11footnotetext: Department of Mathematics, Clark University, 950 Main Street,
Worcester, MA 01610, USA. E-mail address<EMAIL_ADDRESS>Dipartimento di Matematica, Università di Bologna. E-mail address:
<EMAIL_ADDRESS>
# Compactness of Dirac-Einstein spin manifolds and horizontal deformations.

Ali Maalaoui(1) & Vittorio Martino(2)

Abstract In this paper we consider the Hilbert-Einstein-Dirac functional,
whose critical points are pairs, metrics-spinors, that satisfy a system
coupling the Riemannian and the spinorial part. Under some assumptions, on the
sign of the scalar curvature and the diameter, we prove a compactness result
for this class of pairs, in dimension three and four. This can be seen as the
equivalent of the study of compactness of sequences of Einstein manifolds as
in [2, 28]. Indeed, we study the compactness of sequences of critical points
of the Hilbert-Einstein-Dirac functional which is an extension of the Hilbert-
Einstein functional having Einstein manifolds as critical points. Moreover we
will study the second variation of the energy, characterizing the horizontal
deformations for which the second variation vanishes. Finally we will exhibit
some explicit examples.

Keywords: Einstein-Dirac, Hausdorff Compactness, Second Variation

2010 MSC. Primary: 53C21, 53C23. Secondary: 53C27, 58E30

##  1 Introduction and main results

Let $M$ be a closed (compact, without boundary) manifold of dimension $n$. We
recall the energy functional $E$ that couples the gravity with fermionic
interactions:

$E(g,\psi)=\int_{M}R_{g}+\langle
D_{g}\psi,\psi\rangle-\lambda|\psi|^{2}dv_{g}\;,\qquad\lambda\in\mathbb{R}$
(1)

where $g$ is a Riemannian metric on $M$, $\psi$ is a spinor in the spin bundle
$\Sigma M$ on $M$, $R_{g}$ is the scalar curvature, $D_{g}$ is the Dirac
operator and $\langle\cdot,\cdot\rangle$ is the compatible Hermitian metric on
$\Sigma M$; we will give the precise definitions in the next section.
The functional $E$ generalizes the classical Hilbert-Einstein functional and
it is invariant under the group of diffeomorphisms of $M$ as well; we address
the reader to [8, 15, 20] for a detailed exposition on this topic. In
particular, when restricted to a fixed conformal class, this functional gives
rise to equations of Yamabe type, see for instance [16], [24], [25], [26],
[27] and the references therein. We point out here the work in [1], where a
different energy functional coupling spinors and metric was introduced and
studied; its critical points are very restrictive since the spinor component
is either a parallel spinor or a Killing spinor (under the constraint). In
particular, this shrinks the set of critical points and provides flexibility
in the study of their deformations.
In this paper we will consider the case with $\lambda>0$, which will imply
that the scalar curvature is positive, as we will see. The critical points of
this functional are solutions of the system (see next section)

$\left\\{\begin{array}[]{ll}Ric_{g}-\displaystyle\frac{R_{g}}{2}g=T^{g,\psi}\\\
\\\ D_{g}\psi=\lambda\psi\end{array}\right.$ (2)

where $T^{g,\psi}$ denotes the energy-momentum tensor

$T^{g,\psi}(X,Y)=-\frac{1}{4}\langle
X\cdot\nabla_{Y}\psi+Y\cdot\nabla_{X}\psi,\psi\rangle\;,$ (3)

here $\cdot$ and $\nabla$ denote the Clifford multiplication and the metric
connection extended to the spin bundle $\Sigma M$.
A non-trivial spinor $\psi$ is called an Einstein spinor for the eigenvalue
$\lambda$ if it is a solution of the previous system; in addition if $n\geq
3$, by contracting the first equation in (2), it holds

$R_{g}=\frac{\lambda}{n-2}|\psi|^{2}_{g}\;.$ (4)

We will consider the set

$\mathcal{E}(D,c,K)=\\{(g,\psi)\in Crit_{c}(E);\quad diam(M,g)\leq
D,-\Delta_{g}R_{g}\geq-KR_{g}\\}\;,$ (5)

where the constant $c$ introduced in the previous definition is meant to
prevent the collapsing of the manifold (see formula (21) in the sequel). The
constant $K>0$ will be useful for bounding the gradient of the spinor $\psi$.
Besides their technical use, we recall that in the literature regarding the
study of compactness of sets of Riemannian manifolds, several assumptions were
considered. For instance, in the classical theory of Gromov-Cheeger, a bound
on the curvature tensor is necessary and geometrically meaningful (since
without such a bound, conical singularities can form where the curvature blows
up), as well as a bound on the diameter an a lower bound on the volume in
order to avoid collapsing. These type of assumptions were weakened in several
situations, one can check for instance [2, 3, 12]. In our setting, the bound
$c$, on the energy, is natural from a variational point of view, in the spirit
of studying a moduli space of metrics. The diameter bound is again
geometrically natural to avoid a scaling non-compactness. The bound $K$ on the
Laplacian of the curvature is actually technical. Notice that if one assumes
that the metrics are of constant scalar curvature, then we can drop the
dependence on $K$. In fact, one can replace this ”analytic” constraint by a
geometric one by assuming that the $Q$-curvature is bounded from below.
Indeed, the $Q$-curvature $Q$ has the following formula:

$Q=-\frac{1}{2(n-1)}\Delta_{g}R_{g}-\frac{2}{(n-2)^{2}}|Ric|^{2}+\frac{n^{3}-4n^{2}+16n-16}{8(n-1)^{2}(n-2)^{2}}R_{g}^{2}.$

Hence, if $Q$ is bounded from below, then $-\Delta_{g}R_{g}$ is also bounded
below in terms of $R^{2}_{g}$ and this is enough for our analysis. In fact, if
$(g,\psi)\in Crit_{c}(E)$, then the lower bound on the $Q$-curvature is
equivalent to our condition $-\Delta_{g}R_{g}\geq-KR_{g}$.

We want to point out that the study of compactness and convergence of
manifolds with underlying spinors was investigated in [19, 23], also some
cases of collapsing along the limit were studied in [30, 31]. In our case, the
functional provides more control on the spinorial component and this allow us
to keep track of its limit.

Our first theorem is a compactness (up to bubbling) type result, in terms of
convergence in the Gromov-Hausdorff sense, in the spirit of works [3, 2, 6,
28]. We have the following:

######  Theorem 1.1.

Let $n=3$, then the space $\mathcal{E}(D,c,K)$ is compact in the topology
induced by the Hausdorff distance. That is, if
$(g_{k},\psi_{k})\in\mathcal{E}(D,c,K)$ then there exists a subsequence again
denoted by $(g_{k},\psi_{k})$ that converges in $C^{\ell,\alpha}(M)$ to
$(g_{\infty},\psi_{\infty})$ for all $\ell>0$ and $0<\alpha<1$ and
$\psi_{\infty}$ is an Einstein spinor on $(M,g_{\infty})$.

Let $n=4$, then there exist a compact orbifold $(M_{\infty},g_{\infty})$ with
a finite set $S$ of orbifold singularities, a spinor
$\psi_{\infty}\in\Sigma(M_{\infty}\setminus S)$ and a sequence of $C^{\infty}$
embeddings $F_{k}:(M_{\infty}\setminus S,g_{\infty})\to(M,g_{k})$, for $k$
large enough, such that

  * •

$((F_{k})^{*}g_{k},(F_{k})^{*}\psi_{k})$ converges uniformly on compact
subsets in the $C^{\ell,\alpha}$ topology on $M_{\infty}\setminus S$, to
$(g_{\infty},\psi_{\infty})$ for every $\ell>0$ and $0<\alpha<1$ and
$\psi_{\infty}$ is an Einstein spinor on $(M_{\infty}\setminus S,g_{\infty})$.

  * •

For each $p_{i}\in S$, there exist a sequence of real numbers $r_{k}$ and a
sequence of points $x_{k}\in M$ such that $(M,r_{k}g_{k},x_{k})$ converges in
the pointed Gromov-Hausdorff sense to $(Y_{i},\overline{g}_{i},x_{\infty})$,
where $(Y_{i},\overline{g}_{i})$ is a Ricci flat non-flat manifold. That is,
there exists a sequence of smooth diffeomorphisms
$H_{k}:(B(p_{i},r),\overline{g}_{i})\to(M,r_{k}g_{k})$ so that
$(H_{k}^{*}(r_{k}g_{k}))$ converges in the $C^{\ell,\alpha}$ topology in
$B(p_{i},r)\subset Y_{i}$, to $\overline{g}_{i}$ for every $r>0$, $\ell>0$ and
$0<\alpha<1$.

  * •

Moreover, there exists a parallel spinor $\psi_{i,\infty}\in\Gamma(\Sigma
Y_{i})$ such that $H_{k}^{*}\psi_{k}$ converges in the $C^{\ell,\alpha}$
topology in $B(p_{i},r)$, to the spinor $\psi_{i,\infty}$ for every $\ell>0$
and $0<\alpha<1$ and

$\liminf_{k\to\infty}\int_{M}|Rm_{g_{k}}|^{2}dv_{g_{k}}\geq\int_{M}|Rm_{g_{\infty}}|^{2}dv_{g_{\infty}}+\sum_{i=1}^{|S|}\int_{Y_{i}}|Rm_{\overline{g}_{i}}|^{2}dv_{\overline{g}_{i}}.$

Notice that the set $Crit_{c}(E)$ is invariant under the action of the
diffeomorphism group and the previous theorem is dedicated to the study of its
compactness. Hence, in order to have a better understanding of this set, we
would like to know if it has more structure. This is again similar to the
study of the moduli space of Einstein manifolds and Einstein deformations as
in [4, 5, 21]. Hence, we would like to know if transversally to the
diffeomorphisms’ action, $Crit_{c}(E)$ is a finite dimensional manifold. If
this is the case, then combined with the previous theorem, we could have the
structure of a compact manifold with boundary. In general the study of moduli
spaces is not an easy task because of the action of the diffeomorphisms’ group
(we refer the reader to [14, 32, 1]). Our study here is the first step in
analyzing the moduli space. Our second result concerns the Dirac-Einstein
deformations, which are defined by the vanishing of the second variation of
the energy functional at critical points. These are the deformations that does
not affect the set of critical points. In particular we characterize the
horizontal deformations, namely deformations of the metric at a fixed spinor.
This requires actually a carefully procedure since the spin bundle varies with
the metric as well: tracking such variations has been done by Bourguignon-
Gauduchon in [10] (see also [1]). We will recall it in the sequel; here we
first determine the second variation of the functional $E$ at a critical point
$(g,\psi)$:

