1 Introduction
Let
$\mathbb {R}^{n+2}_1$
be the
$(n+2)$
- dimensional Minkowski space, that is, the real vector space
$\mathbb {R}^{n+2}$
endowed with the Lorentz metric
$$\begin{align*}\left<v, w\right>=-v^0w^0+\sum\limits_{i=1}^{n+1}v^iw^i.\end{align*}$$
The one sheeted hyperboloid
$$\begin{align*}\mathbb S^{n+1}_1=\{y\in\mathbb{R}^{n+2}_1: \left<y, y\right>=1, y^0>0\},\end{align*}$$
consisting of all unit spacelike vectors and equipped with the induced metric is called de Sitter space. It is a geodesically complete simply connected Lorentzian manifold with constant curvature one. We say a hypersurface
$M\subset \mathbb S^{n+1}_1$
is spacelike if its induced metric is Riemannian.
Let
$\mathbb S^n$
be the standard round sphere. Then de Sitter space may be parametrized by
$Y: \mathbb S^n\times \mathbb {R}\rightarrow \mathbb S^{n+1}_1$
as follows:
$$\begin{align*}Y(r, \xi)=\sinh(r)E_1+\cosh(r)\xi.\end{align*}$$
In this coordinate system, the induced metric is
$$\begin{align*}\bar g=-dr^2+\phi^2(r)\sigma\end{align*}$$
where
$\sigma $
is the standard metric on
$\mathbb S^n$
and
$\phi =\cosh $
. For a hypersurface
$M\subset \mathbb S^{n+1}_1,$
we define
$$\begin{align*}u=-\left<\phi\frac{\partial}{\partial r}, \nu\right>\end{align*}$$
to be the support function, where
$\nu $
is the future directed unit normal to M and
$\left <\cdot , \cdot \right>$
is the inner product with respect to
$\bar g$
.
In this paper, we prove an optimal isoperimetric inequality for spacelike, compact, star-shaped, and
$2$
-convex hypersurfaces in de Sitter space. Before stating our main results, we need the following definition.
Definition 1.1 A
$C^2$
regular hypersurface
$M\subset \mathbb S^{n+1}_1$
is strictly k-convex, if the principal curvature vector of M at
$X\in M$
satisfies
$\kappa [X]\in \Gamma _k$
for all
$X\in M,$
where
$\Gamma _k$
is the Garding’s cone
$$\begin{align*}\Gamma_k=\{\lambda\in\mathbb R^n: \sigma_j(\lambda)>0, 1\leq j\leq k\},\end{align*}$$
and
$\sigma _j$
is the j-th elementary symmetric polynomial. If the principal curvature vector of M at
$X\in M$
satisfies
$\kappa [X]\in \bar \Gamma _k$
for all
$X\in M,$
then we say M is k-convex.
Theorem 1.2 Let
$M_0\subset \mathbb S^{n+1}_1$
be a spacelike, compact, star-shaped, and strictly
$2$
-convex hypersurface. Then the solution to the following flow equation
$$ \begin{align} \left\{\begin{aligned} X_t&=\left(u-b_{n, 2}\phi'\sigma_2^{-1/2}\right)\nu,\\ X_0&=M_0 \end{aligned} \right. \end{align} $$
exists for all time, where
$b_{n, 2}=(\sigma _2(I))^{1/2}=\left [\frac {n(n-1)}{2}\right ]^{1/2}.$
Moreover, the flow hypersurfaces
$M_t$
converge smoothly to a radial coordinate slice as
$t\rightarrow \infty .$
As a consequence we obtain
Corollary 1.3 Let
$M\subset \mathbb S^{n+1}_1$
be a spacelike, compact, star-shaped, and
$2$
-convex hypersurface. Then there holds
$$ \begin{align} \int_M\sigma_2d\mu_g-(n-1)|M|\leq\xi_{2, 0}(|M|) \end{align} $$
with equality is attained if and only if M is a radial coordinate slice. Here,
$\xi _{2, 0}$
is the associated monotonically increasing function for radial coordinate slices and
$|M|$
denotes the surface area of
$M.$
1.1 Background and motivations
The classical Minkowski inequality [Reference Minkowski16] states that: For a convex hypesurface
$M\subset \mathbb {R}^{n+1}$
we have
$$ \begin{align} \frac{1}{|\mathbb{S}^{n}|}\int_{M}\frac{H}{n}d\mu_g\geq\left(\frac{|M|}{|\mathbb{S}^{n}|}\right)^{\frac{n-1}{n}}, \end{align} $$
with equality holds if and only if M is a sphere. Here, H is the mean curvature of
$M.$
A natural question, raised by several authors (see, for example, [Reference Chang and Wang6, Reference Huisken11, Reference Trudinger21]), is whether the Minkowski inequality stays true for larger classes of domains than just for convex ones. By studying weak solutions of the inverse mean curvature flow in
$\mathbb R^n,$
Huisken–Ilmanen [Reference Huisken and Ilmanen12, Reference Huisken and Ilmanen13] showed that the assumption that M is convex can be replaced by the assumption that M is outward-minimizing. In 2009, by studying a normalized inverse curvature flow, Guan and Li [Reference Guan and Li8] proved (1.3) for the case when M is star-shaped and mean convex. Moreover, they also proved Alexandrov–Fenchel inequalities, which is a general form of Minkowski inequality, for star-shaped and k-convex hypersurfaces in Euclidean space.
There are analogous of Minkowski and Alexandrov–Fenchel inequalities in space form. In hyperbolic space, Wang and Xia [Reference Wang and Xia22] proved the Alexandrov–Fenchel type inequalities for horospherically convex hypersurfaces. Since then, many efforts have been carried out to weaken the condition on the convexity. In particular, for star-shaped and mean convex hypersurfaces, a Minkowski type inequality has been proved in [Reference Brendle, Guan and Li3, Reference Scheuer and Xia20]; for star-shaped and
$2$
-convex hypersurfaces, an Alexandrov–Fenchel type inequality that involving integral of the scalar curvature has been proved in [Reference Brendle, Guan and Li3, Reference Li, Wei and Xiong15]. Due to technical reasons there are much less such results in sphere, even with the convexity assumption the Alexandrov–Fenchel type inequalities are still open. For difficulties in proving the Alexandrov–Fenchel type inequalities in sphere, one may refer to the expository paper [Reference Chen, Guan, Li and Scheuer5]. Some variants of Alexandrov–Fenchel type inequalities for convex hypersurfaces in sphere can be found in [Reference Girão and Pinheiro10, Reference Wei and Xiong23]. In de Sitter space, the Alexandrov–Fenchel type inequalities for convex hypersurfaces were deduced through the well-known duality for strictly convex hypersurfaces of hyperbolic/de Sitter space in [Reference Andrews, Hu and Li1]; while a Minkowski type inequality for spacelike, compact, star-shaped, and mean-convex hypersurfaces was derived in [Reference Scheuer18].
