# Rigidity of complex projective spaces in Ricci shrinkers

Yu Li111Supported by YSBR-001 and a research fund from USTC.  and Wenjia Zhang

###### Abstract

In this paper, we prove that any Ricci shrinker that is sufficiently close to
$(\mathbb{C}\text{P}^{N},g_{FS})$ in the Gromov-Hausdorff sense must itself be
isometric to $(\mathbb{C}\text{P}^{N},g_{FS})$.

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Proof of the main theorem
  4. 4 Further discussion
  5. A The variational formulae

##  1 Introduction

A Ricci shrinker $(M^{n},g,f)$ is a complete Riemannian manifold $(M^{n},g)$
together with a smooth function $f:M\to\mathbb{R}$ such that

$Rc+\nabla^{2}f=\frac{1}{2}g,$ (1.1)

where the potential function $f$ is normalized by adding a constant such that

$\displaystyle(4\pi)^{-\frac{n}{2}}\int e^{-f}\,dV_{g}=1.$ (1.2)

The Ricci shrinkers play essential roles in studying the singularities of the
Ricci flow. In dimension $2$ or $3$, all Ricci shrinkers are completely
classified (cf. [9][23][25][4], etc). It turns out that
$\mathbb{R}^{2},S^{2},\mathbb{R}^{3},S^{3},S^{2}\times\mathbb{R}$, and their
quotients are the only examples.

In higher dimensional cases, much less is known. One strategy is to consider
all Ricci shrinkers as one moduli space $\mathcal{M}$ equipped with pointed-
Gromov-Hausdorff topology, where one can choose a minimum point of the
potential function $f$ as the base point. Under the natural non-collapsing
condition, the moduli space $\mathcal{M}$ has the weak-compactness, in the
sense that any sequence of Ricci shrinkers with uniform entropy bound, by
taking a subsequence, will converge to a Ricci shrinker limit space. In
dimension 4, it was proved by Haslhofer-Müller [10, 11] that the limit is a
smooth Ricci shrinker orbifold. In the general dimension, it was proved by Li-
Li-Wang [20], and Huang-Li-Wang [12] that the limit space is Ricci shrinker
conifold.

One can raise the natural question: What kind of Ricci shrinker is isolated in
$\mathcal{M}$?

In other words, the above question discusses the characterization of the
rigidity of Ricci shrinkers. Here, a Ricci shrinker is rigid if there is no
nearby Ricci shrinker other than itself. Rigid examples include spherical
space-forms $S^{n}/\Gamma$ [13], $\mathbb{C}\text{P}^{2m}$ by Kröncke [17],
and $S^{2}\times S^{2}$ proved by Sun-Zhu [29] recently. For other rigid
compact symmetric spaces, see [1] and those with
$\lambda^{-1}\mu_{\text{fns}}>2$ and H. stable in [5, Table 1, Table 2]. For
noncompact Ricci shrinkers, the rigidity problem is much more involved. Some
noncompact rigid examples include $\mathbb{R}^{n}$, proved by Yokota [30][31]
(see also Li-Wang [21]), $S^{n-1}\times\mathbb{R}$ proved by Li-Wang [22], and
$S^{n-k}\times\mathbb{R}^{k}$ by Colding-Minicozzi [6] lately.

In this paper, we will only consider the rigidity of compact Ricci shrinkers
and prove that the complex projective space $\mathbb{C}\text{P}^{N}$ with
standard metric is rigid.

######  Theorem 1.1 (Main Theorem).

For any integer $N\geq 1$, there exists a small constant
$\epsilon=\epsilon(N)>0$ satisfying the following property.

Suppose $(M^{n},g,f)$ is a Ricci shrinker such that

$\displaystyle
d_{GH}\left\\{(M^{n},g),(\mathbb{C}\emph{P}^{N},g_{FS})\right\\}<\epsilon,$
(1.3)

then $(M^{n},g)$ is isometric to $(\mathbb{C}\emph{P}^{N},g_{FS})$. Here,
$n=2N$ and $g_{FS}$ is the Fubini-Study metric with Einstein constant $1/2$.

The proof of Theorem 1.1 relies on the weak-compactness theory of Ricci
shrinkers developed by Li-Li-Wang [20] and the deformation theory of Ricci
shrinkers by Podestà-Spiro [27] and Kröncke [17]. We sketch the proof as
follows.

It follows from [20, Theorem 1.1] that any Ricci shrinker $(M^{n},g,f)$
satisfying (4.14) for sufficiently small $\epsilon$ is close to
$(\mathbb{C}\text{P}^{N},g_{FS})$ in the Sobolev space $H^{s}$ for large $s$.
In particular, $M$ is diffeomorphic to $\mathbb{C}\text{P}^{N}$ and $g$ is in
a small $H^{s}$ neighborhood of $g_{FS}$. Therefore, one only needs to show
that $g=\phi^{*}g_{FS}$ for a self-diffeomorphism $\phi$ of
$\mathbb{C}\text{P}^{N}$. In other words, $(\mathbb{C}\text{P}^{N},g_{FS})$ is
rigid in the sense of [17, Definition 4.1]. Notice that if we further assume
$(M^{n},g,f)$ is a Kähler Ricci shrinker, then the conclusion already follows
from the classification of Kähler Ricci shrinker with positive bisectional
curvature, see [24, Theorem 3(i)].

From [27, Proposition 2.2], there exists a slice $\mathcal{S}^{s}_{f}$ at
$g_{FS}$ such that for a small $H^{s}$ neighborhood $\mathcal{U}$ of $g_{FS}$,
any $g\in\mathcal{U}$ is isometric to a unique metric
$\tilde{g}\in\mathcal{S}^{s}_{f}$. Moreover, $\mathcal{S}^{s}_{f}$ is a smooth
manifold with tangent space
$T_{g_{FS}}\mathcal{S}^{s}_{f}=\text{ker}(\delta_{g_{FS}})$. In addition, it
follows from [27, Theorem 3.4] that all Ricci shrinkers in
$\mathcal{S}^{s}_{f}$ forms a real-analytic subset of a finitely-dimensional
real-analytic submanifold.

Suppose $(\mathbb{C}\text{P}^{N},g_{FS})$ is not rigid, then there exists a
smooth curve $g_{t}$ of Ricci shrinkers in $\mathcal{S}^{s}_{f}$ starting from
$g_{FS}$. In particular, $g_{t}$ satisfies the Ricci shrinker equation (1.1)
at $t=0$ up to any order. Our strategy is to show that this cannot happen.
More precisely, we will prove that any infinitesimal solitonic deformation of
$(\mathbb{C}\text{P}^{N},g_{FS})$ is not integrable of at most third-order.
Notice that if $N=2m$, Kröncke [17, Theorem 6.1] has already proved that
$(\mathbb{C}\text{P}^{2m},g_{FS})$ is not integrable of second-order. Thus, we
will focus on the case $N=2m-1$, and prove that
$(\mathbb{C}\text{P}^{2m-1},g_{FS})$ is not integrable of third-order and
hence is rigid. We remark that even though $(\mathbb{C}\text{P}^{N},g_{FS})$
is rigid, it is dynamically unstable under the Ricci flow [18, Corollary 1.8].

This paper is organized as follows. Section $2$ recalls Ricci shrinkers’
deformation theory, which will be used throughout the paper. Moreover, we
determine all possible infinitesimal solitonic deformations. In section $3$,
we obtain all possible second-order deformations by solving the second-order
Ricci shrinker equation. Combined with the first-order deformation, we prove
that none is integrable of third-order, and hence the main theorem is proved.
In the last section, we discuss the rigidity of the product Ricci shrinker
with the complex projective space as a factor.

Acknowledgements: Both authors would like to thank Professor Bing Wang for his
interest in this work.

##  2 Preliminaries

Let $(M^{n},\bar{g},\bar{f})$ be a fixed Ricci shrinker on a closed manifold
$M^{n}$ satisfying (1.1) and (1.2) and $s$ is an integer with $s>n/2+3$.

From Sun-Wang [28, Lemma 2.2], there exists a $H^{s}$ neighborhood
$\mathcal{U}$ of $\bar{g}$, such that for any $g\in\mathcal{U}$, the minimizer
of $\boldsymbol{\mu}(g,1)$ is unique and depends real-analytically on $g$.
Here, $\boldsymbol{\mu}(g,1)$ is the celebrated functional defined by Perelman
[26]. In particular, if we denote the minimizer by $f=f(g)$, then $f$
satisfies

$\displaystyle(4\pi)^{-\frac{n}{2}}\int e^{-f}\,dV_{g}=1$

and the Euler-Lagrange equation

$\displaystyle 2\Delta f-|\nabla f|^{2}+R(g)+f-n=\boldsymbol{\mu}(g,1).$

On $\mathcal{U}$, we have the following definition.

######  Definition 2.1.

The Ricci shrinker operator is defined as

$\displaystyle\Phi(g):=\frac{1}{2}g-Rc(g)-\nabla^{2}_{g}f$

for any $g\in\mathcal{U}$. In particular, $\Phi(g)=0$ if and only if $(M,g,f)$
is a Ricci shrinker.

Next, we recall the following notations.

######  Notation 2.2.

Let $(M^{n},g,f)$ be a Riemannian manifold coupled with a smooth function $f$.

  1. (i)

$\mathcal{S}^{p}(M)$ consists of smooth symmetric $p$-forms.

  2. (ii)

The weighted divergence $\delta_{f}:C^{\infty}(\mathcal{S}^{p}(M))\rightarrow
C^{\infty}(\mathcal{S}^{p-1}(M))$ is defined as

$\displaystyle(\delta_{f}T)(X_{1},\cdots,X_{p-1})=\sum_{i=1}^{n}\nabla_{e_{i}}T(e_{i},X_{1},\cdots,X_{p-1})-T(\nabla
f,X_{1},\cdots,X_{p-1})$

for any $T\in\mathcal{S}^{p}(M)$, where $\\{e_{i}\\}$ is an orthonormal basis.

  3. (iii)

$\delta_{f}^{*}$ is the formal adjoint of $\delta_{f}$ with respect to the
form $e^{-f}dV_{g}$. More precisely,

$\displaystyle(\delta_{f}^{*}T)(X_{1},\cdots,X_{p})=-\frac{1}{p}\sum_{i=0}^{p-1}\nabla_{X_{1+i}}T(X_{2+i},\cdots,X_{p+1})$

for any $T\in\mathcal{S}^{p-1}(M)$, where the sums $1+i,\cdots,p+i$ are taken
modulo $p$.

If $f$ is a constant, then we omit $f$ in $\delta_{f}$ and $\delta_{f}^{*}$.

For $(M^{n},\bar{g},\bar{f})$, we recall the decomposition
$S^{2}(M)=\text{ker}(\delta_{\bar{f}})\oplus\text{im}(\delta_{\bar{f}}^{*})$.
Moreover, one has the following slice-theorem proved in [27, Proposition 2.2],
which can be regarded as a generalization of the classic Ebin’s slice theorem
[8].

######  Proposition 2.3.

There exists a submanifold $S_{\bar{f}}^{s}$ of all $H^{s}$ metrics satisfying
the following properties:

  1. a)

There is a small $H^{s}$ neighborhood $\mathcal{U^{\prime}}$ of $\bar{g}$ in
the set of metrics such that any $g\in\mathcal{U^{\prime}}$ is isometric to a
unique metric $\tilde{g}\in\mathcal{S}^{s}_{\bar{f}}$.

  2. b)

$\mathcal{S}^{s}_{\bar{f}}$ is a smooth manifold with tangent space
$T_{\bar{g}}\mathcal{S}^{s}_{\bar{f}}=\emph{ker}(\delta_{\bar{f}})$.

By Proposition 2.3, one may focus on the slice $\mathcal{S}^{s}_{\bar{f}}$ and
all Ricci shrinkers nearby on the slice can be represented as

$\displaystyle\mathscr{S}:=\left\\{g\in\mathcal{S}^{s}_{\bar{f}}\mid\Phi(g)=0\right\\}.$

Moreover, it follows from [27, Theorem 3.4] that $\mathscr{S}$ is a real-
analytic subset of a finitely-dimensional real-analytic submanifold of
$\mathcal{S}^{s}_{\bar{f}}$. Therefore, if $(M,\bar{g},\bar{f})$ is not rigid,
then there exists a nontrivial smooth curve $g_{t}\subset\mathscr{S}$ with
$g_{0}=\bar{g}$.

For any integer $k\geq 1$, from

$\displaystyle\partial_{t^{k}}^{k}\rvert_{t=0}\Phi(g_{t})=0$

we have

$\displaystyle\Phi^{\prime}(g^{(k)})+\sum_{l=2}^{k}\sum_{1\leq
k_{1}\leq\cdots\leq
k_{l},k_{1}+\cdots+k_{l}=k}C(k,l,k_{1},\cdots,k_{l})\Phi^{(l)}(g^{(k_{1})},\cdots,g^{(k_{l})})=0$
(2.1)

where $g^{(i)}=\partial_{t^{i}}^{i}\rvert_{t=0}g_{t}\in S^{2}(M)$ and
$\Phi^{(i)}$ is the $i$-th variation of $\Phi$. Moreover,
$C(k,l,k_{1},\cdots,k_{l})$ are constants depending only on
$k,l,k_{1},\cdots,k_{l}$. In particular, we have

$\displaystyle
g^{\prime}\in\text{ker}(\Phi^{\prime})\cap\text{ker}(\delta_{\bar{f}})=:\text{ISD}$

where the last notation stands for infinitesimal solitonic deformation.

Conversely, we have the following definition.

######  Definition 2.4.

For any nontrivial $h=g^{(1)}\in\emph{ISD}$. We call $h$ integrable up to
order $k$, if there exists a sequence of tensors $g^{(2)},\cdots,g^{(k)}\in
S^{2}(M)$ satisfying (2.1).