######  Theorem 1.2.

Let $(g+th,\psi+t\varphi)$ denote a path of deformations, with
$t\in(-\varepsilon,\varepsilon)$ for a small $\varepsilon>0$, $h$ being
transverse to the diffeomorphism action, that is $\delta h=0$, where $(\delta
h)_{i}=-(divh)_{i}=-\nabla^{j}h_{ij}$. If $(g,\psi)$ is a critical point of
$E$, then we have

$\displaystyle\nabla^{2}E(g,\psi)[(h,\varphi),(h,\varphi)]$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h+\nabla\nabla
tr(h),h\rangle$ $\displaystyle+\frac{1}{2}\left(-\Delta tr(h)-\langle
Ric_{g},h\rangle\right)tr(h)+\frac{R_{g}}{2}|h|^{2}$
$\displaystyle+\frac{1}{2}\langle T^{g,\psi},h\rangle tr(h)+\frac{1}{2}\langle
h\times T^{g,\psi},h\rangle+\frac{1}{2}\langle\nabla
tr(h)\cdot\psi,\varphi\rangle$
$\displaystyle+\langle\mathcal{D}^{h}\varphi,\psi\rangle+2\langle
D_{g}\varphi-\lambda\varphi,\varphi\rangle dv$

where $\Delta_{L}$ is the Lichnerowicz Laplacian acting on symmetric
2-tensors, $\nabla\nabla tr(h)$ is the tensor defined by $(\nabla\nabla
tr(h))_{ij}=\nabla_{i}\nabla_{j}tr(h)$, $(A\times
B)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}$ and
$\mathcal{D}^{h}\varphi=\sum_{i,j}h_{ij}e_{i}\cdot\nabla_{e_{j}}\varphi$.

We recall here that if $h$ is a symmetric 2-tensor, then

$\Delta_{L}h=\nabla^{*}\nabla h+Ric\circ h+h\circ Ric-2\overset{\circ}{R}h,$

with $Ric\circ h+h\circ Ric$ is defined by considering $Ric$ as a
$(1,1)$-tensor, hence

$(Ric\circ h+h\circ Ric)(X,Y)=h(Ric(X),Y)+h(Ric(Y),X)$

and

$(\overset{\circ}{R}h)(X,Y)=\sum_{i,j}R(e_{i},X,Y,e_{j})h(e_{i},e_{j})$

for an orthonormal basis $(e_{1},\cdots,e_{n}).$

As a corollary, one can describe the space of horizontal Einstein-Dirac
deformations. Namely, deformations of the metric with fixed spinor that sits
in the null space of $\nabla^{2}E(g,\psi)$. Roughly speaking, let us denote by
$(g+th,\psi)$ a path of deformations that allows variations only on the metric
$g$ at the given spinor $\psi$, then we have

######  Corollary 1.3.

Let the pair $(g,\psi)$ be a critical point for the functional $E$. Then the
symmetric 2-tensor $h$ belongs to the space of horizontal Dirac-Einstein
deformations, if it solves the following system of equations

$\left\\{\begin{array}[]{ll}\delta h=0\\\ \\\ \left\langle
Ric_{g}-\displaystyle\frac{R_{g}}{2}g,h\right\rangle=\left\langle
T^{g,\psi},h\right\rangle=0\\\ \\\ \Delta h+R_{g}h+T^{g,\psi}\times
h=0\end{array}\right.$

where in the first equation, $\delta h=0$ means that the deformation of $g$ is
transverse to the diffeomorphism action. Moreover, if
$\ker\left(\Delta_{g}+\frac{R_{g}}{2}\right)=0$ or if $g$ is Einstein, then we
can replace the second equation by $tr(h)=\langle Ric_{g},h\rangle=0$; in
particular, the space of horizontal Dirac-Einstein perturbations is finite
dimensional.

In sections 5, we provide some examples of computations of the second
variation. For instance, we consider the example of a manifold with a real
Killing spinor and the example of Sasakian manifolds with Quasi-Killing
spinors.

Acknowledgment: The authors want to extend their thanks and gratitude to the
referee for the comments and suggestions that led to this improved version of
the paper.

##  2 Notations and basic formulas

In this section we are going to introduce some definition about spin geometry
and we recall some basic formulas that we will need in the sequel. Essentially
we will use the same notation as in [20], where we address the reader for the
full detailed exposition.
Let $\Sigma M$ be the canonical spinor bundle associated to $M$, whose
sections are simply called spinors on $M$. This bundle is endowed with a
natural Clifford multiplication

$\text{Cliff}:TM\otimes\Sigma M\longrightarrow\Sigma M,$

which we will denote by $\cdot$ in the sequel; also we denote the canonical
hermitian metric on $\Sigma M$ by $(\cdot,\cdot)$ and the induced metric
connection by $\nabla$. We will also denote by
$\langle\cdot,\cdot\rangle:=Re(\cdot,\cdot)$, the real part of the hermitian
metric which defines an Euclidean dot product on $\Sigma M$. Some useful
relations: for $X,Y\in\Gamma(TM),\psi\in\Gamma(\Sigma M)$, where $\Gamma$ is
used here to denote smooth sections of the given bundles,

$\langle X\cdot\psi,Y\cdot\psi\rangle=g(X,Y)|\psi|^{2},\quad\langle
X\cdot\psi,\psi\rangle=0,$
$\nabla_{X}(Y\cdot\psi)=(\nabla_{X}Y)\cdot\psi+Y\cdot(\nabla_{X}\psi)\;.$

If $(e_{1},\ldots,e_{n})$ is a local orthonormal frame, it holds, for
$k=1,\ldots,n$

$\nabla_{e_{k}}\psi=e_{k}(\psi)-\sum_{i,j=1}^{n}\Gamma_{kj}^{i}e_{i}\cdot
e_{j}\cdot\psi,$

where $\Gamma_{ki}^{j}$ are the coefficients of the connection for the frame
$(e_{1},\ldots,e_{n})$,
$\nabla_{e_{k}}e_{i}=\sum_{j=1}^{n}\Gamma_{ki}^{j}e_{j}.$
We denote by $D:\Sigma M\rightarrow\Sigma M$ the Dirac operator acting on
spinors in the following way

$D\psi=\sum_{i=1}^{n}e_{i}\cdot\nabla_{e_{i}}\psi\;.$

Now, let

$R(X,Y)Z=\nabla_{X}\nabla_{Y}Z-\nabla_{Y}\nabla_{X}Z-\nabla_{[X,Y]}Z$

be the curvature tensor on $M$, in the local orthonormal frame
$(e_{1},\ldots,e_{n})$, we denote

$Rm_{ijkl}=Rm(e_{i},e_{j},e_{k},e_{l})=-g(R(e_{i},e_{j})e_{k},e_{l}),$
$Ric_{jl}=Ric(e_{j},e_{l})=\sum_{i=1}^{n}Rm_{ijil},\quad
R_{g}=\sum_{i=1}^{n}Ric_{ii},$

where $Rm,Ric,R_{g}$ are the Riemann tensor, the Ricci tensor and the scalar
curvature of $M$ respectively. We can extend the curvature tensor on the
spinor bundle $\Sigma M$ by

$R(X,Y)\psi=\nabla_{X}\nabla_{Y}\psi-\nabla_{Y}\nabla_{X}\psi-\nabla_{[X,Y]}\psi,$

and it holds

$R(X,Y)\psi=-\frac{1}{2}R(X,Y)\cdot\psi,$

where in the right hand side we used the notation for the Clifford
multiplication of a 2-form with a spinor, that is:

$R(X,Y)\cdot\psi=\sum_{i<j}Rm(e_{i},e_{j},X,Y)e_{i}\cdot e_{j}\cdot\psi.$

Similarly,

$Ric(X)\cdot\psi=-\sum_{i=1}^{n}e_{i}\cdot R(e_{i},X)\cdot\psi,\qquad
R_{g}\psi=-\sum_{i=1}^{n}e_{i}\cdot Ric(e_{i})\cdot\psi.$

Moreover, if $M$ admits a parallel spinor, that is $\psi\in\Sigma M$ such that
$\nabla\psi=0$, then $M$ is Ricci-flat. Let us also write the following
formula (Lemma 1.2 in [20])

$2Ric(X)\cdot\psi=D(\nabla_{X}\psi)-\nabla_{X}(D\psi)-\sum_{i=1}^{n}e_{i}\cdot\nabla_{(\nabla_{e_{i}}X)}\psi,\quad
X\in\Gamma(TM),\;\psi\in\Gamma(\Sigma M)$

which implies the Schrödinger-Lichnerowicz formula

$D^{2}\psi=-\Delta\psi+\frac{1}{4}R_{g}\psi,$ (6)

where $-\Delta=\nabla^{*}\nabla$ is the Laplacian of the connection. Finally,
we recall the formulas for the first variation of the Dirac operator, the
volume form and the scalar curvature, in order to compute the variation of the
energy functional. The behaviour of the spinor bundle under small changes of
the metric has been done by Bourguignon-Gauduchon in [10], where they
introduced a natural isomorphism to identify spinor bundles related to
different metrics on the same manifold. So, we denote by $Sym(0,2)$ the space
of all symmetric (0,2)-tensor fields on $M$ and we will still denote by
$\langle\cdot,\cdot\rangle$ the induced metric on $Sym(0,2)$. It holds

$\frac{d}{dt}\Big{|}_{t=0}\langle
D_{g+th}\psi_{g+th},\psi_{g+th}\rangle_{g+th}=-\langle
T^{g,\psi},h\rangle,\quad h\in Sym(0,2)$ (7)

with $T^{g,\psi}$ the energy-momentum tensor introduced in (3); here $t$ is a
sufficiently small real parameter and we have set $D_{g+th}$ the Dirac
operator of the metric $g+th$, also $\psi_{g+th}$ is the push forward of
$\psi$ by using the isomorphism defined in [10]. We recall also, for $h\in
Sym(0,2)$

$\frac{d}{dt}\Big{|}_{t=0}dv_{g+th}=\frac{1}{2}\langle g,h\rangle
dv_{g},\quad\frac{d}{dt}\Big{|}_{t=0}R_{g+th}=-\Delta
tr(h)+\delta^{2}(h)-\langle Ric,h\rangle,$ (8)

where $\delta^{2}(h)=\sum_{i,j=1}^{n}\nabla^{i}\nabla^{j}h_{ij}$. Finally,
putting together formulas (7) and (8) we obtain the system (2) for the
critical points of the energy functional $E$.

