1. Introduction and main results
The Yamabe flow
\begin{equation} \dfrac{\partial g}{\partial t}(t)= -(R_{g(t)}-r_{g(t)})g(t),\quad g(0)=g_{0}, \end{equation}
where
$ R_{g(t)}$
is the scalar curvature of
$ g(t)$
and
\begin{equation*} r_{g(t)} = \dfrac { { \displaystyle \int _MR_{g(t)}dv_{g(t)} } }{ { \displaystyle \int _Mdv_{g(t)} } }, \end{equation*}
is the mean value of
$ R_{g(t)}$
along
$ M^{n}$
, it was introduced by R. Hamilton [Reference Hamilton1] and has become one of the standard tools of recent differential geometry. Yamabe solitons arise as self-similar solutions of (1).
Definition 1.
A self-similar solution
$g(t)$
of (1) is a Yamabe soliton if there exists a scalar factor
$\alpha \;:\;[0,\varepsilon )\rightarrow (0,\infty )$
,
$\varepsilon \gt 0$
, and a
$1$
-parameter family
$\{\psi _t\}$
of diffeomorphisms of
$M^{n}$
such that
\begin{equation*} g(t)=\alpha (t)\psi _{t}^{\ast }(g_{0}), \quad \alpha (0)=1\quad \mbox {and}\quad \psi _0=id_M. \end{equation*}
One gets
\begin{equation} \frac{1}{2}\mathcal{L}_{X}g=\left (R_g-\lambda \right )g, \end{equation}
by substituting
$ g(t)=\alpha (t)\psi _t^\ast (g_0)$
into (1) and evaluating the resulting expression at
$ t=0$
, where
$ \mathcal{L}_Xg$
is the Lie derivative of
$ g$
with respect to the field
$ X$
of directions associated with the
$ 1$
-parameter family
$ \{ \psi _t \}$
and
$ \lambda =\alpha '(0)+r_{g}$
. Equation (2) is the fundamental equation of Yamabe solitons. Since their beginning, a lot of results were proved on the nature of Yamabe solitons. For example, Chow [Reference Chow2] proved that compact Yamabe solitons have constant scalar curvature (see also [Reference Di Cerbo and Disconzi3,Reference Hsu4]). Daskalopoulos and Sesum [Reference Daskalopoulos and Sesum5] proved that complete locally conformally flat Yamabe solitons with positive sectional curvature are rotationally symmetric and must belong to the conformal class of flat Euclidean space.
A new notion of soliton is born if one replaces the scalar curvature in (1) by functions of the higher order scalar curvatures. As is the case with any generalization, it is hoped that one recovers the old objects as particular instances of the new ones, while open up room for new and exciting phenomena to happen. In what follows, we give formal definitions and even before we state our main results, we examine a few examples. We included a section containing the lemmas that we have used in the text for the convenience of the reader and a separate section with the proofs of our statements can be found right after it.
The Riemann curvature tensor
$\textrm{Rm}$
of
$ (M^{n},g)$
admits the following decomposition
\begin{equation*} \textrm {Rm}=W_g+A_{g}\bigcirc \,\!\!\!\!\!\!\!\wedge \;g, \end{equation*}
where
$ W_g$
and
$ A_g$
are the tensors of Weyl and Schouten, respectively, and
$\bigcirc \,\!\!\!\!\!\!\wedge\;$
is the Kulkarni–Nomizu product of
$ (M^{n},g)$
. Recall that the Schouten tensor is given by
\begin{equation*} A_g=\frac {1}{n-2}\left (\textrm {Ric}_{g}-\frac {R_g}{2(n-1)}g\right ). \end{equation*}
The
$ \sigma _k$
-curvature of
$ g$
is defined as the
$ k$
th elementary symmetric function of the eigenvalues
$\lambda _{1},\dots,\lambda _{n}$
of the endomorphism
$ g^{-1}A_g$
, that is,
\begin{equation*} \sigma _{k}(g)= \sigma _{k}(g^{-1}A_{g})= \sum _{1\leqslant i_{1}\lt \cdots \lt i_{k}\leqslant n} \lambda _{i_{1}}\cdots \lambda _{i_{k}},\quad 1\leqslant k\leqslant n. \end{equation*}
Here, we set
$\sigma _{0}(g)=1$
for convenience. A simple calculation shows that
$\sigma _{1}(g)=\frac{R_g}{2(n-1)}$
, which indicates that the
$\sigma _{k}$
-curvature is a reasonable substitute for the scalar curvature of
$ (M^{n},g)$
in (1).
Guan and Guofang introduced [Reference Guan and Guofang6] the fully nonlinear flow
\begin{equation} \dfrac{\partial g}{\partial t}(t) = -\left ( \log \frac{\sigma _{k}(g(t))}{\sigma _{l}(g(t))}-\log r_{k,l}(g(t)) \right )g(t),\quad g(0)=g_{0}, \end{equation}
where
\begin{equation*} \log r_{k,l}(g(t))= \dfrac { { \displaystyle \int _{M} \sigma _{l}(g(t))\log \dfrac {\sigma _{k}(g(t))}{\sigma _{l}(g(t))} dv_{g(t)} } }{ {\displaystyle \int _{M}\sigma _{l}(g(t))dv_{g(t)}} }, \end{equation*}
was defined as to make the flow preserve the quantities
\begin{equation*} \mathcal {E}_l(g(t))= \begin {cases} \displaystyle \int _{M}\sigma _{l}(g(t))dv_{g(t)}, &\text {if}\quad l\neq \frac {n}{2}, \\[9pt] -\displaystyle \int _{0}^{1}dt\int _{M}u\sigma _{n/2}(g(t))dv_{g(t)}, & \text {if}\quad l=\frac {n}{2}, \end {cases} \end{equation*}
where
$u\in C^\infty (M)$
,
$g=e^{-2u}g_0$
, and
$g(t)=e^{-2tu}g_0$
. The convergence of the fully nonlinear flow was then proved under certain conditions to be satisfied by the eigenvalues of the Schouten tensor. The authors also provided geometric inequalities such as the Sobolev-type inequality in case
$0\leqslant l\lt k\lt \frac{n}{2}$
, the conformal quasimass-integral-type inequality for
$\frac{n}{2}\leqslant k\leqslant n$
,
$1\leqslant l \lt k$
, and the Moser–Trudinger-type inequality for
$k=\frac{n}{2}$
.
Bo et al. [Reference Bo, Ho and Sheng7] presented quotient Yamabe solitons as self-similar solutions of the flow (3) and stated rigidity results for the existence of such objects on top of locally conformally flat manifolds. For example, it was shown that any compact and locally conformally flat manifold with the structure of a quotient Yamabe soliton, where both
$\sigma _k\gt 0$
and
$\sigma _l\gt 0$
, must have constant quotient curvature
$\frac{\sigma _{k}}{\sigma _{l}}$
. Also, for the so-called gradient
$k$
-Yamabe soliton (
$ l=0$
), they proved that, for
$ k\gt 1$
, any compact gradient
$k$
-Yamabe soliton with negative constant scalar curvature has necessarily constant
$\sigma _{k}$
-curvature. Almost Yamabe solitons were introduced by Barbosa and Ribeiro [Reference Barbosa and Ribeiro8] as generalizations of self-similar solutions of the Yamabe flow. Essentially, they allowed the parameter
$\lambda$
in (2) to be a function on
$M$
. The authors then stated rigidity results for almost Yamabe solitons on compact manifolds. We refer the reader to [Reference Barbosa and Ribeiro8–Reference Seko and Maeta11] for further information.
