

Theorem 8.4 and Corollary 8.2 are useful tools to understand equivalence
classes of paths in nondegenerate dimer models. In particular, if $Q$ is a
consistent (hence also nondegenerate) dimer model in a disk and $e$ is the
boundary idempotent of $Q$, the boundary algebra $eA_{Q}e$ may be used to
obtain an additive Frobenius categorification of the ice quiver of $Q$ [38].
In a future paper, we will use these result to study boundary algebras of
consistent dimer models on disks.

##  9 Reduction of a Dimer Model

There has been some interest, particularly in the disk case, in reducing a
dimer model by removing digon-faces. See [38] and [9, Remark 2.9]. In
particular, if a weakly consistent, Noetherian, boundary-finite dimer model
$Q$ has no digons, our categorification result Corollary 7.8 shows that
$\textup{GP}(Q)$ categorifies the cluster algebra $\mathscr{A}_{Q}$ given by
the underlying quiver of $Q$. On the other hand, if $Q$ has digons, then it
may not be immediately clear what cluster algebra is being categorified.
Reducing a dimer model helps to avoid this issue. In [38], Pressland works
with consistent dimer models on a disk with more than three boundary vertices.
He observes that, in this case, the removal of digon-faces corresponds to
untwisting moves in the strand diagram and hence their removal is
straightforward. Moreover, the resulting reduced dimer models do not have any
digons, and the disk version of Corollary 7.8 can be used to prove results
about categorification. The situation for more general weakly consistent dimer
models is more complicated, as there may be copies similar to Figure 21
contained in the dimer model whose digon-faces may not be removed. However,
such occurrences will force the dimer model to be degenerate. Then in the
nondegenerate case we may freely remove internal digon-faces from a dimer
model.

First, we consider the general situation.

######  Proposition 9.1.

Let $Q$ be a weakly consistent dimer model with a finite number of digon-
faces. There exists a reduced dimer model $Q_{red}$ of $Q$ satisfying the
following:

  1. 1.

$S(Q_{red})=S(Q)$,

  2. 2.

$A_{Q_{red}}\cong A_{Q}$,

  3. 3.

$Q_{red}$ is weakly consistent,

  4. 4.

If $Q$ is nondegenerate, then $Q_{red}$ is nondegenerate, and

  5. 5.

Either $Q_{red}$ is a dimer model on a disk composed of a single digon-face,
or every digon-face of $Q_{red}$ is an internal face incident to only one
other face.

The condition (5) means that, as long as $Q_{red}$ is not composed of a single
digon-face, every digon-face of $Q_{red}$ is in a configuration like that of
Figure 21. See also Figure 14. In contrast, the configuration of Figure 20 is
not possible in a weakly consistent dimer model.

###### Proof.

Note that (3) follows from (2). Suppose that $Q$ is not merely a digon-face
and let $\alpha\beta$ be a digon-face of $Q$ which is either a boundary face
or is incident to two distinct faces. Then $\alpha$ and $\beta$ may not both
be boundary arrows. Suppose $\alpha\beta$ is a clockwise face; the counter-
clockwise case is symmetric.

If $\alpha$ and $\beta$ are both internal arrows, then by assumption the faces
$F_{\alpha}^{cc}$ and $F_{\beta}^{cc}$ are the distinct neighbors of the digon
$\alpha\beta$. Let $Q^{1}=(Q^{1}_{0},Q^{1}_{1},Q^{1}_{2})$ be the dimer model
whose underlying quiver is $Q$ without $\alpha$ and $\beta$ and whose set of
faces is the same as $Q_{2}$, but with the faces
$\\{\alpha\beta,F_{\alpha}^{cc},F_{\beta}^{cc}\\}$ replaced with one face
whose arrows are those of $R_{\alpha}^{cc}$ and $R_{\beta}^{cc}$. Since the
dimer algebra relations of $Q$ give $[\alpha]=[R_{\beta}^{cc}]$ and
$[\beta]=[R_{\alpha}^{cc}]$, the dimer algebra is not changed by this
operation. The surface is also unchanged by this operation. If $\mathcal{M}$
is a perfect matching of $Q$, then $\mathcal{M}$ contains either $\alpha$ or
$\beta$, and the removal of this arrow from $\mathcal{M}$ gives a perfect
matching of $Q$. Hence, if $Q$ is nondegenerate then $Q^{1}$ is nondegenerate.
See Figure 18.

On the other hand, suppose without loss of generality that $\alpha$ is a
boundary arrow and $\beta$ is internal. Then let $Q^{1}$ be the dimer model
obtained by removing the arrow $\alpha$ from $Q_{1}$ and the face
$\alpha\beta$ from $Q_{2}$. As above, $[\alpha]=[R_{\beta}^{cc}]$ in $A_{Q}$
and hence the dimer model is unchanged by this operation. If $\mathcal{M}$ is
a perfect matching of $Q$, then $\mathcal{M}$ contains $\alpha$ or $\beta$. If
$\mathcal{M}$ contains $\alpha$, then removing this arrow gives a perfect
matching of $\mathcal{M}$. If $\mathcal{M}$ contains $\beta$, then
$\mathcal{M}$ is also a perfect matching of $Q$. Then if $Q$ is nondegenerate,
then $Q^{1}$ is nondegenerate. Furthermore, $S(Q^{1})=S(Q)$. See Figure 19.

Figure 18: Removing an internal digon-face. Figure 19: Removing a boundary
digon-face.

In either case, we have defined a quiver $Q^{1}$ such that $S(Q^{1})=S(Q)$ and
$A_{Q^{1}}\cong A_{Q}$. Furthermore, $Q^{1}$ has strictly less digon-faces
than $Q$. We may now apply this process repeatedly to remove all digon-faces
of $Q$. Since $Q$ has a finite number of digon-faces, this process must
terminate with some $Q_{red}$ such that either $Q_{red}$ only has two arrows
or every digon-face of $Q_{red}$ is an internal face incident to only one
other face. In the former case, $Q_{red}$ must be composed of a single digon-
face since if it has two digon-faces then $S(Q_{red})=S(Q)$ is a sphere, hence
$Q$ is not weakly consistent, a contradiction. ∎

Figure 20: One way that that a digon-face could be incident to only one other
face. The strand diagram has a bad digon, hence this configuration is not
possible in a weakly consistent dimer model. Figure 21: A reduced dimer model
on an annulus with a digon-face which may not be removed.

Figure 21 shows a weakly consistent model with an internal digon-face which
may not be removed by the process of the above theorem. Indeed, if the digon-
face is removed then the resulting “dimer model” would have a face which is
not homeomorphic to an open disk. On the level of strand diagrams, removing
the digon-face corresponds to an untwisting move that disconnects the strand
diagram. If $Q$ has an infinite number of digon-faces, a new problem appears.
See Figure 22, which shows an infinite model with an infinite number of
digons. While any finite number of digons may be removed, all of them may not
be removed at once. The universal cover of the dimer model of Figure 21
displays the same behavior.

######  Remark 9.2.

In [37, §3], Pressland outlines a method of removing digon-faces from general
Jacobian ice quivers without changing the completed algebra. If this process
is applied to the dimer model of Figure 21, the resulting ice quiver is not a
dimer model.

Figure 22: Shown is a dimer model on an infinite half-strip. If all digons
were removed, then there would only be one infinite face making up the entire
non-compact surface, which is impossible.

Note that neither Figure 21 nor Figure 22 is nondegenerate. In fact, any
nondegenerate dimer model may be reduced since the condition (5) necessitates

######  Corollary 9.3.

Let $Q$ be a strongly consistent dimer model. There exists a reduced dimer
model $Q_{red}$ of $Q$ satisfying the following:

  1. 1.

$S(Q_{red})=S(Q)$,

  2. 2.

$A_{Q_{red}}\cong A_{Q}$,

  3. 3.

$Q_{red}$ is strongly consistent, and

  4. 4.

Either $Q_{red}$ is a dimer model on a disk composed of a single digon-face,
or $Q_{red}$ has no digon-faces.

###### Proof.

Apply Proposition 9.1. If $Q_{red}$ is not a dimer model on a disk composed of
a single digon-face, then $Q_{red}$ has an internal digon-face $\alpha\beta$
which is incident to a single other face $F$. Let $\gamma$ be an arrow of $F$
which is not $\alpha$ or $\beta$. Any perfect matching $\mathcal{M}$ must
contain $\alpha$ or $\beta$, since $\alpha\beta$ is a face. Then $\mathcal{M}$
cannot contain $\gamma$, since $\gamma$ shares a face with these arrows. We
have shown that no perfect matching contains $\gamma$. Then $Q_{red}$, and by
extension $Q$, is degenerate. ∎

We remark that $Q_{red}$ may have digons, even if it has no digon-faces.
Consider the dimer model on a torus pictured on the left of Figure 23. While
the quiver $Q$ has digons, it has no null-homotopic digons. As we see by
looking at the universal cover model on the right, this means that there are
no digon-faces in $Q$, hence $Q=Q_{red}$ is reduced.

Figure 23: Shown on the left is a dimer model on a torus. Opposite dashed
edges should be identified. Shown on the right is a piece of its universal
cover model.

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# The Anti-Self-Dual Deformation Complex and a conjecture of Singer

A. Rod Gover  Department of Mathematics
University of Auckland
Private Bag 92019, Auckland 1142, New Zealand  and  Matthew J. Gursky
Department of Mathematics
University of Notre Dame
Notre Dame, IN 46556

###### Abstract.

Let $(M^{4},g)$ be a smooth, closed, oriented anti-self-dual (ASD) four-
manifold. $(M^{4},g)$ is said to be unobstructed if the cokernel of the
linearization of the self-dual Weyl tensor is trivial. This condition can also
be characterized as the vanishing of the second cohomology group of the ASD
deformation complex, and is central to understanding the local structure of
the moduli space of ASD conformal structures. It also arises in construction
of ASD manifolds by twistor and gluing methods. In this article we give
conformally invariant conditions which imply an ASD manifold of positive
Yamabe type is unobstructed.

###### 2020 Mathematics Subject Classification:

53C21, 53C07

##  1\. Introduction

Let $M^{4}$ be a smooth, closed, oriented four-manifold. Given a Riemannian
metric $g$ on $M^{4}$, the bundle of two-forms
$\Lambda^{2}=\Lambda^{2}(M^{4})$ splits into the sub-bundles of self-dual and
anti-self=dual two-forms under the action of the Hodge $\star$=operator:

$\displaystyle\Lambda^{2}=\Lambda^{2}_{+}\oplus\Lambda^{2}_{-}.$

By a result of Singer-Thorpe [33], the curvature operator
$Rm:\Lambda^{2}\rightarrow\Lambda^{2}$ has a canonical block decomposition of
the form

$\displaystyle Rm=\left(\begin{array}[]{ll}A^{+}&B\\\
B^{t}&A^{-}\end{array}\right),$

where $A^{\pm}:\Lambda^{2}_{\pm}\rightarrow\Lambda^{2}_{\pm}$ and
$B:\Lambda^{2}_{+}\rightarrow\Lambda^{2}_{-}$. If $W$ denotes the Weyl tensor,
then $W^{\pm}:\Lambda^{2}_{\pm}\rightarrow\Lambda^{2}_{\pm}$, and

$\displaystyle A^{\pm}=W^{\pm}+\frac{1}{12}R\,I,$

where $I$ is the identity and $R$ is the scalar curvature.

######  Definition 1.1.

We say that $(M^{4},g)$ is anti-self-dual (ASD) if $W^{+}_{g}\equiv 0$.

The notion of (anti-)self-duality is conformally invariant: if $W^{+}_{g}=0$
for a metric $g$ and $\tilde{g}=e^{f}g$, then $W^{+}_{\tilde{g}}=0$. This
property will be crucial for the proof of our main result below.