1.2 Outline
In Section 2, we give basic notations and establish fundamental equations for geometric quantities in de Sitter space that will be used in later sections. In Section 3, we introduce the flow equation and prove the monotonicity properties for the quermassintegrals along the flow. In Section 4, we will establish a priori estimates for the flow equation (1.1) and prove that the flow (1.1) exists for all time. In Section 5, we show the flow converges to a radial coordinate slice. This completes the proof of Theorem 1.2 and Corollary 1.3.
2 Preliminary
In this section, we will collect some formulas and lemmas for k-th symmetric functions as well as hypersurfaces in
$\mathbb S^{n+1}_1.$
2.1 Elementary symmetric functions
For any
$k=1, \ldots , n,$
and
$\lambda =(\lambda _1, \ldots , \lambda _n)$
the k-th elementary symmetric function is defined by
$$\begin{align*}\sigma_k(\lambda)=\sum\limits_{1\leq i_1<i_2<\cdots<i_k\leq n}\lambda_{i_1}\lambda_{i_2}\ldots\lambda_{i_k},\end{align*}$$
and we also define
$\sigma _0=1.$
In this paper, we will denote
$\sigma _k(\lambda | i)$
the symmetric function with
$\lambda _i=0.$
The following properties are well known.
Lemma 2.1 Let
$\lambda =(\lambda _1, \ldots , \lambda _n)\in \mathbb R^n$
and
$k=1, \ldots , n,$
then
$$\begin{align*}\sigma_k(\lambda)=\sigma_k(\lambda|i)+\lambda_i\sigma_{k-1}(\lambda|i),\,\,\forall 1\leq i\leq n,\end{align*}$$
$$\begin{align*}\sum\limits_i\lambda_i\sigma_{k-1}(\lambda| i)=k\sigma_k(\lambda),\end{align*}$$
and
$$\begin{align*}\sum\limits_i\sigma_k(\lambda| i)=(n-k)\sigma_k(\lambda).\end{align*}$$
Lemma 2.2 Let
$\lambda \in \Gamma _k$
with
$\lambda _1\geq \ldots \geq \lambda _k\geq \ldots \geq \lambda _n,$
then we have
$$\begin{align*}\sigma_{k-1}(\lambda| n)\geq\sigma_{k-1}(\lambda| n-1)\geq\ldots\geq\sigma_{k-1}(\lambda| 1)>0,\end{align*}$$
$$\begin{align*}\lambda_1\geq \ldots \geq\lambda_k>0, \,\,\sigma_k(\lambda)\leq C_n^k\lambda_1 \ldots \lambda_k,\end{align*}$$
$$\begin{align*}\sum\limits_i\sigma_{k-1}(\lambda|i)\lambda_i^2=\sigma_k\sigma_1-(k+1)\sigma_{k+1},\end{align*}$$
where
$C_n^k=\frac {n!}{k!(n-k)!}.$
The generalized Newton–Maclaurin inequality is as follows, which will be used all the time (see [Reference Spruck19]).
Proposition 2.3 For
$\lambda \in \Gamma _k, k>l\geq 0, r>s\geq 0, k\geq r,$
and
$l\geq s,$
we have
$$\begin{align*}\left[\frac{\sigma_k(\lambda)/C_n^k}{\sigma_l(\lambda)/C_n^l}\right]^{\frac{1}{k-l}}\leq\left[\frac{\sigma_r(\lambda)/C_n^r}{\sigma_s(\lambda)/C_n^s}\right]^{\frac{1}{r-s}}.\end{align*}$$
Moreover, the equality holds if and only if
$\lambda =c(1, \cdots , 1)$
for some
$c> 0.$
Let A be a symmetric matrix and
$\lambda (A)=(\lambda _1, \ldots , \lambda _n)$
be the eigenvalue vector of
$A.$
Let F be the function defined by
$$\begin{align*}F(A)=f(\lambda(A))\end{align*}$$
and denote
$$\begin{align*}F^{ij}(A)=\frac{\partial F}{\partial a_{ij}},\,\,F^{pq, rs}=\frac{\partial^2 F}{\partial a_{pq}\partial a_{rs}}.\end{align*}$$
When A is diagonal, we have
$$\begin{align*}F^{ij}(A)=f^i(\lambda(A))\delta_{ij},\,\,\text{for } f^i=\frac{\partial f}{\partial\lambda_i}.\end{align*}$$
Furthermore, we also have
$$ \begin{align} F^{ij}(A)a_{ij}=\sum\limits_if^i(\lambda(A))\lambda_i, \end{align} $$
$$ \begin{align} F^{ij}(A)a_{ik}a_{kj}=\sum\limits_if^i(\lambda(A))\lambda_i^2. \end{align} $$
In particular, when
$F(A)=\sigma _k(\lambda (A))$
and suppose A is diagonalized at
$p_0,$
then at
$p_0,$
we have
$$ \begin{align}F^{ij}(A)=\sigma_k^{ij}(\lambda(A))=\sigma_{k-1}(\lambda| i)\delta_{ij},\end{align} $$
$$ \begin{align}F^{pq, rs}(A) &=\sigma_k^{pq, rs}(\lambda(A)) \nonumber\\ &=\left\{\begin{aligned} &\frac{\partial^2\sigma_k}{\partial\lambda_p\partial\lambda_r}(\lambda)=\sigma_{k-2}(\lambda|pr), \,\, &p=q, r=s, p\neq r,\\ &-\frac{\partial^2\sigma_k}{\partial\lambda_p\partial\lambda_q}(\lambda)=-\sigma_{k-2}(\lambda|pq), \,\,&p=s, q=r, p\neq q,\\ &0,\,\,&\text{otherwise.} \end{aligned}\right.\end{align} $$
In order to prove the long time existence of the flow (1.1) (see Section 4), we need the following concavity inequality for Hessian operator, which is proved by Siyuan Lu.