Catino et al. [Reference Catino, Mantegazza and Mazzieri12] proposed the study of conformal solitons. A conformal soliton is a Riemannian manifold
$(M^{n},g)$
together with a nonconstant function
$ f\in C^\infty (M)$
satisfying
$\nabla ^2f=\lambda g$
for some
$ \lambda \in \mathbb{R}$
. They provided classification results according to the number of critical points of
$f$
. It should be noticed that solitons of Yamabe,
$k$
-Yamabe, and quotient Yamabe types are examples of conformal solitons.
We introduce almost quotient Yamabe solitons in extension to the quotient Yamabe solitons.
Definition 2.
A solution
$ g(t)$
of (3) is an almost quotient Yamabe soliton if there exist a scalar factor
$ \alpha \;:\;M\times [0,\varepsilon )\rightarrow (0,\infty )$
,
$ \varepsilon \gt 0$
, and a
$ 1$
-parameter family
$ \{\psi _t\}$
of diffeomorphisms of
$ M^{n}$
such that
\begin{equation*} g(t)=\alpha (x,t)\psi _{t}^{\ast }(g_{0}),\quad \alpha (\,\cdot \,,0)\equiv 1\mbox { on }M^{n}\quad \mbox {and}\quad \psi _0=id_{M}. \end{equation*}
Equivalently,
$(M^{n},g)$
is an almost quotient Yamabe soliton if there exists a pair
$X\in \mathfrak{X}(M)$
,
$\lambda \in C^\infty (M)$
satisfying
\begin{equation} \frac{1}{2}\mathcal{L}_{X}g= \left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right )g,\quad \sigma _{k}\cdot \sigma _{l}\gt 0. \end{equation}
We will write the soliton in (4) as
$(M^{n}, g, X, \lambda )$
for the sake of simplicity. Following the terminology already in use with almost Yamabe solitons, a soliton
$ (M^{n},g,X,\lambda )$
will be called:
-
a. expanding if
$ \lambda \lt 0$
, -
b. steady if
$ \lambda =0$
, -
c. shrinking if
$ \lambda \gt 0$
and, finally, -
d. indefinite if
$ \lambda$
change signs on
$ M^{n}$
.
Definition 3.
An almost gradient quotient Yamabe soliton is an almost quotient Yamabe soliton
$(M^{n},g,X,\lambda )$
such that
$ X=\nabla f$
is the gradient field of a function
$ f\in C^\infty (M)$
.
Since
\begin{equation*} \dfrac {1}{2}\mathcal {L}_{\nabla f}g=\nabla ^2 f, \end{equation*}
it follows from (4) that an almost gradient quotient Yamabe soliton
$(M^{n},g,\nabla f,\lambda )$
is characterized by the equation
\begin{equation} \nabla ^2 f= \left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right )g, \quad \sigma _{k}\cdot \sigma _{l}\gt 0. \end{equation}
Almost quotient Yamabe solitons, gradient or not, are regarded as trivial if their defining equation vanishes identically. Thus,
$(M^{n},g,X,\lambda )$
is trivial if
$\mathcal{L}_Xg=0$
and
$(M^{n},g,\nabla f,\lambda )$
if
$\nabla ^2 f=0$
. In either case,
$\log{\frac{\sigma _k}{\sigma _l}}-\lambda =0$
. Let us take a look at a few examples.
Example 1. The product manifold
$(\mathbb{R}\times \mathbb{S}^{n},g=dt^2+g_{\mathbb{R}^n})$
alongside the function
\begin{equation*} f\;:\;\mathbb {R}\times \mathbb {S}^n\to \mathbb {R},\quad (t,x)\mapsto f(t,x)=at+b\quad (a,b\in \mathbb {R}), \end{equation*}
is, for
$ k=l=1$
, a trivial almost gradient quotient Yamabe soliton with
$\lambda =0$
, since
$\sigma _{1}(g^{-1}A_{g})=\frac{n}{2}$
and
$\nabla ^{2}f=0$
.
Example 2. Identities
\begin{equation*} \textrm {Ric}_{g_{\mathbb {S}^{n}}}=(n-1)g_{\mathbb {S}^{n}},\quad R_{g_{\mathbb {S}^{n}}}=n(n-1)\quad \mbox {and}\quad A_{g_{\mathbb {S}^n}}=\frac {1}{2}g_{\mathbb {S}^n}, \end{equation*}
stand for the Ricci tensor, scalar curvature, and Schouten tensor, respectively, of the Euclidean sphere
$ (\mathbb{S}^n,g_{\mathbb{S}^n})$
. Therefore, we have that
\begin{equation*} \sigma _{k}(g_{\mathbb {S}^{n}}^{-1}A_{g_{\mathbb {S}^{n}}})= \frac {1}{2^k}\binom {n}{k},\quad 1\leqslant k\leqslant n. \end{equation*}
Consider the height function
\begin{equation*} h_v\;:\;\mathbb {S}^n\rightarrow \mathbb {R},\quad x\mapsto h_{v}(x)=\langle x,v\rangle, \end{equation*}
on
$ \mathbb{S}^n$
with respect to a given
$ v\in \mathbb{S}^n$
. It then follows that
\begin{equation*} \nabla ^{2}h_{v}=-h_{v}g_{g_{\mathbb {S}^n}}, \end{equation*}
which shows that
$(\mathbb{S}^{n},g_{\mathbb{S}^{n}}, \nabla h_{v},\lambda )$
is a compact almost quotient Yamabe soliton with
\begin{equation*} \lambda \;:\;\mathbb {S}^n\to \mathbb {R},\quad x\mapsto h_v(x)+\log {\frac {\sigma _k}{\sigma _l}}. \end{equation*}
Example 3. On the hyperbolic space
$(\mathbb{H}^{n},g_{\mathbb{H}^{n}})$
, we consider
\begin{equation*} \textrm {Ric}_{g_{\mathbb {H}^{n}}}=-(n-1)g_{\mathbb {H}^{n}},\quad R_{g_{\mathbb {H}^{n}}}=-n(n-1)\quad \mbox {and}\quad A_{g_{\mathbb {H}^{n}}}=-\frac {1}{2}g_{\mathbb {H}^{n}}, \end{equation*}
to denote the Ricci tensor, scalar curvature, and Schouten tensor, respectively. Therefore, we have that
\begin{equation*} \sigma _{k}(g_{\mathbb {H}^{n}}^{-1}A_{g_{\mathbb {H}^{n}}})= \frac {(\!-\!1)^{k}}{2^k}\binom {n}{k},\quad 1\leqslant k\leqslant n. \end{equation*}
We consider the model
$\mathbb{H}^{n}=\{x\in \mathbb{R}^{n,1}\;:\; \langle x,x\rangle _{0}=-1, x_{1}\gt 0\}$
of the hyperbolic space, where
$\mathbb{R}^{n,1}$
denotes the Euclidean space
$\mathbb{R}^{n+1}$
endowed with Lorentzian inner product
$\langle x,x\rangle _{0}=-x_{1}^2+x_{2}^{2}+\dots +x_{n+1}^{2}$
. As in our previous example, we consider the height function
\begin{equation*} h_v\;:\;\mathbb {H}^n\rightarrow \mathbb {R},\quad x\mapsto h_{v}(x)=\langle x,v\rangle _{0}, \end{equation*}
on
$ \mathbb{H}^n$
with respect to a given
$ v\in \mathbb{H}^n$
. Because
\begin{equation*} \nabla ^{2}h_{v}=h_{v}g_{g_{\mathbb {H}^n}}, \end{equation*}
we conclude that
$(\mathbb{H}^{n},g_{\mathbb{H}^{n}}, \nabla h_{v},\lambda )$
is an almost quotient Yamabe soliton with
\begin{equation*} \lambda \;:\;\mathbb {H}^n\to \mathbb {R},\quad x\mapsto -h_v(x)+\log {\frac {\sigma _k}{\sigma _l}}, \end{equation*}
as long as we have
$ k\equiv l\pmod 2$
.