There are topological obstructions to the existence of ASD metrics. By the
Hirzebruch signature formula,

(1.1) $\displaystyle
48\pi^{2}\tau(M^{4})=\int\left(|W^{+}_{g}|^{2}-|W^{-}_{g}|^{2}\right)\,dv_{g},$

where $\tau(M^{4})$ is the signature of the intersection form on
$H^{2}_{dR}(M^{4})$. In particular, we see that if $(M^{4},g)$ is ASD then
$\tau(M^{4})\leq 0$, with equality if and only if $g$ is LCF. If $(M^{4},g)$
is ASD with positive scalar curvature, then the intersection form is actually
definite (see Proposition 1 of [28]). To see this, we first observe that the
splitting of $\Lambda^{2}(M^{4})$ induces a splitting on the space of harmonic
two-forms, hence $H^{2}_{dR}(M^{4})=H^{2}_{+}(M^{4})\oplus H^{2}_{-}(M^{4})$.
If $\omega\in H^{2}_{+}(M^{4})$, then the Weitzenböck formula for the Hodge
laplacian $\Delta_{2}$ is given by

(1.2)
$\displaystyle\Delta_{2}\omega=\Delta\omega+2W^{+}(\omega)-\frac{1}{3}R\omega,$

where $\Delta=g^{ij}\nabla_{i}\nabla_{j}$ is the rough laplacian. If
$(M^{4},g)$ is ASD and the scalar curvature $R>0$, then (1.2) immediately
implies $\omega=0$.

Examples of ASD manifolds include locally conformally flat (LCF) manifolds,
since in dimensions greater than three LCF is equivalent to the vanishing of
the Weyl tensor. In particular, $S^{4}$ endowed with the round metric $g_{c}$
is ASD. Non-simply connected examples include the product metric on
$S^{3}\times S^{1}$, and more generally the metrics constructed via gluing on
the connected sums $S^{3}\times S^{1}\hash\cdots\hash S^{3}\times S^{1}$.

A non-LCF example is given by complex projective space with the Fubini-Study
metric, and we take the opposite of the its natural orientation as a complex
manifold; i.e., $(-\mathbb{CP}^{2},g_{FS})$. Any scalar-flat Kähler metric is
also ASD, since the Kähler condition implies that $W^{+}$ is determined by $R$
(see [12], Section 3). There are many constructions ASD manifolds in the
literature; see for example [31], [32], [29], [13], [16], [34], and the
references in Lecture 6 of [36].

Roughly speaking, the constructions of ASD manifolds are based on either
‘twistor’ or ‘analytic’ methods. The former approach relies on the so-called
Penrose correspondence, which will play no role in our work but is of profound
importance in the study of ASD manifolds. Briefly, the unit sphere bundle
$\mathcal{Z}$ of $\Lambda^{2}_{+}(M^{4})$ carries a canonical complex
structure. As shown in [1], this complex structure is integrable if and only
if the metric is ASD. Therefore, we can associate to any ASD manifold a
complex manifold of (complex) dimension three, called the twistor space of
$(M^{4},g)$. This important observation allows one to use methods of complex
geometry to study the existence and deformation theory of ASD conformal
structures.

Analytic methods involve the construction of an ASD metric on the connected
sum of two manifolds admitting ASD metrics via perturbative methods. As in
other geometric gluing constructions, if $(M_{1},g_{1})$ and $(M_{2},g_{2})$
are ASD manifolds, one first constructs a metric $h$ on the connected sum
$M_{1}\hash M_{2}$ which is “approximately” ASD; i.e., $W^{+}(h)$ is small in
some appropriately defined norm. This reduces the problem to the study of the
mapping properties of the linearized operator, in order to perturb $h$ to
produce an actual ASD metric. To make this more precise, we now introduce the
ASD deformation complex.

###  1.1. The ASD deformation complex

Let $\mathcal{M}(M^{4})$ be the space of smooth Riemannian metrics on $M^{4}$,
and $\mathcal{R}(M^{4})$ the bundle of algebraic curvature tensors. We can
view $W^{+}$ as a mapping

$\displaystyle W^{+}:\mathcal{M}(M^{4})\rightarrow\mathcal{R}(M^{4}).$

Let $g\in\mathcal{M}(M^{4})$ be an ASD metric. We can identify the formal
tangent space of $\mathcal{M}$ at $g$ with sections of the bundle of symmetric
two-tensors, $S^{2}(T^{*}M^{4})$. Let

(1.3)
$\displaystyle\mathcal{D}:\Gamma(S^{2}(T^{*}M^{4}))\rightarrow\Gamma(\mathcal{R}(M^{4}))$

denote the linearization of $W^{+}$ at $g$; i.e., for
$h\in\Gamma(S^{2}(T^{*}M^{4}))$,

$\displaystyle\mathcal{D}_{g}h=\frac{d}{ds}W^{+}(g+sh)\big{|}_{s=0}.$

The choice of a conformal class of metrics $[g]$ determines the bundle of
algebraic Weyl tensors, $\mathcal{W}=\mathcal{W}(M^{4},[g])$, and the sub-
bundles $\mathcal{W}^{\pm}\subset\mathcal{W}\subset
S^{2}_{0}(\Lambda^{2}_{+})$, where $S^{2}_{0}(\Lambda^{2}_{+})$ is the bundle
of symmetric, trace-free endomorphisms of $\Lambda^{2}_{+}$. Note that

(1.4)
$\displaystyle\mathcal{D}_{g}:\Gamma(S^{2}(T^{*}M^{4}))\rightarrow\Gamma(\mathcal{W}^{+})\subset\Gamma(S^{2}_{0}(\Lambda^{2}_{+})).$

In fact, since $g$ is ASD, by conformal invariance $\mathcal{D}_{g}(fg)=0$ for
any $f\in C^{\infty}(M^{4})$, hence

(1.5)
$\displaystyle\mathcal{D}_{g}:\Gamma(S^{2}_{0}(T^{*}M^{4}))\rightarrow\Gamma(\mathcal{W}^{+}).$

We also let

(1.6)
$\displaystyle\mathcal{D}^{*}_{g}:\Gamma(\mathcal{W}^{+})\rightarrow\Gamma(S^{2}_{0}(T^{*}M^{4}))$

denote the $L^{2}$-formal adjoint of $\mathcal{D}_{g}$. Although the formula
for $\mathcal{D}_{g}$ is somewhat involved, the formula for $\mathcal{D}^{*}$
is much more compact (see Proposition A.4 of [23]):

(1.7)
$\displaystyle\mathcal{D}^{*}U_{ij}=2\left(\nabla^{k}\nabla^{\ell}U_{ikj\ell}+P^{k\ell}U_{ikj\ell}\right),$

where $P$ denotes the Schouten tensor (see Section 2).

Let $\mathcal{K}_{g}:\Gamma(T^{*}M^{4})\rightarrow S^{2}_{0}(T^{*}M^{4})$
denote the Killing operator:

$\displaystyle\mathcal{K}_{g}(\omega)_{ij}=\nabla_{i}\omega_{j}+\nabla_{j}\omega_{i}-\frac{1}{2}(\delta_{g}\omega)\,g_{ij},$

where $\delta_{g}\omega=\nabla^{k}\omega_{k}$ is the divergence of $\omega$.
The kernel of $\mathcal{K}$ consists of those one-forms whose dual vector
field are conformal Killing. Moreover, by diffeomorphism and conformal
invariance,

$\displaystyle\operatorname{Im}\mathcal{K}\subset\ker\mathcal{D}.$

The ASD deformation complex is given by

(1.8)
$\displaystyle\Gamma(T^{*}M^{4})\xrightarrow{\mathcal{K}}\Gamma(S^{2}_{0}(T^{*}M^{4}))\xrightarrow{\mathcal{D}}\Gamma(S^{2}_{0}(\Lambda^{2}_{+})).$

This complex is elliptic; see [26], Section 2. The associated cohomology
groups are given by

(1.9) $\displaystyle\begin{split}H^{0}_{ASD}(M^{4},g)&=\ker\mathcal{K}_{g},\\\
H^{1}_{ASD}(M^{4},g)&=\\{h\in\Gamma(S^{2}_{0}(T^{*}M^{4}))\,:\,\delta_{g}h=0,\
\mathcal{D}_{g}h=0\\},\\\
H^{2}_{ASD}(M^{4},g)&=\ker\mathcal{D}^{*}_{g},\end{split}$

where for $h\in\Gamma(S^{2}(T^{*}M^{4}))$,
$(\delta_{g}h)_{j}=g^{ik}\nabla_{k}h_{ij}$ is the divergence. The Atiyah-
Singer index theorem can be used to calculate the index of (1.8)111This was
calculation was done by Singer in unpublished notes, but can be found in the
literature for example in [26]:

$\displaystyle\mbox{Ind}_{ASD}(M^{4},g)=\frac{1}{2}\left(15\chi\left(M^{4}\right)+29\tau\left(M^{4}\right)\right).$

Vanishing of the cohomology groups also provides information on the local
structure of the moduli space of ASD conformal structures:

######  Proposition 1.2.

(See [23], [36]) Suppose $(M^{4},g)$ is ASD with

$\displaystyle H^{0}_{ASD}(M^{4},g)=\\{0\\},\ \ H^{2}_{ASD}(M^{4},g)=\\{0\\}.$

Then the moduli space of anti-self-dual conformal structures near $g$ is a
smooth, finite-dimensional manifold of dimension $\dim H^{1}_{ASD}(M^{4},g)$.

This leads to the following definition:

######  Definition 1.3.

Let $(M^{4},g)$ be ASD. We say that $(M^{4},g)$ is unobstructed if
$H^{2}_{ASD}(M^{4},g)=\\{0\\}$.

By the work of Floer [16] and Donaldson-Friedman [13], if $(M_{1},g_{1})$ and
$(M_{2},g_{2})$ are unobstructed ASD manifolds, then the connected sum
$M_{1}\hash M_{2}$ admits an ASD metric. Thus, we are lead to the question:
under what condition is an ASD manifold unobstructed? The following conjecture
is often attributed to Singer:

######  Conjecture 1.4.

Let $(M^{4},g)$ be ASD. If the Yamabe invariant of $(M^{4},g)$ is positive,
then $(M^{4},g)$ is unobstructed.

For ASD Einstein manifolds the operators $\mathcal{D}\mathcal{D}^{*}$ and
$\mathcal{D}^{*}\mathcal{D}$ were explicitly computed in [24] and [27]. It
follows from these calculations (and can also be seen by more or less direct
calculation) that $(S^{4},g_{0})$ and $(\mathbb{CP}^{2},g_{FS})$ are
unobstructed. We remark that for non-Einstein ASD manifolds, the formulas for
these operators are fairly intractable.

The vanishing of $H^{2}_{ASD}(M^{4},[g])$ can sometimes be verified when the
twistor space is explicitly known. For example, the LCF metrics $k\hash
S^{3}\times S^{1}$ ([30], Theorem 8.2; [14]), and the ASD metrics on
$m\hash\left(-\mathbb{CP}^{2}\right)$ constructed by LeBrun ([29]), are
unobstructed. Our goal in this paper is to provide a criterion for the
vanishing of $H^{2}_{ASD}$ that only involves conformal invariants of the ASD
manifold (and in particular does not depend on verifying any properties of the
twistor space). Our main result is the following:

######  Theorem 1.5.

Suppose $(M^{4},g)$ is ASD with Yamabe invariant $Y(M^{4},[g])>0.$ If

(1.10) $\displaystyle
2\chi(M^{4})+3\tau(M^{4})\geq-\frac{1}{24\pi^{2}}Y(M^{4},[g])^{2},$

then $(M^{4},g)$ is unobstructed.

The proof of Theorem 1.5 relies on two key ideas. The first is that any
element $U\in\ker\mathcal{D}^{*}$ can be associated to a self-dual harmonic
two-form $z=z(U)\in\Lambda^{2}_{+}\left(\mathcal{E}\right)$ taking its values
in the adjoint tractor bundle (see Section 2). Therefore, $z$ satisfies a
twisted version of the usual Weitzenböck formula for self-dual (real-valued)
two-forms. This twisted version provides us with two identities for $U$ (see
Theorem 3.6).

The second key idea is to make a judicious choice of conformal representative
in order to show that the condition (1.10) implies the vanishing of $U$. As
explained in Section 4, we choose a conformal metric whose scalar curvature
satisfies a differential inequality involving the Schouten tensor (see Theorem
4.1), and is adapted to the curvature terms appearing in the Weitzenböck
formula(s) for $U$. Curiously, to prove the existence of this metric we
consider a modification of the functional determinant of an conformally
covariant elliptic operator first computed by Branson and Ørsted [3]. An
robust theory for the existence of critical points of this functional was
developed by Chang and Yang [10], and we are able to show that their ideas
also give existence for our modified functional.

We remark that the lower bound in (1.10) can likely be improved, and it is
possible that the techniques of this paper can be used to ultimately remove
any additional topological assumption.