##  3 Proof of Theorem 1.1

Let $\mathcal{E}(D,c,K)$ the set introduced in (5). For the sake of
simplicity, in the sequel sometimes we will omit the dependance from $g$ and
$\psi$. We will start with the following

######  Lemma 3.1.

There exists a positive constant $C(n,\lambda)$, depending on the dimension
$n$ and the positive real parameter $\lambda$, such that, if
$(g,\psi)\in\mathcal{E}(D,c,K)$ then

$\|\psi\|_{\infty}\leq C(n,\lambda).$

###### Proof.

By formula (4) and since $D^{2}_{g}\psi=\lambda^{2}\psi$ from the Schrödinger-
Lichnerowicz equation (6) we obtain

$-\Delta\psi+\lambda\frac{|\psi|^{2}}{4(n-2)}\psi=\lambda^{2}\psi.$ (9)

Now we let $f=\frac{1}{2}|\psi|^{2}$ and we have

$-\Delta
f=\langle-\Delta\psi,\psi\rangle-|\nabla\psi|^{2}\leq-\frac{R}{2}f+2\lambda^{2}f,$

where the laplacian acting on $f$ is the usual metric laplacian on $M$, the
one acting on $\psi$ is the Laplacian of the connection $\nabla^{*}\nabla$; we
denote them in the same way since it is clearly understood which we use.
Hence,

$-\Delta f+\frac{R}{2}f\leq 2\lambda^{2}f.$

At a maximum point $x_{0}\in M$, we have $\frac{R}{2}\leq 2\lambda^{2}$. Thus
$R\leq 4\lambda^{2}$. But, again by formula (4) we get

$|\psi|^{2}\leq 4(n-2)\lambda.$

∎

Next we prove the following

######  Proposition 3.2.

For any metric $g$ such that $(g,\psi)\in\mathcal{E}(D,c,K)$, we have that
$|Ric_{g}|$ is uniformly bounded in terms of $\lambda$ and $n$ and $K$. In
particular the Sobolev constant of $g$ is uniformly bounded from below and if
$p\in M$ and $r>0$ sufficiently small, then there exists a positive constant
$C$ depending on $\lambda,n,K$ and $D$ such that

$vol(B(p,r))\geq Cr^{n}.$

###### Proof.

Notice that the uniform bound on $\psi$ above, implies a uniform bound on the
scalar curvature $R$. The next step is then to find a bound on $|T^{g,\psi}|$.
So as above, we compute $-\Delta R$:

$\displaystyle-\Delta R$ $\displaystyle=-\frac{\lambda}{n-2}\Delta|\psi|^{2}$
$\displaystyle=\frac{2\lambda}{n-2}\Big{(}\langle-\Delta\psi,\psi\rangle-|\nabla\psi|^{2}\Big{)}$
$\displaystyle=\frac{2\lambda}{n-2}\Big{(}\left[\lambda^{2}-\frac{R}{4}\right]|\psi|^{2}-|\nabla\psi|^{2}\Big{)}.$
(10)

Since $(g,\psi)\in\mathcal{E}(D,c,K)$ we have

$\displaystyle|\nabla\psi|^{2}$ $\displaystyle=\frac{n-2}{2\lambda}\Delta
R+\left[\lambda^{2}-\frac{R}{4}\right]|\psi|^{2}$
$\displaystyle\leq\left[\frac{K}{2}+\lambda^{2}-\frac{R}{4}\right]|\psi|^{2}.$

Therefore, $|\nabla\psi|^{2}$ is uniformly bounded. Thus, we see that
$|T^{g,\psi}|$ is also uniformly bounded, hence $|Ric|$ is also uniformly
bounded in terms of $\lambda$ and $n$ and $K$. Now, the uniform bound on the
Ricci curvature and the diameter of the metric $g$ implies, by the result of
Croke [13], that the Sobolev constant is uniformly bounded from below. Where
here, the Sobolev constant $c_{S}$ is defined as the best constant $c$
satisfying the inequality

$\|u\|_{L^{\frac{2n}{n-2}}(M)}\leq\frac{1}{c}\|\nabla
u\|_{L^{2}(M)}+Vol(M)^{-\frac{2}{n}}\|f\|_{L^{2}(M)},\forall u\in C^{0,1}(M),$

if $M$ is compact and

$\|u\|_{L^{\frac{2n}{n-2}}(M)}\leq\frac{1}{c}\|\nabla u\|_{L^{2}(M)},\forall
u\in C^{0,1}_{c}(M),$

if $M$ is not compact. By using the equivalence between the Sobolev inequality
and the isoperimetric inequality, as in [28], we have that

$vol(\partial B(x,\rho))\geq
C_{1}\Big{(}vol(B(x,\rho))\Big{)}^{\frac{n-1}{n}},$

for all $x\in M$ and $\rho>0$ such that $B(x,\rho)\subset B(p,r)$ and where
$C_{1}$ is the lower bound on the Sobolev constant. Integrating this last
inequality yields the desired result. ∎

We notice that as a byproduct of Proposition 3.2, the total volume of the
metric $g$ is uniformly bounded: this follows from the Bishop-Gromov
inequality and the uniform bound on the diameter.

We focus now on the case $n=4$; in fact, from the Chern-Gauss-Bonnet formula
in dimension 4, we have that

$\chi(M)=\frac{1}{32\pi^{2}}\int_{M}|Rm|^{2}-4|Ric|^{2}+R^{2}\ dv_{g},$

in particular a uniform bound on the Ricci curvature implies automatically a
uniform bound on $\int_{M}|Rm|^{2}dv_{g}$. Therefore we are going to prove a
local $L^{\infty}$ bound on $|Rm|$ provided smallness on its $L^{2}$-norm.
This will be done through a sequence of propositions.

######  Proposition 3.3.

Let $(g,\psi)\in\mathcal{E}(D,K,c)$, $p\in M$ and $r>0$ sufficiently small.
There exists $\varepsilon_{0}(\lambda,K,D)>0$ such that, if

$\int_{B_{2r}}|Rm|^{2}dv<\varepsilon_{0},$

then there exists $C(\lambda,K,D)>0$ such that

$\|Rm\|_{L^{4}(B_{r})}\leq
C\Big{(}\|Rm\|_{L^{2}(B_{2r})}+vol(B_{8r})^{\frac{1}{2}}\Big{)}$ (11)

and

$\|\nabla Rm\|_{L^{2}(B_{r})}\leq
C\Big{[}\|Rm\|_{L^{2}(B_{2r})}^{2}+\|Rm\|_{L^{2}(B_{2r})}+vol(B_{8r})+vol(B_{8r})^{\frac{1}{2}}\Big{]},$
(12)

where we set $B_{r}=B(p,r)$.

###### Proof.

Given two tensors $A$ and $B$, we will use $A*B$ to denote a bilinear
expression of $A$ and $B$. We recall then from [18, Lemma 7.4], that

$\Delta Rm=Rm*Rm+\nabla^{2}Ric,$ (13)

where

$(\nabla^{2}Ric)_{ijkl}=-Ric_{jl,ik}+Ric_{il,jk}+Ric_{jk,il}-Ric_{ik,jl},$

where the comma denotes the covariant derivative in local coordinates
$(e_{1},\ldots,e_{n})$. Since $Ric-\frac{R}{2}g=T^{g,\psi}$, we need to
compute the contribution of $T^{g,\psi}$ in $\nabla^{2}Ric$. By the definition
(3) of $T^{g,\psi}$ we have

$T^{g,\psi}_{ij}=-\frac{1}{4}\Big{(}\langle
e_{i}\cdot\nabla_{e_{j}}\psi,\psi\rangle+\langle
e_{j}\cdot\nabla_{e_{i}}\psi,\psi\rangle\Big{)}.$

We set $S_{ij}=\langle e_{i}\cdot\nabla_{e_{j}}\psi,\psi\rangle$ and we
compute the Laplacian of $S_{ij}$. For simplicity, we can assume that we are
in inertial coordinates at a given point $x$, that is $\Gamma_{ij}^{k}(x)=0$
and $g_{ij}(x)=\delta_{ij}$.

$\displaystyle\nabla_{e_{k}}\nabla_{e_{k}}S_{ij}$
$\displaystyle=\nabla_{e_{k}}[\langle
e_{i}\cdot\nabla_{e_{k}}\nabla_{e_{j}}\psi,\psi\rangle+\langle
e_{i}\cdot\nabla_{e_{j}}\psi,\nabla_{e_{k}}\psi\rangle+\langle\nabla_{e_{k}}e_{i}\cdot\nabla_{e_{j}}\psi,\psi\rangle]$
$\displaystyle=\nabla_{e_{k}}[\langle
e_{i}\cdot[\nabla_{e_{j}}\nabla_{e_{k}}\psi-\frac{1}{2}R(e_{k},e_{j})(\psi)],\psi\rangle]$
$\displaystyle+\langle
e_{i}\cdot\nabla_{e_{k}}\nabla_{e_{j}}\psi,\nabla_{e_{k}}\psi\rangle+\langle
e_{i}\cdot\nabla_{e_{j}}\psi,\nabla_{e_{k}}\nabla_{e_{k}}\psi\rangle$
$\displaystyle+\langle\nabla_{e_{k}}\nabla_{e_{k}}e_{i}\cdot\nabla_{e_{j}}\psi,\psi\rangle+\langle\nabla_{e_{k}}e_{i}\cdot\nabla_{e_{k}}\nabla_{e_{j}}\psi,\psi\rangle+2\langle\nabla_{e_{k}}e_{i}\cdot\nabla_{e_{j}}\psi,\nabla_{e_{k}}\psi\rangle$
$\displaystyle=\langle
e_{i}\cdot\nabla_{e_{k}}\nabla_{e_{j}}\nabla_{e_{k}}\psi,\psi\rangle+\langle
e_{i}\cdot[\nabla_{e_{j}}\nabla_{e_{k}}+\nabla_{e_{k}}\nabla_{e_{j}}]\psi,\nabla_{e_{k}}\psi\rangle$
$\displaystyle+\langle
e_{i}\cdot\nabla_{e_{j}}\psi,\nabla_{e_{k}}\nabla_{e_{k}}\psi\rangle-\frac{1}{2}\langle
e_{i}\cdot\nabla_{e_{k}}R(e_{k},e_{j})(\psi),\psi\rangle+Rm*\nabla\psi*\psi$
$\displaystyle=\langle
e_{i}\cdot\nabla_{e_{j}}\nabla_{e_{k}}\nabla_{e_{k}}\psi,\psi\rangle+Rm*\nabla\psi*\psi+\nabla
Rm*\psi*\psi$ $\displaystyle+\langle
e_{i}\cdot[\nabla_{e_{j}}\nabla_{e_{k}}+\nabla_{e_{k}}\nabla_{e_{j}}]\psi,\nabla_{e_{k}}\psi\rangle+\langle
e_{i}\cdot\nabla_{e_{j}}\psi,\nabla_{e_{k}}\nabla_{e_{k}}\psi\rangle.$