Example 4. Consider
$ \mathbb{R}^n$
endowed with a metric tensor of the form
\begin{equation*} g_{ij}=e^{2u_i}\delta _{ij},\quad 1\leqslant i,j\leqslant n, \end{equation*}
so given in cartesian coordinates
$ x=(x_1,\ldots,x_n)$
of
$ \mathbb{R}^n$
, where
$ u_1,\ldots,u_n\in C^\infty (\mathbb{R}^n)$
. Then, the Ricci tensor of
$ (\mathbb{R}^n,g)$
is given in [Reference Landau, Lifšic and and Landau13] by the formulas
\begin{equation*} \textrm {Ric}_g(\partial _j,\partial _k)= \begin {cases} \sum _{l\neq k,j}\,U_{jk}^l+u_{j,k}u_{l,j}, & \text {if}\quad j\neq k, \\[5pt] \sum _{l\neq k}\,e^{2(u_k-u_l)}U_{ll}^k+U_{kk}^l- \sum _{m\neq k,l}\,e^{2(u_k-u_m)}u_{k,m}u_{l,m}, & \text {if}\quad j=k, \end {cases} \end{equation*}
where
\begin{equation*} u_{i,j}=\dfrac {\partial u_i}{\partial x_j}\quad \mbox {and}\quad u_{i,j,k}=\dfrac {\partial ^2u_i}{\partial x_k\partial x_j}, \end{equation*}
and
\begin{equation*} U_{jk}^l=u_{l,k}(u_k-u_l)_{,j}-u_{l,j,k}, \end{equation*}
for every
$1\leqslant i,j,k,l\leqslant n$
. Assume that
$n\geqslant 4$
. Also, let
$ \tau$
be the
$ n$
-cycle
$ (1,2,3,\ldots,n)$
in the symmetric group
$ S_n$
of degree
$n$
. It turns out that by choosing functions
\begin{equation*} u_i(x_1,\ldots,x_n)= \begin {cases} \log {\cosh {\left (x_{\tau (i)}\right )}}, & \text {if}\quad i\equiv 0\pmod 2, \\[5pt] 0, & \text {if}\quad i\equiv 1\pmod 2, \end {cases} \end{equation*}
we simplify the situation quite a little bit as the Ricci tensor of
$ (\mathbb{R}^n,g)$
ends up being a constant multiple of the metric,
$ \textrm{Ric}_g=-g$
. Therefore,
$ (\mathbb{R}^n,g)$
is a complete Einstein manifold and, as such,
$ A_g=\frac{-1}{2(n-1)}g$
. Then, we have that
\begin{equation*} \sigma _k(g^{-1}A_g)=\dfrac {(\!-\!1)^k}{2^k(n-1)^k}\binom {n}{k},\quad 1\leqslant k\leqslant n. \end{equation*}
Because
$ X=(0,1,\ldots,0,1)$
is a Killing field on
$ (\mathbb{R}^n,g)$
we know that
$ (\mathbb{R}^n,g,X,\lambda )$
is a trivial almost quotient Yamabe soliton whenever
$ k\equiv l\pmod 2$
. It should be noticed that
$ X$
is not a gradient field with respect to the metric
$ g$
.
Any smooth vector field
$ X$
on a compact Riemannian manifold
$ (M^{n},g)$
can be written in the form
\begin{equation} X = \nabla h + Y, \end{equation}
where
$Y\in \mathfrak{X}(M)$
is divergence free and
$h\in C^\infty (M)$
. In fact, by the Hodge-de Rham Theorem [Reference Warner14], we have that
\begin{equation*} X^{\flat }=d\alpha +\delta \beta +\gamma. \end{equation*}
Now, take
$Y = (\delta \beta +\gamma )^{\sharp },\,\nabla h=(d\alpha )^{\sharp }$
and we are done. The function
$ h$
is called the Hodge-de Rham potential of
$ X$
. Our first theorem states the triviality of a compact almost quotient Yamabe soliton under certain integral assumptions.
Theorem 1.
A compact almost quotient Yamabe soliton
$(M^{n},g, X,\lambda )$
is trivial if one of the following assertions holds:
-
a)
$\displaystyle \int _{M}e^{\lambda }\sigma _{l}\langle \nabla \lambda,X\rangle dv_{g}= -\int _{M}e^{\lambda }\langle \nabla \sigma _{l}, X\rangle dv_{g}$
, plus any of these:
-
i.
$\nabla \textrm{Ric}_g=0$
; -
ii.
$div\,{C_g}=0$
, where
$C_g$
is the Cotton tensor of
$(M^{n},g)$
; -
iii.
$X=\nabla f$
is a gradient vector field;
-
-
b)
$\displaystyle \int _{M}\langle \nabla h,X\rangle dv_{g}\leqslant 0$
, where
$h$
is the Hodge-de Rham potential of
$ X$
.
The next two corollaries deal with quotient Yamabe solitons (
$ \lambda$
is a real constant) and constitute direct applications of Theorem 1. In [Reference Bo, Ho and Sheng7], Bo et al. proved that
$\sigma _k/\sigma _l$
must be constant on any compact and locally conformally flat quotient Yamabe soliton. We extend Bo’s result.
Corollary 1.
Let
$(M^{n},g, X,\lambda )$
be any compact quotient Yamabe soliton with a vanishing cotton tensor. Then,
$\sigma _{k}/\sigma _{l}$
is constant and, as such, the soliton is trivial.
In [Reference Catino, Mantegazza and Mazzieri12], Catino et al. proved that any compact gradient
$k$
-Yamabe soliton with a nonnegative Ricci tensor is trivial. Bo et al. [Reference Bo, Ho and Sheng7] also proved that any compact gradient
$k$
-Yamabe soliton with constant negative scalar curvature is trivial. In [Reference Tokura and Batista15], it was shown that any compact gradient
$k$
-Yamabe soliton must be trivial. We extend all these results at once.
Corollary 2.
Let
$(M^{n},g, \nabla f,\lambda )$
be any compact quotient gradient Yamabe soliton. Then,
$\sigma _k/\sigma _l$
is constant and, as such, the soliton is trivial.
Yet another triviality result holds for almost quotient Yamabe solitons if one drops compacity on
$ M^{n}$
in favor of a decay condition on the norm of the soliton field
$ X$
.
Theorem 2.