Lemma 2.4 [Reference Lu17, Lemma 3.1]
Let
$\lambda =(\lambda _1, \ldots , \lambda _n)\in \Gamma _k$
with
$\lambda _1\geq \ldots \geq \lambda _n$
and let
$1\leq l<k.$
For any
$\epsilon , \delta , \delta _0\in (0, 1),$
there exists a constant
$\delta '>0$
depending only on
$\epsilon , \delta , \delta _0, n, k$
and l such that if
$\lambda _l\geq \delta \lambda _1$
and
$\lambda _{l+1}\leq \delta '\lambda _1,$
then we have
$$\begin{align*}-\sum\limits_{p\neq q}\frac{\sigma_k^{pp, qq}\xi_p\xi_q}{\sigma_k}+\frac{(\sum_i\sigma_k^{ii}\xi_i)^2}{\sigma_k^2}\geq(1-\epsilon)\frac{\xi_1^2}{\lambda_1^2}-\delta_0\sum\limits_{i>l}\frac{\sigma_k^{ii}\xi_i^2}{\lambda_1\sigma_k}, \end{align*}$$
where
$\xi =(\xi _1, \ldots , \xi _n)$
is an arbitrary vector in
$\mathbb R^n.$
Note that from the proof in [Reference Lu17], we can see that for fixed
$\delta , \delta _0\in (0, 1), \delta '=\delta '(\epsilon , \delta , \delta _0, n, k)=O(\epsilon )>0$
is a small constant.
2.2 Star-shaped graph in
$\mathbb S^{n+1}_1$
Let
$M=\{(\rho (\xi ), \xi ): \xi \in \mathbb S^n\},$
we will use
$\tilde \nabla $
to denote the standard covariant derivative for the metric
$\sigma $
on
$\mathbb S^n.$
Then the tangent space of the hypersurface at a point
$Y\in M$
is spanned by
$$\begin{align*}Y_j=\rho_j\partial_r+\partial_j,\,\,\text{where } \partial_j:=\tilde\nabla_{\xi_j},\end{align*}$$
and the induced metric on M is given by
$$\begin{align*}g_{ij}=\left<Y_i, Y_j\right>=-\rho_i\rho_j+\cosh^2(\rho)\sigma_{ij}.\end{align*}$$
M is spacelike if
$(g_{ij})$
is positive-definite. A unit normal vector
$\nu $
to M can be obtained by solving the equation
$\left <Y_i, \nu \right>=0$
for
$\forall 1\leq i\leq n.$
Thus we have
$$\begin{align*}\nu=\frac{(\cosh\rho, \tilde\nabla\rho/\cosh\rho)}{\sqrt{\cosh^2(\rho)-|\tilde\nabla\rho|^2}},\end{align*}$$
here
$|\tilde \nabla \rho |^2=\sigma ^{ij}\rho _i\rho _j$
and
$(\sigma ^{ij})$
is the inverse of
$(\sigma _{ij}).$
In the following, for our convenience we will denote
$w:=\sqrt {\cosh ^2(\rho )-|\tilde \nabla \rho |^2},$
then the support function
$$ \begin{align}u:=-\left<\cosh \rho\frac{\partial}{\partial\rho}, \nu\right>=\frac{\cosh^2(\rho)}{w}.\end{align} $$
Moreover, by some routine calculations (for details see [Reference Ballesteros-Chávez, Klingenberg and Lambert4]) we get
$$ \begin{align} g^{ij}=\frac{1}{\cosh^2(\rho)}\left(\sigma^{ij}+\frac{\rho^i\rho^j}{w^2}\right), \end{align} $$
and
$$ \begin{align} h_{ij}=\frac{\cosh\rho}{w}\left(\tilde\nabla_{ij}\rho-2\rho_i\rho_j\tanh\rho+\sinh\rho\cosh\rho\sigma_{ij}\right), \end{align} $$
where
$(g^{ij})$
is the inverse of
$(g_{ij}), \rho ^i=\sigma ^{il}\rho _l,$
and
$h_{ij}$
is the second fundamental form of
$M.$
2.3 Hypersurfaces in
$\mathbb S^{n+1}_1$
In this paper, we will define
$$ \begin{align}\Phi=-\int^r_0\cosh sds=-\sinh r,\end{align} $$
and
$$ \begin{align} V=\cosh r\frac{\partial}{\partial r}. \end{align} $$
We note that
$\Phi =-\phi '.$
Now, let
$M\in \mathbb S^{n+1}_1$
be a spacelike hypersurface with induced metric g. We will use
$\nabla $
to denote the covariant derivative with respect to
$g.$
Then the following fundamental equations are well known:
$$\begin{align*}\begin{aligned} &\nabla_{\tau_i}\tau_j=h_{ij}\nu\,\,&\text{Gauss formula},\\ &\nabla_{\tau_i}\nu=h_i^k\tau_k\,\,&\text{Weingarten equation},\\ &h_{ijk}=h_{ikj}\,\,&\text{Codazzi equation.} \end{aligned}\end{align*}$$
Following the proof of Lemma 2.2 in [Reference Guan and Li9] (see also [Reference Scheuer18, Equation (2.7)]) we have
Lemma 2.5 Let
$M\subset \mathbb S^{n+1}_1$
be a spacelike, compact, connected hypersurfaces with induced metric g. Let
$\Phi $
be defined as in (2.8). Then
$\Phi \mid _{M}$
satisfies,
$$ \begin{align} \nabla_{ij}\Phi=\phi'(\rho)g_{ij}-h_{ij}u, \end{align} $$
where
$\nabla $
is the covariant derivative with respect to
$g, h_{ij}$
is the second fundamental form of
$M,$
and
$u=-\left <V, \nu \right>$
is the support function of
$M.$
Next, following the proof of Lemma 2.6 in [Reference Guan and Li9], we derive the gradient and hessian of the support function u under the induced metric g on
$M.$
Lemma 2.6 The support function u satisfies
$$ \begin{align} \nabla_iu=-h_i^k\nabla_k\Phi, \end{align} $$
$$ \begin{align} \nabla_{ij}u=-g^{kl}\nabla_kh_{ij}\nabla_l\Phi-\phi'h_{ij}+uh^k_ih_{kj}, \end{align} $$
where
$h_i^k=g^{kl}h_{li}.$
3 Curvature flow and monotonicity formula
In this paper, we consider hypersurface flows related to the quermassintegrals. Similar to [Reference Chen, Guan, Li and Scheuer5], let