Summing in $k$, the first term after the last equality yields

$\langle
e_{i}\cdot\nabla_{j}\left[\left(\frac{R}{4}-\lambda^{2}\right)\psi\right],\psi\rangle=\left(\frac{R}{4}-\lambda^{2}\right)\langle
e_{i}\cdot\nabla_{j}\psi,\psi\rangle,$

and the last term yields

$\left(\frac{R}{4}-\lambda^{2}\right)\langle
e_{i}\cdot\nabla_{j}\psi,\psi\rangle.$

Therefore,

$\Delta
S_{ij}=2\left(\frac{R}{4}-\lambda^{2}\right)S_{ij}+Rm*\nabla\psi*\psi+\nabla
Rm*\psi*\psi+\nabla^{2}\psi*\nabla\psi.$

Notice that since $\Delta\psi\in L^{\infty}$ and since $Ric$ is bounded
uniformly, we have from [17], the existence of $C$ depending on $\lambda,K$
and $D$ such that

$\displaystyle\|\nabla^{2}\psi\|_{L^{2}(B_{r})}$ $\displaystyle\leq
C(\|\Delta\psi\|_{L^{2}(B_{2r})}+\|\psi\|_{L^{2}(B_{2r})})\leq
CVol(B_{2r})^{\frac{1}{2}}.$ (14)

Also, recall that $S_{ij}\in L^{\infty}$ uniformly. Hence, if we let $\eta\geq
0$ be a smooth cutoff function such that $\eta=1$ on $B_{r}$ and $\eta=0$
outside $B_{2r}$, there exists $C(\lambda,D)$ such that

$\displaystyle\|\nabla^{2}(\eta S_{ij})\|_{L^{2}(B_{r})}$ $\displaystyle\leq
C\Big{(}\|\Delta(\eta S_{ij})\|_{L^{2}(B_{2r})}+\|\eta
S_{ij}\|_{L^{2}(B_{2r})}\Big{)}$ $\displaystyle\leq C\Big{(}\|\eta
Rm\|_{L^{2}(B_{2r})}+\|\eta\nabla
Rm\|_{L^{2}(B_{2r})}+Vol(B_{4r})^{\frac{1}{2}}\Big{)}.$ (15)

Hence, if we set $H=\nabla^{2}Ric$, we have

$\|\eta H\|_{L^{2}(B_{r})}\leq C\Big{(}\|\nabla^{2}(\eta
S_{ij})\|_{L^{2}(B_{r})}+Vol(B_{2r})^{\frac{1}{2}}\Big{)}.$

Therefore,

$\|\eta H\|_{L^{2}(B_{r})}\leq C(\|\eta Rm\|_{L^{2}(B_{2r})}+\|\eta\nabla
Rm\|_{L^{2}(B_{2r})}+Vol(B_{4r})^{\frac{1}{2}}).$ (16)

We go back now to the equation (13) for $Rm$. We have

$\Delta(\eta Rm)=\eta Rm*Rm+\eta H+\nabla Rm*\nabla\eta.$

Thus, we get

$\displaystyle\|\nabla(\eta Rm)\|_{L^{2}(B_{2r})}^{2}\leq C\Big{(}$
$\displaystyle\|\eta Rm\|_{L^{4}(B_{2r})}\|Rm\|_{L^{2}(B_{2r})}^{2}+\|\eta
Rm\|_{L^{2}(B_{2r})}\|\eta H\|_{L^{2}(B_{2r})}$
$\displaystyle+\|Rm\|_{L^{2}(B_{2r})}\|\eta\nabla
Rm\|_{L^{2}(B_{2r})}\Big{)}.$

Now using estimate $(\ref{est2})$,

$\displaystyle\|\nabla(\eta Rm)\|_{L^{2}(B_{2r})}^{2}\leq C\Big{(}$
$\displaystyle\|Rm\|_{L^{2}(B_{2r})}\|\eta
Rm\|_{L^{4}(B_{2r})}^{2}+\|Rm\|_{L^{2}(B_{2r})}\|\nabla(\eta
Rm)\|_{L^{2}(B_{2r})}$
$\displaystyle+\|Rm\|_{L^{2}(B_{2r})}\Big{[}\|Rm\|_{L^{2}(B_{2r})}+Vol(B_{8r})^{\frac{1}{2}}\Big{]}\Big{)},$

where we used above that $\nabla(\eta Rm)=\eta\nabla Rm+(\nabla\eta)*Rm$ and
$supp(\eta)\subset B_{2r}$. Using the inequality $2ab\leq\alpha
a^{2}+\frac{1}{\alpha}b^{2}$ for a positive $\alpha$ chosen so that
$C\alpha<1$, we have the existence of another constant $C(\lambda,K,D)$ such
that

$\displaystyle\|\nabla(\eta Rm)\|_{L^{2}(B_{2r})}^{2}\leq$ $\displaystyle
C\Big{(}\|Rm\|_{L^{2}(B_{2r})}\|\eta Rm\|_{L^{4}(B_{2r})}^{2}$
$\displaystyle+\|Rm\|_{L^{2}(B_{2r})}\Big{[}\|Rm\|_{L^{2}(B_{2r})}+Vol(B_{8r})^{\frac{1}{2}}\Big{]}\Big{)}.$
(17)

By the Sobolev inequality we have

$\|\eta Rm\|_{L^{4}(B_{2r})}^{2}\leq C\Big{(}\|Rm\|_{L^{2}(B_{2r})}\|\eta
Rm\|_{L^{4}(B_{2r})}^{2}+\|Rm\|_{L^{2}(B_{2r})}\Big{[}\|Rm\|_{L^{2}(B_{2r})}+Vol(B_{8r})^{\frac{1}{2}}\Big{]}\Big{)}.$

In particular, there exists $\varepsilon_{0}=\varepsilon_{0}(\lambda,n,D)>0$
such that if $\|Rm\|_{L^{2}(B_{2r})}<\varepsilon_{0}$, then

$\|Rm\|_{L^{4}(B_{r})}\leq
C\Big{(}\|Rm\|_{L^{2}(B_{2r})}+Vol(B_{8r})^{\frac{1}{2}}\Big{)}.$

Also, from $(\ref{est3})$, we obtain

$\|\nabla Rm\|_{L^{2}(B_{r})}^{2}\leq
C\Big{(}\|Rm\|_{L^{2}(B_{2r})}^{2}+\|Rm\|_{L^{2}(B_{2r})}+Vol(B_{8r})+Vol(B_{8r})^{\frac{1}{2}}\Big{)}.$

∎

Now we recall the following classical lemma (see [6, Lemma 4.6] for a proof)
that we use in the next proposition:

######  Lemma 3.4.

Consider two functions $u\in L^{2}_{loc}$ and $f\in L^{4}_{loc}$ such that
$u\geq 0$ and $f\geq 0$ and

$\Delta u\geq-fu.$

Then there exists $\varepsilon>0$ and $r>0$ sufficiently small, such that if
$\int_{B_{r}}f^{4}<\varepsilon$ then

$\sup_{B_{\frac{r}{2}}}u\leq C\left(\int_{B_{r}}f^{4}\right)^{\frac{1}{4}}.$

We have then

######  Proposition 3.5.

[$\varepsilon$-regularity] There exist $\varepsilon_{1}(\lambda,K,D)>0$ and
$0<r_{0}<1$ such that, if

$\int_{B_{16r}}|Rm|^{2}dv<\varepsilon_{1},\quad r<r_{0},$

then there exists $C(\lambda,K,D)$ such that

$\sup_{B_{\frac{r}{2}}}|Rm|\leq C(\lambda,K,D).$

###### Proof.

The proof is based on the previous Lemma 3.4, considering $u=|Rm|^{2}+1$.
Since

$\Delta u\geq-C(|Rm|+|H|)u,$

we need an estimate for $\||Rm|+|H|\|_{L^{4}(B_{r})}$. Proposition 3.3
provides us with an estimate for $\|Rm\|_{L^{4}(B_{r})}$, so we will estimate
$\|H\|_{L^{4}(B_{r})}$.
First, we use the commutation relation between $\Delta$ and $\nabla_{i}$ to
write

$\Delta\nabla_{i}\psi=\nabla_{i}\Delta\psi+Rm*\nabla\psi+\nabla Rm*\psi.$

Hence, again using [17], we get

$\|\nabla^{3}\psi\|_{L^{2}(B_{r})}\leq
C(\|\Delta\nabla\psi\|_{L^{2}(B_{2r})}+\|\nabla\psi\|_{L^{2}(B_{2r})}),$

therefore, by Sobolev inequality again, we have

$\|\nabla^{2}\psi\|_{L^{4}(B_{r})}^{2}\leq
C\Big{[}Vol(B_{2r})+\|Rm\|_{L^{2}(B_{2r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{2r})}^{2}\Big{]},$ (18)

where we have used the fact that $-\Delta\psi+\frac{R}{4}\psi=\lambda^{2}\psi$
and the uniform boundedness of $\psi,R$ and $\nabla\psi$. This provides an
estimate for the terms of $H$ containing $\nabla^{2}R$. Now we compute
$\Delta\nabla_{k}S_{ij}$. So we start by noticing

$\displaystyle\nabla_{k}\Delta S_{ij}=$ $\displaystyle\nabla R*S_{ij}+\nabla
Rm*\nabla\psi*\psi+Rm*\nabla^{2}\psi*\psi$
$\displaystyle+Rm*\nabla\psi*\nabla\psi+\nabla^{2}Rm*\psi*\psi+\nabla^{3}\psi*\nabla\psi+\nabla^{2}\psi*\nabla^{2}\psi.$

Hence

$\displaystyle\|\nabla_{k}\Delta S_{ij}\|_{L^{2}(B_{r})}\leq$ $\displaystyle
C\Big{(}Vol(B_{r})^{\frac{1}{2}}+\|\nabla
Rm\|_{L^{2}(B_{r})}+\|Rm\|_{L^{4}(B_{r})}\|\nabla^{2}\psi\|_{L^{4}(B_{r})}$
$\displaystyle+\|\nabla^{2}Rm\|_{L^{2}(B_{r})}+\|\nabla^{3}\psi\|_{L^{2}(B_{r})}+\|\nabla^{2}\psi\|_{L^{4}(B_{r})}^{2}\Big{)}.$