Let
$(M^{n},g,X,\lambda )$
be a complete and noncompact almost quotient Yamabe soliton satisfying
\begin{equation*} \int _{M^{n}\setminus B_{r}(x_{0})}\dfrac {|X|}{d(x,x_{0})}dv_{g}\lt \infty \quad {\textit{and}}\quad \mathcal {L}_Xg\geqslant 0, \end{equation*}
where
$d$
is the distance function with respect to
$g$
and
$B_{r}(x_{0})$
is the ball of radius
$r\gt 0$
centered at
$x_{0}$
. Then,
$(M^{n},g,X,\lambda )$
is trivial.
Next, we give a sufficient condition for a compact almost quotient gradient Yamabe soliton to be isometric with an Euclidean sphere.
Theorem 3.
Let
$(M^{n},g, \nabla f,\lambda )$
be a nontrivial compact quotient gradient almost Yamabe soliton with constant scalar curvature
$R_g=R\gt 0$
. Then
$(M^{n},g)$
is isometric to the Euclidean sphere
$\mathbb{S}^{n}(\sqrt{r})$
,
$ r=R/n(n-1)$
. Moreover, up to a rescaling, the potential
$f$
is given by
$f=h_{v}+c$
where
$h_{v}$
is the height function on the sphere and
$c$
is a real constant.
Another situation in which an almost gradient quotient Yamabe soliton must be isometric with an Euclidean sphere is described below.
Theorem 4.
Let
$(M^{n},g, \nabla f,\lambda )$
be a nontrivial compact quotient gradient almost Yamabe soliton with constant
$\sigma _{k}$
-curvature, for some
$k=2,\dots,n$
, and
$A_g\gt 0$
. Then,
$(M^{n},g)$
is isometric with an Euclidean sphere
$\mathbb{S}^{n}$
.
Remark 1. A similar result concerning almost Ricci solitons can be found in [Reference Barros and Evangelista16].
Finally, we investigate the structure of noncompact almost quotient gradient Yamabe solitons satisfying reasonable conditions on its potential function and both Ricci and scalar curvatures.
Theorem 5.
Let
$(M^{n},g, \nabla f,\lambda )$
be a nontrivial and noncompact almost quotient gradient Yamabe soliton. Assume that
\begin{equation*} \mathcal {L}_{\nabla f^2}R\geqslant 0, \quad \overset {\circ }{\textrm {Ric}_g}(\nabla f,\nabla f)\geqslant {0} \quad {\textit{and}}\quad |\overset {\circ }{\textrm {Ric}}(\nabla f^2)|\in L^{1}(M). \end{equation*}
Then,
$(M^{n},g)$
has constant scalar curvature
$R_g=R\leqslant 0$
and
$f$
has at most one critical point. Moreover, we have that:
-
a) If
$R=0$
, then
$(M^{n},g)$
is isometric with a Riemannian product manifold
$(\mathbb{R}\times \mathbb{F}^{n-1},dt^2+g_{\mathbb{F}})$
; -
b) If
$R\lt 0$
and
$f$
has no critical points, then
$(M^{n},g)$
is isometric with a warped product manifold
$(\mathbb{R}\times \mathbb{F}^{n-1},dt^2+\xi (t)^2g_{\mathbb{F}})$
such that
\begin{equation*} \xi ''+\frac {R}{n(n-1)}\xi =0; \end{equation*}
-
c) If
$R\lt 0$
and
$f$
has only one critical point, then
$(M^{n},g)$
is isometric with a hyperbolic space.
Remark 2. Einstein manifolds satisfy the hypothesis of Theorem (5) quite naturally for if
\begin{equation*} \textrm {Ric}=\rho g, \end{equation*}
for some
$\rho \in \mathbb{R}$
, then
$R$
is constant over
$M$
and, as such, we have that
\begin{equation*} \mathcal {L}_{\nabla f^{2}}R\equiv 0. \end{equation*}
Furthermore, the traceless Ricci tensor
\begin{equation*} \overset {\circ }{\textrm {Ric}}=\textrm {Ric}-\frac {R}{n}g\equiv 0, \end{equation*}
vanishes identically, thus giving
$|\overset{\circ }{Ric}(\nabla f^{2})|\in L^{1}(M)$
.
2. Key Lemmas
In this section, we collect some useful lemmas that will used in the proof of the main results.
Lemma 1. ([Reference Barbosa and Freitas17,Reference Gover and Ørsted18]). Let
$ (M^{n},g)$
be a compact Riemannian manifold with a possibly empty boundary
$ \partial M$
. Then,
\begin{equation*} \int _{M}X(\textrm {tr}\,T)dv_{g}= n\int _{M}\textrm {div}\,{T(X)}dv_{g}+ \frac {n}{2} \int _{M}\langle \overset {\circ }{T},\mathcal {L}_{X}g\rangle dv_{g}- n\int _{\partial M}\overset {\circ }{T}(X,\nu )ds_{g}, \end{equation*}
for every symmetric
$ (0,2)$
-tensor
$ T$
and every vector field
$ X$
on
$M$
, where
\begin{equation*} \textrm {tr}\,T=g^{ij}T_{ij}\quad {\textit{and}}\quad \overset {\circ }{T}=T-\frac {\textrm {tr}\,T}{n}g, \end{equation*}
and
$\nu$
is the outward unit normal field on
$ \partial M$
.
Proof. First, notice that integration by parts yields
\begin{equation*} \int _{\partial M}T(X,\nu )dA_{g}=\int _{M}\nabla ^{i}(T_{ij}X^{j})dv_{g}, \end{equation*}
and because
\begin{align*} \nabla ^{i}(T_{ij}X^{j}) & = \nabla ^{i}T_{ij}X^{j}+T_{ij}\nabla ^{i}X^{j} \\[5pt] &= \nabla ^{i}T_{ij}X^{j}+\frac{1}{2}T_{ij}(\nabla ^{i}X^{j}+\nabla ^{j}X^{i}) \\[5pt] &= \textrm{div}\,{T(X)}+\frac{1}{2}\langle T,\mathcal{L}_{X}g\rangle, \end{align*}
we get that
\begin{equation} \begin{split} \int _{\partial M}T(X,\nu )dA_{g} &= \int _{M}\nabla ^{i}(T_{ij}X^{j})dv_{g} \\[5pt] &= \int _{M} \textrm{div}\,{T(X)} dv_{g}+ \frac{1}{2}\int _{M}\langle T,\mathcal{L}_{X}g\rangle dv_{g} \\[5pt] &= \int _{M} \textrm{div}\,{T(X)} dv_{g}+ \frac{1}{2} \int _{M}\langle \overset{\circ }{T},\mathcal{L}_{X}g\rangle dv_{g}+ \frac{1}{2} \int _{M}\frac{\textrm{tr}\,T}{n}\langle g,\mathcal{L}_{X}g\rangle dv_{g} \\[5pt] &= \int _{M} \textrm{div}\,{T(X)} dv_{g}+ \frac{1}{2} \int _{M}\langle \overset{\circ }{T},\mathcal{L}_{X}g\rangle dv_{g}+ \frac{1}{n}\int _{M}\textrm{tr}\,T\cdot \textrm{div}\,Xdv_{g}. \end{split} \end{equation}
On the other hand, we have that
\begin{equation} \int _{M}\textrm{tr}\,T\cdot \textrm{div}\,Xdv_{g}= \int _{\partial M} \textrm{tr}\,T\cdot \langle X,\nu \rangle dA_{g}-\int _{M}X(\textrm{tr}\,T) dv_{g}. \end{equation}
The result now follows from (7) and (8) above.