$M=\partial \Omega ,$
set
$$ \begin{align} \begin{aligned} \mathcal{A}_{-1}&=\text{Vol}(\Omega),\,\,\mathcal A_0=\int_Md\mu_g\\ \mathcal A_1&=\int_M\sigma_1d\mu_g-n\text{Vol}(\Omega)\\ \mathcal A_m&=\int_M\sigma_md\mu_g-\frac{n-m+1}{m-1}\mathcal A_{m-2}, \end{aligned} \end{align} $$
where
$2\leq m\leq n, g$
is the induced metric on
$M,$
and
$d\mu _g$
is the associated volume element. Let
$M_t$
be a smooth family of spacelike, compact, connected hypersurfaces in
$\mathbb S^{n+1}_1$
evolving along the flow
$$ \begin{align} X_t=S\nu. \end{align} $$
From [Reference Gerhardt7] we get
$$ \begin{align} \partial_tg_{ij}=2Sh_{ij,} \end{align} $$
and
$$ \begin{align} \partial_t\nu=\nabla S. \end{align} $$
Moreover, by [Reference Scheuer18, Lemma 3.1] we have
$$ \begin{align} \partial_th_i^j=\nabla^j\nabla_i S-Sh^k_ih^j_k+S\delta_i^j. \end{align} $$
By [Reference Scheuer18, Lemma 3.2] we also have
$$ \begin{align} \begin{aligned} \partial_t\mathcal A_{-1}&=\int_{M_t}Sd\mu_g,\\ \partial_t\mathcal A_{0}&=\int_{M_t}\sigma_1Sd\mu_g. \end{aligned} \end{align} $$
Combining (3.5), (3.6) with an induction argument, we derive for
$0\leq l\leq n-1$
$$ \begin{align} \partial_t\mathcal A_l=(l+1)\int_{M_t}S\sigma_{l+1}d\mu_g. \end{align} $$
3.1 A specific flow equation.
In order to obtain an isoperimetric type inequality, we will consider the following curvature flow
$$ \begin{align} X_t=\left(u-b_{n, k}\phi'\sigma_k^{-1/k}\right)\nu, \end{align} $$
where
$b_{n, k}=\left (C_n^k\right )^{1/k}$
and
$1\leq k\leq n.$
Then from now on, our normal velocity
$S=u-b_{n, k}\phi '\sigma _k^{-1/k}.$
Recall the Hsiung–Minkowski identities (see [Reference Scheuer18, Equation (2.8)] or [Reference Chen, Guan, Li and Scheuer5, Equation (1.4)])
$$ \begin{align} \int_Mu\sigma_{m+1}d\mu_g=C_{n, m}\int_M\phi'\sigma_md\mu_g,\,\,0\leq m\leq n-1, \end{align} $$
for
$C_{n, m}=\frac {n-m}{m+1}=\frac {C_n^{m+1}}{C_n^m}$
we obtain the following lemma.
Lemma 3.1 Let
$M_t$
be a smooth family of spacelike, compact, connected, strictly k-convex hypersurfaces in
$\mathbb S^{n+1}_1$
evolving along the flow (3.8). Then the surface area
$\mathcal A_0$
is non-increasing and the quantity
$$\begin{align*}\mathcal A_{k}(M_t)=\left\{\begin{aligned} &\int_{M_t}\sigma_1d\mu_g-n\text{Vol}(\Omega_t),\,\,k=1\\ &\int_{M_t}\sigma_kd\mu_g-\frac{n-k+1}{k-1}\mathcal A_{k-2}(M_t),\,\,2\leq k\leq n-1, \end{aligned}\right.\end{align*}$$
is non-decreasing. Moreover,
$\mathcal A_k$
is strictly increasing unless
$M_t$
is totally umbilic.
Proof In view of (3.7) and (3.9) we get, along the flow (3.8)
$$ \begin{align} \begin{aligned} \partial_t \mathcal A_0&=\int_{M_t}u\sigma_1-b_{n, k}\phi'\sigma_k^{-1/k}\sigma_1d\mu_g\\ &=\int_{M_t}n\phi'-b_{n, k}\phi'\sigma_k^{-1/k}\sigma_1d\mu_g. \end{aligned} \end{align} $$
By the Newton–Maclaurin inequality (see Proposition 2.3) we know that when
$M_t$
is strictly k-convex,
$$\begin{align*}\frac{\sigma_1}{n}\geq\frac{\sigma_k^{1/k}}{b_{n, k}}.\end{align*}$$
Thus,we conclude that
$\partial _t\mathcal A_0\leq 0.$
Similarly, we can also obtain for
$1\leq k\leq n-1$
$$ \begin{align} \begin{aligned} \partial_t \mathcal A_k&=(k+1)\int_{M_t}\sigma_{k+1}(u-b_{n, k}\phi'\sigma_k^{-1/k})d\mu_g\\ &=(k+1)\int_{M_t}C_{n, k}\phi'\sigma_k-b_{n, k}\phi'\sigma_k^{-1/k}\sigma_{k+1} d\mu_g. \end{aligned} \end{align} $$
It’s easy to see that at the point where
$\sigma _{k+1}\leq 0$
we have
$C_{n, k}\phi '\sigma _k-b_{n, k}\phi '\sigma _k^{-1/k}\sigma _{k+1}>0;$
while at the point where
$\sigma _{k+1}>0,$
applying Newton–Maclaurin inequality we still have
$C_{n, k}\phi '\sigma _{k}-b_{n, k}\phi '\sigma _k^{-1/k}\sigma _{k+1}\geq 0.$
Moreover, the equality holds if and only if at this point the principal curvature vector of
$M_t$
is
$\kappa =c(1, \cdots , 1)$
for some
$c>0.$
Therefore, the lemma is proved.
We want to point out that a straightforward calculation yields
$\partial _t\mathcal A_n\equiv 0.$
Hence, if one can prove the flow (3.8) moves an arbitrary k-convex hypersurface to a round sphere, then the following conjecture would turn into a theorem:
Conjecture 3.2 Let
$M\subset \mathbb S^{n+1}_1$
be a spacelike, compact, star-shaped, and k-convex hypersurface. Then there holds
$$ \begin{align} \mathcal A_k\leq\xi_{k, 0}(\mathcal A_0),\,\,1\leq k\leq n, \end{align} $$
with equality holds if and only if M is a radial coordinate slice. Here,
$\xi _{k, 0}$
is the associated monotonically increasing function for radial coordinate slices.