But

$\displaystyle\|\nabla^{2}Rm\|_{L^{2}(B_{r})}$ $\displaystyle\leq
C(\|Rm\|_{L^{4}(B_{2r})}^{2}+\|H\|_{L^{2}(B_{2r})})$ $\displaystyle\leq
C(\|Rm\|_{L^{2}(B_{4r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{4r})}+\|Rm\|_{L^{2}(B_{4r})}+Vol(B_{8r})^{\frac{1}{2}}+Vol(B_{16r})).$

Therefore

$\displaystyle\|\nabla_{k}\Delta S_{ij}\|_{L^{2}(B_{r})}\leq$ $\displaystyle
C\Big{[}\|\nabla Rm\|_{L^{2}(B_{4r})}+\|Rm\|_{L^{2}(B_{4r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{4r})}^{2}$
$\displaystyle+\|Rm\|_{L^{4}(B_{r})}\Big{(}vol(B_{2r})^{\frac{1}{2}}+\|Rm\|_{L^{2}(B_{2r})}+\|\nabla
Rm\|_{L^{2}(B_{2r})}\Big{)}$
$\displaystyle+\|Rm\|_{L^{2}(B_{4r})}+Vol(B_{16r})^{\frac{1}{2}}\Big{]}.$

Using,

$\Delta\nabla_{k}S_{ij}=\nabla_{k}\Delta S_{ij}+Rm*\nabla S_{ij}+\nabla
Rm*\psi,$

we get that

$\displaystyle\|\Delta\nabla_{k}S_{ij}\|_{L^{2}(B_{r})}\leq$ $\displaystyle
C\Big{[}\|\nabla Rm\|_{L^{2}(B_{4r})}+\|Rm\|_{L^{2}(B_{4r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{4r})}^{2}$
$\displaystyle+\|Rm\|_{L^{4}(B_{r})}\Big{(}vol(B_{2r})^{\frac{1}{2}}+\|Rm\|_{L^{2}(B_{2r})}+\|\nabla
Rm\|_{L^{2}(B_{2r})}\Big{)}$
$\displaystyle+\|Rm\|_{L^{2}(B_{4r})}+Vol(B_{16r})^{\frac{1}{2}}\Big{]}.$

Therefore, we have

$\displaystyle\|\nabla^{3}S_{ij}\|_{L^{2}(B_{r})}\leq$ $\displaystyle
C\Big{[}\|\nabla Rm\|_{L^{2}(B_{8r})}+\|Rm\|_{L^{2}(B_{8r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{8r})}^{2}$
$\displaystyle+\|Rm\|_{L^{4}(B_{2r})}\Big{(}vol(B_{4r})^{\frac{1}{2}}+\|Rm\|_{L^{2}(B_{4r})}+\|\nabla
Rm\|_{L^{2}(B_{4r})}\Big{)}$
$\displaystyle+\|Rm\|_{L^{2}(B_{8r})}+Vol(B_{32r})^{\frac{1}{2}}\Big{]}.$

Thus, Sobolev inequality yields

$\displaystyle\|\nabla^{2}S_{ij}\|_{L^{4}(B_{r})}\leq$ $\displaystyle
C\Big{[}\|\nabla Rm\|_{L^{2}(B_{8r})}+\|Rm\|_{L^{2}(B_{8r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{8r})}^{2}$
$\displaystyle+\|Rm\|_{L^{4}(B_{2r})}\Big{(}vol(B_{4r})^{\frac{1}{2}}+\|Rm\|_{L^{2}(B_{4r})}+\|\nabla
Rm\|_{L^{2}(B_{4r})}\Big{)}$
$\displaystyle+\|Rm\|_{L^{2}(B_{8r})}+Vol(B_{32r})^{\frac{1}{2}}\Big{]},$ (19)

and combining $(\ref{est2s})$ and $(\ref{est2psi})$ we obtain

$\displaystyle\|H\|_{L^{4}(B_{r})}\leq$ $\displaystyle C\Big{[}\|\nabla
Rm\|_{L^{2}(B_{8r})}+\|Rm\|_{L^{2}(B_{8r})}^{2}+\|\nabla
Rm\|_{L^{2}(B_{8r})}^{2}$
$\displaystyle+\|Rm\|_{L^{4}(B_{2r})}\Big{(}vol(B_{4r})^{\frac{1}{2}}+\|Rm\|_{L^{2}(B_{4r})}+\|\nabla
Rm\|_{L^{2}(B_{4r})}\Big{)}$
$\displaystyle+\|Rm\|_{L^{2}(B_{8r})}+Vol(B_{32r})^{\frac{1}{2}}\Big{]}.$

Now using $(\ref{ineq1})$ and $(\ref{ineq2})$ of Proposition 3.3, assuming
that $\|Rm\|_{L^{2}(B_{2r})}<\varepsilon_{0}$, we have

$\displaystyle\|H\|_{L^{4}(B_{r})}\leq$ $\displaystyle
C\Big{(}Vol(B_{32r})^{\frac{1}{2}}+Vol(B_{32r})+Vol(B_{32r})^{2}+\|Rm\|_{L^{2}(B_{16r})}+\|Rm\|^{2}_{L^{2}(B_{16r})}$
$\displaystyle+\|Rm\|_{L^{2}(B_{16r})}^{3}+\|Rm\|_{L^{2}(B_{16r})}^{4}\Big{)}.$
(20)

Hence, there exist $\varepsilon_{1}(\lambda,K,D)$ and $r_{0}(\lambda,K.D)$
such that if $\|Rm\|_{L^{2}(B_{16r})}<\varepsilon_{1}$ and $r<r_{0}$, then
$\||Rm|+|H|\|_{L^{4}(B_{r})}<\varepsilon$. Thus, it follows from Lemma 3.4
that

$\sup_{B_{\frac{r}{2}}}u\leq C(\lambda,K,D).$

∎

Let us recall now the following definition:

######  Definition 3.6.

Let $(M,g)$ be a Riemannian manifold. We say that $(M,g)$ has an adapted
$(r,N,C^{\ell,\alpha})$ (resp. $(r,N,W^{k,p})$) harmonic atlas, if there is a
covering $(B(x_{k},r))_{1\leq k\leq N}$ of $M$ made of geodesic balls of
radius $r$, for which the balls $B(x_{k},\frac{r}{2})$ cover $M$ and
$B(x_{k},\frac{r}{4})$ are disjoint, such that each $B(x_{k},10r)$ has a
harmonic coordinate chart $(U_{k},\phi_{k})$, such that the metric tensor in
these coordinates is $C^{\ell,\alpha}$ (resp. $W^{k,p}$) and on
$B(x_{k},10r)$, there exists a constant $C>1$ such that

$C^{-1}\delta_{ij}\leq g_{ij}\leq C\delta_{ij}$

and

$r^{\ell+\alpha}\|g_{ij}\|_{C^{\ell,\alpha}}\leq C,\quad(\text{ resp.
}r^{k-\frac{p}{n}}\|g_{ij}\|_{W^{k,p}}\leq C).$

The $C^{\ell,\alpha}$ (resp. $W^{k,p}$)- harmonic radius at $x$, $r_{h}(x)$ is
the maximum radius $r$ for which $B_{x}(r)$ has harmonic coordinates.
Now we will complete the proof of Theorem 1.1.

###### Proof.

(of Theorem 1.1, case $n=4$) First we notice that if
$(g,\psi)\in\mathcal{E}(D,c,K)$ then

$c=\int_{M}R_{g}dv_{g}\leq\frac{\lambda}{2}\|\psi\|_{\infty}^{2}vol(M,g)\leq
4\lambda^{2}vol(M,g),$ (21)

therefore the volume is uniformly bounded from below. Now given a sequence
$(g_{i},\psi_{i})\in\mathcal{E}(D,c,K)$, then, combining the lower bound on
the volume with the uniform bound on $Ric_{g}$, $\|Rm\|_{L^{2}}$ and the
diameter assumption, we have from [3, Theorem 2.6] the convergence of a
subsequence of $(M,g_{i})$ in the Gromov-Hausdorff topology to a four
dimensional orbifold $(M_{\infty},g_{\infty})$ with finitely many
singularities. We can assume without loss of generality that
$M_{\infty}\setminus S\hookrightarrow M$. Moreover the convergence to
$g_{\infty}$ is in $C^{1,\alpha}$ away from the singularities in the Cheeger-
Gromov sense. Also, around the singularity the metric blows up to an ALE
(Asymptotically Locally Euclidean) manifold (see [3, Theorem 1.2] and the
definition therein). We will denote by $S$ the set of singularities and
$M^{0}=M_{\infty}\setminus S$ that we identify with $M\setminus S$.

Let $r<\frac{r_{0}}{4}$ as defined in Proposition 3.5 and let us consider a
covering of $(M,g_{i})$ by balls $B(x_{k},r)$ such that $B(x_{k},\frac{r}{2})$
are disjoint. We let

$I=\left\\{k\in\mathbb{N};\int_{B(x_{k},16r)}|Rm_{g_{i}}|^{2}dv_{g_{i}}<\varepsilon_{1}\right\\},$

and we define

$G_{i}(r)=\bigcup_{k\in I}B(x_{k},r).$

Similarly, we let

$H_{i}(r)=\bigcup\left\\{B(x_{k},r);\int_{B(x_{k},16r)}|Rm_{g_{i}}|_{g_{i}}^{2}\
dv_{g_{i}}\geq\varepsilon_{1}\right\\}.$

Notice that $M_{i}=H_{i}(r)\cup G_{i}(r)$. This is a splitting of $M_{i}$ into
a good set where one can control the curvature and a bad set where there is no
$L^{\infty}$ control on the curvature. In fact, following [2, 28], one can
show that the number of balls in the definition of $H_{i}(r)$ is finite and
uniformly bounded in $i$ and $r<\frac{r_{0}}{4}$.
By Propositions 3.2 and 3.5, one has a uniform lower bound on the injectivity
radius of $G_{i}$ (see for instance [12, Section 3]). In particular
$(G_{i}(r),g_{i})$ has an adapted $(r,N,W^{2,p})$ harmonic atlas with a
uniform lower bound on the injectivity radius. Now in a harmonic coordinate
patch, we recall that the metric satisfies the quasilinear elliptic equation