We now recall a useful result established in [Reference Barros and Gomes19].
Lemma 2. ([Reference Barros and Gomes19]) Let
$ (M^{n},g)$
be a Riemannian manifold and
$ T$
be a symmetric
$ (0,2)$
-tensor field on
$ M^{n}$
. Then
\begin{equation*} \textrm {div}\,(T(\varphi X))= \varphi (\textrm {div}\,T)(X)+ \varphi \langle \nabla X, T\rangle + T(\nabla \varphi,X), \end{equation*}
for any
$X\in \mathfrak{X}(M)$
and
$\varphi \in C^\infty (M)$
where
$T(X)$
is the vector field
$ g$
-equivalent to
$ T$
.
For locally conformally flat manifolds, a proposition similar to the next one can be found in [Reference Han20]. Recall that a vector field
$ X$
on a Riemannian manifold
$ (M^{n},g)$
is a conformal field in case
\begin{equation*} \frac {1}{2}\mathcal {L}_Xg=\varphi g, \end{equation*}
for some
$ \varphi \in C^\infty (M)$
.
Recall that the
$k$
-Newton tensor field associated with
$g^{-1}A_g$
is defined by
\begin{equation*} T_{k}(g^{-1}A_g)=\sum _{j=0}^k\,(\!-\!1)^j\sigma _{k-j}(g)(g^{-1}A_g)^j, \quad 1\leqslant k\leqslant n. \end{equation*}
Among the identities satisfied by
$ T_k(g^{-1}A_g)$
one finds (see [Reference Barros and Evangelista16])
\begin{equation*} \textrm {tr}\,{T_k}(g^{-1}A_g)=(n-k)\sigma _k(g) \quad \mbox {and}\quad \textrm {div}\,{T_k}(g^{-1}A_g)=0, \end{equation*}
for every
$1\leqslant k\leqslant n$
.
Proposition 1.
If
$ X$
is a conformal vector field on a compact Riemannian manifold
$(M^{n},g)$
with null Cotton tensor, then
\begin{equation*} \int _{M^{n}}\langle X,\nabla \sigma _{k}\rangle dv_{g}=0, \end{equation*}
for every
$ k=1,2,\ldots,n$
.
Proof. Let
$ \varphi \in C^\infty (M)$
be such that
\begin{equation*} \frac {1}{2}\mathcal {L}_Xg=\varphi g, \end{equation*}
and take
$ T_k=T_k(g^{-1}A_g)$
where
$ k\in \{1,2,\ldots,n-1\}$
. Now a direct application of Lemma 1 yields
\begin{equation} \int _{M}X(\textrm{tr}\,{T_k}) dv_{g}= n\int _{M} \textrm{div}\,{T_k(X)}dv_{g}+ n\int _{M}\varphi \langle \overset{\circ }{T_k},g\rangle dv_{g}. \end{equation}
It follows from Corollary 1 of [Reference Barros and Evangelista16] that
$ \textrm{div}\,{T_k}=0$
and because
\begin{equation*} \overset {\circ }{T_k} = T_k- \frac {\textrm {tr}\,{T_k}}{n}g = T_k- \frac {n-k}{n}\sigma _{k}g, \end{equation*}
Equation (9) can rewritten in the simpler form
\begin{equation*} (n-k)\int _{M}\langle X,\nabla \sigma _{k}\rangle dv_{g}=0, \end{equation*}
which proves the proposition in case
$k\neq n$
. As for the remaining case, it follows from [Reference Han20] that
\begin{equation*} n\langle X,\nabla \sigma _{n}\rangle = \nabla _{a} \left [ T_{b}^{a}\nabla ^{b}(\textrm {div}\,X)+2n\sigma _{n}X^a \right ], \end{equation*}
where
$T_{b}^{a}$
are the components of
$T_{n-1}(g^{-1}A_g)$
. Therefore, if we go there and write
\begin{equation*} Y^{a}=T_{b}^{a}\nabla ^{b}(\textrm {div}\,X)+2n\sigma _{n}X^a, \end{equation*}
we get that
\begin{equation*} n\int _{M}\langle X,\nabla \sigma _{k}\rangle dv_{g}= \int _{M}\nabla _{a}Y^{a}dv_{g}=0, \end{equation*}
which proves the proposition also for
$k=n$
.
Our next lemma states some structural equations for almost quotient gradient Yamabe solitons.
Lemma 3.
Let
$(M^{n},g,\nabla f,\lambda )$
be an almost gradient quotient Yamabe soliton. Then, we have that:
-
a)
$\Delta f= n\left (\log \dfrac{\sigma _{k}}{\sigma _{l}}- \lambda \right )$
; -
b)
$(n-1)\nabla \left (\log \dfrac{\sigma _{k}}{\sigma _{l}}- \lambda \right )+\textrm{Ric}(\nabla f)=0$
; -
c)
$(n-1)\Delta \left (\log \dfrac{\sigma _{k}}{\sigma _{l}}- \lambda \right )+ \dfrac{1}{2}\langle \nabla R,\nabla f\rangle + \left (\log \dfrac{\sigma _{k}}{\sigma _{l}}- \lambda \right )R=0$
.
Proof.
-
a) The first assertion is obtained by tracing (5);
-
b) Next, we differentiate (5) to get
from what we see that
\begin{equation*} \nabla _{j}\nabla _{r}\nabla _{i}f= \nabla _{j}\left (\log \frac {\sigma _{k}}{\sigma _{l}}- \lambda \right )g_{ri}, \end{equation*}
with the help of the Ricci identity that can be found in ([Reference Aubin21], pg. 4). Now, we only need to contract this equation on the indices
\begin{equation*} \nabla _{i}\nabla _{j}\nabla _{r}f+ \sum _{s}R_{rijs}\nabla _{s}f= \nabla _{j}\left (\log \frac {\sigma _{k}}{\sigma _{l}}- \lambda \right )g_{ri}, \end{equation*}
$j,r$
in order to getthen yielding
\begin{equation*} \nabla _{i}\Delta f+ \sum _{s}Ric_{is}\nabla _{s}f= \nabla _{i}\left (\log \frac {\sigma _{k}}{\sigma _{l}}- \lambda \right ), \end{equation*}
(10)by
\begin{equation} (n-1)\nabla _{i}\left (\log \frac{\sigma _{k}}{\sigma _{l}}- \lambda \right )+\sum _{s}Ric_{is}\nabla _{s}f=0, \end{equation}
$ \textrm{a)}$
, which proves the second assertion;
-
c) Now, we deal with the third one. We apply the divergence operator on both sides of (10) and use the twice contracted second Bianchi’s identity to obtain
which is equivalent to
\begin{equation*} (n-1)\Delta \left ( \log \frac {\sigma _{k}}{\sigma _{l}}-\lambda \right )+ \frac {1}{2}\langle \nabla R,\nabla f\rangle + \sum _{sl}Ric_{sl}\nabla _{s}\nabla _{l}f=0, \end{equation*}
by using the fundamental equation (5), one concludes the asserted result.
\begin{equation*} (n-1)\Delta \left ( \log \frac {\sigma _{k}}{\sigma _{l}}-\lambda \right )+ \frac {1}{2}\langle \nabla R,\nabla f\rangle + \left ( \log \frac {\sigma _{k}}{\sigma _{l}}-\lambda \right )R =0, \end{equation*}
3. Proofs of the main results
This section contains proofs for the main results in this paper.