###  1.2. Organization

In Section 2 we provide the necessary background on the tractor bundle and
associated connection and metric. The main result (for our purposes) is
Proposition 2.2, in which we associate to $U\in\ker\mathcal{D}^{*}$ a twisted
SD harmonic two-form $z=z(U)\in\Lambda^{2}_{+}\left(\mathcal{E}\right)$. In
Section 3 we compute two Weitzenböck formulas for $U$ that follow from this
correspondence. In Section 4 we give the proof of Theorem 1.5, and relegate
the PDE aspects of our work to the Appendix.

###  1.3. Acknowledgements

The authors would like to express their sincere thanks to Claude LeBrun for
initially suggesting this problem, and for being an invaluable resource during
our work.

The first author is supported by the Royal Society of New Zealand Marsden
grant 19-UOA-008. The second author is supported by NSF grant DMS-2105460 and
a Simons Foundation Fellowship in Mathematics, award 923208.

This project began during a visit of the second author to the Department of
Mathematics at the University of Auckland, in September 2022\. The author
would like to thank the department for its support and hospitality.

##  2\. Background and the interpretation via tractor calculus

###  2.1. Some conventions for Riemannian geometry

For index calculations we will use Penrose’s abstract index notation, unless
otherwise indicated. In this we write $\mathcal{E}_{a}$ and $\mathcal{E}^{a}$
as alternative notations for, respectively the cotangent and tangent bundles
and the contraction $\omega(v)$ of a 1-form $\omega$ with a tangent vector
$v^{a}$ is written with a repeated index $\omega_{a}v^{a}$ (mimicking the
Einstein summation convention). Tensor bundles are denoted then by adorning
the symbol $\mathcal{E}$ with appropriate indices and sometimes also
indicating symmetries. For example $\mathcal{E}_{(ab)}$ is the notation for
$S^{2}T^{*}M$, the subbundle of symmetric tensors in $T^{*}M\otimes T^{*}M$.

Then our convention for the Riemann tensor $R_{ab}{}^{c}{}_{d}$ is such that

(2.1)
$\displaystyle\begin{split}\left[\nabla_{a},\nabla_{b}\right]v^{c}&=R_{ab}{}^{c}{}_{d}v^{d},\\\
\left[\nabla_{a},\nabla_{b}\right]\omega_{c}&=R_{abc}{}^{d}\omega_{d},\end{split}$

where $\nabla_{a}$ is the Levi-Civita connection of a metric $g_{ab}$, $v$ any
tangent vector field, and $\omega$ is any one-form. Using the metric to raise
and lower indices we have, for example $R_{abcd}=g_{ce}R_{ab}{}^{e}{}_{d}$.
This may be decomposed:

(2.2) $R_{abcd}=W_{abcd}+2\left(g_{c[a}P_{b]d}+g_{d[b}P_{a]c}\right),$

where the completely trace-free part $W_{abcd}$ is the Weyl tensor and
$P_{ab}$ is the Schouten tensor. It follows that

(2.3)
$P_{ab}:=\frac{1}{n-2}\left(R_{ab}-\frac{R}{2\left(n-1\right)}g_{ab}\right)$

where $R_{bc}=R_{ab}{}^{a}{}_{c}$ is the Ricci tensor and its metric trace
$R=g^{ab}R_{ab}$ is scalar curvature. We will use $J$ to denote the metric
trace of Schouten, i.e. $J:=g^{ab}P_{ab}$. Lastly, we have

(2.4) $\displaystyle C_{abc}$ $\displaystyle:=2\nabla_{[b}P_{c]a},$
$\displaystyle B_{ab}$ $\displaystyle:=-\nabla^{c}C_{abc}+P^{dc}C_{dacb},$

where $C_{abc}$ and $B_{ab}$ are Cotton and Bach tensors, respectively. It
should be noted that the Bianchi identities imply

(2.5) $\left(n-3\right)C_{abc}=\nabla^{d}W_{dabc}.$

###  2.2. The tractor bundle and connection

To treat and work with objects that are conformally invariant it is natural
work, at least partly, in the setting of conformal manifolds. Here by a
conformal manifold $(M,\boldsymbol{c})$ we mean a smooth manifold of dimension
$n\geq 3$ equipped with an equivalence class $\boldsymbol{c}$ of Riemanian
metrics, where $g_{ab}$, $\widehat{g}_{ab}\in\boldsymbol{c}$ means that
$\widehat{g}_{ab}=\Omega^{2}g_{ab}$ for some smooth positive function
$\Omega$. On a general conformal manifold $(M,\boldsymbol{c})$, there is no
distinguished connection on $TM$. But there is an invariant and canonical
connection on a closely related bundle, namely the conformal tractor
connection on the standard tractor bundle, see [2, 7].

Here we review the basic conformal tractor calculus on Riemannian and
conformal manifolds. See [2, 11, 18] for more details. Unless stated
otherwise, calculations will be done with the use of generic
$g\in\boldsymbol{c}$.

On an $n$-manifold $M$ the top exterior power of the tangent bundle
$\Lambda^{n}TM$ is a line bundle. Thus its square
$\mathcal{K}:=(\Lambda^{n}TM)^{\otimes 2}$ is canonically oriented and so one
can take oriented roots of it: Given $w\in\mathbb{R}$ we set

(2.6) $\mathcal{E}[w]:=\mathcal{K}^{\frac{w}{2n}},$

and refer to this as the bundle of conformal densities. For any vector bundle
$\mathcal{V}$, we write $\mathcal{V}[w]$ to mean
$\mathcal{V}[w]:=\mathcal{V}\otimes\mathcal{E}[w]$. For example,
$\mathcal{E}_{(ab)}[w]$ denotes the symmetric second tensor power of the
cotangent bundle tensored with $\mathcal{E}[w]$, i.e.
$S^{2}T^{*}M\otimes\mathcal{E}[w]$ on $M$. On a fixed Riemannian manifold
$\mathcal{K}$ is canonically trivialised by the square of the volume form, and
so $\mathcal{K}$ and its roots are not usually needed explicitly. However if
we wish to change the metric conformally, or work on a conformal structure
then these objects become important.

Since each metric in a conformal class determines a trivialisation of
$\mathcal{K}$, it follows easily that on a conformal structure there is a
canonical section $\boldsymbol{g}_{ab}\in\Gamma(\mathcal{E}_{(ab)})[2]$. This
has the property that for each positive section
$\sigma\in\Gamma(\mathcal{E}_{+}[1])$ (called a scale)
$g_{ab}:=\sigma^{-2}\boldsymbol{g}_{ab}$ is a metric in $\boldsymbol{c}$.
Moreover, the Levi-Civita connection of $g_{ab}$ preserves $\sigma$ and
therefore $\boldsymbol{g}_{ab}$. Thus it makes sense to use the conformal
metric to raise and lower indices, even when we are chosing a particular
metric $g_{ab}\in\boldsymbol{c}$ and its Levi-Civita connection for
calculations. It turns out that this simplifies many computations, and so in
this section we will do that without further mention. (In the subsequent
sections we will work with a fixed metric and use that to trivialise density
bundles – so indices will be raised and lowered using the metric.)

Considering Taylor series for sections of $\mathcal{E}[1]$ one recovers the
jet exact sequence at 2-jets,

(2.7)
$0\to\mathcal{E}_{(ab)}[1]\stackrel{{\scriptstyle\iota}}{{\to}}J^{2}\mathcal{E}[1]\to
J^{1}\mathcal{E}[1]\to 0.$

Note that $J^{2}\mathcal{E}[1]$ and its sequence (2.7) are canonical objects
on any smooth manifold. But with a conformal structure $\boldsymbol{c}$ we
have the orthogonal decomposition of $\mathcal{E}_{ab}[1]$ into trace-free and
trace parts

(2.8)
$\mathcal{E}_{ab}[1]=\mathcal{E}_{(ab)_{0}}[1]\oplus\boldsymbol{g}_{ab}\cdot\mathcal{E}[-1].$

Thus we can canonically quotient $J^{2}\mathcal{E}[1]$ by the image of
$\mathcal{E}_{(ab)_{0}}[1]$ under $\iota$ (in (2.7)). The resulting quotient
bundle is denoted $\mathcal{T}^{*}$ (or $\mathcal{E}_{A}$ in abstract
indices), and called the conformal cotractor bundle. Observing that the jet
exact sequence at 1-jets (of $\mathcal{E}[1]$),

$0\to\mathcal{E}_{b}[1]\stackrel{{\scriptstyle\iota}}{{\to}}J^{1}\mathcal{E}[1]\to\mathcal{E}[1]\to
0,$

we see at once that $\mathcal{T}^{*}$ has a composition series

(2.9)
$\mathcal{T}^{*}=\mathcal{E}[1]\mbox{$\begin{picture}(12.7,8.0)(-0.5,-1.0)\put(2.0,0.2){$+$}
\put(6.2,2.8){\oval(8.0,8.0)[l]}
\end{picture}$}\mathcal{E}_{a}[1]\mbox{$\begin{picture}(12.7,8.0)(-0.5,-1.0)\put(2.0,0.2){$+$}
\put(6.2,2.8){\oval(8.0,8.0)[l]} \end{picture}$}\mathcal{E}[-1],$

meaning that $\mathcal{E}[-1]$ is a subbundle of $\mathcal{T}^{*}$ and the
quotient of $\mathcal{T}^{*}$ by this (which is $J^{1}\mathcal{E}[1]$) has
$\mathcal{E}_{a}[1]$ as a subbundle, whereas there is a canonical projection
$X:\mathcal{T}^{*}\to\mathcal{E}[1]$. In abstact indices we write $X^{A}$ for
this map and call it the canonical tractor.

Given a choice of metric $g\in\boldsymbol{c}$, the formula

(2.10) $\sigma\mapsto\frac{1}{n}{[D_{A}\sigma]}_{g}:=\begin{pmatrix}\sigma\\\
\nabla_{a}\sigma\\\ -\frac{1}{n}\left(\Delta+J\right)\sigma\end{pmatrix}$

(where $\Delta$ is the Laplacian $\nabla^{a}\nabla_{a}$) gives a second-order
differential operator on $\mathcal{E}[1]$ which is a linear map
$J^{2}\mathcal{E}[1]\to\mathcal{E}[1]\oplus\mathcal{E}_{a}[1]\oplus\mathcal{E}[-1]$
that clearly factors through $\mathcal{T}^{*}$ and so determines an
isomorphism

(2.11)
$\mathcal{T}^{*}\stackrel{{\scriptstyle\sim}}{{\longrightarrow}}{[\mathcal{T}^{*}]}_{g}=\mathcal{E}[1]\oplus\mathcal{E}_{a}[1]\oplus\mathcal{E}[-1].$

In subsequent discussions, we will use (2.11) to split the tractor bundles
without further comment. Thus, given $g\in\boldsymbol{c}$, an element $V_{A}$
of $\mathcal{E}_{A}$ may be represented by a triple $(\sigma,\mu_{a},\rho)$,
or equivalently by

(2.12) $V_{A}=\sigma Y_{A}+\mu_{a}Z_{A}{}^{a}+\rho X_{A}.$

The last display defines the algebraic splitting operators
$Y:\mathcal{E}[1]\to\mathcal{T}^{*}$ and $Z:T^{*}M[1]\to\mathcal{T}^{*}$
(determined by the choice $g_{ab}\in\boldsymbol{c}$) which may be viewed as
sections $Y_{A}\in\Gamma(\mathcal{E}_{A}[-1])$ and
$Z_{A}{}^{a}\in\Gamma(\mathcal{E}_{A}{}^{a}[-1])$. We call these sections
$X_{A},Y_{A}$ and $Z_{A}{}^{a}$ _tractor projectors_.