Note that when
$k=1$
the above conjecture has been proved in [Reference Scheuer18]. In this paper, we solve the case when
$k=2.$
4 Long time existence of (1.1)
In this section, we will establish a priori estimates for the flow equation (1.1) and prove the long time existence theory. For greater generality, we will start with the study of (3.8) instead. In the rest of this section, we will assume the initial hypersurface
$M_0$
is spacelike, compact, star-shaped, and strictly k-convex. Then by the short time existence theorem we know there exists
$T^*>0$
such that the flow (3.8) has a unique solution
$M_t$
for
$t\in [0, T^*).$
Moreover, the flow hypersurface
$M_t$
is also spacelike, compact, star-shaped, and strictly k-convex.
4.1 Estimates up to first order
In this section, we will derive the
$C^0$
and
$C^1$
estimates for the solution of (3.8).
Lemma 4.1 Along the flow (3.8) there holds for all
$(\xi , t)\in \mathbb S^n\times (0, T^*)$
we have
$$\begin{align*}\min\limits_{\mathbb S^n}\rho(\cdot, 0)\leq\rho(\xi, t)\leq\max\limits_{\mathbb S^n}\rho(\cdot, 0).\end{align*}$$
Proof The proof is the same as the one in [Reference Scheuer18], for completeness, we include it here.
The radial function
$\rho $
satisfies
$$\begin{align*}\rho_t=\left(u-b_{n,k}\phi'\sigma_k^{-1/k}\right)\frac{\cosh\rho}{w}.\end{align*}$$
At the critical point of
$\rho ,$
we get
$$\begin{align*}\tilde\nabla\rho=0\,\,\text{and } w=\cosh\rho=u.\end{align*}$$
In view of (2.6) and (2.7) we can see that at the the critical point,
$$\begin{align*}h^i_j=g^{ik}h_{kj}=\frac{1}{\cosh^2(\rho)}\left(\tilde\nabla_{ij}\rho+\sinh\rho\cosh\rho\delta_{ij}\right).\end{align*}$$
Therefore, we obtain that at the critical point
$$\begin{align*}\rho_t=\cosh\rho-\frac{b_{n,k}\sinh\rho}{\sigma_k^{1/k}\left(\frac{\tilde\nabla_{ij}\rho}{\cosh^2(\rho)}+\frac{\sinh\rho}{\cosh\rho}\delta_{ij}\right)}.\end{align*}$$
Note that at the spacial maximal points of
$\rho $
we have
$(\tilde \nabla _{ij}\rho )\leq 0$
, which implies that
$$\begin{align*}\sigma_k^{1/k}\left(\frac{\tilde\nabla_{ij}\rho}{\cosh^2(\rho)}+\tanh\rho\delta_{ij}\right)\leq b_{n,k}\tanh\rho.\end{align*}$$
Thus we have
$\max \rho $
is non-increasing. Similarly, we can show that
$\min \rho $
is non-decreasing.
Define
$L:=\partial _t-b_{n,k}\phi 'F^{-2}F^{ij}\nabla ^j\nabla _i$
for
$F=\sigma _k^{1/k}$
and
$F^{ij}=\frac {\partial F}{\partial h^i_j},$
denote
$$ \begin{align*}\hat L: =L+\left<V, \nabla\cdot\right>,\end{align*} $$
we will use Lemmas 2.5 and 2.6 to derive the evolution equations for
$\Phi $
and
$u.$
Lemma 4.2 Along the flow (3.8),
$\Phi $
and u evolve as follows
$$ \begin{align} \hat L\Phi=2b_{n,k} F^{-1}\phi'u-b_{n,k}(\phi')^2F^{-2}\sum f^i-\cosh^2(\rho), \end{align} $$
and
$$ \begin{align} \hat Lu=\phi'u\left(1-b_{n,k}F^{-2}\sum f^i\kappa_i^2\right)-b_{n, k} F^{-1}|\nabla\Phi|^2, \end{align} $$
where
$F(A)=f(\kappa [A])$
and
$\sum f^i=\sum F^{ii}.$
Proof Since the values of
$\hat L\Phi $
and
$\hat Lu$
are independent of the choice of coordinates, we may always choose an orthonormal frame
$\{\tau _1, \ldots , \tau _n\}$
such that
$(h_{ij})$
is diagonalized. Then at the point of consideration, we have
$F^{ij}=f^i\delta _{ij}$
and
$h_{ij}=\kappa _i\delta _{ij}.$
In view of Lemma 2.5, we get
$$ \begin{align*} \hat L\Phi&=\partial_t\Phi-b_{n,k}\phi'F^{-2}F^{ii}\nabla_i\nabla_i\Phi+\left<V, \nabla\Phi\right>\\ &=\left<V, S\nu\right>-b_{n,k}\phi'F^{-2}F^{ii}(\phi'\delta_{ii}-h_{ii}u)+|\nabla\Phi|^2\\ &=-(u-b_{n,k} F^{-1}\phi')u-b_{n,k}(\phi')^2F^{-2}\sum f^i+b_{n,k}\phi'F^{-1}u+|\nabla\Phi|^2\\ &=2b_{n,k} F^{-1}\phi'u-b_{n,k}(\phi')^2F^{-2}\sum f^i-\cosh^2(\rho). \end{align*} $$
Here, we have used
$\sum f^i \kappa _i=f$
and
$$ \begin{align*}\left<V, \nabla\Phi\right>=\sum\left<V, \left<V, \tau_i\right>\tau_i\right>=|V|^2+u^2=u^2-\cosh^2(\rho)=|\nabla\Phi|^2.\end{align*} $$
Similarly, by Lemma 2.6 and (3.4) we have
$$ \begin{align*} \hat Lu&=\partial_tu-b_{n,k}\phi'F^{-2}F^{ii}\nabla_i\nabla_iu+\left<V, \nabla u\right>\\ &=\phi'S-\left<V, \nabla S\right>+\left<V, \nabla u\right>-b_{n,k}\phi'F^{-2}F^{ii}(-h_{iik}\nabla_k\Phi-\phi'h_{ii}+uh^l_ih_{li})\\ &=\phi'S+b_{n,k}\left<V, (\nabla\phi')F^{-1}\right>+b_{n,k}\left<V, \phi'\nabla F^{-1}\right>+b_{n,k}\phi'F^{-2}\left<\nabla F, V\right>\\ &\quad +b_{n,k}(\phi')^2F^{-1}-ub_{n,k}\phi'F^{-2}F^{ii}h^l_ih_{li}\\ &=\phi'u-b_{n,k} F^{-1}|\nabla\Phi|^2-ub_{n,k}\phi'F^{-2}\sum f^i\kappa_i^2\\ &=\phi'u\left(1-b_{n,k} F^{-2}\sum f^i\kappa_i^2\right)-b_{n,k} F^{-1}|\nabla\Phi|^2. \end{align*} $$
From Lemma 4.2 we obtain the
$C^1$
estimate.