Proof of Theorem 1.
-
a) Integrating by parts, one sees that
if either
\begin{equation*} \int _{M}\textrm {Ric}_{jk}\nabla _{i}C_{ijk}dv_{g}= -\int _{M}\nabla _{i}\textrm {Ric}_{jk}C_{ijk}dv_{g}=0, \end{equation*}
$\nabla \textrm{Ric}_g=0$
or
$\textrm{div}\,{C_g}=0$
and because(11)we conclude that
\begin{equation} \begin{split} \int _{M}&\nabla _{i}\textrm{Ric}_{jk}C_{ijk}dv_{g} = \\[5pt] &=\int _{M} \left [ C_{ijk}+ \frac{1}{2(n-1)} \left ( g_{jk}\nabla _{i}R_g-g_{ij}\nabla _{j}R_g \right ) \right ]C_{ijk}dv_{g} \\[5pt] &= \int _{M}|C_g|^{2}dv_{g}+ \frac{1}{2(n-1)} \int _{M} \left ( C_{ijk}g_{jk}\nabla _{i}R_g- C_{ijk}g_{ij}\nabla _{j}R_g \right )dv_{g} \\[5pt] &=\int _{M}|C_g|^{2}dv_{g}, \end{split} \end{equation}
$C_g=0$
. Equation (5) implies that
$ X$
is a conformal field and so we can apply Proposition 1 to conclude thatTherefore, we have
\begin{equation*} \int _{M^{n}} \sigma _{k}\left (\log \frac {\sigma _{k}}{\sigma _{l}}-\lambda \right ) dv_{g} = -\dfrac {1}{n} \int _{M^{n}} \langle \nabla \sigma _{k},X\rangle dv_{g} =0. \end{equation*}
(12)by our hypothesis on the nullity of the integral at the right hand of (12). Since
\begin{equation} \begin{split} \int _{M^{n}} \frac{\sigma _{l}}{n} &\left ( \frac{\sigma _{k}}{\sigma _{l}}-e^\lambda \right ) \left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right ) dv_{g} = \\[5pt] &= -\int _{M^{n}} \frac{e^{\lambda }\sigma _{l}}{n} \left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right ) dv_{g} \\[5pt] &= \int _{M^{n}} e^{\lambda }\sigma _{l}\langle \nabla \lambda, X\rangle dv_{g}+ \int _{M^{n}} e^{\lambda }\langle \nabla \sigma _{l}, X\rangle dv_{g}=0, \end{split} \end{equation}
$\sigma _{l}\neq 0$
does not change sign on
$ M^{n}$
, we then admit that
$\sigma _{k}/\sigma _{l}=e^\lambda$
, which proves our assertion in case of parallel Ricci curvature or divergence free Cotton tensor. On the other hand, if
$ X=\nabla f$
, we argue by contradiction to show that
$ f$
is a constant function. Should
$f$
not be constant on
$ M^{n}$
, the manifold
$(M^{n},g)$
could not lie in any conformal class other than that of the Euclidean sphere
$(\mathbb{S}^n,g_{\mathbb{S}^n})$
, by Theorem 1.1 of [Reference Catino, Mantegazza and Mazzieri12]. So, just as it happens with any locally conformally flat manifold, the Cotton tensor of
$(M^{n},g)$
would then vanish identically and by what has been said above
$(M^{n},g,\nabla{f},\lambda )$
ought to be trivial. This contradiction shows that
$ f$
is indeed a constant function, now concluding a);
-
b) Because the fields
$ \nabla h,Y$
in the Hodge-de Rham decomposition
$ X=\nabla h+Y$
of
$ X$
are orthogonal to one another in
$ L^2(M)$
, we get thatthe inequality is part of the hypothesis. Then,
\begin{equation*} \int _{M^{n}}|\nabla h|^{2}dv_{g}= \int _{M^{n}}\langle \nabla h,\nabla h+Y\rangle dv_{g}= \int _{M^{n}}\langle \nabla h, X\rangle dv_{g}\leqslant 0, \end{equation*}
$ \nabla h=0$
and
$ X=Y$
. Since
$Y$
is divergence free, we conclude thatand, as such, the soliton is trivial. So, the proof of the theorem is complete.
\begin{equation*} n\left (\log \frac {\sigma _{k}}{\sigma _{l}}-\lambda \right )= \textrm {div}\,X=0, \end{equation*}
Proof of Theorem 2. As we already know, the fundamental equation
\begin{equation*} \dfrac {1}{2}\mathcal {L}_Xg= \left (\log {\dfrac {\sigma _k}{\sigma _l}}-\lambda \right )g, \end{equation*}
leads to
\begin{equation} \textrm{div}\,X=n\left (\log \dfrac{\sigma _{k}}{\sigma _{l}}-\lambda \right ), \end{equation}
and because we suppose that
$ \mathcal{L}_Xg\geqslant 0$
we must then admit that
$ \log{\frac{\sigma _k}{\sigma _l}}-\lambda \geqslant 0$
. So, if we now take a cutoff function
$\psi \;:\;M\rightarrow \mathbb{R}$
satisfying
\begin{equation*} 0\leqslant \psi \leqslant 1\mbox { on } M,\quad \psi \equiv 1\hspace {0,2cm}\text {in}\hspace {0,2cm} B_{r}(x_{0}),\quad \textrm {supp}{(\psi )}\subset B_{2r}(x_{0})\quad \text {and}\quad |\nabla \psi |\leqslant \frac {K}{r}, \end{equation*}
where
$ K\gt 0$
is a real constant, we are in place to conclude that
\begin{equation*} \begin{split} n\int _{B_{r}(x_{0})} \left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right ) dv_{g} &= \int _{B_r(x_0)} n\psi \left (\log{\frac{\sigma _k}{\sigma _l}}-\lambda \right ) dv_g \\[5pt] &\leqslant \int _{B_{2r}(x_0)} n\psi \left (\log{\frac{\sigma _k}{\sigma _l}}-\lambda \right ) dv_g \\[5pt] &= \int _{B_{2r}(x_{0})}\psi \,\textrm{div}\,Xdv_{g} \\[5pt] &= -\int _{B_{2r}(x_{0})}g(\nabla \psi,X)dv_{g} \\[5pt] & \leqslant \int _{B_{2r}(x_{0})\setminus B_{r}(x_{0})}|-\nabla \psi ||X|dv_{g} \\[5pt] & \leqslant K\int _{B_{2r}(x_{0})\setminus B_{r}(x_{0})}\frac{|X|}{r}dv_{g}, \\[5pt] & \leqslant 2K\int _{M\setminus B_r(x_0)}\dfrac{|X|}{d(x,x_0)}dv_g, \end{split} \end{equation*}
from what it follows that
\begin{align*} 0\leqslant \int _M{\left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right )} dv_{g} & = \lim _{r\to \infty } \int _{B_{r}(x_{0})}{\left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right )} dv_{g} \\[5pt] & \leqslant \frac{2K}{n}\lim _{r\to \infty } \int _{M\setminus B_r(x_0)}\frac{|X|}{d(x,x_0)}dv_g=0. \end{align*}
Henceforth, we have that
$ \mathcal{L}_Xg=\log{\frac{\sigma _k}{\sigma _l}}-\lambda =0$
which proves the Theorem.