By construction the tractor bundle is conformally invariant, i.e. determined
by $(M,\boldsymbol{c})$ and indpenedent of any choice of $g\in\boldsymbol{c}$.
However the splitting (2.12) is not. Considering the transformation of the
operator (2.10) determining the splitting we see that if
$\widehat{g}=\Omega^{2}f$ the components of an invariant section of
$\mathcal{T}^{*}$ should transform according to:

(2.13)
$[\mathcal{T}^{*}]_{\widehat{g}}\ni\left(\begin{array}[]{c}\widehat{\sigma}\\\
\widehat{\mu}_{b}\\\
\widehat{\rho}\end{array}\right)=\left(\begin{array}[]{ccc}1&0&0\\\
\Upsilon_{b}&\delta^{c}_{b}&0\\\
-\frac{1}{2}\Upsilon^{2}&-\Upsilon^{c}&1\end{array}\right)\left(\begin{array}[]{c}\sigma\\\
\mu_{c}\\\ \rho\end{array}\right)~{}\sim~{}\left(\begin{array}[]{c}\sigma\\\
\mu_{b}\\\ \rho\end{array}\right)\in[\mathcal{T}^{*}]_{g},$

where $\Upsilon_{a}=\Omega^{-1}\nabla_{a}\Omega$, and converesely this
transformation of triples is the hallmark of an invariant tractor section.
Equivalent to the last display is the rule for how the algebraic splitting
operators transform

(2.14)
$\textstyle\widehat{X}_{A}=X_{A},\quad\widehat{Z}_{A}{}^{b}=Z_{A}{}^{b}+\Upsilon^{b}X_{A},\quad\widehat{Y}_{A}=Y_{A}-\Upsilon_{b}Z_{A}{}^{b}-\frac{1}{2}\Upsilon_{b}\Upsilon^{b}X_{A}\,.$

Given a metric $g\in\boldsymbol{c}$, and the corresponding splittings, as
above, the tractor connection is given by the formula

(2.15) $\nabla_{a}^{\mathcal{T}}\begin{pmatrix}\sigma\\\ \mu_{b}\\\
\rho\end{pmatrix}:=\begin{pmatrix}\nabla_{a}\sigma-\mu_{a}\\\
\nabla_{a}\mu_{b}+P_{ab}\sigma+\boldsymbol{g}_{ab}\rho\\\ \nabla_{a}\rho-
P_{ac}\mu^{c}\end{pmatrix},$

where on the right hand side the $\nabla$s are the Levi-Civita connection of
$g$. Using the transformation of components, as in (2.13), and also the
conformal transformation of the Schouten tensor,

(2.16)
$P^{\widehat{g}}_{\phantom{\widehat{g}}ab}=P_{ab}-\nabla_{a}\Upsilon_{b}+\Upsilon_{a}\Upsilon_{b}-\frac{1}{2}g_{ab}\Upsilon_{c}\Upsilon^{c},\quad\widehat{g}=\Omega^{2}g,$

reveals that the triple on the right hand side transforms as a 1-form taking
values in $\mathcal{T}^{*}$ – i.e. again by (2.13) except twisted by
$\mathcal{E}_{a}$. Thus the right hand side of (2.15) is the splitting into
slots of a conformally invariant connection $\nabla^{\mathcal{T}}$ on (section
of) the bundle $\mathcal{T}^{*}$.

There is a nice conceptual origin for the connection (2.15). Using (2.16) and
the transformation of the Levi-Civita connection it is straightforward to
verify that the equation

(2.17) $\nabla_{(a}\nabla_{b)_{0}}\sigma+P_{(ab)_{0}}\sigma=0$

on conformal densities $\sigma\in\Gamma(\mathcal{E}[1])$ is conformally
invariant. As this is an overdetermined PDE, solutions in general do not
exist. Overdetermined linear PDE are typically studied by prolongation and it
is quickly verified that the tractor parallel transport given by (2.15) is
exactly the closed system that arises from prolonging (2.17) (see [2, 11]).
From this observation, and the formula (2.15), it follows that non-trivial
solutions of (2.17) are non-vanishing on an open dense set (for $M$ connected)
on which the metric $\sigma^{-2}\boldsymbol{g}_{ab}$ is Einstein – an
observation that has a number of applications, see [11, 17] and references
therein.

The tractor bundle is also equipped with a conformally invariant signature
$(n+1,1)$ metric $h_{AB}\in\Gamma\left(\mathcal{E}_{(AB)}\right)$, defined as
quadratic form by the mapping

(2.18) $[V_{A}]_{g}=\begin{pmatrix}\sigma\\\ \mu_{a}\\\
\rho\end{pmatrix}\mapsto\mu_{a}\mu^{a}+2\sigma\rho=:h\left(V,V\right),and$

with the polarization identity. This is important not only by dint of its
conformal invariance, but it is easily checked that this tractor metric $h$ is
preserved by $\nabla_{a}^{\mathcal{T}}$, i.e.
$\nabla_{a}^{\mathcal{T}}h_{AB}=0$. Thus it makes sense to use $h_{AB}$ (and
its inverse) to raise and lower tractor indices, and we do this henceforth
without further comment. In particular $X^{A}=h^{AB}X_{B}$ is the canonical
tractor (and hence our use of the same kernel symbol). For computations the
table of Figure 1 is useful.

$\begin{array}[]{l|ccc}&Y^{A}&Z^{Ac}&X^{A}\\\ \hline\cr Y_{A}&0&0&1\\\
Z_{Ab}&0&\delta_{b}{}^{c}&0\\\ X_{A}&1&0&0\end{array}$ Figure 1. Tractor inner
product

We see that $h$ may be decomposed into a sum of projections

$h_{AB}=Z_{A}{}^{a}Z_{B}{}^{b}\boldsymbol{g}_{ab}+X_{A}Y_{B}+Y_{A}X_{B}\,.$

Finally for this section we note that, of course, the curvature of the tractor
connection $\kappa_{abCD}$ is determined by

$2\nabla_{[a}^{\mathcal{T}}\nabla_{b]}^{\mathcal{T}}V^{C}=\kappa_{ab}{}^{C}{}_{D}V^{D}\quad\mathrm{for\
all}\quad V^{C}\in\Gamma\left(\mathcal{E}^{A}\right),$

and can be written in terms of tractor projectors as

(2.19)
$\kappa_{abCD}=W_{abcd}Z_{C}{}^{c}Z_{D}{}^{d}+C_{cab}Z_{C}{}^{c}X_{D}-C_{cab}X_{C}Z_{D}{}^{c}\,.$

The bundle $\Lambda^{2}\mathcal{T}$ is often termed the adjoint tractor
bundle, as it is a vector bundle modelled on the Lie algebra of the conformal
group $SO(n+1,1)$. So, as expected, the tractor curvature is a 2-form taking
values in this bundle.

###  2.3. The differential splitting operator

In dimensions $n\geq 4$ the tractor curvature is the image of a conformally
invariant operator $z$ acting on the Weyl curvature. The operator is as
follows:

######  Lemma 2.1.

In any dimension $n\geq 4$ there is a conformally invariant differential map

(2.20)
$z:\Gamma(\mathcal{W})\to\Gamma(\Lambda^{2}\otimes\Lambda^{2}\mathcal{T}),$

given by

(2.21) $\displaystyle\begin{split}U_{ij}{}^{k}{}_{\ell}\mapsto
z_{ijAB}&=U_{ij}^{\ \ k\ell}Z_{kA}Z_{\ell B}-\frac{1}{n-3}(\delta U)^{k}_{\
ij}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)\\\ &=\frac{1}{2}U_{ij}^{\ \
k\ell}\left(Z_{kA}Z_{\ell B}-Z_{\ell A}Z_{kB}\right)-\frac{1}{n-3}(\delta
U)^{k}_{\ ij}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right),\end{split}$

where

$\displaystyle(\delta U)_{kij}=\nabla^{m}U_{mkij}.$

The conformal invariance is easily verified directly using (2.14) and the
conformal transformation of the Levi-Civita connection. It can also be deduced
from the conformal invariance of the tractor curvature $\kappa_{abCD}$. In
fact the map (2.20) is a standard “BGG-splitting operator”, as in the theory
[8, 5], and the conformal deformation sequence can de understood as arising
from a twisting by $\Lambda^{2}\mathcal{T}$ of the de Rham complex [6, 19].

We are, in particular, interested in the case of dimension $n=4$. Then it is
evident, from the formula (2.21), that if $U$ is SD (or ASD) then so is $z(U)$
as a tractor-twisted 2-form, as the Hodge-$\star$ commutes with the Levi-
Civita connection. We obtain more if also $U\in\ker\mathcal{D}^{*}$.

######  Proposition 2.2.

Suppose $(M^{4},g)$ is ASD, and $U$ is SD. Then $U\in\ker\mathcal{D}^{*}$ iff
$z=z(U)\in\Lambda^{2}_{+}(\mathcal{E})$ is a harmonic, self-dual two form.

###### Proof.

First note that by standard BGG theory [8, 5, 15] this is true in the
conformally flat setting. Now consider the curved case.

Certainly $\delta z$ is conformally invariant as it is a twisting of the usual
divergence of 2-forms (in dimension 4) with the conformally invariant tractor
connection. Thus, starting from the top (meaning from the left in the
filtration on the adjoint tractor bundle that is induced from (2.9)), the
first non-zero slot of $\delta z$ must be conformally invariant and
constructed from $U$, its covariant derivatives and possibly curvature
contracted with $U$. From order considerations the last of these can only
happen in the bottom slot. The very top slot is rank 2 and involves no
derivatives of $U$. Thus it is zero, as $U$ is trace-free. At the next level
we have one derivative of $U$, but it is well known that in dimension 4 there
is no such conformal invariant.

Thus the image of $\delta z$ lies in the bottom slot which contains a rank 2
tensor. Considering the knowledge of conformally flat case, it follows that
this conformally invariant object must be a multiple of $\mathcal{D}^{*}U$,
plus possibly another conformally invariant rank 2 tensor constructed by
contracting curvature into $U$. It is easily checked that it is not possible
to construct a rank 2 conformal invariant by contracting curvature into $U$,
as $U$ is SD while the Weyl curvature is, by assumption, ASD. Thus the only
possibility is that the bottom slot is a non-zero multiple of
$\mathcal{D}^{*}U$. ∎

We will give another proof of this result (by direct calculation) in the next
section; see Corollary 3.3.

######  Corollary 2.3.

(See [4], Theorem 3.10; [22], Lemma 2.1) Suppose $(M^{4},g)$ is ASD and
$U\in\ker\mathcal{D}^{*}$. Let $z$ be defined as in (2.21). Then

(2.22) $\displaystyle\Delta z=\frac{1}{3}Rz=2Jz.$

###### Proof.

Since $z$ satisfies a twisted version of the Weitzenböck formula (1.2), when
$(M^{4},g)$ is ASD the only non-zero curvature term is given by $\frac{1}{3}R$
in both versions. ∎

##  3\. Weitzenböck formula(s)

In this section we use Proposition 2.2 and Corollary 2.3 to prove a
Weitzenböck formula for $z=z(U)\in\Lambda^{2}_{+}(\mathcal{E})$ when
$U\in\ker\mathcal{D}^{*}$. We begin with some more general calculations.

######  Proposition 3.1.

Let $U\in\Gamma(\mathcal{W}^{+})$ and let $z=z(U)$ be given by (2.21). Then
the covariant derivative of $z$ is given by

(3.1) $\displaystyle\begin{split}\nabla_{m}&z_{ijAB}=\left(\delta
U\right)_{mij}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\left(-\nabla_{m}\left(\delta
U\right)^{\ell}_{\ ij}-U_{ij}^{\ \ k\ell}P_{mk}\right)\left(X_{A}Z_{\ell
B}-Z_{\ell A}X_{B}\right)\\\ &\ \ \ \ +U_{ij\ m}^{\ \
k}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+\left(\frac{1}{2}\nabla_{m}U_{ij}^{\ \
k\ell}+\left(\delta U\right)^{k}_{\
ij}\,\delta^{\ell}_{m}\right)\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right).\end{split}$

For the proof of this and the following proposition we will use the following
formulas (see (6) from [18]):

(3.2) $\displaystyle\begin{split}\nabla_{k}X_{A}&=Z_{kA},\\\
\nabla_{k}Z_{A\ell}&=-P_{k\ell}X_{A}-Y_{A}g_{k\ell},\\\
\nabla_{k}Y_{A}&=P_{k\ell}Z_{A}^{\ell}.\end{split}$

######  Lemma 3.2.

The following formulas hold:

(3.3)
$\displaystyle\nabla_{m}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)=-\left(Y_{A}Z_{mB}-Z_{mA}Y_{B}\right)+\left(P_{m}^{q}X_{A}Z_{qA}-P_{m}^{q}Z_{qA}X_{B}\right),$
(3.4)
$\displaystyle\nabla_{m}\left(X_{A}Z_{kB}-Z_{kA}Z_{B}\right)=-g_{km}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)-\left(Z_{kA}Z_{mB}-Z_{mA}Z_{kB}\right),$
(3.5)
$\displaystyle\nabla_{m}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)=P_{mk}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+P_{m}^{q}\left(Z_{qA}Z_{kB}-Z_{qB}Z_{kA}\right),$
(3.6) $\displaystyle\begin{split}\nabla_{m}\left(Z_{kA}Z_{\ell
B}-Z_{kB}Z_{\ell A}\right)&=-P_{mk}\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+P_{m\ell}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)\\\ &\ \
-g_{mk}\left(Y_{A}Z_{\ell B}-Y_{B}Z_{\ell
A}\right)+g_{m\ell}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right).\end{split}$