Lemma 4.3 Along the flow (3.8) there holds for all
$(\xi , t)\in \mathbb S^n\times (0, T^*)$
we have
$$\begin{align*}u(\xi, t)\leq\max\limits_{\mathbb S^n}u(\cdot, 0).\end{align*}$$
Proof Let
$\kappa =(\kappa _1, \ldots , \kappa _n)$
be the principal curvature vector of
$M_t,$
then
$\kappa \in \Gamma _k.$
In view of Lemma 2.2 and Proposition 2.3 we get
$$ \begin{align*} \sum f^i\kappa_i^2 &=\frac{1}{k}\sigma_k^{1/k-1}\sigma_{k-1}(\kappa|i)\kappa_i^2\\ &=\frac{1}{k}\sigma_k^{1/k-1}[\sigma_k\sigma_1-(k+1)\sigma_{k+1}]\\ &\geq\frac{1}{k}\sigma_k^{1/k}\left[\frac{n}{b_{n, k}}\sigma_k^{1/k}-(k+1)C_{n, k}\frac{\sigma_k^{1/k}}{b_{n, k}}\right]. \end{align*} $$
Therefore, we have
$$ \begin{align}\sum f^i\kappa_i^2\geq\frac{f^2}{b_{n, k}}.\end{align} $$
Combining with Equation (4.2) we obtain, along the flow (3.8)
$\hat Lu\leq 0$
. Then the lemma follows from the standard maximum principle.
Recall Equation (2.5), we can see that Lemma 4.3 implies that along the flow (3.8),
$\frac {|\tilde \nabla \rho |^2}{\cosh ^2(\rho )}$
is uniformly bounded away from 1.
4.2 Uniform bounds of F
In this section, we will prove that F is uniformly bounded from below along the flow (3.8). However, due to technical reasons, to obtain the uniform upper bound of F we have to restrict ourselves to the case when
$k=2,$
that is, the flow (1.1).
Lemma 4.4 Along the flow (3.8) there holds for all
$(\xi , t)\in \mathbb S^n\times (0, T^*)$
we have
$$\begin{align*}F(\xi, t)\geq\min\limits_{\mathbb S^n}F(\cdot, 0).\end{align*}$$
Proof At the critical point of F we may choose an orthonormal frame
$\{\tau _1, \ldots , \tau _n\}$
such that
$h_{ij}=\kappa _i\delta _{ij}$
is diagonalized. By virtue of (3.5) we have
$$ \begin{align*} \partial_tF&=F^{ii}\left\{u_{ii}+b_{n, k} F^{-1}\Phi_{ii}+2b_{n,k} F^{-1}F_i\left(\frac{\phi'}{F}\right)_i+b_{n,k}\phi'F^{-2}\nabla_{ii}F\right\}\\ &\quad -(u-b_{n,k}\phi'F^{-1})\sum f^i\kappa_i^2+(u-b_{n,k}\phi'F^{-1})\sum f^i. \end{align*} $$
Applying Lemmas 2.5 and 2.6, we obtain
$$ \begin{align} \begin{aligned} \partial_tF&=b_{n,k}\phi'F^{-2}F^{ii}\nabla_{ii}F+F^{ii}(-h_{iil}\nabla_l\Phi-\phi'h_{ii}+uh^l_ih_{li}) \\ & \quad +b_{n,k} F^{-1}F^{ii}(\phi'\delta_{ii}-h_{ii}u)\\ &\quad +2b_{n,k} F^{-1}F^{ii}F_i\left(\frac{\phi'}{F}\right)_i-(u-b_{n,k}\phi'F^{-1})\sum f^i\kappa_i^2\\ &\quad +(u-b_{n, k}\phi'F^{-1})\sum f^i. \end{aligned} \end{align} $$
Let
$F_{\min }(t)=\min \limits _{\xi \in \mathbb S^n}F(\xi , t),$
then
$F_{\min }$
satisfies
$$ \begin{align*} \frac{d}{dt}F_{\min}&\geq-\phi'F+u\sum f^i\kappa_i^2+b_{n,k} F^{-1}\phi'\sum f^i-b_{n,k} u\\ &\quad -u\sum f^i\kappa_i^2+b_{n,k}\phi'F^{-1}\sum f^i\kappa_i^2+u\sum f^i-b_{n,k}\phi'F^{-1}\sum f^i\\ &=u(\sum f^i-b_{n,k})+\phi'F^{-1}\left(b_{n,k}\sum f^i\kappa_i^2-F^2\right). \end{align*} $$
By the concavity of F we get
$$ \begin{align}\sum f^i\geq f(1, \cdots, 1)=b_{n,k}.\end{align} $$
In conjunction with inequality (4.3), we conclude
$$\begin{align*}\frac{d}{dt}F_{\min}\geq 0,\end{align*}$$
which yields this lemma.
In the proofs of Lemmas 4.5 and 4.6 below, we will explicitly use the fact that
$k=2.$
Therefore, from now on, we will restrict ourselves to the locally constrained inverse scalar curvature flow
$$ \begin{align} \left\{\begin{aligned} X_t&=\left(u-b_{n, 2}\phi'\sigma_2^{-1/2}\right)\nu,\\ X_0&=M_0. \end{aligned}\right. \end{align} $$
Lemma 4.5 Along the flow (4.6) there holds for all
$(\xi , t)\in \mathbb S^n\times (0, T^*)$
we have
$$\begin{align*}F(\xi, t)\leq C,\end{align*}$$
where
$C>0$
is a constant depending on
$M_0, n, \rho ,$
and u.