Notice that by [17, Lemma 5.2], if $h$ is integrable up to any order, then
there exists a smooth curve $g_{t}\in\mathscr{S}$ with
$\partial_{t}\rvert_{t=0}g_{t}=h$. In particular, $(M,\bar{g},\bar{f})$ is not
rigid.

In this paper, we are only interested in the special case that
$(M,\bar{g},\bar{f})$ is Einstein, that is, $\bar{f}$ is constant. We first
recall the following first variation of $\Phi$.

######  Lemma 2.5.

For any $h\in S^{2}(M)$

$2\Phi^{\prime}(h)=\Delta h+2Rm(h)+2\delta^{*}\delta
h+\nabla^{2}(H-2f^{\prime})$ (2.2)

where $H=\emph{Tr}(h)$ and $f^{\prime}$ is determined by

$(\Delta+\frac{1}{2})(H-2f^{\prime})=\delta^{2}h.$ (2.3)

Moreover, $\Phi^{\prime}$ is a self-adjoint operator on $S^{2}(M)$ preserving
the decomposition $S^{2}(M)=\emph{ker}(\delta)\oplus\emph{im}(\delta^{*})$. In
particular, for $h\in\emph{ker}(\delta)$, we have

$2\Phi^{\prime}(h)=\Delta h+2Rm(h)\quad\text{and}\quad H=2f^{\prime}.$

###### Proof.

The proofs of formulae (2.2) and (2.3) can be found, e.g., in [7, Lemma 2.3]
and [7, (2.15)] respectively. Notice that $H-2f^{\prime}$ is uniquely
determined since the first eigenvalue of $\Delta$ is greater than $n/2(n-1)$
by Lichnerowicz [19]. Moreover, it is immediately known from (2.2) and (2.3)
that $\Phi^{\prime}$ is a self-adjoint operator.

Since $\Phi(\bar{g})=0$, we conclude that $\Phi(\phi^{*}\bar{g})=0$ for any
self-diffeomorphism $\phi$ of $M$. In particular,
$\text{im}(\delta^{*})\subset\text{ker}(\Phi^{\prime})$. In addition, if
$h\in\text{ker}(\delta)$, one has $2\Phi^{\prime}(h)=\Delta h+2Rm(h)$ and
hence

$\displaystyle 2\delta(\Phi^{\prime}(h))=\delta(\Delta
h+2Rm(h))=(\Delta+\frac{1}{2})(\delta h)=0.$

In sum, the proof is complete. ∎

From Lemma 2.5, it is easy to obtain the following decomposition, see [15,
Lemma 6.2] for details.

######  Lemma 2.6.

We have

$\displaystyle\emph{ISD}=\emph{IED}\oplus\\{u\bar{g}+2\nabla^{2}u\mid u\in
C^{\infty}(M),\Delta u+u=0\\},$

where _IED_ consists of all TT _(traceless-transverse)_ $h\in S^{2}(M)$ such
that $\Delta h+2Rm(h)=0$.

Now, we focus on the case $(M,\bar{g})=(\mathbb{C}\text{P}^{N},g_{FS})$. It
follows immediately from [14] that there is no nontrivial infinitesimal
Einstein deformation. In other words, $\text{IED}=0$. Therefore, we only need
to consider the conformal variation. Next, we recall the following result of
Kröncke [17, Theorem 5.7].

######  Proposition 2.7.

Let $(M,\bar{g})$ be an Einstein manifold with Einstein constant $1/2$. Let
$u\in C^{\infty}(M)$ be such that $\Delta u+u=0$. Then
$u\bar{g}+2\nabla^{2}u\in\emph{ISD}$ is not integrable of second-order if
there exists another function $w\in C^{\infty}(M)$ with $\Delta w+w=0$ such
that

$\int_{M}u^{2}w\neq 0.$

######  Remark 2.8.

In Proposition 2.7, if we further assume $\emph{IED}=0$, then
$u\bar{g}+2\nabla^{2}u$ is not integrable of second-order only if there exists
another function $w$ with $\Delta w+w=0$ such that

$\int_{M}u^{2}w\neq 0.$

By using Proposition 2.7, Kröncke [17, Theorem 6.1] proved the rigidity of
$(\mathbb{C}\text{P}^{2m},g_{FS})$:

######  Theorem 2.9.

All infinitesimal solitonic deformations of $(\mathbb{C}\emph{P}^{2m},g_{FS})$
are not integrable of second-order. Therefore,
$(\mathbb{C}\emph{P}^{2m},g_{FS})$ is rigid.

Notice that if $N=2m-1$, there indeed exists $u$ with $\Delta u+u=0$ such that
$\int u^{2}w=0$ for any $w$ with $\Delta w+w=0$. In fact, we have the
following precise characterization of $u$.

######  Lemma 2.10.

On $(\mathbb{C}\text{P}^{2m-1},g_{FS})$, if there exists a function $u$ with
$\Delta u+u=0$ such that

$\displaystyle\int u^{2}w=0$

for any $w$ with $\Delta w+w=0$. Then, after a possible change of coordinates,

$\displaystyle
u([z_{0},z_{1},\cdots,z_{2m-1}])=\lambda\left(\frac{|z_{0}|^{2}+\cdots+|z_{m-1}|^{2}-|z_{m}|^{2}-\cdots-|z_{2m-1}|^{2}}{|z_{0}|^{2}+\cdots+|z_{2m-1}|^{2}}\right)$
(2.4)

for a constant $\lambda$.

###### Proof.

Let $P_{k,k}$ be the space of polynomials on $\mathbb{C}^{2m}$ which are
homogeneous of degree k in $z$ and $\bar{z}$, and let $H_{k,k}$ the subspace
of harmonic polynomials in $P_{k,k}$. It is well-known that
$P_{k,k}=H_{k,k}\oplus r^{2}P_{k-1,k-1}$ and any function $u$ with $\Delta
u+u=0$ on $\mathbb{C}\text{P}^{2m-1}$ can be lifted to a function $f\in
H_{1,1}$.

By a change of coordinates, we may assume $f=\sum\lambda_{i}|z_{i}|^{2}$ and
$\sum\lambda_{i}=0$. Since

$\displaystyle f^{2}\in P_{2,2}=H_{2,2}\oplus
r^{2}H_{1,1}\oplus\mathbb{R}\cdot r^{4},$

it follows from our assumption that $f^{2}\in H_{2,2}\oplus\mathbb{R}\cdot
r^{4}$ and hence $\Delta_{E}f^{2}\in\mathbb{R}\cdot\Delta_{E}(r^{4})$. On the
one hand,

$\displaystyle\Delta_{E}f^{2}=8\sum\lambda_{i}^{2}|z_{i}|^{2}.$ (2.5)

On the other hand, we have

$\displaystyle\Delta_{E}r^{4}=8(m+1)r^{2}.$ (2.6)

Comparing (2.5) with (2.6), we conclude that $|\lambda_{i}|$ are the same and
the conclusion follows. ∎

######  Convention 2.11.

For later computations, we require

  1. a)

The scalar curvature $R$, the operator $\Delta$, etc., are the concepts in
real Riemannian geometry.

  2. b)

The curvature operator in local coordinates are given by
$R_{abcd}=\langle[\nabla_{\partial_{a}},\nabla_{\partial_{b}}]\partial_{c}-\nabla_{[\partial_{a},\partial_{b}]}\partial_{c},\partial_{d}\rangle$.
Therefore, for any $h\in S^{2}$, $Rm(h)_{ac}=-R_{abcd}h_{bd}$.

  3. c)

$i,j,k,\cdots$ and $\bar{i},\bar{j},\bar{k},\cdots$ denote the local
holomorphic and antiholomorphic coordinates respectively, unless otherwise
stated.

##  3 Proof of the main theorem

Throughout this section, we consider
$(M^{n},g)=(\mathbb{C}\text{P}^{2m-1},g_{FS})$ for $m\geq 2$. Notice that
$\mathbb{C}\text{P}^{1}=S^{2}$ and its rigidity is obvious. We will do most
calculations in the domain
$U_{0}=\\{z_{0}=1|z\in\mathbb{C}\text{P}^{2m-1}\\}$, i.e.,
$(z_{1},\cdots,z_{2m-1})$ are the local coordinates. In addition, the function
$u$ is defined as in (2.4).

### Basic equations for $u$

######  Notation 3.1.

For simplicity, we define

$\displaystyle A:=1+|z_{1}|^{2}+\cdots+|z_{m-1}|^{2},\quad
B:=|z_{m}|^{2}+\cdots+|z_{2m-1}|^{2},\quad S:=A+B.$

Under the local coordinates, the Fubini-Study metric and its inverse are
represented as

$g_{i\bar{j}}=4m\left(\frac{\delta_{ij}}{S}-\frac{\bar{z}_{i}z_{j}}{S^{2}}\right)\quad\text{and}\quad
g^{i\bar{j}}=\frac{S}{4m}\left(\delta_{ij}+z_{i}\bar{z}_{j}\right).$ (3.1)

Moreover, its Christoffel coefficients are

$\displaystyle\Gamma_{ij}^{k}=-\frac{1}{S}\left(\delta_{ik}\bar{z}_{j}+\delta_{jk}\bar{z}_{i}\right).$
(3.2)

Next, we have the following results by direct computation

######  Lemma 3.2.

For the function $u$ defined in (2.4), we have

$\displaystyle\lambda^{-1}u_{i}=\begin{cases}2S^{-2}B\bar{z}_{i},\quad&\forall
i<m,\\\ -2S^{-2}A\bar{z}_{i},\quad&\forall i\geq
m,\end{cases}\quad\text{and}\quad u_{\bar{i}}=\overline{u_{i}}.$

###### Proof.

From our definition, $\lambda^{-1}u=2S^{-1}A-1$ and hence

$\displaystyle\lambda^{-1}u_{i}=2S^{-1}A_{i}-2S^{-2}A\bar{z}_{i}.$

From this, the conclusion follows. ∎

Next, we compute the Hessian of $u$.

######  Lemma 3.3.

We have

$\displaystyle\lambda^{-1}u_{i\bar{j}}=\begin{cases}2S^{-2}B(\delta_{ij}-2S^{-1}\bar{z}_{i}z_{j}),\quad&\forall
i,j<m,\\\ &\\\ -2S^{-2}A(\delta_{ij}-2S^{-1}\bar{z}_{i}z_{j}),\quad&\forall
i,j\geq m,\\\ &\\\
2S^{-3}(A-B)\bar{z}_{i}z_{j},\quad&\text{otherwise}.\end{cases}$

Morever, $u_{ij}=u_{\bar{i}\bar{j}}=0$.

###### Proof.

We may assume $\lambda=1$. By direct calculations,

$\displaystyle\partial_{j}\partial_{i}u=\partial_{j}(2S^{-1}A_{i}-2S^{-2}A\bar{z}_{i})=4S^{-3}A\bar{z}_{i}z_{j}-2S^{-2}A_{i}\bar{z}_{j}-2S^{-2}A_{j}\bar{z}_{i}.$

Therefore, by (3.2),

$\displaystyle
u_{ij}=\partial_{j}\partial_{i}u-\Gamma_{ij}^{k}u_{k}=\partial_{j}\partial_{i}u-S^{-1}(\delta_{ik}\bar{z}_{j}+\delta_{jk}\bar{z}_{i})(2S^{-1}A_{k}-2S^{-2}A\bar{z}_{k})=0.$

Since $u$ is a real function, $u_{\bar{i}\bar{j}}=0$ as well. In addition,

$\displaystyle u_{i\bar{j}}=$
$\displaystyle\partial_{\bar{j}}\partial_{i}u=\partial_{\bar{j}}(2S^{-1}A_{i}-2S^{-2}A\bar{z}_{i})$
$\displaystyle=$
$\displaystyle-2S^{-2}S_{\bar{j}}A_{i}+2S^{-1}A_{i\bar{j}}+4S^{-3}AS_{\bar{j}}\bar{z}_{i}-2S^{-2}A_{\bar{j}}\bar{z}_{i}-2S^{-2}A\delta_{ij}$
$\displaystyle=$
$\displaystyle-2S^{-2}A_{i}z_{j}+2S^{-1}A_{i\bar{j}}+4S^{-3}A\bar{z}_{i}z_{j}-2S^{-2}A_{\bar{j}}\bar{z}_{i}-2S^{-2}A\delta_{ij}.$

From this, the conclusion follows immediately. ∎

######  Lemma 3.4.

$u$ satisfies the following identities:

  1. (a)

$|\nabla u|^{2}=\dfrac{1}{2m}(\lambda^{2}-u^{2})$.

  2. (b)

$\Delta u^{2}=\dfrac{\lambda^{2}}{m}-(2+\dfrac{1}{m})u^{2}$.

  3. (c)

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int u^{2}=\dfrac{\lambda^{2}}{2m+1}$, where
$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int=\frac{1}{\emph{Vol}(M)}\int$.

  4. (d)

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int u^{4}=\dfrac{3\lambda^{4}}{(2m+1)(2m+3)}$.

###### Proof.

Without loss of generality, we assume $\lambda=1$.

(a) We compute from (3.1) and Lemma 3.2 that

$\displaystyle|\nabla u|^{2}=2g^{i\bar{j}}u_{i}u_{\bar{j}}$ $\displaystyle=$
$\displaystyle\frac{S}{2m}\left(\sum_{i,j<m}(\delta_{ij}+z_{i}\bar{z}_{j})4S^{-4}B^{2}\bar{z}_{i}z_{j}+\sum_{i,j\geq
m}(\delta_{ij}+z_{i}\bar{z}_{j})4S^{-4}A^{2}\bar{z}_{i}z_{j}+\sum_{\text{otherwise}}z_{i}\bar{z}_{j}(-4S^{-4}AB\bar{z}_{i}z_{j})\right)$
$\displaystyle=$
$\displaystyle\frac{2S^{-3}}{m}\left(B^{2}A(A-1)+A^{2}(B^{2}+B)-2A(A-1)B^{2}\right)=\frac{2S^{-2}AB}{m}=\frac{1}{2m}(1-u^{2}).$

(b) From (a), we have

$\Delta u^{2}=2u\Delta u+2|\nabla
u|^{2}=-2u^{2}+\frac{1}{m}(1-u^{2})=\frac{1}{m}-(2+\frac{1}{m})u^{2}.$

(c) Integrating (b), we get

$\displaystyle(2+\frac{1}{m})\int u^{2}=\frac{1}{m}\text{Vol}(M),$

and the identity follows.