###### Proof.

First, by the Leibniz rule,

(3.7)
$\displaystyle\nabla_{m}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)=\left(\nabla_{m}X_{A}\right)Y_{B}+X_{A}\left(\nabla_{m}Y_{B}\right)-\left(\nabla_{m}Y_{A}\right)X_{B}-Y_{A}\left(\nabla_{m}X_{B}\right).$

By (3.2) we find

(3.8)
$\displaystyle\begin{split}\nabla_{m}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)&=Z_{mA}Y_{B}+P_{mq}X_{A}Z_{qA}-P_{mq}Z_{qA}X_{B}-Y_{A}Z_{mB}\\\
&=-\left(Y_{A}Z_{mB}-Z_{mA}Y_{B}\right)+\left(P_{mq}X_{A}Z_{qA}-P_{mq}Z_{qA}X_{B}\right).\end{split}$

Similarly,

(3.9)
$\displaystyle\begin{split}\nabla_{m}&\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)=\left(\nabla_{m}X_{A}\right)Z_{kB}+X_{A}\left(\nabla_{m}Z_{kB}\right)-\left(\nabla_{m}Z_{kA}\right)X_{B}-Z_{kA}\left(\nabla_{m}X_{B}\right)\\\
&=Z_{mA}Z_{kB}+X_{A}\left(-P_{mk}X_{B}-Y_{B}g_{mk}\right)-\left(-P_{mk}X_{A}-Y_{A}g_{mk}\right)X_{B}-Z_{kA}Z_{mB}\\\
&=-g_{km}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)-\left(Z_{kA}Z_{mB}-Z_{mA}Z_{kB}\right),\end{split}$
(3.10)
$\displaystyle\begin{split}\nabla_{m}&\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)=\left(\nabla_{m}Y_{A}\right)Z_{kB}+Y_{A}\left(\nabla_{m}Z_{kB}\right)-\left(\nabla_{m}Z_{kA}\right)Y_{B}-Z_{kA}\left(\nabla_{m}Y_{B}\right)\\\
&=P_{mq}Z_{qA}Z_{kB}+Y_{A}\left(-P_{mk}X_{B}-Y_{B}g_{mk}\right)-\left(-P_{mk}X_{A}-Y_{A}g_{mk}\right)Y_{B}-P_{mq}Z_{qB}Z_{kA}\\\
&=P_{mk}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+P_{mq}\left(Z_{qA}Z_{kB}-Z_{qB}Z_{kA}\right),\end{split}$
(3.11) $\displaystyle\begin{split}\nabla_{m}\left(Z_{kA}Z_{\ell
B}-Z_{kB}Z_{\ell A}\right)&=\left(\nabla_{m}Z_{kA}\right)Z_{\ell
B}+Z_{kA}\left(\nabla_{m}Z_{\ell
B}\right)-\left(\nabla_{m}Z_{kB}\right)Z_{\ell
A}-Z_{kB}\left(\nabla_{m}Z_{\ell A}\right)\\\
&=\left(-P_{mk}X_{A}-Y_{A}g_{mk}\right)Z_{\ell
B}+Z_{kA}\left(-P_{m\ell}X_{B}-Y_{B}g_{m\ell}\right)\\\ &\ \ \ \
-\left(-P_{mk}X_{B}-Y_{B}\delta_{mk}\right)Z_{\ell
A}-Z_{kB}\left(P_{m\ell}X_{A}-Y_{A}\delta_{m\ell}\right)\\\
&=-P_{mk}\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+P_{m\ell}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)\\\ &\ \
-g_{mk}\left(Y_{A}Z_{\ell B}-Y_{B}Z_{\ell
A}\right)+g_{m\ell}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right).\end{split}$

∎

###### Proof of Proposition 3.1.

By (2.21),

(3.12)
$\displaystyle\begin{split}\nabla_{m}z_{ijAB}&=\nabla_{m}\Big{\\{}\frac{1}{2}U_{ij}^{\
\ k\ell}\left(Z_{kA}Z_{\ell B}-Z_{\ell A}Z_{kB}\right)-(\delta U)^{k}_{\
ij}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)\Big{\\}}\\\
&=\frac{1}{2}\nabla_{m}U_{ij}^{\ \ k\ell}\left(Z_{kA}Z_{\ell B}-Z_{\ell
A}Z_{kB}\right)+\frac{1}{2}U_{ij}^{\ \ k\ell}\nabla_{m}\left(Z_{kA}Z_{\ell
B}-Z_{\ell A}Z_{kB}\right)\\\ &\ \ -\nabla_{m}\left(\delta U\right)^{k}_{\
ij}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)-\left(\delta U\right)^{k}_{\
ij}\nabla_{m}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)\\\
&=\frac{1}{2}\nabla_{m}U_{ij}^{\ \ k\ell}\left(Z_{kA}Z_{\ell B}-Z_{\ell
A}Z_{kB}\right)+\frac{1}{2}U_{ij}^{\ \
k\ell}\Big{\\{}-P_{mk}\left(X_{A}Z_{\ell B}-Z_{\ell A}X_{B}\right)\\\ &\ \
+P_{m\ell}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)-g_{mk}\left(Y_{A}Z_{\ell
B}-Y_{B}Z_{\ell
A}\right)+g_{m\ell}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)\Big{\\}}\\\ &\ \ \ \
-\nabla_{m}\left(\delta U\right)^{k}_{\
ij}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)-\left(\delta U\right)^{k}_{\
ij}\Big{\\{}-g_{km}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)\\\ &\ \ \ \ \ \ \
-\left(Z_{kA}Z_{mB}-Z_{mA}Z_{kB}\right)\Big{\\}}\\\ &=\left(\delta
U\right)_{mij}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\left(-\nabla_{m}\left(\delta
U\right)^{\ell}_{\ ij}-U_{ij}^{\ \ k\ell}P_{mk}\right)\left(X_{A}Z_{\ell
B}-Z_{\ell A}X_{B}\right)\\\ &\ \ \ \ +U_{ij\ m}^{\ \
k}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+\left(\frac{1}{2}\nabla_{m}U_{ij}^{\ \
k\ell}+\left(\delta U\right)^{k}_{\
ij}\,\delta^{\ell}_{m}\right)\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right).\end{split}$

∎

We can now give another proof of Proposition 2.2:

######  Corollary 3.3.

Suppose $(M^{4},g)$ is ASD, and $U$ is SD. Then $U\in\ker\mathcal{D}^{*}$ iff
$z=z(U)\in\Lambda^{2}_{+}(\mathcal{E})$ is a harmonic, self-dual two form.

###### Proof.

Since $z$ is self-dual, it suffices to show that $\delta z=0$. By (3.1),

(3.13) $\displaystyle\begin{split}\left(\delta
z\right)_{jAB}&=g^{im}\nabla_{m}z_{ijAB}\\\ &=g^{im}\left(\delta
U\right)_{mij}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+g^{im}\left(-\nabla_{m}\left(\delta
U\right)^{\ell}_{\ ij}-U_{ij}^{\ \ k\ell}P_{mk}\right)\left(X_{A}Z_{\ell
B}-Z_{\ell A}X_{B}\right)\\\ &\ \ \ \ +g^{im}U_{ij\ m}^{\ \
k}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+g^{im}\left(\frac{1}{2}\nabla_{m}U_{ij}^{\
\ k\ell}+\left(\delta U\right)^{k}_{\
ij}\,\delta^{\ell}_{m}\right)\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right).\end{split}$

Since $U$ is a curvature-type tensor for which all contractions vanish, it
follows that

(3.14) $\displaystyle\left(\delta
z\right)_{jAB}=-\frac{1}{2}\left(\mathcal{D}^{*}U\right)_{j\ell}\left(X_{A}Z_{\ell
B}-Z_{\ell A}X_{B}\right).$

∎

######  Proposition 3.4.

Same assumptions as Proposition 3.1. Then the rough laplacian of $z$, i.e.,
$\Delta z_{ijAB}=g^{k\ell}\nabla_{k}\nabla_{\ell}z_{ijAB}$, is given by

(3.15)
$\displaystyle\begin{split}\Delta&z_{ijAB}=\Big{\\{}\left(W^{+}\right)_{pqi}^{\
\ \ k}U^{qp}_{\ \ kj}+\left(W^{+}\right)_{pqj}^{\ \ \ k}U^{pq}_{\ \
ik}\Big{\\}}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)\\\
&+\Big{\\{}-\Delta\left(\delta U\right)^{\ell}_{\ ij}-2\nabla^{m}U_{ij}^{\ \
k\ell}P_{mk}-U_{ij}^{\ \ k\ell}\nabla_{k}J+J\left(\delta U\right)^{\ell}_{\
ij}\Big{\\}}\left(X_{A}Z_{\ell B}-Z_{\ell A}X_{B}\right)\\\
&+\Big{\\{}\frac{1}{2}\Delta U_{ij}^{\ \ k\ell}-\nabla^{k}\left(\delta
U\right)^{\ell}_{\ ij}+\nabla^{\ell}\left(\delta U\right)^{k}_{\ ij}-U_{ij}^{\
\ km}P^{\ell}_{m}+U_{ij}^{\ \ \ell
m}P^{k}_{m}\Big{\\}}\times\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right).\end{split}$

###### Proof.

By (3.1),

(3.16) $\displaystyle\begin{split}\Delta
z_{ijAB}&=\nabla^{m}\left(\nabla_{m}z_{ijAB}\right)\\\
&=\nabla^{m}\Big{\\{}\left(\delta
U\right)_{mij}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\left(-\nabla_{m}\left(\delta
U\right)^{\ell}_{\ ij}-U_{ij}^{\ \ k\ell}P_{mk}\right)\left(X_{A}Z_{\ell
B}-Z_{\ell A}X_{B}\right)\\\ &\ \ \ \ +U_{ij\ m}^{\ \
k}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+\left(\frac{1}{2}\nabla_{m}U_{ij}^{\ \
k\ell}+\left(\delta U\right)^{k}_{\
ij}\,\delta^{\ell}_{m}\right)\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right)\Big{\\}}\\\ &=\left(\nabla^{m}\left(\delta
U\right)_{mij}\right)\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\left(-\Delta\left(\delta
U\right)^{\ell}_{\ ij}-\nabla^{m}U_{ij}^{\ \ k\ell}P_{mk}-U_{ij}^{\ \
k\ell}\nabla^{m}P_{mk}\right)\\\ &\times\,\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+\left(\nabla^{m}U_{ij\ m}^{\ \
k}\right)\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+\left(\frac{1}{2}\Delta
U_{ij}^{\ \ k\ell}+\nabla^{\ell}\left(\delta U\right)^{k}_{\ ij}\right)\\\
&\times\,\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell A}\right)+\left(\delta
U\right)_{mij}\nabla^{m}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)\\\
&+\left(-\nabla_{m}\left(\delta U\right)^{\ell}_{\ ij}-U_{ij}^{\ \
k\ell}P_{mk}\right)\nabla^{m}\left(X_{A}Z_{\ell B}-Z_{\ell A}X_{B}\right)\\\
&\ \ +U_{ij\ m}^{\ \
k}\nabla^{m}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+\left(\frac{1}{2}\nabla_{m}U_{ij}^{\
\ k\ell}+\left(\delta U\right)^{k}_{\
ij}\,\delta^{\ell}_{m}\right)\nabla^{m}\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right).\end{split}$

Then applying the formulas from Lemma 3.2 and using the Bianchi identity to
write $\nabla^{m}P_{mk}=\nabla_{k}J$, we get