Proof of Theorem 3. It follows from Lemma 3
$\textrm{(c)}$
that if the scalar curvature of
$ (M^{n},g,\nabla f,\lambda )$
is a constant function on
$ M^{n}$
, then
\begin{equation} \Delta \left ( \log \dfrac{\sigma _{k}}{\sigma _{l}}-\lambda \right )+ \frac{R}{n-1} \left ( \log \dfrac{\sigma _{k}}{\sigma _{l}}-\lambda \right )=0, \end{equation}
and, by the min-max principle, we must have
$ R\gt 0$
. By using that
\begin{equation} \Delta f=n \left ( \log \dfrac{\sigma _{k}}{\sigma _{l}}-\lambda \right ), \end{equation}
we then get
\begin{equation*} \Delta \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda +\frac {R}{n(n-1)}f \right ) =0, \end{equation*}
and since
$ (M^{n},g)$
is a compact Riemannian manifold, one see that
\begin{equation*} \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda +\frac {R}{n(n-1)}f=c \quad \mbox {on}\quad M^{n}, \end{equation*}
for a certain
$ c\in \mathbb{R}$
, by the maximum principle. Hence,
\begin{equation*} \nabla \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right ) + \frac {R}{n(n-1)}\nabla f =0, \end{equation*}
and so
\begin{equation*} \nabla _{X}\nabla \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right ) = -\frac {R}{n(n-1)}\nabla _{X}\nabla f = -\frac {R}{n(n-1)} \left ( \log \frac {\sigma _{k}}{\sigma _{l}}-\lambda \right )X. \end{equation*}
We can now apply Obata’s theorem ([Reference Obata22], Theorem 1) to conclude that
$(M^{n},g)$
is isometric with an Euclidean sphere of radius
$ \sqrt{r}$
,
$r=R/n(n-1)$
. To prove our last claim, we notice that we can assume that
$R=n(n-1)$
possibly at the cost of rescaling the metric
$ g$
. From equations (14) and (15), it is seen that
$\frac{\Delta f}{n}$
is an eigenfunction of the Laplacian on
$(\mathbb{S}^n,g)$
and so there must exist a
$v\in \mathbb{S}^n$
such that
$\frac{1}{n}\Delta f=h_{v}=-\frac{1}{n}\Delta h_{v}$
. Hence,
$\Delta (f+ h_{v})=0$
but then
$f=h_{v}+c$
for some real
$c$
.
Proof of Theorem 4. By Theorem 1.1 of [Reference Catino, Mantegazza and Mazzieri12] the only nontrivial compact almost gradient quotient Yamabe solitons reside in the conformal class of the Euclidean sphere and because of that we can assume that
\begin{equation*} M^{n}=\mathbb {S}^n\quad \mbox {and}\quad \varphi ^{-2}g=g_{\mathbb {S}^n}, \end{equation*}
where
$ \varphi \in C^\infty (\mathbb{S}^n)$
is strictly positive. Then, the Ricci tensors of
$ g$
and
$ g_{\mathbb{S}^n}$
are correlated by the equation [Reference Besse23]
\begin{equation*} \textrm {Ric}_{\mathbb {S}^n} = \textrm {Ric}_{g} + \frac {1}{\varphi ^2} \big {\{} (n-2)\varphi \nabla ^{2}\varphi + [\varphi \Delta \varphi -(n-1)|\nabla \varphi |^2]g \big {\}}, \end{equation*}
which we algebraically manipulate in order to get the similar equation
\begin{equation} A_{g_{\mathbb{S}^n}} = A_{g} + \frac{\nabla ^2\varphi }{\varphi } - \frac{1}{2}\frac{|\nabla \varphi |^2}{\varphi ^2}g, \end{equation}
for the Schouten tensors. But then we have
\begin{equation*} \frac {1}{2} \left ( \varphi ^2+\frac {|\nabla \varphi |^2}{\varphi ^2} \right )g= A_{g}+\frac {\nabla ^2\varphi }{\varphi }, \end{equation*}
from what it follows that
\begin{equation} \nabla ^2\varphi = \varphi \left [ -A_g+ \frac{1}{n}\left (\sigma _1(g)+\frac{\Delta \varphi }{\varphi }\right )g \right ]. \end{equation}
Notice that Lemma 1 applied to
$ T=T_k(g^{-1}A_g)$
and
$ X=\nabla \varphi$
gives
\begin{equation} \int _{M}\langle T_{k}(g^{-1}A_{g}),\nabla ^{2}\varphi \rangle dv_{g}=0, \end{equation}
because
$ \textrm{tr}\,{T_k(g^{-1}A_g)}=(n-k)\sigma _k(g)$
is constant on
$ \mathbb{S}^n$
by hypothesis and
$ \textrm{div}\,{T_k(g^{-1}A_g)}=0$
. A combination of (18) and (17) above leads to
\begin{align*} 0 & = \int _{M} \langle T_{k}(g^{-1}A_{g}), -\varphi A_{g}+ \frac{\sigma _1(g)\varphi +\Delta \varphi }{n}g \rangle dv_{g}=0 \\[5pt] & = \int _M \left [ -\varphi \langle T_k(g^{-1}A_g),A_g\rangle + \frac{\sigma _1(g)\varphi +\Delta \varphi }{n} \langle T_k(g^{-1}A_g),g\rangle \right ] dv_g \\[5pt] & = \int _M \varphi \left [ \left (\frac{n-k}{n}\right )\sigma _1(g)\sigma _k(g)- (k+1)\sigma _{k+1}(g) \right ] dv_g \end{align*}
where we have used the identity
$\textrm{tr}\,{T_k(g^{-1}A_g\circ A_g)}=(k+1)\sigma _{k+1}(g)$
[Reference Reilly24]. By Lemma 23 of [Reference Viaclovsky25], we conclude that
\begin{equation*} \left (\frac {n-k}{n}\right )\sigma _{1}\sigma _{k}=(k+1)\sigma _{k+1}, \end{equation*}
implying that
$(\mathbb{S}^n,g)$
is an Einstein manifold. In particular, the scalar curvature of
$ g$
is constant on
$ \mathbb{S}^n$
and by Theorem 3 there is even an isometry between
$(\mathbb{S}^n,g)$
and
$ (\mathbb{S}^n,g_{\mathbb{S}^n})$
which proves the Theorem.