(d) We compute

$\displaystyle\Delta u^{4}=$ $\displaystyle 2(\Delta u^{2})u^{2}+2|\nabla
u^{2}|^{2}$ $\displaystyle=$ $\displaystyle
2(\frac{1}{m}-(2+\frac{1}{m})u^{2})u^{2}+8u^{2}(\frac{1}{2m}(1-u^{2}))$
$\displaystyle=$ $\displaystyle\frac{6}{m}u^{2}+(-4-\frac{6}{m})u^{4}.$

By integration, we have

$\displaystyle\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int
u^{4}=\frac{3}{2m+3}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int u^{2}=\frac{3}{(2m+1)(2m+3)}.$

∎

From now on, we will only consider the infinitesimal deformation $ug$ by Lemma
2.6 and Lemma 2.10. Notice that we consider here $ug$ instead of
$ug+2\nabla^{2}u$ as one can compose with a family of diffeomorphisms
generated by $-\nabla u$.

Our strategy goes as the following:

Step 1: Solve all possible second variation $g^{(2)}$ from

$\Phi^{(2)}(ug,ug)+\Phi^{\prime}(g^{(2)})=0.$ (3.3)

Step 2: There exists a symmetric 2-tensor $\eta\in\text{ker}(\Phi^{\prime})$
such that for any $g^{(2)}$ obtained in Step 1,

$\int_{M}\langle\Phi^{(3)}(ug,ug,ug)+3\Phi^{(2)}(ug,g^{(2)}),\eta\rangle\neq
0.$ (3.4)

Combining these two steps, one can prove that $ug$ is not integrable of third-
order.

### Second Variation

Now, we set $g(t)=(1+tu)g$ and $f(t)=f(g(t))$. Moreover, we denote the partial
derivative at $t=0$ by the subscript $t$.

By Lemma A.2(i),

$\Phi^{(2)}(ug,ug)=-\nabla^{2}f_{tt}-2(m-1)du\otimes du+(u^{2}-|\nabla
u|^{2})g-4(m-1)u\nabla^{2}u,$ (3.5)

where $f_{tt}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{tt}=$
$\displaystyle(2m-1)u^{2}-(3m-2)|\nabla u|^{2}$ $\displaystyle=$
$\displaystyle(2m-\frac{1}{m}+\frac{1}{2})u^{2}+(\frac{1}{m}-\frac{3}{2})\lambda^{2}.$
(3.6)

Here, we have used Lemma 3.4 (a).

######  Lemma 3.5.

The solution $f_{tt}$ of (3.6) is

$f_{tt}=-\frac{4m^{2}+m-2}{3m+2}u^{2}-\frac{m-2}{3m+2}\lambda^{2}.$

###### Proof.

We choose two functions $u^{2},\lambda^{2}$ as basis to form a linear space
and calculate from Lemma 3.4 (b) that

$\Delta\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}=\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}\begin{bmatrix}-(2+\frac{1}{m})&0\\\
\frac{1}{m}&0\end{bmatrix}.$

and hence

$(\Delta+\frac{1}{2})\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}=\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}\begin{bmatrix}-(\frac{3}{2}+\frac{1}{m})&0\\\
\frac{1}{m}&\frac{1}{2}\end{bmatrix}.$

In particular, the operator $\Delta+1/2$ preserves the linear space $\langle
u^{2},\lambda^{2}\rangle$. Since the kernel of $\Delta+1/2$ is trivial, we
have from (3.6)

$\displaystyle f_{tt}$
$\displaystyle=\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}\begin{bmatrix}-(\frac{3}{2}+\frac{1}{m})&0\\\
\frac{1}{m}&\frac{1}{2}\end{bmatrix}^{-1}\begin{bmatrix}2m-\frac{1}{m}+\frac{1}{2}\\\
\frac{1}{m}-\frac{3}{2}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}\begin{bmatrix}-\frac{2m}{3m+2}&0\\\
\frac{4}{3m+2}&2\end{bmatrix}\begin{bmatrix}2m-\frac{1}{m}+\frac{1}{2}\\\
\frac{1}{m}-\frac{3}{2}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}u^{2}&\lambda^{2}\end{bmatrix}\begin{bmatrix}-\frac{4m^{2}+m-2}{3m+2}\\\
-\frac{m-2}{3m+2}\end{bmatrix}.$

∎

From (3.3) and (3.5), we try to solve $h$ in

$\Phi^{\prime}(h)=\nabla^{2}f_{tt}+2(m-1)du\otimes du+(|\nabla
u|^{2}-u^{2})g+4(m-1)u\nabla^{2}u$

Since $\text{im}(\delta^{*})\subset\text{ker}(\Phi^{\prime})$ and
$\Phi^{\prime}$ preserves $\text{ker}(\delta)$, we may further assume $\delta
h=0$. In particular, from Lemma 2.5, Lemma 3.4 (a) and Lemma 3.5, one has

$\displaystyle\frac{1}{2}Lh$ $\displaystyle=\nabla^{2}f_{tt}+2(m-1)du\otimes
du+(|\nabla u|^{2}-u^{2})g+4(m-1)u\nabla^{2}u$ (3.7)
$\displaystyle=\frac{\lambda^{2}}{2m}g-\frac{2m+1}{2m}u^{2}g-\frac{2m(m+2)}{3m+2}du\otimes
du+\frac{4m^{2}-6m-4}{3m+2}u\nabla^{2}u,$

where, for simplicity, we denote the operator $\Delta+2Rm$ by $L$.

Our next goal is to obtain a particular solution $h_{0}$ of (3.7). To achieve
this, we first recall a general result for Kähler manifolds.

######  Lemma 3.6.

On any Kähler manifold, the operator $L$ satisfies

$L(S^{(1,1)})\subset S^{(1,1)}\quad\text{and}\quad L(S^{(0,2)}\oplus
S^{(2,0)})\subset S^{(0,2)}\oplus S^{(2,0)},$ (3.8)

where we denote the space of symmetric $(p,q)$-form by $S^{(p,q)}$.

###### Proof.

Recall the fact that $h\in S^{(1,1)}$ if and only if $h(X,Y)=h(JX,JY)$, and
$h\in S^{(0,2)}\oplus S^{(2,0)}$ if and only if $h(X,Y)=-h(JX,JY)$. By
slightly abusing the notation, we define an operator $J:S^{2}\rightarrow
S^{2}$,where $J(h)(X,Y)=h(JX,JY)$.

To show (3.8), one only needs to prove

$L(Jh)=J(Lh).$ (3.9)

Under an orthonormal basis $\\{e_{i}\\}$, we compute

$\displaystyle Rm(Jh)(e_{a},e_{b})=$ $\displaystyle-
Rm(e_{a},e_{i},e_{b},e_{i})h(Je_{i},Je_{i})$ $\displaystyle=$ $\displaystyle-
Rm(e_{a},Je_{i},e_{b},Je_{i})h(e_{i},e_{i})$ $\displaystyle=$ $\displaystyle-
Rm(Je_{a},e_{i},Je_{b},e_{i})h(e_{i},e_{i})=J(Rm(h))(e_{a},e_{b}).$

Combined with the fact that $J$ is parallel, we readily conclude (3.9) and
thus (3.8) is proved. ∎

For later calculation, we define the following auxiliary tensor.

######  Definition 3.7.

We define a symmetric $(1,1)$-form $\xi$, where

$\xi_{i\bar{j}}=\xi_{\bar{j}i}=g^{\bar{k}l}u_{i\bar{k}}u_{l\bar{j}}$

One important property of $\xi$ is that it can be represented as a linear
combination of $\\{\lambda^{2}g,u^{2}g,\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial u,u\nabla^{2}u\\}$.

######  Lemma 3.8.

We have

$\displaystyle\xi_{i\bar{j}}=\frac{\lambda^{2}-u^{2}}{16m^{2}}g_{i\bar{j}}-\frac{1}{4m}u_{i}u_{\bar{j}}-\frac{1}{2m}uu_{i\bar{j}}.$

###### Proof.

Without loss of generality, we assume $\lambda=1$.

For $i,j<m$, it follows from Lemma 3.3 that

$\displaystyle\xi_{i\bar{j}}=$ $\displaystyle
g^{\bar{k}l}u_{i\bar{k}}u_{l\bar{j}}$ $\displaystyle=$
$\displaystyle\sum_{k,l<m}4S^{-4}B^{2}g^{\bar{k}l}(\delta_{ik}-2S^{-1}\bar{z}_{i}z_{k})(\delta_{lj}-2S^{-1}\bar{z}_{l}z_{j})$
$\displaystyle+\sum_{k<m,l\geq
m}4S^{-5}B(A-B)g^{\bar{k}l}(\delta_{ik}-2S^{-1}\bar{z}_{i}z_{k})\bar{z}_{l}z_{j}$
$\displaystyle+\sum_{k\geq
m,l<m}4S^{-5}B(A-B)g^{\bar{k}l}(\delta_{lj}-2S^{-1}\bar{z}_{l}z_{j})\bar{z}_{i}z_{k}$
$\displaystyle+\sum_{k,l\geq
m}4S^{-6}(A-B)^{2}g^{\bar{k}l}\bar{z}_{i}z_{k}\bar{z}_{l}z_{j}$
$\displaystyle=:I_{11}+I_{12}+I_{13}+I_{14}.$

We compute

$\displaystyle I_{11}=$ $\displaystyle
m^{-1}B^{2}S^{-3}\sum_{k,l<m}(\delta_{kl}+\bar{z}_{k}z_{l})(\delta_{ik}-2S^{-1}\bar{z}_{i}z_{k})(\delta_{lj}-2S^{-1}\bar{z}_{l}z_{j})$
$\displaystyle=$ $\displaystyle
m^{-1}B^{2}S^{-3}\sum_{l<m}\left(\delta_{il}+(1-2S^{-1}A)\bar{z}_{i}z_{l}\right)(\delta_{lj}-2S^{-1}\bar{z}_{l}z_{j})$
$\displaystyle=$ $\displaystyle
m^{-1}B^{2}S^{-3}\left(\delta_{ij}+(1-4S^{-1}A+4S^{-2}A^{2}-4S^{-2}A)\bar{z}_{i}z_{j}\right).$

Moreover,

$\displaystyle I_{12}=I_{13}=$ $\displaystyle m^{-1}(A-B)BS^{-4}\sum_{k\geq
m,l<m}z_{l}\bar{z}_{k}\bar{z}_{i}z_{k}(\delta_{lj}-2S^{-1}\bar{z}_{l}z_{j})$
$\displaystyle=$ $\displaystyle
m^{-1}(A-B)B^{2}S^{-4}\sum_{l<m}z_{l}\bar{z}_{i}(\delta_{lj}-2S^{-1}\bar{z}_{l}z_{j})$
$\displaystyle=$ $\displaystyle
m^{-1}(A-B)B^{2}S^{-4}(1+2S^{-1}-2S^{-1}A)\bar{z}_{i}z_{j}$

and we have

$\displaystyle I_{14}=$ $\displaystyle m^{-1}(A-B)^{2}S^{-5}\sum_{k,l\geq
m}(\delta_{kl}+z_{l}\bar{z}_{k})\bar{z}_{i}z_{k}\bar{z}_{l}z_{j}=m^{-1}(A-B)^{2}B(1+B)S^{-5}\bar{z}_{i}z_{j}.$

Combining all terms, we obtain for $i,j<m$,

$\displaystyle\xi_{i\bar{j}}=m^{-1}B^{2}S^{-3}\left(\delta_{ij}+(B^{-1}-4S^{-1})\bar{z}_{i}z_{j}\right).$

In the same way, we can calculate the other cases and obtain

$\displaystyle\xi_{i\bar{j}}=\begin{cases}m^{-1}B^{2}S^{-3}\left(\delta_{ij}+(B^{-1}-4S^{-1})\bar{z}_{i}z_{j}\right),\quad&\forall
i,j<m,\\\ &\\\
m^{-1}A^{2}S^{-3}\left(\delta_{ij}+(A^{-1}-4S^{-1})\bar{z}_{i}z_{j}\right),\quad&\forall
i,j\geq m,\\\ &\\\
-m^{-1}(A-B)^{2}S^{-4}\bar{z}_{i}z_{j},\quad&\text{otherwise}.\end{cases}$

By using Lemma 3.2 and Lemma 3.3, one can easily check

$\displaystyle\xi_{i\bar{j}}=\frac{\lambda^{2}-u^{2}}{16m^{2}}g_{i\bar{j}}-\frac{1}{4m}u_{i}u_{\bar{j}}-\frac{1}{2m}uu_{i\bar{j}}$

and we omit the detailed calculations. ∎

Now we can completely solve $h$ in (3.7).

######  Theorem 3.9.

The solution set of (3.7), denoted by $\mathcal{H}$, can be represented as

$\displaystyle\mathcal{H}=h_{0}+K_{0}\oplus\emph{im}(\delta^{*}),$

where $K_{0}:=\\{vg+2\nabla^{2}v\mid v\in C^{\infty}(M),\Delta v+v=0\\}$ and
$h_{0}$ is a particular solution in $\emph{ker}(\delta)$ defined as

$\displaystyle h_{0}:=$
$\displaystyle-\frac{2}{m+1}\lambda^{2}g+\frac{2m}{m+1}u^{2}g+\frac{4m(m^{2}+5m+2)}{(m+1)(3m+2)}(\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial u)$
$\displaystyle-\frac{8m^{3}}{(m+1)(3m+2)}u\nabla^{2}u+\frac{4m^{2}(m+2)}{(m+1)(3m+2)}(\partial
u\otimes\partial u+\bar{\partial}u\otimes\bar{\partial}u).$ (3.10)

###### Proof.

From Lemma 2.5 and Lemma 2.6, we only need to prove $h_{0}$ is indeed a
particular solution in $\text{ker}(\delta)$.