Proof Recall the flow equation (4.1) of
$\Phi $
we have
$$ \begin{align*} \hat L\Phi&=2b_{n, 2} F^{-1}\phi'u-b_{n, 2}(\phi')^2F^{-2}\sum f^i-\cosh^2(\rho)\\ &\leq-\left(u-\frac{b_{n, 2}\phi'}{F}\right)^2+u^2-\cosh^2(\rho), \end{align*} $$
where we have used
$\sum f^i\geq b_{n, 2}.$
Moreover, by Lemma 4.3 we get
$$\begin{align*}u^2-\cosh^2(\rho)=\frac{\cosh^2(\rho)}{1-|\tilde\nabla\rho|^2/\cosh^2(\rho)}-\cosh^2(\rho)\leq\beta_0 u^2,\end{align*}$$
for some
$0<\beta _0=\beta _0(|\tilde \nabla \rho |/\cosh \rho )<1.$
Therefore, we can see that there exists some uniform constant
$0<\beta _1=\beta _1(n, \beta _0)<1$
such that whenever
$F\geq C_0(n, \rho , u)$
$$ \begin{align}\hat L\Phi\leq-\beta_1 u^2.\end{align} $$
Here,
$C_0(n, \rho , u)$
is a large constant that only depends on
$n, \rho ,$
and
$u.$
Now, consider
$\Psi =\log F+\lambda u+\alpha \Phi ,$
where
$\lambda , \alpha>0$
to be determined. Assume
$\Psi $
achieves its maximum at an interior point
$X_0\in M_{t_0}$
for some
$t_0\in (0, T^*).$
Then at
$X_0$
we can choose an orthonormal frame
$\{\tau _1, \ldots , \tau _n\}$
such that
$h_{ij}=\kappa _i\delta _{ij}.$
We have, at
$X_0$
$$ \begin{align} \Psi_i=\frac{F_i}{F}+\lambda u_i+\alpha\Phi_i=0 \end{align} $$
and
$$ \begin{align} \begin{aligned} 0&\leq\hat L\Psi=\frac{\hat LF}{F}+b_{n, 2}\phi'F^{-2}F^{ii}\left(\frac{F_i}{F}\right)^2+\lambda\hat Lu+\alpha\hat L\Phi\\&\leq\frac{1}{F} \left[u(\sum f^i-b_{n, 2})+\phi'F^{-1}(b_{n, 2}\sum f^i\kappa_i^2-F^2)\right.\\&\quad \left. +\ 2b_{n, 2} F^{-1}F^{ii}\left(\frac{\phi^{\prime}_iF_i}{F}-\frac{\phi'F_i^2}{F^2}\right)\right]+b_{n, 2}\phi'F^{-4}F^{ii}F_i^2\\&\quad +\lambda\left(\phi'u-b_{n, 2}\phi'uF^{-2}\sum f^i\kappa_i^2-b_{n, 2} F^{-1}|\nabla\Phi|^2\right)+\alpha\hat L\Phi, \end{aligned} \end{align} $$
where we have used Equations (4.2) and (4.4). We can see that (4.9) implies
$$ \begin{align} \begin{aligned} 0&\leq\frac{u}{F}\sum f^i+\phi'F^{-2}b_{n, 2}\sum f^i\kappa_i^2\left(1-\lambda u\right)\\ &\quad +2b_{n, 2} F^{-3}F^{ii}\phi^{\prime}_iF_i-b_{n, 2}\phi'F^{-4}F^{ii}F_i^2+\lambda\phi'u+\alpha\hat L\Phi. \end{aligned} \end{align} $$
Since
$$\begin{align*}\begin{aligned} 2b_{n, 2} F^{-3}F^{ii}\phi^{\prime}_iF_i&\leq b_{n, 2}\phi'F^{-4}F^{ii}F_i^2+\frac{b_{n, 2}}{\phi'}F^{-2}F^{ii}(\phi^{\prime}_i)^2\\ &\leq b_{n, 2}\phi'F^{-4}F^{ii}F_i^2+\frac{b_{n, 2}}{\phi'}F^{-2}|\nabla\Phi|^2\sum f^i,\end{aligned}\end{align*}$$
if at the critical point
$F\geq C_1=C_1(u, n, \rho )$
very large, applying (4.7) we can see that (4.10) becomes
$$ \begin{align} \begin{aligned} 0&\leq\hat L\Psi\leq\sum f^i+\phi'F^{-2}b_{n, 2}\sum f^i\kappa_i^2\left(1-\lambda u\right) +\lambda\phi'u-\alpha\beta_1u^2. \end{aligned} \end{align} $$
From Equation (2.3), Lemmas 2.1 2.2, and Proposition 2.3 we obtain
$$ \begin{align}\sum f^i=\frac{1}{2}F^{-1}(n-1)\sigma_1\end{align} $$
and
$$ \begin{align}\sum f^i\kappa_i^2=\frac{1}{2}F\sigma_1-\frac{3}{2}F^{-1}\sigma_3\geq\frac{1}{n}\sigma_1 F.\end{align} $$
Thus we may choose
$\lambda =\lambda (\rho , n, u)>0$
such that
$$\begin{align*}\sum f^i+\phi'F^{-2}b_{n, 2}\sum f^i\kappa_i^2\left(1-\lambda u\right)\leq-\sum f^i.\end{align*}$$
After we fix the value of
$\lambda ,$
we can choose
$\alpha =\alpha (\rho , u, \lambda , \beta _1)>0$
such that
$$\begin{align*}\lambda\phi'u-\alpha\beta_1u^2<0.\end{align*}$$
We conclude that if
$F>C_1$
large at
$X_0$
we would have the right hand side of (4.11) is negative. This leads to a contradiction. Therefore,
$\Psi $
is uniformly bounded on
$[0, T^*).$
4.3 Uniform bounds for principal curvatures
In this section, we will show that the principal curvatures of
$M_t$
remain bounded along the flow (4.6). More precisely, we will prove the following lemma
Lemma 4.6 Let
$M_t$
be the solution of (4.6) on
$[0, T^*),$
then there exists a constant C depending on
$M_0, n, \rho , u,$
and F such that