$-\Delta_{g_{i}}g_{i}+\partial g_{i}*g_{i}=Ric_{g_{i}}.$

From the first equation in the system (2), any $g_{i}$ is a solution of

$Ric_{g_{i}}=\frac{R_{g_{i}}}{2}\psi_{i}+T^{g_{i},\psi_{i}},$

and since $\psi_{i}$ satisfies

$-\Delta_{g_{i}}\psi_{i}+\frac{R_{g_{i}}}{2}\psi_{i}=\lambda^{2}\psi_{i},$

we have a uniform $C^{1,\alpha}$ bound on $\psi_{i}$ leading to a
$C^{0,\alpha}$ bound on $T^{g_{i},\psi_{i}}$. Therefore, we have a uniform
$C^{2,\alpha}$ bound on $g_{i}$. This can be naturally iterated to get a
$C^{\ell,\alpha}$ bound on $g_{i}$. Therefore, there exists a subsequence of
$(G_{i}(r),g_{i})$ that converges in the $C^{\infty}$ topology to an open
manifold $(G_{r},g_{\infty}^{r})$ and $\psi_{i}\to\psi_{\infty}^{r}$ in the
$C^{\infty}$ topology.
We choose now a sequence $r_{j}\to 0$ such that $r_{j+1}\leq\frac{1}{2}r_{j}$
and we let

$G_{i}^{j}=\\{x\in M;x\in G_{i}(r_{m})\text{ for some }m\leq j\\}.$

Then we have

$G_{i}^{1}\subset G_{i}^{2}\subset\cdots\subset(M,g_{i}).$

We also have that $G_{r_{j}}\subset G_{r_{j+1}}$. So we set
$M^{0}=\cup_{j=1}^{\infty}G_{r_{j}}\subset M$. Now using a diagonal argument,
we can extract a subsequence of $(G_{i}^{j},g_{i},\psi_{i})$ that converges in
$C^{\ell,\alpha}$ to $(M^{0},g_{\infty},\psi_{\infty})$ for all $\ell$. This
proves the convergence away from a finite set of points, namely, the singular
set. It remains then to study the convergence in the neighborhood of points in
this latter set. The singular set $S$ can then be defined by
$S=M_{\infty}\setminus M^{0}$. In particular, we can characterize $S$ by

$S=\left\\{\begin{array}[]{ll}p\in M,\text{ such that there exists }x_{k}\in
M,r>0,\varepsilon_{1}>0\text{ with}\\\ \\\ x_{k}\to p\text{
and}\displaystyle\liminf_{r\to
0}\liminf_{k\to\infty}\int_{B(x_{k},r)}|Rm_{g_{k}}|^{2}dv_{g_{k}}\geq\varepsilon_{1}\end{array}\right\\}.$

Now let us take a point $p_{0}\in S$. We know that there exists
$x_{k}\in(M,g_{k})$ such that $x_{k}\to p_{0}$. Since $S$ is finite, we pick
$\delta>0$ so that $B(p_{0},2\delta)\cap S=\\{p_{0}\\}$ and let

$r_{k}=\sup_{x\in B(x_{k},\delta)}|Rm_{g_{k}}|,$

and we notice that $r_{k}\to\infty$. We can then, consider a sequence of
pointed Riemannian manifolds $(M,\tilde{g}_{k},x_{k})$ where
$\tilde{g}_{k}=r_{k}g_{k}$. Under this new rescaled metric $\tilde{g}_{k}$ we
have:

$\left\\{\begin{array}[]{ll}Ric_{\tilde{g}_{k}}-\displaystyle\frac{R_{\tilde{g}_{k}}}{2}\tilde{g}_{k}=\sqrt{r_{k}}T^{\tilde{g}_{k},\psi}\\\
\\\
D_{\tilde{g}_{k}}\psi=\displaystyle\frac{\lambda}{\sqrt{r_{k}}}\psi.\end{array}\right.$
(22)

Using the same procedure as above, knowing that now we have $Rm_{\tilde{g}}$
uniformly bounded in every ball $B_{r}\subset B(x_{k},r_{k}\delta)$ we have
the convergence of the space $(M,\tilde{g}_{k},x_{k})$ to
$(Y,\overline{g},x_{\infty})$ in the pointed Hausdorff topology. In fact, as
in [28], one can show that for every ball $B(x_{\infty},r)\subset Y$, there
exists a smooth diffeomorphism $\Phi_{k}:B(x_{\infty},r)\to M$, so that
$\Phi_{k}^{*}\tilde{g}_{k}$ converges to $\overline{g}$ in
$C^{\ell,\alpha}(B(x_{\infty},r))$ for all $\ell\in\mathbb{N}$. Moreover,
since $|Ric_{\tilde{g}_{k}}|^{2}=\frac{1}{r_{k}^{2}}|Ric_{g_{k}}|^{2}$, we see
that $\overline{g}$ is indeed a Ricci-flat, non-flat metric and from equation
$(\ref{equalim})$, $\psi_{k}$ converges to a harmonic spinor $\psi_{\infty}$
in $C^{\ell,\alpha}(B(x_{\infty},r))$ for every $r>0$. That is on
$(Y,\overline{g})$ we have that

$D_{\overline{g}}\psi_{\infty}=0.$

Using a result of [22, Lemma 5.1], we know that there $\psi_{\infty}$ is
asymptotic to a parallel spinor at infinity. Next, using [6, Theorem 1.5], we
have that the mass of $(Y,\overline{g})$ is zero. Thus, we see that
$\psi_{\infty}$ is actually parallel.
To prove the last point of Theorem 1.1, we use again the splitting
$M=G_{i}(r)\cup H_{i}(r)$. Thus we have

$\int_{M}|Rm_{g_{i}}|^{2}_{g_{i}}\
dv_{g_{i}}=\int_{G_{i}(r)}|Rm_{g_{i}}|^{2}_{g_{i}}\
dv_{g_{i}}+\int_{H_{i}(r)}|Rm_{g_{i}}|^{2}_{g_{i}}\ dv_{g_{i}}.$

First we notice that since the convergence of the metric is in the
$C^{k,\alpha}$ sense in $G_{i}(r)$,

$\liminf_{i\to\infty}\int_{G_{i}(r)}|Rm_{g_{i}}|^{2}_{g_{i}}\
dv_{g_{i}}=\int_{G_{r}}|Rm_{g_{\infty}^{r}}|^{2}_{g_{\infty}^{r}}\
dv_{g_{\infty}^{r}}.$

On the other hand, since $r$ is arbitrary and by scale invariance,

$\liminf_{i\to\infty}\int_{H_{i}(r)}|Rm_{g_{i}}|^{2}_{g_{i}}\
dv_{g_{i}}=\liminf_{i\to\infty}\int_{H_{i}(r_{i}r)}|Rm_{\tilde{g}_{i}}|^{2}_{\tilde{g}_{i}}\
dv_{\tilde{g}_{i}}\geq\sum_{k=1}^{|S|}\int_{Y_{k}}|Rm_{\overline{g}_{k}}|^{2}_{\overline{g}_{k}}dv_{\overline{g}_{k}},$

which leads to the desired inequality. ∎

We notice that in the case $n=3$ we have the compactness result as a corollary
from the uniform boundedness of the Ricci curvature proved in Proposition 3.2,
that is

######  Corollary 3.7.

The space $\mathcal{E}(D,c,K)$ is compact in dimension 3.

###### Proof.

The uniform bound on the Ricci curvature induces a uniform bound on the
curvature tensor in dimension 3. Hence, one has automatically a global version
of Proposition 3.5. Now combining this uniform bound with the diameter and the
volume bound, provides a uniform lower bound on the injectivity radius.The
limiting process is then similar to the one in dimension 4 and the desired
result follows (see for instance [3], [6], [28]). ∎

######  Remark 3.8.

Let us make the following remarks:

  * i)

our proof does not require the manifold $M$ to be fixed. Indeed, one can take
a sequence $(M_{k},g_{k},\psi_{k})$ and the limiting procedure still works
under the same kind of bounds, with an additional upper bound on
$b_{2}(M_{k})$, the second Betti number of $M_{k}$. The bound on $b_{2}$
allows to have a uniform upper bound on the $L^{2}$ norm of the curvature
tensor. The rest of the proof follows from Cheeger’s finiteness theorem (see
[11, 29]). That is $\mathcal{E}(D,c,K)$ contains only a finite number of
diffeomorphism classes;

  * ii)

we can also take the $\lambda$ to be varying in a compact set $[a,b]$ with
$a>0$. The case when $a=0$ is different in nature and was investigated in
[19];

  * iii)

the constant $c$ in the definition of $\mathcal{E}$ is of a variational nature
but it can be replaced by a non-collapsing geometric condition. That is
$Vol_{g}M\geq c$;

  * iv)

the introduction of the constant $K$, was necessary to get a bound on the
gradient of the spinor and hence a uniform bound on the Ricci. But notice that
one can impose another condition that would replace that bound. For instance
if $P^{g}$ is the Penrose operator defined by
$P^{g}_{X}\psi=\nabla_{X}\psi+\frac{1}{n}X\cdot D\psi$, one can impose that
$|P^{g}\psi|\leq K$. This condition combined with the equation of the spinor
(9) and the uniform bound of $\|\psi\|_{L^{\infty}}$ will lead to a uniform
bound on $\|\nabla\psi\|_{L^{\infty}}$. In fact, one can rewrite the first
equation of $(\ref{eq1})$ as follows:

$Ric_{g}-\frac{R_{g}}{n}g=\tilde{T}^{g,\psi}_{ij},$

where

$\tilde{T}^{g,\psi}_{ij}=-\frac{1}{4}[\langle e_{i}\cdot
P^{g}_{e_{j}}\psi,\psi\rangle+\langle e_{j}\cdot
P^{g}_{e_{i}}\psi,\psi\rangle];$

  * v)

As we mentioned above, the bound $-\Delta_{g}R_{g}\geq-KR_{g}$ can be replaced
by a more geometric condition in dimension 4. Namely, the $Q$-curvature
$Q_{g}$ is bounded from below by $-K$. Indeed, recall that
$Q_{g}=\frac{1}{6}(-\Delta_{g}R+R^{2}-3|Ric_{g}|^{2})$. Therefore, if
$Q_{g}\geq-K$, then

$-\Delta_{g}R\geq-6K-R^{2},$

and this is enough to carry out the estimates needed in Proposition 3.2.