(3.17) $\displaystyle\begin{split}\Delta&z_{ijAB}=\left(\nabla^{m}\left(\delta
U\right)_{mij}\right)\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\left(-\Delta\left(\delta
U\right)^{\ell}_{\ ij}-\nabla^{m}U_{ij}^{\ \ k\ell}P_{mk}-U_{ij}^{\ \
k\ell}\nabla_{k}J\right)\\\ &\times\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+\left(\nabla^{m}U_{ij\ m}^{\ \
k}\right)\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)+\left(\frac{1}{2}\Delta
U_{ij}^{\ \ k\ell}+\nabla^{\ell}\left(\delta U\right)^{k}_{\ ij}\right)\\\
&\times\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell A}\right)+\left(\delta
U\right)_{mij}\Big{\\{}-\left(Y_{A}Z^{m}_{B}-Z^{m}_{A}Y_{B}\right)+\left(P^{m}_{q}X_{A}Z^{q}_{A}-P^{m}_{q}Z^{q}_{A}X_{B}\right)\Big{\\}}\\\
&+\left(-\nabla_{m}\left(\delta U\right)^{\ell}_{\ ij}-U_{ij}^{\ \
k\ell}P_{mk}\right)\Big{\\{}-\delta_{\ell}^{m}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)-\left(Z_{\ell
A}Z^{m}_{B}-Z^{m}_{A}Z_{\ell B}\right)\Big{\\}}\\\ &\ \ +U_{ij\ m}^{\ \
k}\Big{\\{}P^{m}_{k}\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+P^{m}_{q}\left(Z_{qA}Z_{kB}-Z_{qB}Z_{kA}\right)\Big{\\}}\\\
&+\left(\frac{1}{2}\nabla_{m}U_{ij}^{\ \ k\ell}+\left(\delta U\right)^{k}_{\
ij}\,\delta^{\ell}_{m}\right)\Big{\\{}-P^{m}_{k}\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+P^{m}_{\ell}\left(X_{A}Z_{kB}-Z_{kA}X_{B}\right)\\\ &\ \
-\delta^{m}_{k}\left(Y_{A}Z_{\ell B}-Y_{B}Z_{\ell
A}\right)+\delta^{m}_{\ell}\left(Y_{A}Z_{kB}-Z_{kA}Y_{B}\right)\Big{\\}}\\\
&=2\left(\nabla^{m}\left(\delta
U\right)_{mij}\right)\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\Big{\\{}-\Delta\left(\delta
U\right)^{\ell}_{\ ij}-2\nabla^{m}U_{ij}^{\ \ k\ell}P_{mk}-U_{ij}^{\ \
k\ell}\nabla_{k}J\\\ &\ +J\left(\delta U\right)^{\ell}_{\
ij}\Big{\\}}\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+\Big{\\{}\frac{1}{2}\Delta U_{ij}^{\ \
k\ell}+2\nabla^{\ell}\left(\delta U\right)^{k}_{\ ij}+U_{ij}^{\ \
mk}P^{\ell}_{m}+U_{ij}^{\ \ \ell m}P^{k}_{m}\Big{\\}}\\\
&\times\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell A}\right).\end{split}$

Since the very last term is skew-symmetric in $k,\ell$, we can rewrite this as

(3.18)
$\displaystyle\begin{split}\Delta&z_{ijAB}=2\left(\nabla^{m}\left(\delta
U\right)_{mij}\right)\left(X_{A}Y_{B}-Y_{A}X_{B}\right)+\Big{\\{}-\Delta\left(\delta
U\right)^{\ell}_{\ ij}-2\nabla^{m}U_{ij}^{\ \ k\ell}P_{mk}-U_{ij}^{\ \
k\ell}\nabla_{k}J\\\ &\ +J\left(\delta U\right)^{\ell}_{\
ij}\Big{\\}}\left(X_{A}Z_{\ell B}-Z_{\ell
A}X_{B}\right)+\Big{\\{}\frac{1}{2}\Delta U_{ij}^{\ \
k\ell}-\nabla^{k}\left(\delta U\right)^{\ell}_{\ ij}+\nabla^{\ell}\left(\delta
U\right)^{k}_{\ ij}\\\ &\ \ -U_{ij}^{\ \ km}P^{\ell}_{m}+U_{ij}^{\ \ \ell
m}P^{k}_{m}\Big{\\}}\times\left(Z_{kA}Z_{\ell B}-Z_{kB}Z_{\ell
A}\right).\end{split}$

The proposition will follow, once we prove

######  Lemma 3.5.

(3.19) $\displaystyle\nabla^{m}\left(\delta
U\right)_{mij}=\frac{1}{2}\left(W^{+}\right)_{pqi}^{\ \ \ k}U^{qp}_{\ \
kj}-\frac{1}{2}\left(W^{+}\right)_{pqj}^{\ \ \ k}U^{qp}_{\ \ ki}.$

In particular, if $(M^{4},g)$ is ASD, then

(3.20) $\displaystyle\nabla^{m}\left(\delta U\right)_{mij}=0.$

###### Proof.

By definition of the divergence and the fact that $U_{\ell mij}=-U_{m\ell
ij}$,

$\displaystyle\nabla^{m}\left(\delta U\right)_{mij}$
$\displaystyle=\nabla^{m}\nabla^{\ell}U_{\ell mij}$
$\displaystyle=g^{mp}g^{\ell q}\nabla_{p}\nabla_{q}U_{\ell mij}$
$\displaystyle=\frac{1}{2}g^{mp}g^{\ell
q}\left(\nabla_{p}\nabla_{q}-\nabla_{q}\nabla_{p}\right)U_{\ell mij}$
$\displaystyle=\frac{1}{2}g^{mp}g^{\ell q}\left(R_{pq\ell}^{\ \ \
k}U_{kmij}+R_{pqm}^{\ \ \ k}U_{\ell kij}+R_{pqi}^{\ \ \ k}U_{\ell
mkj}+R_{pqj}^{\ \ \ k}U_{\ell mik}\right)$
$\displaystyle=\frac{1}{2}\left(-R^{mk}U_{kmij}+R^{\ell k}U_{\ell
kij}\right)+\frac{1}{2}g^{mp}g^{\ell q}\left(R_{pqi}^{\ \ \ k}U_{\ell
mkj}+R_{pqj}^{\ \ \ k}U_{\ell mik}\right)$
$\displaystyle=\frac{1}{2}g^{mp}g^{\ell q}\left(R_{pqi}^{\ \ \ k}U_{\ell
mkj}+R_{pqj}^{\ \ \ k}U_{\ell mik}\right).$

Using the decomposition of the curvature tensor as in (2.2), it follows that

$\displaystyle\nabla^{m}\left(\delta U\right)_{mij}$
$\displaystyle=\frac{1}{2}g^{mp}g^{\ell q}\left(W_{pqi}^{\ \ \
k}+g_{ip}P_{q}^{k}-\delta_{p}^{k}P_{iq}-g_{iq}P_{p}^{k}+\delta_{q}^{k}P_{ip}\right)U_{\ell
mkj}$ $\displaystyle\ \ \ \ +\frac{1}{2}g^{mp}g^{\ell q}\left(W_{pqj}^{\ \ \
k}+g_{jp}P_{q}^{k}-\delta_{p}^{k}P_{jq}-g_{jq}P_{p}^{k}+\delta_{q}^{k}P_{jp}\right)U_{\ell
mik}$ $\displaystyle=\frac{1}{2}W_{pqi}^{\ \ \ k}U^{qp}_{\ \
kj}+\frac{1}{2}\left(\delta_{i}^{m}P^{k\ell}-g^{km}P_{i}^{\ell}-\delta_{i}^{\ell}P^{km}+g^{k\ell}P_{i}^{m}\right)U_{\ell
mkj}$ $\displaystyle\ \ \ \ +\frac{1}{2}W_{pqj}^{\ \ \ k}U^{qp}_{\ \
ik}+\frac{1}{2}\left(\delta_{j}^{m}P^{k\ell}-g^{km}P_{j}^{\ell}-\delta_{j}^{\ell}P^{km}+g^{k\ell}P_{j}^{m}\right)U_{\ell
mik}.$

Using the fact that all contractions of $U$ vanish, the above simplifies to

$\displaystyle\nabla^{m}\left(\delta U\right)_{mij}$
$\displaystyle=\frac{1}{2}W_{pqi}^{\ \ \ k}U^{qp}_{\ \
kj}+\frac{1}{2}W_{pqj}^{\ \ \ k}U^{qp}_{\ \
ik}+\frac{1}{2}\left(P^{k\ell}U_{\ell ikj}-P^{km}U_{imkj}+P^{k\ell}U_{\ell
jik}-P^{km}U_{jmik}\right)$ $\displaystyle=\frac{1}{2}W_{pqi}^{\ \ \
k}U^{qp}_{\ \ kj}+\frac{1}{2}W_{pqj}^{\ \ \ k}U^{qp}_{\ \ ik}$
$\displaystyle=\frac{1}{2}W_{pqi}^{\ \ \ k}U^{qp}_{\ \
kj}-\frac{1}{2}W_{pqj}^{\ \ \ k}U^{qp}_{\ \ ki}$

(note that all terms involving the Schouten tensor can be seen to cancel after
re-indexing). Finally, since $U\in\Gamma(\mathcal{W}^{+})$, (3.19) follows. ∎

∎

######  Theorem 3.6.

If $(M^{4},g)$ is ASD and $U\in\ker\mathcal{D}^{*}$, then

(3.21) $\displaystyle\Delta\left(\delta U\right)_{\ell
ij}=-2\nabla^{m}U_{ijk\ell}P^{k}_{m}-U_{ijk\ell}\nabla^{k}J+3J\left(\delta
U\right)_{\ell ij},$ (3.22) $\displaystyle\frac{1}{2}\Delta
U_{ijk\ell}=\nabla_{k}\left(\delta U\right)_{\ell
ij}-\nabla_{\ell}\left(\delta U\right)_{kij}+U_{ijkm}P_{\ell}^{m}-U_{ij\ell
m}P^{m}_{k}+JU_{ijk\ell}.$

###### Proof.

If $U\in\ker\mathcal{D}^{*}$ then $z\in\Lambda^{2}_{+}(\mathcal{E})$ is
harmonic. Therefore, (3.21) and (3.22) follow from Proposition 3.4 and
Corollary 2.3. ∎

######  Remark 3.7.

Although we will not provide a proof, it is not difficult to show that (3.22)
holds for any section $U\in\Gamma\left(\mathcal{W}^{+}\right)$; i.e., the
condition $U\in\ker\mathcal{D}^{*}$ is not necessary. However, this is not the
case for (3.21).

##  4\. Proof of Theorem 1.5

Let $Q_{g}$ denote the $Q$-curvature of $(M^{4},g)$:

(4.1)
$\displaystyle\begin{split}Q_{g}&=\frac{1}{12}\left(-\Delta_{g}R+R_{g}^{2}-3|Ric_{g}|^{2}\right)\\\
&=-\frac{1}{2}\Delta_{g}J+J^{2}-|P|^{2}.\end{split}$

The total $Q$-curvature is a conformal invariant, and we can rewrite (1.10) as
a condition on the total $Q$-curvature and the Yamabe invariant as follows. By
the Chern-Gauss-Bonnet formula,

(4.2) $\displaystyle\int
Q_{g}\,dv_{g}=4\pi^{2}\chi(M^{4})-\frac{1}{8}\int|W_{g}|^{2}\,dv_{g}.$

Since $g$ is ASD, (4.2) becomes

(4.3) $\displaystyle\int
Q_{g}\,dv_{g}=4\pi^{2}\chi(M^{4})-\frac{1}{8}\int|W^{-}_{g}|^{2}\,dv_{g}.$

By the Hirzebruch signature formula,

(4.4) $\displaystyle
48\pi^{2}\tau(M^{4})=\int\left(|W^{+}_{g}|^{2}-|W^{-}_{g}|^{2}\right)\,dv_{g}=-\int|W^{-}_{g}|^{2}\,dv_{g}.$

Combining this with (4.3) we see that

(4.5) $\displaystyle\int
Q_{g}\,dv_{g}=2\pi^{2}\left(2\chi(M^{4})+3\tau(M^{4})\right).$

Therefore, the assumption (1.10) is equivalent to

(4.6) $\displaystyle\int Q_{g}\,dv_{g}\geq-\frac{1}{12}Y(M^{4},[g])^{2}.$

###### The proof of Theorem 1.5.

Since our assumptions are conformally invariant, it suffices to prove the
result when $g$ is replaced by any metric in the same conformal class. The
metric we will use is a solution of a variational problem related to the
regularized determinant of an elliptic operator. The precise formulation is
contained in Theorem 5.2 in the Appendix, while the following consequence will
suffice for our purposes:

######  Theorem 4.1.

Let $(M^{4},g_{0})$ be a closed Riemannian four-manifold with

$(i)$ $Y(M^{4},[g_{0}])>0$,

$(ii)$
$\int_{M^{4}}Q_{g_{0}}\,dv_{g_{0}}\geq-\frac{1}{12}Y(M^{4},[g_{0}])^{2}$.