Observe that the right side of $\eqref{E307}$ is in the linear space $V$
spanned by $\\{\lambda^{2}g,u^{2}g,\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial u,u\nabla^{2}u,\partial
u\otimes\partial u+\bar{\partial}u\otimes\bar{\partial}u\\}$. So we consider
solving an explicit $h_{0}$ in this linear space. We will show that $L$
actually preserves $V$, and hence one can solve a particular $h_{0}$ in $V$ by
inverting $L$.

In the following, we do calculations of $L$ on each basis.

  1. (a)

We compute

$\displaystyle L(g)=\Delta(g)+2Rm(g)=2Rc=g.$

  2. (b)

From Lemma 3.4 (b), we have

$\displaystyle L(u^{2}g)=(\Delta
u^{2})g+2u^{2}Rm(g)=\left(\frac{\lambda^{2}}{m}-(2+\frac{1}{m})u^{2}\right)g+u^{2}g=\frac{\lambda^{2}}{m}g-(1+\frac{1}{m})u^{2}g.$

  3. (c)

Notice that under general coordinates $a,b,c$, etc., we have by Bochner’s
formula

$\displaystyle\Delta u_{a}=(\Delta
u)_{a}+g^{bc}R_{ab}u_{c}=-u_{a}+\frac{1}{2}u_{a}=-\frac{1}{2}u_{a}.$

Therefore,

$\displaystyle\Delta(u_{a}u_{b})=(\Delta u_{a})u_{b}+(\Delta
u_{b})u_{a}+2g^{cd}u_{ac}u_{bd}=-u_{a}u_{b}+2\xi_{ab}.$ (3.11)

From Lemma 3.6, $Rm(\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial u)$ is of type $(1,1)$.
More precisely, we recall

$R_{i\bar{j}k\bar{l}}=\frac{1}{4m}(g_{i\bar{j}}g_{k\bar{l}}+g_{i\bar{l}}g_{k\bar{j}})$

and hence

$\displaystyle
Rm(u_{i}u_{\bar{j}})_{k\bar{l}}=-R_{k\bar{t}\bar{l}s}u_{i}u_{\bar{j}}g^{i\bar{t}}g^{\bar{j}s}=\frac{1}{4m}(u_{k}u_{\bar{l}}+\frac{1}{2}|\nabla
u|^{2}g_{k\bar{l}})=\frac{1}{4m}u_{k}u_{\bar{l}}+\frac{\lambda^{2}-u^{2}}{16m^{2}}g_{k\bar{l}},$
(3.12)

where we have used Lemma 3.4 (a).

Combined with (3.11), we obtain

$\displaystyle L(u_{i}u_{\bar{j}})_{k\bar{l}}$
$\displaystyle=\Delta(u_{i}u_{\bar{j}})_{k\bar{l}}+2Rm(u_{i}u_{\bar{j}})_{k\bar{l}}$
$\displaystyle=-u_{k}u_{\bar{l}}+2\xi_{k\bar{l}}+\frac{1}{2m}u_{k}u_{\bar{l}}+\frac{\lambda^{2}-u^{2}}{8m^{2}}g_{k\bar{l}}$
$\displaystyle=-u_{k}u_{\bar{l}}-\frac{1}{m}uu_{k\bar{l}}+\frac{\lambda^{2}-u^{2}}{4m^{2}}g_{k\bar{l}},$

where for the last line we have used Lemma 3.8.

  4. (d)

Similarly, we have

$Rm(u_{i}u_{j})_{kl}=-R_{\bar{t}k\bar{s}l}u_{i}u_{j}g^{i\bar{t}}g^{j\bar{l}}=-\frac{1}{2m}u_{k}u_{l}$

and hence

$\displaystyle L(u_{i}u_{j})_{kl}$
$\displaystyle=\Delta(u_{i}u_{j})_{kl}+2Rm(u_{i}u_{j})_{kl}=-(1+\frac{1}{m})u_{k}u_{l}.$

  5. (e)

Next, we compute

$\displaystyle L(u\nabla^{2}u)=\Delta u\nabla^{2}u+2\langle\nabla
u,\nabla(\nabla^{2}u)\rangle+uL(\nabla^{2}u)=2\langle\nabla
u,\nabla(\nabla^{2}u)\rangle-u\nabla^{2}u,$ (3.13)

where we have used the fact that $L(\nabla^{2}u)=\nabla^{2}(\Delta u+u)=0$.
Moreover,

$\displaystyle 2g^{cd}u_{d}(\nabla_{c}\nabla_{a}\nabla_{b}u)=$ $\displaystyle
2g^{cd}u_{d}(\nabla_{a}\nabla_{b}\nabla_{c}u)-2g^{cd}g^{ef}R_{cabe}u_{f}u_{d}$
$\displaystyle=$
$\displaystyle\nabla_{a}\nabla_{b}(g^{cd}u_{c}u_{d})-2g^{cd}\nabla_{a}\nabla_{c}u\nabla_{b}\nabla_{d}u-2g^{cd}g^{ef}R_{cabe}u_{f}u_{d}.$

In other words,

$\displaystyle 2\langle\nabla u,\nabla(\nabla^{2}u)\rangle=\nabla^{2}|\nabla
u|^{2}-2\xi-2Rm(du\otimes du).$

Combined with (3.13), we obtain

$\displaystyle L(u\nabla^{2}u)_{k\bar{l}}=$ $\displaystyle|\nabla
u|^{2}_{k\bar{l}}-2\xi_{k\bar{l}}-2Rm(du\otimes du)_{k\bar{l}}-uu_{k\bar{l}}$
$\displaystyle=$
$\displaystyle-\frac{\lambda^{2}-u^{2}}{4m^{2}}g_{k\bar{l}}-\frac{1}{m}u_{k}u_{\bar{l}}-uu_{k\bar{l}},$

where we have used Lemma 3.8, Lemma 3.4 (a) and (3.12).

To summarize, the matrix of $L$ acts on the basis
$\\{\lambda^{2}g,u^{2}g,\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial u,u\nabla^{2}u,\partial
u\otimes\partial u+\bar{\partial}u\otimes\bar{\partial}u\\}$ by

$\begin{aligned}
L&\begin{bmatrix}\lambda^{2}g_{i\bar{j}}&u^{2}g_{i\bar{j}}&u_{i}u_{\bar{j}}&uu_{i\bar{j}}&u_{i}u_{j}\end{bmatrix}\\\
&=\begin{bmatrix}\lambda^{2}g_{i\bar{j}}&u^{2}g_{i\bar{j}}&u_{i}u_{\bar{j}}&uu_{i\bar{j}}&u_{i}u_{j}\end{bmatrix}\begin{bmatrix}1&\frac{1}{m}&\frac{1}{4m^{2}}&-\frac{1}{4m^{2}}&0\\\
0&-1-\frac{1}{m}&-\frac{1}{4m^{2}}&\frac{1}{4m^{2}}&0\\\
0&0&-1&-\frac{1}{m}&0\\\ 0&0&-\frac{1}{m}&-1&0\\\
0&0&0&0&-\frac{1}{m}-1\end{bmatrix}\end{aligned}.$

The matrix of $L^{-1}$ is

$\begin{aligned}
\begin{bmatrix}1&\frac{1}{m+1}&\frac{1}{4(m^{2}-1)}&-\frac{1}{4(m^{2}-1)}&0\\\
0&-\frac{m}{m+1}&\frac{1}{4(m^{2}-1)}&-\frac{1}{4(m^{2}-1)}&0\\\
0&0&\frac{m^{2}}{1-m^{2}}&\frac{m}{m^{2}-1}&0\\\
0&0&\frac{m}{m^{2}-1}&\frac{m^{2}}{1-m^{2}}&0\\\
0&0&0&0&-\frac{m}{m+1}\end{bmatrix}\end{aligned}.$

Therefore, we can solve one explicit $h_{0}$ from (3.7) by

$\displaystyle\frac{1}{2}h_{0}$
$\displaystyle=\begin{bmatrix}\lambda^{2}g_{i\bar{j}}&u^{2}g_{i\bar{j}}&u_{i}u_{\bar{j}}&uu_{i\bar{j}}&u_{i}u_{j}\end{bmatrix}\begin{bmatrix}1&\frac{1}{m+1}&\frac{1}{4(m^{2}-1)}&-\frac{1}{4(m^{2}-1)}&0\\\
0&-\frac{m}{m+1}&\frac{1}{4(m^{2}-1)}&-\frac{1}{4(m^{2}-1)}&0\\\
0&0&\frac{m^{2}}{1-m^{2}}&\frac{m}{m^{2}-1}&0\\\
0&0&\frac{m}{m^{2}-1}&\frac{m^{2}}{1-m^{2}}&0\\\
0&0&0&0&-\frac{m}{m+1}\end{bmatrix}\begin{bmatrix}\frac{1}{2m}\\\
-1-\frac{1}{2m}\\\ -\frac{2m(m+2)}{3m+2}\\\ \frac{4m^{2}-6m-4}{3m+2}\\\
-\frac{2m(m+2)}{3m+2}\end{bmatrix}$
$\displaystyle=\begin{bmatrix}\lambda^{2}g_{i\bar{j}}&u^{2}g_{i\bar{j}}&u_{i}u_{\bar{j}}&uu_{i\bar{j}}&u_{i}u_{j}\end{bmatrix}\begin{bmatrix}-\frac{1}{1+m}\\\
\frac{m}{m+1}\\\ \frac{2m(m^{2}+5m+2)}{(1+m)(3m+2)}\\\
-\frac{4m^{3}}{(m+1)(3m+2)}\\\ \frac{2m^{2}(m+2)}{(m+1)(3m+2)}\end{bmatrix}.$

It remains to check that $h_{0}\in\text{ker}(\delta)$. Indeed, one can easily
compute

$\displaystyle\begin{cases}&\delta(\lambda^{2}g)=0,\\\
&\delta(u^{2}g)=2u\nabla u,\\\ &\delta\left(\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial
u\right)=-\frac{1}{2}u\nabla u,\\\
&\delta(u\nabla^{2}u)=\frac{1}{2}\nabla|\nabla
u|^{2}+u\delta(\nabla^{2}u)=-(\frac{1}{2m}+\frac{1}{2})u\nabla u,\\\
&\delta\left(\partial u\otimes\partial
u+\bar{\partial}u\otimes\bar{\partial}u\right)=-(\frac{1}{2m}+\frac{1}{2})u\nabla
u.\end{cases}$

Therefore, it is easy to see from (3.10) that

$\displaystyle\delta(h_{0})=\left(\frac{4m}{m+1}-\frac{2m(m^{2}+5m+2)}{(1+m)(3m+2)}+\frac{4m^{2}}{3m+2}-\frac{2m(m+2)}{3m+2}\right)u\nabla
u=0.$

In sum, the proof is complete. ∎

### Third-order obstruction

In the subsection, we prove that

$I(h):=\int_{M}\langle\Phi^{(3)}(ug,ug,ug)+3\Phi^{(2)}(h,ug),ug\rangle\neq 0$
(3.14)

for any $h\in\mathcal{H}$. In other words, we choose $\eta=ug$ in (3.4).

Recall that $\mathcal{H}=h_{0}+K_{0}\oplus\text{im}(\delta^{*})$, where
$h_{0}$ is defined in (3.10). We next show that $I(h)$ depends only on
$I(h_{0})$.

######  Proposition 3.10.

With above definitions, for any $h\in\mathcal{H}$,

$\displaystyle I(h)=I(h_{0}).$

The proof relies on the fact that $\int\langle 3\Phi^{(2)}(h,ug),ug\rangle$
depends only on $h_{0}$, which the following two lemmas can show.

######  Lemma 3.11.

For any $1$-form $\alpha$, we have

$\int\langle\Phi^{(2)}(ug,\delta^{*}\alpha),ug\rangle=0.$

###### Proof.

Let $X$ be the vector field dual to $\alpha$ and $\phi_{s}$ be a family of
diffeomorphisms generated by $X$.

Evaluated at $t=s=0$, we have

$\displaystyle\partial^{2}_{st}\Phi\left(\phi_{s}^{*}((1+tu)g)\right)=\partial^{2}_{st}\Phi(g+tug+sL_{X}g)+\Phi^{\prime}(L_{X}(ug))=\Phi^{(2)}(ug,L_{X}g)+\Phi^{\prime}\left(L_{X}(ug)\right)$
(3.15)

since

$\displaystyle\phi_{s}^{*}((1+tu)g)=(1+tu)g+sL_{X}\left((1+tu)g\right)+R_{1}=(1+tu)g+sL_{X}g+stL_{X}(ug)+R_{2},$

where $R_{1}$ and $R_{2}$ are remainders containing only terms of $t^{2}$ or
$s^{2}$ or higher-order.

On the other hand, we have

$\displaystyle\partial^{2}_{st}\int\langle\Phi\left(\phi_{s}^{*}((1+tu)g)\right),ug\rangle_{g}\,dV_{g}=\partial^{2}_{st}\int\langle\Phi((1+tu)g),\phi_{-s}^{*}(ug)\rangle_{\phi_{-s}^{*}(g)}\,dV_{\phi_{-s}^{*}(g)}=0$
(3.16)

since $\Phi^{\prime}(ug)=0$. Combining (3.15) and (3.16), we obtain

$\displaystyle-2\int\langle\Phi^{(2)}(ug,\delta^{*}\alpha),ug\rangle=\int\langle\Phi^{(2)}(ug,L_{X}g),ug\rangle$
$\displaystyle=$
$\displaystyle-\int\langle\Phi^{\prime}(L_{X}(ug)),ug\rangle=-\int\langle
L_{X}(ug),\Phi^{\prime}(ug)\rangle=0,$

where we have used the fact that $\Phi^{\prime}$ is self-adjoint.