##  4 Second Variation

In this section we want to compute the second variation of $E$ at a critical
point $(g,\psi)$. For the sake of simplicity, in the sequel sometimes we will
omit the dependance from $g$ and $\psi$ whenever there is no confusion. We
first recall the variation of some geometric quantities under the perturbation
of the metric. So we consider a variation of metrics $g(t)$ such that
$\frac{\partial g(t)}{\partial t}_{|t=0}=h$. Then one has:

$\left\\{\begin{array}[]{ll}\displaystyle\frac{\partial Ric_{ij}}{\partial
t}_{|t=0}=-\frac{1}{2}\Big{(}\Delta_{L}h_{ij}+\nabla_{i}\nabla_{j}tr(h)+\nabla_{i}(\delta
h)_{j}+\nabla_{j}(\delta h)_{i}\Big{)}\\\ \\\ \displaystyle\frac{\partial
R}{\partial t}_{|t=0}=-\Delta(tr(h))+\delta^{2}h-\langle
Ric,h\rangle,\end{array}\right.$

where $(\delta h)_{i}=-(divh)_{i}=-\nabla^{j}h_{ij}$,
$\delta^{2}(h)=\sum_{i,j=1}^{n}\nabla^{i}\nabla^{j}h_{ij}$ and $\Delta_{L}$ is
the Lichnerowicz Laplacia on 2-tensors. Now since we are dealing with
variations involving spinors, we need an effective way of differentiation
spinors. We state here the results developed in [7] (we also refer the reader
to [1, Section 4.2]). Consider the manifold $\tilde{M}=I\times M$, where $I$
is a small interval centered at zero, where the variation $g(t)$ is defined.
On $\tilde{M}$ we define the metric $\tilde{g}=dt^{2}+g(t)$. We denote by
$\nabla^{C}$ the Levi-Civita connection on $(\tilde{M},\tilde{g})$ and
”$\bullet$” the Clifford multiplication on $\Sigma\tilde{M}$. Then we have

$X\cdot\psi=\nu\bullet X\bullet\psi,$ (23)

where $\nu$ is the normal vector to the $M$ as a submanifold of $\tilde{M}$.
We can take $\nu=\frac{\partial}{\partial t}$. Therefore, if $X$ and $Y\in
TM$, we have

$\nabla^{C}_{X}Y=\nabla_{X}Y+\tilde{g}(W(X),Y)\nu,$ (24)

where $W$ is the Weingarten map, i.e. $W(x)=-\nabla_{X}^{C}\nu$. Now on the
spinor bundle the covariant derivative can be expressed as follow for
$\varphi\in\Sigma\tilde{M}_{|M}$ and $X\in TM$:

$\nabla^{C}_{X}\varphi=\nabla_{X}\varphi-\frac{1}{2}\nu\bullet
W(X)\bullet\varphi.$ (25)

In our case, as shown in [7], we have

$\left\\{\begin{array}[]{ll}\tilde{g}(W(X),Y)=-\frac{1}{2}h(X,Y)\\\ \\\
Rm_{\tilde{g}}(X,Y,U,\nu)=\frac{1}{2}\Big{(}(\nabla_{Y}h)(X,U)-(\nabla_{X}h)(Y,U)\Big{)}.\end{array}\right.$
(26)

Now, we are ready to compute the second variation of $E$. Let us call

$E_{1}(g,\psi)=Ric_{g}-\frac{R_{g}}{2}g-T^{g,\psi},$

and

$E_{2}(g,\psi)=D_{g}\psi-\lambda\psi.$

We first compute the variation with respect to $\psi$ since it is easier. So
we assume that $g$ is fixed and $\frac{\partial\psi}{\partial
t}_{|t=0}=\varphi$. Then we have

$\frac{\partial E_{1}}{\partial t}_{|t=0}=F(\psi,\varphi)\;,$

where the tensor $F$ is defined by

$F(X,Y)=\frac{1}{4}\Big{(}\langle
X\cdot\nabla_{Y}\varphi+Y\cdot\nabla_{X}\varphi,\psi\rangle+\langle
X\cdot\nabla_{Y}\psi+Y\cdot\nabla_{X}\psi,\varphi\rangle\Big{)}.$

Also,

$\frac{\partial E_{2}}{\partial t}_{|t=0}=D_{g}\varphi-\lambda\varphi.$

Now we move to the variation in the $g$ direction as described above. We start
by the term $\frac{\partial T_{ij}}{\partial t}$. So let us consider again the
quantity $S_{ij}=\langle e_{i}\cdot\nabla_{e_{j}}\psi,\psi\rangle$ and let us
compute its variation. Then using the metric $\tilde{g}$ and assuming that the
spinor $\psi$ is parallel transported along the variation and using the fact
that $\nabla_{\nu}^{C}\nu=0$ and $(\ref{bull})$, we have

$\displaystyle\frac{\partial S_{ij}}{\partial t}_{|t=0}$
$\displaystyle=\nabla_{\nu}^{C}\langle
e_{i}\cdot\nabla_{e_{j}}\psi,\psi\rangle_{g_{t}}=\langle
e_{i}\cdot\nabla_{\nu}^{C}\nabla_{e_{j}}\psi,\psi\rangle_{g_{t}}.$

But from $(\ref{spinbull})$ we have

$\displaystyle\nabla_{\nu}^{C}\nabla_{e_{j}}\psi$
$\displaystyle=\nabla_{\nu}^{C}\Big{(}\nabla_{e_{j}}^{C}\psi+\frac{1}{2}\nu\bullet
W(e_{j})\bullet\psi\Big{)}$
$\displaystyle=Rm_{\tilde{g}}(\nu,e_{j})\psi+\nabla^{C}_{[\nu,e_{j}]}\psi+\frac{1}{2}\nu\bullet\nabla^{C}_{\nu}W(e_{j})\bullet\psi$
$\displaystyle=Rm_{\tilde{g}}(\nu,e_{j})\psi+\nabla_{W(e_{j})}^{C}\psi+\frac{1}{2}\nu\bullet\nabla^{C}_{\nu}W(e_{j})\bullet\psi$
(27)
$\displaystyle=Rm_{\tilde{g}}(\nu,e_{j})\psi+\nabla_{W(e_{j})}\psi+\frac{1}{2}\nu\bullet[\nabla^{C}_{\nu}W(e_{j})-W^{2}(e_{j})]\bullet\psi$
$\displaystyle=Rm_{\tilde{g}}(\nu,e_{j})\psi+\nabla_{W(e_{j})}\psi+\frac{1}{2}\nu\bullet[Rm_{\tilde{g}}(e_{j},\nu)\nu]\bullet\psi$
(28)

where here we use the Riccati equation for the Weingarten map
$(\nabla_{\nu}^{C}W)(X)=Rm_{\tilde{g}}(X,\nu)\nu+W^{2}(X)$. Now we have (with
the convention that $e_{0}=\nu$),

$Rm_{\tilde{g}}(\nu,e_{j})\psi=-\frac{1}{2}\sum_{0\leq p<\ell\leq
n}Rm_{\tilde{g}}(e_{p},e_{\ell},\nu,e_{j})e_{p}\bullet e_{\ell}\bullet\psi$

and

$Rm_{\tilde{g}}(e_{j},\nu)\nu\bullet\psi=Rm_{\tilde{g}}(e_{j},\nu,\nu,e_{k})e_{k}\bullet\psi.$

Hence

$Rm_{\tilde{g}}(\nu,e_{j})\psi-\frac{1}{2}\nu\bullet[Rm_{\tilde{g}}(e_{j},\nu)\nu]\bullet\psi=-\frac{1}{2}\sum_{1\leq
p<\ell\leq n}Rm_{\tilde{g}}(e_{p},e_{\ell},\nu,e_{j})e_{p}\bullet
e_{\ell}\bullet\psi.$

But using $(\ref{meth})$, we have

$Rm_{\tilde{g}}(e_{p},e_{\ell},\nu,e_{j})=-\frac{1}{2}(\nabla_{\ell}(h)_{pj}-\nabla_{p}(h)_{\ell
j}).$

and $W(e_{j})_{k}=-\frac{1}{2}h_{jk}$. Therefore,

$\nabla_{\nu}^{C}\nabla_{e_{j}}\psi=\frac{1}{4}\sum_{1\leq p<\ell\leq
n}\Big{(}(\nabla_{\ell}h)_{pj}-(\nabla_{p}h)_{\ell j}\Big{)}e_{p}\cdot
e_{\ell}\cdot\psi-\frac{1}{2}h_{jk}\nabla_{e_{k}}\psi.$

We finally notice that

$\langle e_{i}\cdot\nabla_{W(e_{j})}\psi,\psi\rangle=-\frac{1}{2}h_{jk}\langle
e_{i}\cdot\nabla_{e_{k}}\psi,\psi\rangle=-\frac{1}{2}(h\times S)_{ij},$

where here we set $(A\times B)_{ij}=\sum_{k=1}^{n}A_{ik}B_{kj}$. Also,
rewriting the term below

$\frac{1}{4}\langle e_{i}\cdot\sum_{1\leq p<q\leq
n}\Big{(}(\nabla_{p}h)_{qj}-(\nabla_{q}h)_{pj}\Big{)}e_{p}\cdot
e_{q}\cdot\psi,\psi\rangle=\frac{1}{4}\sum_{p\not=q\not=i}(\nabla_{p}h)_{qj}\langle
e_{i}\cdot e_{p}\cdot e_{q}\cdot\psi,\psi\rangle,$

We can define a $3$-tensor $Q$ by $Q(X,Y,Z)=\langle(X\wedge Y\wedge
Z)\cdot\psi,\psi\rangle$ in a way that, we can write the later term as
$\frac{1}{4}(\nabla h)\times Q$. Therefore, we have

$\frac{\partial T_{ij}}{\partial t}_{|t=0}=-\frac{1}{2}(h\times
T)_{ij}-\frac{1}{16}(\nabla h\times\tilde{Q})_{ij},$

where $\tilde{Q}$ is the symmetrization of $Q$. But since $Q$ is totally skew-
symmetric, we have that $\tilde{Q}=0$. Now, given a symmetric 2-tensor $h$, we
define the operator $\mathcal{D}^{h}$ by

$\mathcal{D}^{h}\varphi=\sum_{i,j=1}^{n}h_{ij}e_{i}\cdot\nabla_{e_{j}}\varphi.$

With this notation we have

$\frac{\partial D_{g}\varphi}{\partial
t}_{|t=0}=-\frac{1}{2}\mathcal{D}^{h}\varphi+\frac{1}{4}\nabla(tr(h))\cdot\varphi-\frac{1}{4}\delta(h)\cdot\varphi.$

To summarize the previous computations, if we denote by $\nabla\nabla tr(h)$
the tensor defined by $(\nabla\nabla tr(h))_{ij}=\nabla_{i}\nabla_{j}tr(h)$
and $sym(\nabla(\delta h))$ the symmetric tensor defined by
$(sym(\nabla(\delta h)))_{ij}=\nabla_{i}(\delta h)_{j}+\nabla_{j}(\delta
h)_{i}$, we write the second variation of $E$:

$\displaystyle\nabla^{2}E(g,\psi)[(h,\varphi),(h,\varphi)]$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h+\nabla\nabla
tr(h)+sym(\nabla(\delta h)),h\rangle$ $\displaystyle+\frac{1}{2}\left(-\Delta
tr(h)+\delta^{2}h-\langle Ric,h\rangle\right)tr(h)+\frac{R}{2}|h|^{2}$
$\displaystyle-\frac{1}{4}\langle\mathcal{D}^{h}\psi,\psi\rangle
tr(h)+\frac{1}{2}\langle h\times T,h\rangle$
$\displaystyle+2\langle-\frac{1}{2}\mathcal{D}^{h}\psi+\frac{1}{4}\nabla
tr(h)\cdot\psi-\frac{1}{4}\delta h\cdot\psi,\varphi\rangle$
$\displaystyle+2\langle F(\varphi,\psi),h\rangle+2\langle
D\varphi-\lambda\varphi,\varphi\rangle\ dv.$

We have

$\langle
F(\varphi,\psi),h\rangle=\frac{1}{2}\langle\mathcal{D}^{h}\psi,\varphi\rangle+\frac{1}{2}\langle\mathcal{D}^{h}\varphi,\psi\rangle$

and

$-\frac{1}{2}\langle\mathcal{D}^{h}\psi,\psi\rangle=\langle T,h\rangle.$

This will allow us to finalize the proof of our result.