Then there is a smooth, unit volume conformal metric $g\in[g_{0}]$ with
$J_{g}=\operatorname{tr}P_{g}>0$ satisfying

(4.7) $\displaystyle\Delta
J_{g}\leq-|\mathring{P}_{g}|^{2}+\frac{15}{4}J_{g}^{2},$

where $\mathring{P}_{g}=P-\frac{1}{4}Jg$ is the trace-free Schouten tensor.

The proof of this result is somewhat involved, and will also be given in the
Appendix. In the following we will show how Theorem 1.5 follows from Theroem
4.1 and the Weitzenböck formulas of the preceding section. For the rest of the
proof the metric $g$ is assumed to satisfy the conclusions of Theorem 4.1, and
we will usually suppress the subscript $g$.

Assume $U\in\ker\mathcal{D}^{*}$. We will record two integral identities that
follow from Theorem 3.6 and the inequality (4.7). First, by (3.22),

(4.8) $\displaystyle\begin{split}\frac{1}{2}\Delta|U|^{2}&=\langle U,\Delta
U\rangle+|\nabla U|^{2}\\\ &=U^{ijk\ell}\Delta U_{ijk\ell}+|\nabla U|^{2}\\\
&=U^{ijk\ell}\Big{\\{}2\nabla_{k}\left(\delta U\right)_{\ell
ij}-2\nabla_{\ell}\left(\delta U\right)_{kij}+2U_{ijkm}P_{\ell}^{m}-2U_{ij\ell
m}P^{m}_{k}+2JU_{ijk\ell}\Big{\\}}+|\nabla U|^{2}\\\
&=4U^{ijk\ell}\nabla_{k}\left(\delta U\right)_{\ell
ij}+4U^{ijk\ell}\,U_{ijkm}P_{\ell}^{m}+2J|U|^{2}+|\nabla U|^{2}.\end{split}$

Recalling that

$\displaystyle U^{ijk\ell}U_{ijkm}=\frac{1}{4}|U|^{2}\delta^{\ell}_{m},$

(4.8) implies

(4.9)
$\displaystyle\frac{1}{2}\Delta|U|^{2}=4U^{ijk\ell}\nabla_{k}\left(\delta
U\right)_{\ell ij}+3J|U|^{2}+|\nabla U|^{2}.$

If we multiply both sides by $J$ and integrate over $M^{4}$, then

(4.10)
$\displaystyle\frac{1}{2}\int_{M^{4}}J\Delta|U|^{2}\,dv=\int_{M^{4}}\big{\\{}4JU^{ijk\ell}\nabla_{k}\left(\delta
U\right)_{\ell ij}+3J^{2}|U|^{2}+J|\nabla U|^{2}\big{\\}}\,dv.$

Integrating by parts on the left and using (4.7), we find

(4.11)
$\displaystyle\begin{split}\frac{1}{2}\int_{M^{4}}J\Delta|U|^{2}\,dv&=\frac{1}{2}\int_{M^{4}}\left(\Delta
J\right)|U|^{2}\,dv\leq\int_{M^{4}}\left(-\frac{1}{2}|\mathring{P}|^{2}+\frac{15}{8}J^{2}\right)|U|^{2}\,dv.\end{split}$

Combining (4.9) and (4.10),

(4.12) $\displaystyle
0\geq\int_{M^{4}}\big{\\{}4JU^{ijk\ell}\nabla_{k}\left(\delta U\right)_{\ell
ij}+\frac{1}{2}|\mathring{P}|^{2}|U|^{2}+\frac{9}{8}J^{2}|U|^{2}+J|\nabla
U|^{2}\big{\\}}\,dv.$

We now use (3.21):

(4.13) $\displaystyle\begin{split}\frac{1}{2}\Delta|\delta
U|^{2}&=|\nabla\delta U|^{2}+\langle\delta U,\Delta\left(\delta
U\right)\rangle\\\ &=|\nabla\delta U|^{2}+\left(\delta U\right)^{\ell
ij}\Delta\left(\delta U\right)_{\ell ij}\\\ &=|\nabla\delta
U|^{2}+\left(\delta U\right)^{\ell
ij}\Big{\\{}-2\nabla^{m}U_{ijk\ell}P^{k}_{m}-U_{ijk\ell}\nabla^{k}J+3J\left(\delta
U\right)_{\ell ij}\Big{\\}}\\\ &=|\nabla\delta U|^{2}-2\left(\delta
U\right)^{\ell ij}\nabla^{m}U_{ijk\ell}P_{m}^{k}-\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\nabla^{k}J+3J|\delta U|^{2}.\end{split}$

Integrating this over $M^{4}$ gives

(4.14) $\displaystyle 0=\int_{M^{4}}\Big{\\{}|\nabla\delta
U|^{2}-2\left(\delta U\right)^{\ell
ij}\nabla^{m}U_{ijk\ell}P_{m}^{k}-\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\nabla^{k}J+3J|\delta U|^{2}\Big{\\}}\,dv.$

If we integrate by parts in the second term and use the contracted second
Bianchi identity $\nabla^{m}P_{m}^{k}=\nabla^{k}J$, then

(4.15) $\displaystyle\begin{split}\int_{M^{4}}-2\left(\delta U\right)^{\ell
ij}\nabla^{m}U_{ijk\ell}P_{m}^{k}\,dv_{g}&=\int_{M^{4}}\Big{\\{}2\nabla^{m}\left(\delta
U\right)^{\ell ij}U_{ijk\ell}P_{m}^{k}+2\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\nabla^{m}P_{m}^{k}\Big{\\}}\,dv_{g}\\\
&=\int_{M^{4}}\Big{\\{}2\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}P_{m}^{k}+2\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\nabla^{k}J\Big{\\}}\,dv_{g}.\end{split}$

Substituting this result back into (4.14) gives

(4.16) $\displaystyle 0=\int_{M^{4}}\Big{\\{}|\nabla\delta
U|^{2}+2\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}P_{m}^{k}+U_{ijk\ell}\left(\delta U\right)^{\ell
ij}\nabla^{k}J+3J|\delta U|^{2}\Big{\\}}\,dv.$

Next, integrate by parts in the second-to-last term in (4.16):

(4.17) $\displaystyle\begin{split}\int_{M^{4}}U_{ijk\ell}\left(\delta
U\right)^{\ell
ij}\nabla^{k}J\,dv&=\int_{M^{4}}\Big{\\{}-U_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}J-\nabla^{k}U_{ijk\ell}\left(\delta U\right)^{\ell
ij}J\Big{\\}}\,dv\\\ &=\int_{M^{4}}\Big{\\{}-U_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}J-J|\delta U|^{2}\Big{\\}}\,dv.\end{split}$

Substituting this back into (4.16) gives

(4.18) $\displaystyle 0=\int_{M^{4}}\Big{\\{}|\nabla\delta
U|^{2}+2\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}P_{m}^{k}-JU_{ijk\ell}\nabla^{k}\left(\delta U\right)^{\ell
ij}+2J|\delta U|^{2}\Big{\\}}\,dv.$

Finally, we rewrite the curvature terms above in terms of $\mathring{P}$ and
$J$:

(4.19) $\displaystyle 0=\int_{M^{4}}\Big{\\{}|\nabla\delta
U|^{2}+2\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\mathring{P}_{mk}-\frac{1}{2}JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}+2J|\delta U|^{2}\Big{\\}}\,dv.$

Multiplying by two and rearranging terms, we find

(4.20) $\displaystyle\int_{M^{4}}JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}\,dv=\int_{M^{4}}\Big{\\{}2|\nabla\delta
U|^{2}+4\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\mathring{P}_{mk}+4J|\delta U|^{2}\Big{\\}}\,dv.$

We want to combine (4.20) with (4.12). To do so, we first use (4.20) to write

(4.21)
$\displaystyle\begin{split}\int_{M^{4}}&4JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}\,dv=3\int_{M^{4}}JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}\,dv+\int_{M^{4}}JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}\,dv\\\ &=\int_{M^{4}}3JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}\,dv+\int_{M^{4}}\Big{\\{}2|\nabla\delta
U|^{2}+4\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\mathring{P}_{m}^{k}+4J|\delta U|^{2}\Big{\\}}\,dv\\\
&=\int_{M^{4}}\Big{\\{}2|\nabla\delta U|^{2}+4\nabla^{m}\left(\delta
U\right)^{\ell
ij}U_{ijk\ell}\mathring{P}_{m}^{k}+3JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}+4J|\delta U|^{2}\Big{\\}}\,dv\end{split}$

Substituting this into (4.12), we have

(4.22) $\displaystyle\begin{split}0&\geq\int_{M^{4}}\Big{\\{}2|\nabla\delta
U|^{2}+4\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\mathring{P}_{m}^{k}+3JU_{ijk\ell}\nabla^{k}\left(\delta
U\right)^{\ell ij}+4J|\delta U|^{2}\\\ &\ \ \ \ \ \ \ \ \
+\frac{1}{2}|\mathring{P}|^{2}|U|^{2}+\frac{9}{8}J^{2}|U|^{2}+J|\nabla
U|^{2}\Big{\\}}\,dv\\\ &=\int_{M^{4}}\Big{\\{}2|\nabla\delta
U|^{2}+4T_{m}^{k}\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}+4J|\delta
U|^{2}+\frac{1}{2}|\mathring{P}|^{2}|U|^{2}+\frac{9}{8}J^{2}|U|^{2}+J|\nabla
U|^{2}\Big{\\}}\,dv,\end{split}$

where

(4.23) $\displaystyle T_{km}=\mathring{P}_{mk}+\frac{3}{4}Jg_{km}.$

If we define the tensor

(4.24) $\displaystyle V_{m\ell ij}=T_{m}^{k}U_{ijk\ell},$

then the term involving $T$ in (4.22) can be estimated (via Cauchy-Schwarz) by

$\displaystyle\left|4T_{m}^{k}\nabla^{m}\left(\delta U\right)^{\ell
ij}U_{ijk\ell}\right|$ $\displaystyle=4\left|\nabla^{m}\left(\delta
U\right)^{\ell ij}V_{m\ell ij}\right|$ $\displaystyle\leq 4\left|\nabla\delta
U\right|\left|V\right|.$

By the arithmetic-geometric mean inequality,

(4.25) $\displaystyle\left|4T_{km}\nabla_{m}\left(\delta U\right)_{\ell
ij}U_{ijk\ell}\right|\leq 2\left|\nabla\delta
U\right|^{2}+2\left|V\right|^{2}.$

By the definition of $V$,

$\displaystyle\left|V\right|^{2}$ $\displaystyle=V^{m\ell ij}V_{m\ell ij}$
$\displaystyle=T_{p}^{m}U^{ijp\ell}T_{m}^{k}U_{ijk\ell}$
$\displaystyle=T_{p}^{m}T_{m}^{k}U^{ij\ell p}U_{ij\ell k}$
$\displaystyle=T_{p}^{m}T_{m}^{k}\left(\frac{1}{4}|U|^{2}\delta_{k}^{p}\right)$
$\displaystyle=\frac{1}{4}|T|^{2}|U|^{2}$
$\displaystyle=\frac{1}{4}\left(|\mathring{P}|^{2}+\frac{9}{4}J^{2}\right)|U|^{2}.$

Consequently, (4.25) implies

$\displaystyle\left|4T_{km}\nabla_{m}\left(\delta U\right)_{\ell
ij}U_{ijk\ell}\right|\leq 2\left|\nabla\delta
U\right|^{2}+\frac{1}{2}|\mathring{P}|^{2}|U|^{2}+\frac{9}{8}J^{2}|U|^{2},$

hence

(4.26) $\displaystyle 4T_{km}\nabla_{m}\left(\delta U\right)_{\ell
ij}U_{ijk\ell}\geq-2\left|\nabla\delta
U\right|^{2}-\frac{1}{2}|\mathring{P}|^{2}|U|^{2}-\frac{9}{8}J^{2}|U|^{2}.$

Substituting this into (4.22), we conclude

(4.27) $\displaystyle 0\geq\int_{M^{4}}\Bigg{\\{}4J|\delta U|^{2}+J|\nabla
U|^{2}\Bigg{\\}}\,dv.$

Since $J>0$, it follows that $\nabla U=0$. However, it is then immediate from
(4.12) that $U\equiv 0$. ∎

##  5\. Appendix: The proof of Theorem 4.1

In this Appendix we give the proof of Theorem 4.1. The material in this
section is an extension of the existence work of Chang-Yang [10] for critical
points of the regularized determinant of conformally covariant operators. We
begin with a brief overview of their work, omitting the motivation from
spectral geometry and limiting ourselves to the underlying variational
problem.