∎

######  Lemma 3.12.

For any smooth function $v$ with $\Delta v+v=0$, we have

$\displaystyle\int\langle\Phi^{(2)}(ug,vg),ug\rangle=0.$

###### Proof.

We set $g(t,s)=(1+tu+sv)g$. Then it follows from Lemma A.3 (A.7), (A.8) that

$\displaystyle\langle\Phi_{st},g\rangle=(4-6m)\langle\nabla u,\nabla
v\rangle+2(4m-3)uv-\Delta f_{st},$ (3.17)

where $f_{st}$ satisfies

$\displaystyle(\Delta+\frac{1}{2})f_{st}=(2m-1)uv-(3m-2)\langle\nabla u,\nabla
v\rangle.$ (3.18)

Multiplying (3.18) by $u$ on both sides and integrating, we obtain

$\displaystyle\int(2m-1)u^{2}v-\frac{3m-2}{2}\langle\nabla u^{2},\nabla
v\rangle=\int u(\Delta+\frac{1}{2})f_{st}=-\frac{1}{2}\int uf_{st}.$ (3.19)

From (3.17) and (3.19), we have

$\displaystyle\int\langle\Phi_{st},ug\rangle=$
$\displaystyle\int(2-3m)\langle\nabla u^{2},\nabla
v\rangle+2(4m-3)u^{2}v-u\Delta f_{st}$ $\displaystyle=$
$\displaystyle\int(2-3m)u^{2}v+2(4m-3)u^{2}v-(4m-2)u^{2}v+(3m-2)\langle\nabla
u^{2},\nabla v\rangle$ $\displaystyle=$ $\displaystyle 4(m-1)\int u^{2}v=0,$

where the last equality follows from our choice of $u$ in Lemma 2.10. ∎

Combining Lemma 3.11 and Lemma 3.12, Proposition 3.10 follows immediately.

To obtain (3.14), we only need to show $I(h_{0})\neq 0$. For simplicity, we
define

$\displaystyle
I_{1}:=\int_{M}\langle\Phi^{(3)}(ug,ug,ug),ug\rangle\quad\text{and}\quad
I_{2}:=\int_{M}\langle 3\Phi^{(2)}(h_{0},ug),ug\rangle.$

######  Proposition 3.13.

With above definitions, we have

$\displaystyle
I_{1}=-\frac{6(20m^{3}-15m^{2}-9m+6)}{(2m+1)(2m+3)(3m+2)}\emph{Vol}(M)\lambda^{4}.$

###### Proof.

We define $g(t)=(1+tu)g$. By Lemma A.2(ii), we have

$\displaystyle\Phi_{ttt}=$
$\displaystyle-\nabla^{2}f_{ttt}+\frac{3}{2}\left(du\otimes
df_{tt}+df_{tt}\otimes du-\langle\nabla f_{tt},\nabla u\rangle g\right)$
$\displaystyle+24(m-1)udu\otimes
du+12(m-1)u^{2}\nabla^{2}u-3u^{3}g-6(m-2)u|\nabla u|^{2}g,$

where $f_{ttt}$ satisfies

$\displaystyle(\Delta+\frac{1}{2})f_{ttt}=$ $\displaystyle 3u\Delta
f_{tt}-6(m-1)\langle\nabla f_{tt},\nabla u\rangle-6(3m-2)u^{3}+9(3m-2)u|\nabla
u|^{2}$ $\displaystyle=$ $\displaystyle 3u\Delta
f_{tt}+\frac{6(m-1)(4m^{2}+m-2)}{3m+2}\langle\nabla u^{2},\nabla
u\rangle-6(3m-2)u^{3}+9(3m-2)u|\nabla u|^{2},$

where we have used Lemma 3.5.

Now, we denote

$\displaystyle I_{11}:=-\int\langle\nabla^{2}f_{ttt},ug\rangle=-\int u\Delta
f_{ttt}=\int uf_{ttt}.$

From $\int u(\Delta+1)f_{ttt}=0$, Lemma 3.4 (a) (b) and Lemma 3.5, we conclude

$\displaystyle I_{11}=$ $\displaystyle-2\int u(\Delta+\frac{1}{2})f_{ttt}$
$\displaystyle=$ $\displaystyle-6\int u^{2}\Delta
f_{tt}+\frac{4(m-1)(4m^{2}+m-2)}{3m+2}u^{2}|\nabla
u|^{2}-2(3m-2)u^{4}+3(3m-2)u^{2}|\nabla u|^{2}$ $\displaystyle=$
$\displaystyle-6\int\Delta
u^{2}\left(-\frac{4m^{2}+m-2}{3m+2}u^{2}\right)+\frac{16m^{3}+15m^{2}-12m-4}{2m(3m+2)}u^{2}(\lambda^{2}-u^{2})-2(3m-2)u^{4}$
$\displaystyle=$
$\displaystyle-6\int\left(\frac{\lambda^{2}}{m}-(2+\frac{1}{m})u^{2}\right)\left(-\frac{4m^{2}+m-2}{3m+2}u^{2}\right)$
$\displaystyle-6\int\frac{16m^{3}+15m^{2}-12m-4}{2m(3m+2)}u^{2}(\lambda^{2}-u^{2})-2(3m-2)u^{4}$
$\displaystyle=$ $\displaystyle\frac{3(36m^{2}+3m-22)}{3m+2}\int
u^{4}-\frac{3(16m^{2}+7m-14)}{3m+2}\lambda^{2}\int u^{2}.$ (3.20)

Similarly, we compute

$\displaystyle I_{12}:=$ $\displaystyle\frac{3}{2}\int\langle du\otimes
df_{tt}+df_{tt}\otimes du-\langle\nabla f_{tt},\nabla u\rangle g,ug\rangle$
$\displaystyle+\int\langle 24(m-1)udu\otimes
du+12(m-1)u^{2}\nabla^{2}u-3u^{3}g-6(m-2)u|\nabla u|^{2}g,ug\rangle$
$\displaystyle=$ $\displaystyle\int 6(1-m)\langle\nabla u,\nabla f_{tt}\rangle
u-12(2m^{2}-7m+4)u^{2}|\nabla u|^{2}-6(4m-3)u^{4}$ $\displaystyle=$
$\displaystyle\int-\frac{6(2m^{3}-14m^{2}+m+6)}{m(3m+2)}u^{2}(\lambda^{2}-u^{2})-6(4m-3)u^{4}$
$\displaystyle=$ $\displaystyle-\frac{6(2m+3)(5m^{2}-m-2)}{m(3m+2)}\int
u^{4}-\frac{6(2m^{3}-14m^{2}+m+6)}{m(3m+2)}\lambda^{2}\int u^{2}.$ (3.21)

Combining (3.20) and (3.21), we have

$\displaystyle I_{1}=$ $\displaystyle
I_{11}+I_{12}=\frac{3(16m^{3}-23m^{2}-8m+12)}{m(3m+2)}\int
u^{4}-\frac{3(20m^{3}-21m^{2}-12m+12)}{m(3m+2)}\lambda^{2}\int u^{2}$
$\displaystyle=$
$\displaystyle\left(\frac{9(16m^{3}-23m^{2}-8m+12)}{m(2m+1)(2m+3)(3m+2)}-\frac{3(20m^{3}-21m^{2}-12m+12)}{m(2m+1)(3m+2)}\right)\text{Vol}(M)\lambda^{4},$
$\displaystyle=$
$\displaystyle-\frac{6(20m^{3}-15m^{2}-9m+6)}{(2m+1)(2m+3)(3m+2)}\text{Vol}(M)\lambda^{4},$

where we have used Lemma 3.4 (c) (d). ∎

Next, we compute $I_{2}$.

######  Proposition 3.14.

With above definitions, we have

$\displaystyle
I_{2}=\frac{6(4m^{4}+25m^{3}-32m^{2}-7m+14)}{(m+1)(2m+1)(2m+3)(3m+2)}\emph{Vol}(M)\lambda^{4}.$

###### Proof.

We define $g(t,s)=(1+tu)g+sh_{0}$ and $H_{0}=\text{Tr}(h_{0})$. Then it
follows from (3.10) and Lemma 3.4 (a) that

$\displaystyle H_{0}=$
$\displaystyle-\frac{2(4m-2)}{m+1}\lambda^{2}+\frac{2m(4m-2)}{m+1}u^{2}+\frac{4m(m^{2}+5m+2)}{(m+1)(3m+2)}|\nabla
u|^{2}+\frac{8m^{3}}{(m+1)(3m+2)}u^{2}$ $\displaystyle=$
$\displaystyle-\frac{2(11m^{2}-3m-6)}{(m+1)(3m+2)}\lambda^{2}+\frac{2(16m^{3}+m^{2}-9m-2)}{(m+1)(3m+2)}u^{2}.$
(3.22)

Now it follows from Lemma A.4 (A.12) that

$\displaystyle\langle\Phi_{st},g\rangle=-2(m-1)\langle
h_{0},\nabla^{2}u\rangle-\frac{3}{2}\langle\nabla H_{0},\nabla
u\rangle-u\Delta H_{0}+\frac{uH_{0}}{2}-\Delta f_{st},$

where $f_{st}$ satisfies

$\displaystyle(\Delta+\frac{1}{2})f_{st}=-\frac{1}{2}u\Delta
H_{0}-\frac{3}{4}\langle\nabla u,\nabla H_{0}\rangle.$

By our definition,

$\displaystyle\frac{1}{3}I_{2}$
$\displaystyle=\int\langle\Phi_{st},ug\rangle=\int-2(m-1)u\langle
h_{0},\nabla^{2}u\rangle-\frac{3}{2}u\langle\nabla H_{0},\nabla
u\rangle-u^{2}\Delta H_{0}+\frac{u^{2}H_{0}}{2}-u\Delta f_{st}$
$\displaystyle=\int-2(m-1)u\langle
h_{0},\nabla^{2}u\rangle-\frac{3}{4}\langle\nabla H_{0},\nabla u^{2}\rangle-
H_{0}\Delta u^{2}+\frac{u^{2}H_{0}}{2}+uf_{st}$
$\displaystyle=\int-2(m-1)u\langle
h_{0},\nabla^{2}u\rangle+\frac{u^{2}H_{0}}{2},$ (3.23)

where we have used Lemma 3.4 (b) and the identity

$\displaystyle\int uf_{st}=-2\int u(\Delta+\frac{1}{2})f_{st}=\int u^{2}\Delta
H_{0}+\frac{3}{4}\langle\nabla u^{2},\nabla H_{0}\rangle.$

Now, we define and compute

$\displaystyle I_{21}:=$ $\displaystyle\int u\langle h_{0},\nabla^{2}u\rangle$
$\displaystyle=$
$\displaystyle\int\frac{2}{m+1}\lambda^{2}u^{2}-\frac{2m}{m+1}u^{4}+\frac{4m(m^{2}+5m+2)}{(m+1)(3m+2)}u\langle\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial u,\nabla^{2}u\rangle$
$\displaystyle+\int-\frac{8m^{3}}{(m+1)(3m+2)}u^{2}|\nabla^{2}u|^{2}+\frac{4m^{2}(m+2)}{(m+1)(3m+2)}u\langle\partial
u\otimes\partial u+\bar{\partial}u\otimes\bar{\partial}u,\nabla^{2}u\rangle.$

Since $\nabla^{2}u$ is of type $(1,1)$, we have

$\displaystyle\langle\partial u\otimes\partial
u+\bar{\partial}u\otimes\bar{\partial}u,\nabla^{2}u\rangle=0$ (3.24)

and

$\displaystyle\langle\partial
u\otimes\bar{\partial}u+\bar{\partial}u\otimes\partial
u,\nabla^{2}u\rangle=2g^{l\bar{j}}g^{i\bar{k}}u_{i}u_{\bar{j}}u_{\bar{k}l}=\frac{1}{2}\langle\nabla|\nabla
u|^{2},\nabla u\rangle=-\frac{1}{2m}u|\nabla
u|^{2}=-\frac{1}{4m^{2}}u(\lambda^{2}-u^{2}).$ (3.25)

Moreover, it follows from Lemma 3.8 and Lemma 3.4 (a) that

$\displaystyle|\nabla^{2}u|^{2}=$ $\displaystyle
2g^{i\bar{j}}\xi_{i\bar{j}}=\frac{2m-1}{8m^{2}}(\lambda^{2}-u^{2})-\frac{1}{4m}|\nabla
u|^{2}+\frac{1}{2m}u^{2}=\frac{m-1}{4m^{2}}\lambda^{2}+\frac{m+1}{4m^{2}}u^{2}.$
(3.26)

Combining (3.24), (3.25) and (3.26), we have

$\displaystyle I_{21}=$
$\displaystyle\int\frac{2}{m+1}\lambda^{2}u^{2}-\frac{2m}{m+1}u^{4}-\frac{m^{2}+5m+2}{m(m+1)(3m+2)}u^{2}(\lambda^{2}-u^{2})$
$\displaystyle-\int\frac{8m^{3}}{(m+1)(3m+2)}u^{2}\left(\frac{m-1}{4m^{2}}\lambda^{2}+\frac{m+1}{4m^{2}}u^{2}\right)$
$\displaystyle=$ $\displaystyle-\frac{8m^{2}-3m-2}{m(3m+2)}\int
u^{4}-\frac{2m^{3}-7m^{2}+m+2}{m(m+1)(3m+2)}\lambda^{2}\int u^{2}$

Therefore, it follows from (3.22) and (3.23) that

$\displaystyle\frac{1}{3}I_{2}=$
$\displaystyle-2(m-1)I_{21}+\int\frac{u^{2}H_{0}}{2}$ $\displaystyle=$
$\displaystyle\frac{32m^{4}-5m^{3}-29m^{2}+4m+4}{m(m+1)(3m+2)}\int
u^{4}+\frac{4m^{4}-29m^{3}+19m^{2}+8m-4}{m(m+1)(3m+2)}\lambda^{2}\int u^{2}$
$\displaystyle=$
$\displaystyle\frac{2(4m^{4}+25m^{3}-32m^{2}-7m+14)}{(m+1)(2m+1)(2m+3)(3m+2)}\text{Vol}(M)\lambda^{4}$

∎

Finally, we can prove the main theorem of the section.