###### Proof.

(of Theorem 1.2) Let us take now a deformation of $g$ that is transverse to
the diffeomorphism action, that is $\delta h=0$. Then in the previous formula
we have

$\displaystyle\nabla^{2}E(g,\psi)[(h,\varphi),(h,\varphi)]$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h+\nabla\nabla
tr(h),h\rangle$ $\displaystyle+\frac{1}{2}\Big{(}-\Delta tr(h)-\langle
Ric_{g},h\rangle\Big{)}tr(h)+\frac{R_{g}}{2}|h|^{2}$
$\displaystyle+\frac{1}{2}\langle T^{g,\psi},h\rangle tr(h)+\frac{1}{2}\langle
h\times T^{g,\psi},h\rangle+\frac{1}{2}\langle\nabla
tr(h)\cdot\psi,\varphi\rangle$
$\displaystyle+\langle\mathcal{D}^{h}\varphi,\psi\rangle+2\langle
D_{g}\varphi-\lambda\varphi,\varphi\rangle\ dv.$

∎

Now we move to the case of horizontal deformations.

###### Proof.

(of Corollary 1.3)

A horizontal deformation is a deformation of the metric while the spinorial
part is fixed. Namely, $\varphi=0$. First recall that

$R_{g}=\frac{\lambda}{n-2}|\psi|^{2}.$

Since the right hand side is invariant under the deformation of the metric, we
have that the variation of $R_{g}$ is zero. Hence

$-\Delta tr(h)-\langle Ric_{g},h\rangle=0.$

On the other hand, an Einstein-Dirac deformation, is defined by the vanishing
of $\nabla^{2}E$. Hence,

$\Delta_{L}h+\nabla\nabla tr(h)+R_{g}h+\langle T^{g,\psi},h\rangle g+h\times
T^{g,\psi}=0$

and if we take the trace of this last equation, we get

$2\Delta tr(h)+R_{g}tr(h)+n\langle T^{g,\psi},h\rangle+\langle
T^{g,\psi},h\rangle=0.$ (29)

But

$\langle T^{g,\psi},h\rangle=\langle Ric_{g}-\frac{R_{g}}{2}g,h\rangle=-\Delta
tr(h)-\frac{R_{g}}{2}tr(h).$

Thus, equation (29) becomes

$(1-n)\Delta tr(h)+\frac{R_{g}}{2}(1-n)tr(h)=0,$

or equivalently

$-\Delta tr(h)-\frac{R_{g}}{2}tr(h)=0.$

Since $\langle Ric_{g},h\rangle=-\Delta tr(h)$, the conclusion follows. ∎

######  Remark 4.1.

Some comments are in order:

  * i)

We notice that a conformal variation of the metric cannot be an Einstein-Dirac
deformation. Indeed, if $h=ug$ then the second condition yields

$\lambda u|\psi|^{2}=0,$

hence if $u$ is smooth, then it must be zero.

  * ii)

The functional $E$ that we are studying has $\lambda$ as a fixed parameter and
the Euler-Lagrange equation asserts that $\lambda$ must be an eigenvalue for
$D_{g}$ for all critical metrics $g$. In particular if one studies Einstein-
Dirac deformations, one needs to take into consideration the stability of the
eigenvalue $\lambda$. We recall from [10, Theorem 24] that the variation of an
eigenvalue of the Dirac operator under the metric reads as

$\frac{\partial\lambda}{\partial t}_{|t=0}=-\frac{1}{2}\int_{M}\langle
T^{g,\psi},h\rangle\ dv.$

Hence, our perturbations $h$ need to satisfy the integral version of $\langle
T^{g,\psi},h\rangle=0$.

  * iii)

Let us go back to the case of general deformation and focus on the ones
satisfying $\delta h=tr(h)=0$. This is motivated by the following: if $M$ is
not diffeomorphic to the sphere then $T_{g}S^{2}(M)=C^{\infty}(M)g\oplus
Im(\delta_{g}^{*})\oplus(\ker(\delta_{g})\cap tr_{g}^{-1}(0))$. The first
factor corresponds to conformal changes of the metric and in this setting the
problem was investigated in [27, 16, 9]. The second factor corresponds to the
tangent space of the orbits of the group $Diff_{0}(M)$. So we want to focus on
the third factor. Then we have

$\displaystyle\nabla^{2}E(g,\psi)[(h,\varphi),(h,\varphi)]$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h,h\rangle+\frac{R_{g}}{2}|h|^{2}$
$\displaystyle+\frac{1}{2}\langle h\times
T^{g,\psi},h\rangle+\langle\mathcal{D}^{h}\varphi,\psi\rangle+2\langle
D_{g}\varphi-\lambda\varphi,\varphi\rangle\ dv.$

##  5 Some examples

###  5.1 Example 1: real Killing spinors

We consider a real Killing spinor $\psi$, that is $\nabla_{X}\psi=-\mu
X\cdot\psi$ for a real constant $\mu>0$, for all tangent vector $X$. In this
case $g$ has to be Einstein and

$D\psi=-\mu\sum_{i}e_{i}\cdot e_{i}\cdot\psi=n\mu\psi.$

In particular $\lambda=n\mu$. Also

$T_{ij}=-\frac{1}{4}[\langle-\mu e_{i}\cdot e_{j}\cdot\psi-\mu e_{j}\cdot
e_{i}\cdot\psi,\psi\rangle]=\frac{1}{2}\mu|\psi|^{2}g_{ij},$
$R=4n(n-1)\mu^{2}=\frac{n\mu}{n-2}|\psi|^{2}.$

Thus $|\psi|^{2}=4(n-1)(n-2)\mu$. Now we compute the terms in the second
variation:

$\langle T\times h,h\rangle=\frac{\mu|\psi|^{2}}{2}\langle
g_{ik}h_{jk}.h\rangle=\frac{\mu|\psi|^{2}}{2}|h|^{2}=\frac{n-2}{2n}R|h|^{2}=2(n-1)(n-2)\mu^{2}|h|^{2},$

and

$\mathcal{D}^{h}\psi=-\mu\sum_{i,j}h_{ij}e_{i}\cdot e_{j}\cdot\psi=\mu
tr(h)\psi.$

In particular if $tr(h)=0$ we have $\mathcal{D}^{h}\psi=0$. So the second
variation reads

$\displaystyle\nabla^{2}E(g,\psi)[(h,\varphi),(h,\varphi)]$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h,h\rangle+\frac{R}{2}|h|^{2}+\frac{n-2}{4n}R|h|^{2}+2\langle
D\varphi-\lambda\varphi,\varphi\rangle dv$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h,h\rangle+\frac{3n-2}{4n}R|h|^{2}+2\langle
D\varphi-\lambda\varphi,\varphi\rangle dv.$

###  5.2 Example 2: Quasi-Killing spinors

Now we consider a Sasakian spin-manifold $(M^{2m+1},g,\phi,\eta)$. A quasi-
Killing spinor satisfies

$\nabla_{X}\psi=aX\cdot\psi+b\eta(X)\xi\cdot\psi,$

where $\xi$ is the Reeb vector field of $\eta$. We recall from [20, Theorem
6.6.] that if $M$ is simply connected with

$Ric=\frac{2-m}{m-1}g+\frac{2m^{2}-m-2}{m-1}\eta\otimes\eta,$

then it does carry a WK-spinor (weak Killing spinor) built from quasi-Killing
spinors with

$a=\pm\frac{1}{2},\quad b=\mp\frac{2m^{2}-m-2}{4(m-1)}.$

So let us compute the second variation at such solutions. For the sake of
simplicity we will do the computations for $a=\frac{1}{2}$ and
$b=-\frac{2m^{2}-m-2}{4(m-1)}$. Notice that in this case we have
$R=\frac{2m}{m-1}$. Also, if $\psi$ is an $(a,b)$-quasi-Killing spinor then it
is an eigenspinor of the Dirac operator with eigenvalue $\lambda=-(2m+1)a-b$.
We calculate now the values of the tensor $T$:

$T_{ij}=-2g_{ij}+\frac{2m^{2}-m-2}{m-1}\eta\oplus\eta.$

Also:

$\mathcal{D}^{h}\psi=\sum_{i,j}h_{ij}e_{i}\cdot\nabla_{j}\psi=b\langle
h(\xi)\cdot\xi\cdot\psi,\psi\rangle,$

where we considered $h$ as an endomorphism, that is
$h(\xi)=\sum_{i=1}^{2m+1}h_{i,2m+1}e_{i}$. Now, we compute

$\langle h\times T,h\rangle=-2|h|^{2}+\frac{2m-m-2}{m-1}|h(\xi)|^{2}.$

Replacing it in the formula for the second variation of $E$, it yields

$\displaystyle\nabla^{2}E(g,\psi)[(h,\varphi),(h,\varphi)]$
$\displaystyle=\int_{M}\frac{1}{2}\langle\Delta_{L}h,h\rangle+\frac{m+1}{m-1}|h|^{2}+\frac{2m^{2}-m-2}{2(m-1)}|h(\xi)|^{2}$
$\displaystyle-\frac{2m^{2}-m-1}{4(m-1)}\langle
h(\xi)\cdot\xi\cdot\psi,\varphi\rangle+2\langle
D\varphi-\lambda\varphi,\varphi\rangle\ dv.$

One might notice the difference here compared to the case of real Killing
spinors: if one wants to consider variations that preserves the associate
metric to the contact structure $\eta$, then we necessarily have $h(\xi)=0$
and the second variation takes a similar form as the one of a Killing spinor.

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