Proof of Theorem 5. Lemma 2 applied to the data
$T=\overset{\circ }{\textrm{Ric}_g}$
,
$X=\nabla f$
, and
$\varphi =f$
gives
\begin{equation} \textrm{div}\,{\overset{\circ }{\textrm{Ric}_g}(f\nabla f)} = f(\textrm{div}\,{\overset{\circ }{\textrm{Ric}_g})(\nabla f)} + f \langle \nabla ^2 f, \overset{\circ }{\textrm{Ric}_g}\rangle + \overset{\circ }{\textrm{Ric}_g}(\nabla f,\nabla f), \end{equation}
and it then follows from the second contracted Bianchi identity that
\begin{equation} (\textrm{div}\,{\overset{\circ }{\textrm{Ric}_g})(\nabla f)} = \frac{n-2}{2n}\langle \nabla f,\nabla R\rangle. \end{equation}
A straightforward computation shows that
\begin{equation} f \langle \nabla ^2 f,\overset{\circ }{\textrm{Ric}_g}\rangle = f\left ( \log \frac{\sigma _{k}}{\sigma _{l}}-\lambda \right ) \langle g,\overset{\circ }{\textrm{Ric}_g}\rangle =0, \end{equation}
and equations (19), (20), and (21) together give
\begin{equation} \frac{1}{2}\textrm{div}\,{\overset{\circ }{\textrm{Ric}_g}(\nabla f^2)} = \frac{n-2}{4n}\langle \nabla R_g,\nabla f^2\rangle + \overset{\circ }{\textrm{Ric}_g}(\nabla f,\nabla f). \end{equation}
Proposition 1 of [Reference Caminha, Souza and Camargo26] tell us that
$\textrm{div}\,{\overset{\circ }{\textrm{Ric}_g}(\nabla f^2)}=0$
because
$|\overset{\circ }{\textrm{Ric}_g}(\nabla f^2)|\in L^{1}(M)$
. Consequently,
\begin{equation*} \langle \nabla R_g,\nabla f^2\rangle =0\quad \mbox {and}\quad \overset {\circ }{\textrm {Ric}_g}(\nabla f,\nabla f)=0. \end{equation*}
As
$(M^{n},g,\nabla f,\lambda )$
is a nontrivial almost quotient gradient Yamabe soliton, any regular level set
$\Sigma$
of the potential function
$f$
admits a maximal open neighborhood
$U\subset M$
in which
$ g$
can be written like
\begin{equation} g=dr\otimes dr+(f'(r))^2g^{\Sigma }, \end{equation}
where
$g^{\Sigma }$
is the restriction of
$g$
to
$\Sigma$
(see [Reference Catino, Mantegazza and Mazzieri12]). Since
$M$
is noncompact,
$f$
has at most one critical point. As the Ricci tensor of a warped product metric,
$ \textrm{Ric}_g$
now admits the following decomposition
\begin{equation} \textrm{Ric}_g= \textrm{Ric}^{\Sigma } - (n-1)\frac{f^{'''}}{f^{'}}dr\otimes dr - [(n-2)(f'')^2+f'f''']g^{\Sigma }, \end{equation}
thus giving
$\frac{R_g}{n}=-(n-1)\frac{f'''}{f'}$
because
$\overset{\circ }{\textrm{Ric}_g}(\nabla f,\nabla f)=0$
. Equation (24) can also be manipulated to show that
\begin{equation*} \textrm {Ric}_g(\nabla f)=\frac {R_g}{n}\nabla f, \end{equation*}
of which
\begin{equation} \nabla \left ( \log \dfrac{\sigma _{k}}{\sigma _{l}}-\lambda \right ) + \frac{R_g}{n(n-1)}\nabla f =0, \end{equation}
is a consequence by Lemma 3
$ \textrm{b)}$
. The divergence of equation (25) is
\begin{equation} \Delta \left ( \log \dfrac{\sigma _{k}}{\sigma _{l}} -\lambda \right ) + \frac{1}{n(n-1)} \langle \nabla R_g,\nabla f\rangle + \frac{R_g}{n-1} \left ( \log \dfrac{\sigma _{k}}{\sigma _{l}} - \lambda \right ) =0. \end{equation}
which we compare with the expression in Lemma 3
$ \textrm{c)}$
to see that
$\langle \nabla R_g,\nabla f\rangle =0$
. Since
$R_g$
only depends on
$r$
we get that
\begin{equation*} f'R'_{\!\!g} = f'\langle \nabla R_g,\partial r\rangle = \langle \nabla R_g,\nabla f\rangle =0, \end{equation*}
implying that the scalar curvature
$R_g=R$
is constant. We claim that
$R\leqslant 0$
. As a matter of fact, if we had
$R\gt 0$
, then from (25) we would then have that
$\log \frac{\sigma _{k}}{\sigma _{l}}-\lambda$
is not constant on
$ M^{n}$
and satisfies
\begin{equation*} \nabla _{X}\nabla \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right ) = -\frac {R}{n(n-1)}\nabla _{X}\nabla f = -\frac {R}{n(n-1)} \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right )X. \end{equation*}
From Obata’s theorem [Reference Obata22], the manifold
$ M^{n}$
would then be compact, which is absurd. Therefore,
$ R\leqslant 0$
.
-
a) It follows from (25) that
$ \log{\frac{\sigma _k}{\sigma _l}}-\lambda =c$
for some
$ c\in \mathbb{R}$
because we now have
$ R=0$
. By Theorem 2 of [Reference Tashiro27]
$ (M^{n},g)$
must be isometric with flat Euclidean space
$ \mathbb{R}^n$
in case
$ c\neq 0$
. Since this would leave us with
$ \sigma _1(g)=\sigma _2(g)=\cdots =\sigma _n(g)=0$
, the function
$ \log{\frac{\sigma _k}{\sigma _l}}$
could not be defined. Then,
$ c=0$
and so
$ \nabla ^2 f=0$
by the fundamental equation (5). Theorem B of Kanai [Reference Kanai28] then implies that
$ (M^{n},g)$
is isometric with a Riemannian product manifold
$ \mathbb{R}\times \mathbb{F}^{n-1}$
; -
b) If
$f$
has no critical points and
$ R\lt 0$
, then once more by (25) we get that
$\log \frac{\sigma _{k}}{\sigma _{l}}-\lambda$
is not constant on
$ M^{n}$
and satisfieson
\begin{equation*} \nabla _{X}\nabla \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right ) = -\frac {R}{n(n-1)}\nabla _{X}\nabla f = -\frac {R}{n(n-1)} \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right )X, \end{equation*}
$ M^{n}$
for every
$ X\in \mathfrak{X}(M)$
. In virtue of Theorem D in [Reference Kanai28], the manifold
$(M^{n},g)$
is isometric with a warped product manifold
$(\mathbb{R}\times \mathbb{F}^{n-1},dr^2+\xi (r)^2g_{\mathbb{F}})$
in which the warping function
$\xi$
solves the second-order linear ODE with constant coefficients
$\xi ''+\frac{R}{n(n-1)}\xi =0$
;
-
c) In our last call to equation (25), we observe that if
$f$
has exactly one critical point and
$ R\lt 0$
then
$\log \frac{\sigma _{k}}{\sigma _{l}}-\lambda$
is not constant on
$ M^{n}$
and must satisfyon
\begin{equation*} \nabla _{X}\nabla \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right ) = -\frac {R}{n(n-1)}\nabla _{X}\nabla f = -\frac {R}{n(n-1)} \left ( \log \dfrac {\sigma _{k}}{\sigma _{l}}-\lambda \right )X, \end{equation*}
$ M^{n}$
for every
$ X\in \mathfrak{X}(M)$
. We then apply Theorem C in [Reference Kanai28] to conclude that
$(M^{n},g)$
is isometric with a hyperbolic space.
Competing interests
The authors declare that there is no competing interests.
Data availability statement
Data supporting this manuscript is provided in the bibliography.