Let $(M^{4},g)$ be a closed, four-dimensional Riemannian manifold, and
$W^{2,2}(M)$ the Sobolev space of functions whose weak derivatives up to order
two are in $L^{2}$. Consider the following functionals on $W^{2,2}(M)$:

(5.1) $\displaystyle I^{\pm}[w]=4\int
w|W^{\pm}_{g}|^{2}\,dv_{g}-\big{(}\int|W^{\pm}_{g}|^{2}\,dv_{g}\big{)}\log\fint
e^{4w}\,dv_{g},$ (5.2) $\displaystyle II[w]=\int wP_{g}w\,dv_{g}+4\int
Q_{g}w\,dv_{g}-\big{(}\int Q_{g}\,dv_{g}\big{)}\log\fint e^{4w}\,dv_{g},$
(5.3) $\displaystyle III[w]=12\int(\Delta w+|\nabla
w|^{2})^{2}\,dv_{g}-4\int(w\Delta R_{g}+R_{g}|\nabla w|^{2})\,dv_{g},$

where $P_{g}$ denotes the Paneitz operator, $Q_{g}$ the scalar curvature, and
$\fint$ denotes the normalized integral (i.e., divided by the volume of $g$).
Let $\gamma_{1}^{\pm},\gamma_{2},\gamma_{3}$ be constants and define
$F:W^{2,2}(M)\rightarrow\mathbb{R}$ be given by

(5.4) $\displaystyle
F[w]=\gamma_{1}^{+}I^{+}[w]+\gamma_{1}^{-}I^{-}[w]+\gamma_{2}II[w]+\gamma_{3}III[w].$

We also define the associated conformal invariant:

(5.5)
$\displaystyle\kappa_{g}=-\gamma_{1}^{+}\int|W^{+}|^{2}\,dv_{g}-\gamma_{1}^{-}\int|W^{-}|^{2}\,dv_{g}-\gamma_{2}\int
Q_{g}\,dv_{g}.$

As explained in [10], critical points of $F$ determine a conformal metric
satisfying a fourth order curvature condition. More precisely, if we define
the $U$-curvature of $g$ by

(5.6) $\displaystyle
U=U(g)=\gamma_{1}^{+}|W^{+}_{g}|^{2}+\gamma_{1}^{-}|W^{-}_{g}|^{2}+\gamma_{2}Q_{g}-\gamma_{3}\Delta_{g}R_{g},$

then $w$ is a smooth critical point of $F$ if and only if the conformal metric
$g_{F}=e^{2w}g$ satisfies

(5.7) $\displaystyle U(g_{F})\equiv\mu$

for some constant $\mu$. A general existence result for critical points of $F$
was proved in [10]:

######  Theorem 5.1.

(See [10], Theorem 1.1; also [20], Corollary 1.1) Assume:

$(i)$ $\gamma_{2}<0$ and $\gamma_{3}<0$,

$(ii)$ $\kappa_{g}<(-\gamma_{2})8\pi^{2}$.

Then $\sup_{w\in W^{2,2}(M)}F[w]$ is attained by some $w\in W^{2,2}$.

Regularity of extremals was proved by the second author in joint work with
Chang-Yang [9]; later Uhlenbeck-Viaclovsky proved a more general regularity
result for arbitrary critical points of $F$ (see [35]).

To prove Theorem 4.1 we need to introduce another functional:

(5.8) $\displaystyle IV[w]=\dfrac{\int\left(R_{g}+6|\nabla
w|^{2}\right)e^{2w}\,dv_{g}}{\left(\int e^{4w}\,dv_{g}\right)^{1/2}}.$

This is just the Yamabe functional, written in a slightly non-standard form:
if $\widetilde{g}=e^{2w}g$, then

(5.9) $\displaystyle IV[w]=\dfrac{\int
R_{\widetilde{g}}\,dv_{\widetilde{g}}}{\mbox{Vol}(\widetilde{g})^{1/2}}.$

Given a constant $\gamma_{4}$, we define
$\Phi:W^{2,2}(M)\rightarrow\mathbb{R}$ by

(5.10)
$\displaystyle\Phi[w]=\gamma_{1}^{+}I^{+}[w]+\gamma_{1}^{-}I^{-}[w]+\gamma_{2}II[w]+\gamma_{3}III[w]+\gamma_{4}IV[w].$

A trivial modification of the existence result of Chang-Yang and the
regularity results of Chang-Gursky-Yang and Uhenbeck-Viaclovsky gives the
following:

######  Theorem 5.2.

Assume:

$(i)$ $\gamma_{2}<0,\gamma_{3}<0,\gamma_{4}\leq 0$.

$(ii)$ $\kappa_{g}<(-\gamma_{2})8\pi^{2}$.

Then $\sup_{w\in W^{2,2}(M)}\Phi[w]$ is attained by some $w\in W^{2,2}(M)$.
Moreover, $w\in C^{\infty}$, and the conformal metric $\widetilde{g}=e^{2w}g$
satisfies

(5.11)
$\displaystyle\gamma_{1}^{+}|W^{+}_{\widetilde{g}}|^{2}+\gamma_{1}^{-}|W^{-}_{\widetilde{g}}|^{2}+\gamma_{2}Q_{\widetilde{g}}-\gamma_{3}\Delta_{\widetilde{g}}R_{\widetilde{g}}+\frac{1}{2}\gamma_{4}R_{\widetilde{g}}=\mu,$

for some constant $\mu$.

###### Proof.

Note that the functional $\Phi$ is scale-invariant:

$\displaystyle\Phi[w+c]=\Phi[w]$

for any constant $c$. Therefore, we may normalize a maximizing sequence
$\\{w_{k}\\}$ for $\Phi$ so that

$\displaystyle\int e^{4w_{k}}\,dv_{g}=1.$

Assuming $R_{g}\geq-C$, it follows from the Schwartz inequality that

$\displaystyle IV[w]$ $\displaystyle=\int\left(R_{g}+6|\nabla
w_{k}|^{2}\right)e^{2w_{k}}\,dv_{g}$ $\displaystyle\geq\int
R_{g}e^{2w_{k}}\,dv_{g}$ $\displaystyle\geq-C\left(\int
e^{4w_{k}}\,dv_{g}\right)^{1/2}$ $\displaystyle\geq-C.$

Consequently, if $\gamma_{4}\leq 0$ then $\gamma_{4}IV[w]$ is bounded above.
By (5.9), it is also bounded below (by the Yamabe invariant). Therefore, the
addition of this term has no effect on the estimates in the existence proof of
Chang-Yang. ∎

We are now ready to prove Theorem 4.1:

###### The proof of Theorem 4.1.

We first remark that if $(M^{4},g_{0})$ is conformally equivalent to the round
sphere (suitably normalized), then $g=g_{c}$ satisfies the conclusions of the
Theorem. Therefore, we may assume $(M^{4},g_{0})$ is not conformally the round
sphere.

Taking

$\displaystyle\gamma_{1}^{\pm}$ $\displaystyle=0,$ $\displaystyle\gamma_{2}$
$\displaystyle=-6,$ $\displaystyle\gamma_{3}$ $\displaystyle=-\frac{1}{2},$
$\displaystyle\gamma_{4}$ $\displaystyle=-2Y(M^{4},[g_{0}]),$

then

$\displaystyle\kappa_{g}=6\int Q_{g_{0}}\,dv_{g_{0}}.$

To use Theorem 5.2 we need to verify assumption $(ii)$; i.e.,

(5.12) $\displaystyle\int Q_{g_{0}}\,dv_{g_{0}}<8\pi^{2}.$

By Theorem B of [21], (5.12) holds as long as $(M^{4},g_{0})$ is not
conformally equivalent to the round sphere. Therefore, by Theorem 5.2 there is
a smooth conformal metric $g=e^{2w}g_{0}$ (which we can normalize to have unit
volume) satisfying

(5.13)
$\displaystyle-6Q_{g}+\frac{1}{2}\Delta_{g}R_{g}-Y(M^{4},[g_{0}])R_{g}=\mu.$

For the rest of the proof we will omit the subscript $g$. If
$E=Ric-\frac{1}{4}Rg$ denotes the trace-free Ricci tensor of $g$, then we may
use the definition of the $Q$-curvature to rewrite (5.13) as

(5.14) $\displaystyle\Delta
R=\frac{1}{8}R^{2}-\frac{3}{2}|E|^{2}+Y(M^{4},[g_{0}])\,R+\mu.$

By the arithmetic-geometric mean inequality,

$\displaystyle
Y(M^{4},[g_{0}])\,R\leq\frac{1}{2}R^{2}+\frac{1}{2}Y(M^{4},[g_{0}])^{2}.$

Therefore,

(5.15) $\displaystyle\Delta
R\leq\frac{5}{8}R^{2}-\frac{3}{2}|E|^{2}+\frac{1}{2}Y(M^{4},[g_{0}])^{2}+\mu.$

######  Claim 5.3.

(5.16) $\displaystyle\mu+\frac{1}{2}Y(M^{4},[g_{0}])^{2}\leq 0.$

For now let us assume the claim and see how the theorem follows.

From (5.16) and (5.15) it follows that

(5.17) $\displaystyle\Delta R\leq\frac{5}{8}R^{2}-\frac{3}{2}|E|^{2}.$

Using the fact that $J=\frac{1}{6}R$ and $\mathring{P}=\frac{1}{2}E$, this
inequality can also be written

$\displaystyle\Delta J\leq-|\mathring{P}|^{2}+\frac{15}{4}J^{2},$

hence (4.7) holds.

To see that $R>0$, we use (5.14) and the fact that $\mu<0$ to write

(5.18) $\displaystyle\begin{split}\Delta
R&\leq\frac{1}{8}R^{2}+Y(M^{4},[g_{0}])\,R\\\
&\leq\frac{1}{6}R^{2}+Y(M^{4},[g_{0}])\,R.\end{split}$

Let $\phi>0$ denote the eigenfunction associated to the first eigenvalue
$\lambda_{1}(L)$ of the conformal laplacian:

(5.19) $\displaystyle L\phi:=\left(-6\Delta+R\right)\phi=\lambda_{1}(L).$

Since $Y(M^{4},[g])>0$, it follows that $\lambda_{1}(L)>0$. An easy
calculation using (5.18) and (5.19) gives

$\displaystyle\Delta\dfrac{R}{\phi}\leq-2\langle\nabla\left(\frac{R}{\phi}\right),\frac{\nabla\phi}{\phi}\rangle+\left(Y\left(M^{4},[g_{0}]\right)+\frac{1}{6}\lambda_{1}(L)\right)\frac{R}{\phi}.$

It follows from the strong maximum principle that $R/\phi>0$ on $M$, hence
$R>0$. This completes the proof of the theorem, once we prove Claim 5.3.

###### Proof of Claim 5.3.

If integrate (5.13) over $M$ and use the fact that $g$ has unit volume, we
obtain

(5.20) $\displaystyle\mu=-6\int Q_{g}\,dv_{g}-Y(M^{4},[g_{0}])\int
R_{g}\,dv_{g}.$

By definition of the Yamabe invariant (again using the fact that $g$ has unit
volume) and the fact that $Y(M^{4},[g_{0}])>0$,

$\displaystyle Y(M^{4},[g_{0}])\int R_{g}\,dv_{g}\geq Y(M^{4},[g_{0}])^{2}.$

Therefore, by (5.20),

(5.21) $\displaystyle\mu\leq-6\int Q_{g}\,dv_{g}-Y(M^{4},[g_{0}])^{2}.$

Since the total $Q$-curvature is a conformal invariant, using assumption
$(ii)$ of the theorem we see that

$\displaystyle\mu$ $\displaystyle\leq-6\int
Q_{g}\,dv_{g}-Y(M^{4},[g_{0}])^{2}$ $\displaystyle=-6\int
Q_{g_{0}}\,dv_{g_{0}}-Y(M^{4},[g_{0}])^{2}$
$\displaystyle\leq\frac{1}{2}Y(M^{4},[g_{0}])^{2}-Y(M^{4},[g_{0}])^{2}$
$\displaystyle=-\frac{1}{2}Y(M^{4},[g_{0}])^{2},$

which proves (5.16). ∎

∎

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