######  Theorem 3.15.

For every $h\in\mathcal{H}$,

$\int\langle\Phi^{(3)}(ug,ug,ug)+3\Phi^{(2)}(h,ug),ug\rangle=-\frac{24(m-1)(4m^{3}-m^{2}+m+2)}{(m+1)(2m+1)(2m+3)(3m+2)}\emph{Vol}(M)\lambda^{4}.$

###### Proof.

From our calculations above, we have

$\displaystyle I(h_{0})=I_{1}+I_{2}$ $\displaystyle=$
$\displaystyle\left(-\frac{6(20m^{3}-15m^{2}-9m+6)}{(2m+1)(2m+3)(3m+2)}+\frac{6(4m^{4}+25m^{3}-32m^{2}-7m+14)}{(m+1)(2m+1)(2m+3)(3m+2)}\right)\text{Vol}(M)\lambda^{4}$
$\displaystyle=$
$\displaystyle-\frac{24(m-1)(4m^{3}-m^{2}+m+2)}{(m+1)(2m+1)(2m+3)(3m+2)}\text{Vol}(M)\lambda^{4}.$

∎

Now, Theorem 3.15 implies that $ug$ is not integrable of the third-order and
hence $\mathbb{C}\text{P}^{2m-1}$ is rigid. Combined with Theorem 2.9, Theorem
1.1 is proved.

##  4 Further discussion

In this section, we consider the rigidity of the product of a complex
projective space of complex even dimension and an Einstein manifold $M_{2}$.
For simplicity, we assume $(M_{1},g_{1})=(\mathbb{C}\text{P}^{2m},g_{FS})$ and
$(M_{2},g_{2})$ is an Einstein manifold with Einstein constant $1/2$
satisfying the following conditions:

$\displaystyle\begin{cases}1\notin\text{spec}(\Delta_{2}),\\\
(-\infty,-1]\cup\\{0\\}\notin\text{spec}(L_{2}|_{\text{TT}}).\end{cases}$
($\dagger$)

Here, we use the subscript $2$ to denote operators concerning $g_{2}$.
Moreover, the fact that $L_{2}=\Delta_{2}+2Rm_{2}$ preserves the TT-subspace
follows from the next lemma.

######  Lemma 4.1.

Let $(M,g)$ be a compact Einstein manifold with $Rc=\lambda g$ for a constant
$\lambda$. Then the operator $L=\Delta+2Rm$ preserves the decomposition
$S^{2}(M)=\emph{ker}(\delta)\oplus\emph{im}(\delta^{*})$ and the TT-subspace.
Moreover, the smallest eigenvalue of $L$ restricted on $\emph{im}(\delta^{*})$
is at least $-\lambda$.

###### Proof.

The conclusions follow immediately from the following identities, which can be
verified by direct calculations.

$\displaystyle\delta L(h)=(\Delta+\lambda)(\delta h),\quad
L(\delta^{*}\alpha)=\delta^{*}((\Delta+\lambda)\alpha)\quad\text{and}\quad\text{Tr}(Lh)=(\Delta+2\lambda)(\text{Tr}(h))$

for any $h\in S^{2}(M)$ and $1$-form $\alpha$. ∎

######  Remark 4.2.

Many compact symmetric spaces including $S^{n}$ for $n\geq 3$ satisfy
($\dagger$ ‣ 4). We refer the readers to _[5, Table 1, Table 2]_.

Next, we find all infinitesimal solitonic deformations on the product.

######  Lemma 4.3.

For $(M,g)=(M_{1}\times M_{2},g_{1}\times g_{2})$ with condition ($\dagger$ ‣
4), we have

$\displaystyle\emph{ISD}=\\{ug_{1}+2\nabla_{1}^{2}u+vg_{2}\,|\,u,v\in
C^{\infty}(M_{1}),\Delta_{1}u+u=\Delta_{1}v+v=0\\}.$

###### Proof.

From Lemma 2.5, $2\Phi^{\prime}_{g}=L_{g}$ on $\text{ker}(\delta_{g})$ and
hence we only need consider the kernel of $L_{g}$ on $S^{2}(M)$. It follows
from [16, Proposition 4.1] that

$\displaystyle\text{spec}(L_{g})=(\text{spec}(L_{1})+\text{spec}(\Delta_{2}))\cup(\text{spec}(L_{2})+\text{spec}(\Delta_{1}))\cup(\text{spec}(\Delta_{1}^{1})+\text{spec}(\Delta_{2}^{1})),$

where $\Delta^{1}$ denotes the Laplacian on the $1$-form. In addition, the
eigentensor of $L_{g}$ can be expressed as the product of the corresponding
eigensections. Notice the any tensor in the kernel of $L_{g}$ can only be the
products of the first two types since $\text{spec}(\Delta_{i}^{1})>0$ for
$i=1,2$. Indeed, if $\text{spec}(\Delta_{i}^{1})=0$, it implies that the
universal cover of $M_{i}$ splits a line, which is impossible.

It follows from [3] that $\text{spec}(L_{1})=\\{-1,0,\cdots\\}$. Moreover, if
$L_{1}(h_{1})-h_{1}=0$, then $h_{1}=cg_{1}$ for some constant $c$. If
$L_{1}(h_{1})=0$, then $h_{1}=ug_{1}+2\nabla_{1}^{2}u$ for some function $u$
with $\Delta_{1}u+u=0$. By our assumption that
$1\notin\text{spec}(\Delta_{2})$, we conclude any tensor in the kernel of
$L_{g}$ of the first type can be represented as $ug_{1}+2\nabla_{1}^{2}u$ with
$\Delta_{1}u+u=0$.

Next, we take any $h=vh_{2}$ in the kernel of $L_{g}$, where $v$ is an
eigenfunction of $\Delta_{1}$ and $h_{2}$ is an eigentensor of $L_{2}$. Since
$\text{spec}(\Delta_{1})=\\{0,1,\cdots\\}$, see [3], we assume
$\Delta_{1}v+\lambda_{1}v=0$ for $\lambda_{1}=0,1$ or $\lambda_{1}>1$.

If $\lambda_{1}=0$, then $v$ is a constant and $L_{2}(h_{2})=0$. We decompose
$h_{2}=\tilde{h}_{2}+\delta^{*}_{2}\alpha\in\text{ker}(\delta_{2})\oplus\text{im}(\delta_{2}^{*})$,
where $\alpha$ is a $1$-form on $M_{2}$. From Lemma 4.1, we conclude that
$L_{2}(\tilde{h}_{2})=0$. By taking the trace of the last equation, we obtain
$\Delta_{2}\tilde{H}_{2}+\tilde{H}_{2}=0$, where
$\tilde{H}_{2}=\text{Tr}_{g_{2}}\tilde{h}_{2}$. It follows from our assumption
($\dagger$ ‣ 4) that $\tilde{H}_{2}=0$ and hence $\tilde{h}_{2}=0$. Therefore,
$h=\delta^{*}_{2}\alpha$ with $L_{2}(\delta^{*}_{2}\alpha)=0$. In particular,
$h\in\text{im}(\delta^{*}_{g})$ as one can regard $\alpha$ as a $1$-form on
$M$.

If $\lambda_{1}=1$, then $\Delta_{1}v+v=0$ and $L_{2}(h_{2})-h_{2}=0$. It
follows from Lemma 4.1 that $h_{2}\in\text{ker}(\delta_{2})$. Moreover, by
taking the trace, $\Delta_{2}H_{2}=0$, where $H_{2}=\text{Tr}_{g_{2}}(h_{2})$
and hence $H_{2}$ is a constant. In other words,
$L_{2}(h_{2}-H_{2}g_{2}/n_{2})-(h_{2}-H_{2}g_{2}/n_{2})=0$, where
$n_{2}=\text{dim}(M_{2})$. Thus, by ($\dagger$ ‣ 4), we conclude that
$h_{2}=H_{2}g_{2}/n_{2}$.

If $\lambda_{1}>1$, then $L_{2}(h_{2})-\lambda_{1}h_{2}=0$. Again, it follows
from Lemma 4.1 that $h_{2}\in\text{ker}(\delta_{2})$. Moreover, by taking the
trace, $\Delta_{2}H_{2}+(1-\lambda_{1})H_{2}=0$ and hence $H_{2}=0$. In other
words, $h_{2}$ belongs to the TT-subspace and by ($\dagger$ ‣ 4), $h_{2}=0$
since $-\lambda_{1}\notin\text{spec}(L_{2}|_{\text{TT}})$.

From the above discussion, we conclude that any $h\in\text{ker}(L_{g})$ can be
written as

$\displaystyle h=ug_{1}+2\nabla_{1}^{2}u+vg_{2}+\delta_{g}^{*}\alpha,$

where $u,v$ are functions on $M_{1}$ with $\Delta_{1}u+u=\Delta_{1}v+v=0$ and
$\alpha$ is a $1$-form on $M$. From this, the conclusion immediately follows.
∎

Next, we prove

######  Lemma 4.4.

Let $(M^{n},g)=(M^{n_{1}}_{1}\times M^{n_{2}}_{2},g_{1}\times g_{2})$ such
that $(M_{i},g_{i})$ _( $i=1,2$)_ is an Einstein manifold with Einstein
constant $1/2$. Then

$\displaystyle\Phi^{(2)}(ug,vg_{1})=-\nabla^{2}f_{st}-\frac{n_{1}-2}{4}(du\otimes
dv+dv\otimes du)-g_{1}\langle\nabla u,\nabla
v\rangle+\frac{g+g_{1}}{2}uv-\frac{n_{1}-2}{2}(u\nabla^{2}v+v\nabla^{2}u),$

where $u,v$ are functions on $M_{1}$ with $\Delta_{1}u+u=\Delta_{1}v+v=0$ and
$f_{st}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{st}=\frac{n_{1}}{2}uv-\frac{3n_{1}-2}{4}\langle\nabla
u,\nabla v\rangle.$

In particular, we have

$\displaystyle\langle\Phi^{(2)}(ug,vg_{1}),g\rangle=-\Delta
f_{st}+\frac{2-3n_{1}}{2}\langle\nabla u,\nabla
v\rangle+\frac{n+3n_{1}-4}{2}uv$ (4.1)

and

$\displaystyle\langle\Phi^{(2)}(ug,vg_{1}),g_{1}\rangle=-\Delta
f_{st}+\frac{2-3n_{1}}{2}\langle\nabla u,\nabla v\rangle+2(n_{1}-1)uv.$ (4.2)

###### Proof.

We define $\tilde{g}(t,s)=(1+tu)(g+svg_{1})$ and
$\widetilde{Rc}=Rc(\tilde{g})$.

From Lemma A.1(1), we have at $t=0$,

$\displaystyle\widetilde{Rc}_{t}=-\frac{n-2}{2}\nabla^{2}_{g+svg_{1}}u-\frac{1}{2}(\Delta_{g+svg_{1}}u)(g+svg_{1}).$

By taking the derivative for $s$ and using Lemma A.1 (7) (9), we obtain

$\displaystyle\widetilde{Rc}_{st}=$ $\displaystyle\frac{n-2}{4}(du\otimes
dv+dv\otimes du-\langle\nabla u,\nabla v\rangle
g_{1})+\frac{uv}{2}g_{1}-\frac{g}{2}\left(uv+\frac{n_{1}-2}{2}\langle\nabla
u,\nabla v\rangle\right)$ $\displaystyle=$
$\displaystyle\frac{n-2}{4}(du\otimes dv+dv\otimes
du)-\frac{g_{2}}{2}uv-\left(\frac{n-2}{4}g_{1}+\frac{n_{1}-2}{4}g\right)\langle\nabla
u,\nabla v\rangle,$

since $u,v$ depend only on $M_{1}$.

Now, we have

$\displaystyle Rc_{st}=\widetilde{Rc}_{st}-Rc^{\prime}(uvg_{1}),$

which by Lemma A.1 (1) can be simplified as

$\displaystyle Rc_{st}=$
$\displaystyle\widetilde{Rc}_{st}+\frac{n_{1}-2}{2}\nabla^{2}(uv)+\frac{1}{2}\Delta(uv)g_{1}$
$\displaystyle=$ $\displaystyle\frac{n+2n_{1}-6}{4}(du\otimes dv+dv\otimes
du)+\left(\frac{6-n}{4}g_{1}-\frac{n_{1}-2}{4}g\right)\langle\nabla u,\nabla
v\rangle$
$\displaystyle-\frac{g+g_{1}}{2}uv+\frac{n_{1}-2}{2}(u\nabla^{2}v+v\nabla^{2}u).$

Similarly, we define $\widetilde{R}=R(\tilde{g})$ and compute from Lemma A.1
(4) that

$\displaystyle\widetilde{R}_{t}=-uR_{g+svg_{1}}-(n-1)\Delta_{g+svg_{1}}u$

and hence

$\displaystyle\widetilde{R}_{st}=\frac{4-2n-n_{1}}{2}uv-\frac{(n-1)(n_{1}-2)}{2}\langle\nabla
u,\nabla v\rangle.$

Therefore, we have

$\displaystyle R_{st}=$
$\displaystyle\widetilde{R}_{st}-R^{\prime}(uvg_{1})=\widetilde{R}_{st}+\frac{n_{1}}{2}uv+(n_{1}-1)\Delta(uv)$
$\displaystyle=$
$\displaystyle(4-n-2n_{1})uv+\frac{5n_{1}+2n-6-nn_{1}}{2}\langle\nabla
u,\nabla v\rangle.$

Differentiating $2\Delta f-|\nabla f|^{2}+R+f-n=\boldsymbol{\mu}$, we obtain

$\displaystyle 2\Delta f_{st}+2\Delta_{t}f_{s}+2\Delta_{s}f_{t}-2\langle\nabla
f_{s},\nabla f_{t}\rangle+R_{st}+f_{st}=\boldsymbol{\mu}_{st}=0,$

where $f_{s}=\frac{n_{1}-2}{2}v$ and $f_{t}=\frac{n-2}{2}u$. Therefore, we
obtain

$\displaystyle(\Delta+\frac{1}{2})f_{st}=\frac{n_{1}}{2}uv-\frac{3n_{1}-2}{4}\langle\nabla
u,\nabla v\rangle.$

In addition, we compute

$\displaystyle(\nabla^{2}f)_{st}=$
$\displaystyle\nabla^{2}f_{st}+\nabla^{2}_{t}f_{s}+\nabla^{2}_{s}f_{t}$
$\displaystyle=$ $\displaystyle\nabla^{2}f_{st}-\frac{n+n_{1}-4}{4}(du\otimes
dv+dv\otimes
du)+\left(\frac{n_{1}-2}{4}g+\frac{n-2}{4}g_{1}\right)\langle\nabla u,\nabla
v\rangle.$

Therefore, we have

$\displaystyle\Phi_{st}=$ $\displaystyle-R_{st}-(\nabla^{2}f)_{st}$
$\displaystyle=$ $\displaystyle-\nabla^{2}f_{st}-\frac{n_{1}-2}{4}(du\otimes
dv+dv\otimes du)-g_{1}\langle\nabla u,\nabla
v\rangle+\frac{g+g_{1}}{2}uv-\frac{n_{1}-2}{2}(u\nabla^{2}v+v\nabla^{2}u).$

By taking the trace, we immediately obtain (4.1) and (4.2). ∎

######  Proposition 4.5.

Let $(M^{n},g)=(M_{1}\times M_{2},g_{1}\times g_{2})$ with condition
($\dagger$ ‣ 4). Then any $h\in\emph{ISD}$ is not integrable of second-order.

###### Proof.

From Lemma 4.3, one can represent $h$ as

$\displaystyle h=u^{\prime}g_{1}+2\nabla_{1}^{2}u^{\prime}+v^{\prime}g_{2},$

for functions $u^{\prime},v^{\prime}$ on $M_{1}$ with
$\Delta_{1}u^{\prime}+u^{\prime}=\Delta_{1}v^{\prime}+v^{\prime}=0$. By
composing with a family of diffeomorphisms generated by $-\nabla u^{\prime}$,
we may assume $h=u^{\prime}g_{1}+v^{\prime}g_{2}=ug+vg_{1}$, where
$u:=v^{\prime}$ and $v:=u^{\prime}-v^{\prime}$.

Suppose $h$ is integrable of second-order, then
$\Phi^{(2)}(h,h)\perp\text{ker}(\Phi^{\prime})$. Notice that by Lemma 4.3,
$wg,wg_{1}\in\text{ker}(\Phi^{\prime})$ for any $w\in C^{\infty}(M_{1})$ with
$\Delta_{1}w+w=0$.

Now, we decompose

$\displaystyle\int_{M}\langle\Phi^{(2)}(h,h),wg\rangle$
$\displaystyle=\int_{M}\langle\Phi^{(2)}(ug,ug),wg\rangle+\int_{M}\langle\Phi^{(2)}(vg_{1},vg_{1}),wg\rangle+2\int_{M}\langle\Phi^{(2)}(ug,vg_{1}),wg\rangle$
$\displaystyle=:I_{1}+I_{2}+I_{3}.$

It follows from Lemma A.2 that

$\displaystyle\langle\Phi^{(2)}(ug,ug),g\rangle=-\Delta
f_{1}+2(n-1)u^{2}-\frac{3n-2}{2}|\nabla u|^{2},$

where $f_{1}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{1}=\frac{n}{2}u^{2}-\frac{3n-2}{4}|\nabla
u|^{2}.$

Since $\Delta w+w=0$, we have

$\displaystyle\int_{M}wf_{1}=-2\int_{M}w(\Delta+\frac{1}{2})f_{1}=\int_{M}-nu^{2}w+\frac{3n-2}{2}|\nabla
u|^{2}w.$

Therefore, we compute

$\displaystyle I_{1}=\int_{M}wf_{1}+2(n-1)u^{2}w-\frac{3n-2}{2}|\nabla
u|^{2}w=(n-2)\int_{M}u^{2}w.$ (4.3)

By similar calculations, we have

$\displaystyle
I_{2}=\text{Vol}(M_{2})\int_{M_{1}}\langle\Phi_{1}^{(2)}(vg_{1},vg_{1}),wg_{1}\rangle=(4m-2)\int_{M}v^{2}w.$
(4.4)

Now, it follows from (4.1) that

$\displaystyle\langle\Phi^{(2)}(ug,vg_{1}),g\rangle=-\Delta
f_{2}-(6m-1)\langle\nabla u,\nabla v\rangle+(n/2+6m-2)uvw,$

where $f_{2}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{2}=2muv-\frac{6m-1}{2}\langle\nabla
u,\nabla v\rangle.$

Therefore, one obtains

$\displaystyle I_{3}=2\int_{M}wf_{2}-(6m-1)\langle\nabla u,\nabla v\rangle
w+(n/2+6m-2)uvw=(n+4m-4)\int uvw.$ (4.5)

Combing (4.3), (4.4) and (4.5), we conclude that

$\displaystyle\int_{M}\langle\Phi^{(2)}(h,h),wg\rangle=(n-2)\int_{M}u^{2}w+2(2m-1)\int_{M}v^{2}w+(n+4m-4)\int_{M}uvw.$
(4.6)

Next, we decompose

$\displaystyle\int_{M}\langle\Phi^{(2)}(h,h),wg_{1}\rangle$
$\displaystyle=\int_{M}\langle\Phi^{(2)}(ug,ug),wg_{1}\rangle+\int_{M}\langle\Phi^{(2)}(vg_{1},vg_{1}),wg_{1}\rangle+2\int_{M}\langle\Phi^{(2)}(ug,vg_{1}),wg_{1}\rangle$
$\displaystyle=:I^{\prime}_{1}+I^{\prime}_{2}+I^{\prime}_{3}.$

It follows from Lemma A.2 that

$\displaystyle\langle\Phi^{(2)}(ug,ug),g_{1}\rangle=-\Delta
f_{1}+(n+4m-2)u^{2}-\frac{8m+n-2}{2}|\nabla u|^{2},$

where we have used the fact that $f_{1}$ depends only on $M_{1}$.

Therefore, we have

$\displaystyle I^{\prime}_{1}=$
$\displaystyle\int_{M}wf_{1}+(n+4m-2)u^{2}w-\frac{8m+n-2}{2}|\nabla u|^{2}w$
$\displaystyle=$ $\displaystyle\int_{M}(4m-2)u^{2}w+(n-4m)|\nabla
u|^{2}w=\frac{4m+n-4}{2}\int_{M}u^{2}w,$ (4.7)

since $2\int_{M}|\nabla u|^{2}w=\int_{M}u^{2}w$. Next, we observe that

$\displaystyle I_{2}=I_{2}^{\prime}=(4m-2)\int_{M}v^{2}w.$ (4.8)

Now, it follows from (4.2) that

$\displaystyle I^{\prime}_{3}=2\int_{M}wf_{3}-(6m-1)\langle\nabla u,\nabla
v\rangle w+2(4m-1)uvw=4(2m-1)\int_{M}uvw.$ (4.9)

Combing (4.7), (4.8) and (4.9), we conclude that

$\displaystyle\int_{M}\langle\Phi^{(2)}(h,h),wg_{1}\rangle=\frac{4m+n-4}{2}\int_{M}u^{2}w+2(2m-1)\int_{M}v^{2}w+4(2m-1)\int_{M}uvw.$
(4.10)

From (4.6), (4.10) and the fact that $u,v,w$ depend only on $M_{1}$, we
conclude that

$\displaystyle\int_{M_{1}}\psi_{1}(u,v)w=\int_{M_{1}}\psi_{2}(u,v)w=0,$ (4.11)

where $\psi_{1}$ and $\psi_{2}$ are quadratic functions given by

$\displaystyle\psi_{1}(u,v)=\frac{n-2}{2(2m-1)}u^{2}+v^{2}+\frac{n+4m-4}{2(2m-1)}uv,\quad\psi_{2}(u,v)=\frac{4m+n-4}{4(2m-1)}u^{2}+v^{2}+2uv.$

Now, we define constants

$\displaystyle\lambda=\frac{n-4m}{4(2m-1)},\quad
x=\frac{1+\sqrt{1+4\lambda}}{2},\quad y=\frac{1-\sqrt{1+4\lambda}}{2}.$

Then it is clear that

$\displaystyle(xu+v)^{2}=-\frac{y}{\lambda}\psi_{1}(u,v)+(1+\frac{y}{\lambda})\psi_{2}(u,v)$

and

$\displaystyle(yu+v)^{2}=-\frac{x}{\lambda}\psi_{1}(u,v)+(1+\frac{x}{\lambda})\psi_{2}(u,v).$

Therefore, it follows from (4.11) that for any $w\in C^{\infty}(M_{1})$ with
$\Delta_{1}w+w=0$,

$\displaystyle\int_{M_{1}}(xu+v)^{2}w=\int_{M_{1}}(yu+v)^{2}w=0$

and hence by Proposition 2.7 and Lemma 2.10 that

$\displaystyle xu+v=yu+v=0.$

Clearly, it implies that $h=0$, which is a contradiction.

In sum, the proof is complete. ∎

Combining Proposition 4.5 with [20, Theorem 1.1], we have the following
rigidity result.

######  Theorem 4.6.

For any compact Einstein manifold $(M_{2},g_{2})$ satisfying ($\dagger$ ‣ 4),
there exists a small constant $\epsilon>0$ satisfying the following property.

Suppose $(M^{n},g,f)$ is a Ricci shrinker such that

$\displaystyle d_{GH}\left\\{(M^{n},g),(\mathbb{C}\emph{P}^{2m}\times
M_{2}^{n-4m},g_{FS}\times g_{2})\right\\}<\epsilon,$ (4.12)

then $(M,g)$ is isometric to $(\mathbb{C}\emph{P}^{2m}\times
M_{2}^{n-4m},g_{FS}\times g_{2})$.

In general, we conjecture that the product of two complex projective spaces is
also rigid.

######  Conjecture 4.7.

For any integers $N_{1},N_{2}\geq 1$, there exists a small constant
$\epsilon=\epsilon(N_{1},N_{2})>0$ satisfying the following property.

Suppose $(M^{n},g,f)$ is a Ricci shrinker such that

$\displaystyle
d_{GH}\left\\{(M^{n},g),(\mathbb{C}\emph{P}^{N_{1}}\times\mathbb{C}\emph{P}^{N_{2}},g_{FS}\times
g_{FS})\right\\}<\epsilon,$ (4.13)

then $(M^{n},g)$ is isometric to
$(\mathbb{C}\emph{P}^{N_{1}}\times\mathbb{C}\emph{P}^{N_{2}},g_{FS}\times
g_{FS})$. Here, $n=2(N_{1}+N_{2})$ and $g_{FS}\times g_{FS}$ is the product of
the Fubini-Study metrics.

In addition, motivated by [6], we also make the following conjecture for the
noncompact case.

######  Conjecture 4.8.

For any integers $N,k\geq 1$, there exists a small constant
$\epsilon=\epsilon(N,k)>0$ satisfying the following property.

Suppose $(M^{n},g,f)$ is a Ricci shrinker such that

$\displaystyle
d_{PGH}\left\\{(M^{n},g,p),(\mathbb{C}\emph{P}^{N}\times\mathbb{R}^{k},g_{FS}\times
g_{E})\right\\}<\epsilon,$ (4.14)

then $(M^{n},g)$ is isometric to
$(\mathbb{C}\emph{P}^{N}\times\mathbb{R}^{k},g_{FS}\times g_{E})$. Here,
$n=2N+k$ and $g_{FS}\times g_{E}$ is the product of the Fubini-Study and
Euclidean metrics.

##  Appendix A The variational formulae

We first recall the following formulae of the geometric quantities along
conformal deformation, whose proof can be found in [29, Lemma A.2]. Notice
that the sign of the last term in $Rc_{ttt}$ here is different from that in
[29, Lemma A.2].

######  Lemma A.1.

Let $(M^{n},g)$ be a compact Riemannian manifold and $u$ a smooth function on
$M$. For $g(t)=(1+tu)g$ and any smooth function $\psi$ on $M$, we have at
$t=0$,

  1. (1)

$\text{Rc}_{t}=-\dfrac{n-2}{2}\nabla^{2}u-\dfrac{1}{2}(\Delta u)g$.

  2. (2)

$\text{Rc}_{tt}=(n-2)u\nabla^{2}u+\dfrac{3(n-2)}{2}du\otimes du+\left(u\Delta
u-\dfrac{n-4}{2}|\nabla u|^{2}\right)u$.

  3. (3)

$\text{Rc}_{ttt}=-3(n-2)u^{2}\nabla^{2}u-9(n-2)udu\otimes
du-\left(3u^{2}\Delta u-3(n-4)u|\nabla u|^{2}\right)g$.

  4. (4)

$R_{t}=-uR-(n-1)\Delta u$.

  5. (5)

$R_{tt}=2u^{2}R+4(n-1)u\Delta u-\dfrac{(n-1)(n-6)}{2}|\nabla u|^{2}$.

  6. (6)

$R_{ttt}=-6u^{3}R-18(n-1)u^{2}\Delta u+\dfrac{9(n-1)(n-6)}{2}u|\nabla u|^{2}$.

  7. (7)

$\nabla^{2}_{t}\psi=\dfrac{1}{2}\left(-du\otimes d\psi-d\psi\otimes
du+\langle\nabla\psi,\nabla u\rangle g\right)$.

  8. (8)

$\nabla^{2}_{tt}\psi=u\left(du\otimes d\psi+d\psi\otimes
du-\langle\nabla\psi,\nabla u\rangle g\right)$.

  9. (9)

$\Delta_{t}\psi=-u\Delta\psi+\dfrac{n-2}{2}\langle\nabla u,\nabla\psi\rangle$.

  10. (10)

$\Delta_{tt}\psi=2u^{2}\Delta\psi-2(n-2)u\langle\nabla u,\nabla\psi\rangle$.

Let $(M^{n},g)$ be a compact Einstein manifold with Einstein constant $1/2$.

######  Lemma A.2.

For any smooth function $u$ with $\Delta u+u=0$, we set $g(t)=(1+tu)g$ and
$f(t)=f(g(t))$. At $t=0$, we have the following variational formulae.

  1. (i)

(The second-order variational formula)

$\displaystyle\Phi_{tt}=-\nabla^{2}f_{tt}-\frac{n-2}{2}du\otimes
du+(u^{2}-|\nabla u|^{2})g-(n-2)u\nabla^{2}u,$ (A.1)

where $f_{tt}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{tt}=\frac{n}{2}u^{2}-\frac{3n-2}{4}|\nabla
u|^{2}.$ (A.2)

  2. (ii)

(The third-order variational formula)

$\displaystyle\Phi_{ttt}$
$\displaystyle=-\nabla^{2}f_{ttt}+\frac{3}{2}\left(du\otimes
df_{tt}+df_{tt}\otimes du-\langle\nabla f_{tt},\nabla u\rangle g\right)$
$\displaystyle+6(n-2)udu\otimes
du+3(n-2)u^{2}\nabla^{2}u-3u^{3}g-\frac{3(n-6)}{2}u|\nabla u|^{2}g,$ (A.3)

where $f_{ttt}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{ttt}=3u\Delta
f_{tt}-\frac{3(n-2)}{2}\langle\nabla f_{tt},\nabla
u\rangle-\frac{3(3n-2)}{2}u^{3}+\frac{9(3n-2)}{4}u|\nabla
u|^{2}+\frac{n-2}{2}\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int u^{3}.$ (A.4)

###### Proof.

(i) By our assumption of $u$, we know that $\Phi^{\prime}(ug)=0$ and hence
$\Phi_{t}=\boldsymbol{\mu}_{tt}=0$, where
$\boldsymbol{\mu}=\boldsymbol{\mu}(g(t),1)$. By differentiating twice for the
Euler-Lagrange equation $2\Delta f-|\nabla f|^{2}+R+f-n=\boldsymbol{\mu}$, we
obtain

$\displaystyle 2\Delta f_{tt}+4\Delta_{t}f_{t}-2|\nabla
f_{t}|^{2}+R_{tt}+f_{tt}=0,$ (A.5)

where we have used the fact that $f(0)$ is a constant. On the one hand, it
follows from (2.3) that

$\displaystyle(\Delta+\frac{1}{2})(nu-2f_{t})=\delta^{2}(ug)=\Delta u=-u$

and hence $f_{t}=\frac{n-2}{2}u$. On the other hand, it follows from Lemma A.1
that

$\displaystyle R_{tt}=(4-3n)u^{2}-\frac{(n-1)(n-6)}{2}|\nabla u|^{2}$

and

$\displaystyle\Delta_{t}f_{t}=2u^{2}\Delta f_{t}-2(n-2)u\langle\nabla u,\nabla
f_{t}\rangle=-(n-2)u^{3}-(n-2)^{2}u|\nabla u|^{2}.$

From (A.5), we immediately obtain (A.2). Next, we compute

$\displaystyle\Phi_{tt}=-Rc_{tt}-(\nabla^{2}f)_{tt}=-Rc_{tt}-\nabla^{2}f_{tt}-2\nabla_{t}^{2}f_{t}.$

From Lemma A.1(2)(7), we obtain (A.1) after simplification.

(ii) By differentiating three times for the Euler-Lagrange equation $2\Delta
f-|\nabla f|^{2}+R+f-n=\boldsymbol{\mu}$, we obtain

$\displaystyle 2\Delta f_{ttt}+6\Delta_{t}f_{tt}+6\Delta_{tt}f_{t}+6u|\nabla
f_{t}|^{2}-6\langle\nabla f_{t},\nabla
f_{tt}\rangle+R_{ttt}+f_{ttt}=\boldsymbol{\mu}_{ttt}.$ (A.6)

From the definition of $\boldsymbol{\mu}$, we have

$\displaystyle\boldsymbol{\mu}_{ttt}=\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int\langle\Phi_{tt},ug\rangle=(n-2)\mathchoice{{\vbox{\hbox{$\textstyle-$
}}\kern-7.18056pt}}{{\vbox{\hbox{$\scriptstyle-$
}}\kern-5.40836pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-4.35417pt}}{{\vbox{\hbox{$\scriptscriptstyle-$
}}\kern-3.89584pt}}\\!\int u^{3},$

where the last equality can be calculated similarly as [17, Theorem 5.7] by
using (A.1) and (A.2), see also [18, Proposition 9.1]. From Lemma
A.1(6)(9)(10), we obtain (A.4).

Similarly, we compute

$\displaystyle\Phi_{ttt}=-Rc_{ttt}-(\nabla^{2}f)_{ttt}=-Rc_{ttt}-\nabla^{2}f_{ttt}-3\nabla_{t}^{2}f_{tt}-3\nabla_{tt}^{2}f_{t}.$

Therefore, (A.6) follows from Lemma A.1(3)(7)(8) with a routined
simplification. ∎

######  Lemma A.3.

For any smooth functions $u,v$ with $\Delta u+u=\Delta v+v=0$, we set
$g(t,s)=(1+tu+sv)g$ and $f(t,s)=f(g(t,s))$. At $t=s=0$, we have the following
variational formula.

$\displaystyle\Phi_{st}=$
$\displaystyle-\nabla^{2}f_{st}-\frac{n-2}{4}(du\otimes dv+dv\otimes
du)+(uv-\langle\nabla u,\nabla
v\rangle)g-\frac{n-2}{2}(u\nabla^{2}v+v\nabla^{2}u),$

where $f_{st}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{st}=\frac{n}{2}uv-\frac{3n-2}{4}\langle\nabla
u,\nabla v\rangle.$ (A.7)

In particular,

$\displaystyle\langle\Phi_{st},g\rangle=-\Delta
f_{st}+\frac{2-3n}{4}\langle\nabla u,\nabla v\rangle+2(n-1)uv.$ (A.8)

###### Proof.

The proof follows from Lemma A.2 (i) by polarization. ∎

######  Lemma A.4.

For any smooth function $u$ with $\Delta u+u=0$ and symmetric $2$-tensor $h$
with $\delta h=0$, we set $g(t,s)=(1+tu)g+sh$ and $f(t,s)=f(g(t,s))$. At
$t=s=0$, we have the following variational formulae.

  1. (i)

(Cross second-order variational formula for $Rc$)

$\displaystyle Rc_{st}=$
$\displaystyle\frac{n-2}{2}C_{ij}^{k}u_{k}+\frac{u}{2}h+\langle
h,\nabla^{2}u\rangle\frac{g}{2}-\langle\nabla H,\nabla u\rangle\frac{g}{4}$
$\displaystyle+\frac{1}{2}\left(\Delta(uh)+2Rm(uh)-uh+\nabla^{2}(uH)+2\delta^{*}\delta(uh)\right).$
(A.9)

  2. (ii)

(Cross second-order variational formula for $\Phi$)

$\displaystyle\Phi_{st}=-Rc_{st}-\nabla^{2}f_{st}+\frac{n-2}{2}C_{ij}^{k}u_{k}+\frac{1}{4}\left(du\otimes
dH+dH\otimes du-\langle\nabla u,\nabla H\rangle g\right),$ (A.10)

where and $f_{st}$ is determined by

$\displaystyle(\Delta+\frac{1}{2})f_{st}=-\frac{1}{2}u\Delta
H-\frac{3}{4}\langle\nabla u,\nabla H\rangle.$ (A.11)

Here, $H=\emph{Tr}(h)$ and
$C_{ij}^{k}=\frac{1}{2}g^{kl}(\nabla_{i}h_{jl}+\nabla_{j}h_{il}-\nabla_{l}h_{ij})$.
In particular,

$\displaystyle\langle\Phi_{st},g\rangle=-\frac{n-2}{2}\langle
h,\nabla^{2}u\rangle-\frac{3}{2}\langle\nabla H,\nabla u\rangle-u\Delta
H+\frac{uH}{2}-\Delta f_{st}.$ (A.12)

###### Proof.

We define $\tilde{g}(t,s)=(1+tu)(g+sh)$ and $\widetilde{Rc}=Rc(\tilde{g})$.

From Lemma A.1(1), we have at $t=0$,

$\displaystyle\widetilde{Rc}_{t}=-\frac{n-2}{2}\nabla^{2}_{g+sh}u-\frac{1}{2}(\Delta_{g+sh}u)(g+sh).$

By taking the derivative for $s$, we obtain

$\displaystyle\widetilde{Rc}_{st}=\frac{n-2}{2}C_{ij}^{k}u_{k}+\frac{1}{2}uh+\langle
h,\nabla^{2}u\rangle\frac{g}{2}-\langle\nabla H,\nabla u\rangle\frac{g}{4}$

where we have used

$\displaystyle(\nabla_{i}\nabla_{j}u)_{s}=-(\Gamma_{ij}^{k})_{s}u_{k}=-C_{ij}^{k}u_{k}.$

and

$\displaystyle(\Delta_{g+sh}u)_{s}=-\langle h,\nabla^{2}u\rangle-\langle\delta
h,\nabla u\rangle+\frac{1}{2}\langle\nabla H,\nabla u\rangle=-\langle
h,\nabla^{2}u\rangle+\frac{1}{2}\langle\nabla H,\nabla u\rangle.$

Therefore, (A.9) follows from the fact that

$\displaystyle Rc_{st}=\widetilde{Rc}_{st}-Rc^{\prime}(uh).$

and by [2, Theorem 1.174(d)],

$\displaystyle-2Rc^{\prime}(uh)=\Delta(uh)+2Rm(uh)-uh+\nabla^{2}(uH)+2\delta^{*}\delta(uh).$

Similarly, we define $\widetilde{R}=R(\tilde{g})$ and compute from Lemma A.1
(4) that

$\displaystyle\widetilde{R}_{t}=-uR_{g+sh}-(n-1)\Delta_{g+sh}u$

and hence

$\displaystyle\widetilde{R}_{st}=u(\Delta
H+\frac{H}{2})+(n-1)\langle\nabla^{2}u,h\rangle-\frac{n-1}{2}\langle\nabla
H,\nabla u\rangle,$

where we have used [2, Theorem 1.174(e)].

Therefore, we have

$\displaystyle R_{st}=$
$\displaystyle\widetilde{R}_{st}-R^{\prime}(uh)=\widetilde{R}_{st}+\Delta(uH)-\delta^{2}(uh)+\frac{uH}{2}$
$\displaystyle=$ $\displaystyle\widetilde{R}_{st}+u\Delta H+2\langle\nabla
H,\nabla u\rangle-\langle h,\nabla^{2}u\rangle-\frac{uH}{2}$ $\displaystyle=$
$\displaystyle(n-2)\langle h,\nabla^{2}u\rangle+\frac{5-n}{2}\langle\nabla
H,\nabla u\rangle+2u\Delta H.$

Differentiating $2\Delta f-|\nabla f|^{2}+R+f-n=\boldsymbol{\mu}$, where
$\boldsymbol{\mu}=\boldsymbol{\mu}(g(t,s),1)$, we obtain

$\displaystyle 2\Delta f_{st}+2\Delta_{t}f_{s}+2\Delta_{s}f_{t}-2\langle\nabla
f_{s},\nabla f_{t}\rangle+R_{st}+f_{st}=\boldsymbol{\mu}_{st}=0,$

where the last equality holds since $ug\in\text{ker}(\Phi^{\prime})$. From
Lemma 2.5, $f_{t}=\frac{n-2}{2}u$, $f_{s}=H/2$ and hence by direct
calculations we obtain (A.11).

In addition, we compute

$\displaystyle(\nabla^{2}f)_{st}=$
$\displaystyle\nabla^{2}f_{st}+\nabla^{2}_{t}f_{s}+\nabla^{2}_{s}f_{t}$
$\displaystyle=$ $\displaystyle\nabla^{2}f_{st}+\frac{1}{4}(-du\otimes dH-
dH\otimes du+\langle\nabla H,\nabla u\rangle g)-\frac{n-2}{2}C^{k}_{ij}u_{k}.$

From $\Phi_{st}=-Rc_{st}-(\nabla^{2}f)_{st}$, we immediately obtain (A.10) and
hence (A.12) follows by taking the trace and noting that
$\langle\delta^{*}\delta(uh),g\rangle=-\langle h,\nabla^{2}u\rangle$ since
$\delta h=0$. ∎

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Yu Li, Institute of Geometry and Physics, University of Science and Technology
of China, No. 96 Jinzhai Road, Hefei, Anhui Province, 230026, China;
<EMAIL_ADDRESS>

Wenjia Zhang, School of Mathematical Sciences, University of Science and
Technology of China, No. 96 Jinzhai Road, Hefei, Anhui Province, 230026,
China<EMAIL_ADDRESS>