$$ \begin{align} \left|\kappa[M_t]\right|\leq C. \end{align} $$
Proof Let us consider the test function
$$\begin{align*}G=\log\kappa_1+\lambda\Phi,\end{align*}$$
where
$\kappa _1$
is the largest principal curvature and
$\lambda>0$
is a large constant to be determined. Assume G achieves its maximum at an interior point
$P_0\in M_{t_0}$
for some
$t_0\in (0, T^*).$
Then at this point, we can choose an orthonormal frame such that
$h_{ij}=\kappa _i\delta _{ij}$
is diagonalized. Without loss of generality, we may assume
$\kappa _1$
has multiplicity
$m,$
i.e.,
$$\begin{align*}\kappa_1=\ldots=\kappa_m>\kappa_{m+1}\geq\ldots\geq\kappa_n.\end{align*}$$
By [Reference Brendle, Choi and Daskalopoulos2] (see also [Reference Lu17, Lemma 5]) we have at
$P_0,$
$$ \begin{align} \delta_{kl}\kappa_{1i}=h_{kli},\text{for } 1\leq k, l\leq m, \end{align} $$
and
$$ \begin{align} \kappa_{1ii}\geq h_{11ii}+2\sum\limits_{p>m}\frac{h^2_{1pi}}{\kappa_1-\kappa_p} \end{align} $$
in the viscosity sense. We want to point out that (4.15) yields that when
$m\geq 2, h_{11i}=0$
for
$1<i\leq m.$
We will start with computing the evolution equation of
$\kappa _1$
at
$P_0.$
Recall (3.5) we have
$$ \begin{align*} \partial_th_i^i&=\nabla_{ii}S-S\kappa_i^2+S\\ &=u_{ii}-b_{n, 2}\phi^{\prime}_{ii}F^{-1}+2b_{n, 2}\phi^{\prime}_iF^{-2}F_i-2b_{n, 2}\phi'F^{-3}F^2_i\\ &\quad +b_{n, 2}\phi'F^{-2}\left(F^{kk}h_{kkii}+F^{pq, rs}h_{pqi}h_{rsi}\right)-S\kappa_i^2+S. \end{align*} $$
Combining with the following commuting formula in de Sitter space (for details see [Reference Ballesteros-Chávez, Klingenberg and Lambert4, p. 10])
$$\begin{align*}h_{kkii}=h_{iikk}+\kappa_i^2\kappa_k-\kappa_i\delta_{kk}-\kappa_i\kappa_k^2+\kappa_k\delta_{ii},\end{align*}$$
we deduce
$$ \begin{align*} \partial_th^i_i&=(-h_{iik}\nabla_k\Phi-\phi'\kappa_i+u\kappa_i^2)-b_{n, 2} F^{-1}(-\phi'\delta_{ii}+\kappa_iu)\\ &\quad +2b_{n, 2} F^{-2}\phi^{\prime}_iF_i-2b_{n, 2}\phi'F^{-3}F^2_i\\ &\quad +b_{n, 2}\phi' F^{-2}F^{kk}(h_{iikk}+\kappa_i^2\kappa_k-\kappa_i\delta_{kk}-\kappa_i\kappa_k^2+\kappa_k\delta_{ii})\\ &\quad -(u-b_{n, 2}\phi'F^{-1})\kappa_i^2+(u-b_{n, 2}\phi'F^{-1})\\ &\quad +b_{n, 2}\phi'F^{-2}F^{pq, rs}h_{pqi}h_{rsi}. \end{align*} $$
Therefore, we obtain
$$ \begin{align} \begin{aligned} \hat Lh^i_i&=\kappa_i\left(-\phi'-b_{n, 2} F^{-1}u-b_{n, 2}\phi'F^{-2}\sum f^k-b_{n, 2}\phi'F^{-2}\sum f^k\kappa_k^2\right)\\ &\quad +\left(b_{n, 2}\phi'F^{-1}+u\right)+2b_{n, 2}\phi'F^{-1}\kappa_i^2\\ &\quad +\left(2b_{n, 2} F^{-2}\phi^{\prime}_iF_i-2b_{n, 2}\phi'F^{-3}F^2_i+b_{n, 2}\phi'F^{-2}F^{pq, rs}h_{pqi}h_{rsi}\right). \end{aligned} \end{align} $$
From (4.16) we get, at
$P_0$
$$\begin{align*}\hat L\kappa_1\leq\hat Lh^1_1-2b_{n, 2}\phi'F^{-2}\sum\limits_i\sum\limits_{p>m}F^{ii}\frac{h^2_{1pi}}{\kappa_1-\kappa_p}.\end{align*}$$
By our assumption, we have, at
$P_0$
$$ \begin{align} G_i=\frac{\kappa_{1i}}{\kappa_1}+\lambda\Phi_i=0. \end{align} $$
Moreover, in view of Lemma 4.2 and Equation (4.17) we derive
$$ \begin{align} \begin{aligned} 0&\leq\hat LG=\frac{\hat L\kappa_1}{\kappa_1}+\frac{b_{n, 2}\phi'F^{-2}F^{ii}}{\kappa_1^2}h_{11i}^2+\lambda\hat L\Phi\\&\leq\frac{1}{\kappa_1}\bigg[\kappa_1\left(-\phi'-b_{n, 2} F^{-1}u-b_{n, 2}\phi'F^{-2}\sum f^k-b_{n, 2}\phi'F^{-2}\sum f^k\kappa_k^2\right)\\&\quad +(b_{n, 2}\phi'F^{-1}+u)+2b_{n, 2}\phi'F^{-1}\kappa_1^2 +\left(2b_{n, 2} F^{-2}\phi^{\prime}_1F_1-2b_{n, 2}\phi'F^{-3}F_1^2\right.\\&\quad \left.+\ b_{n, 2}\phi'F^{-2}F^{pq, rs}h_{pq1}h_{rs1}\right) -2b_{n, 2}\phi'F^{-2}\sum\limits_i\sum\limits_{p>m}F^{ii}\frac{h^2_{1pi}}{\kappa_1-\kappa_p}\bigg]\\&\quad +\frac{b_{n, 2}\phi'F^{-2}}{\kappa_1^2}F^{ii}h^2_{11i}+\lambda\left(2b_{n, 2} F^{-1}\phi'u-b_{n, 2}(\phi')^2F^{-2}\sum f^k-\cosh^2(\rho)\right). \end{aligned} \end{align} $$
Since
$$\begin{align*}2b_{n, 2} F^{-2}\phi^{\prime}_1F_1\leq\frac{1}{2}b_{n, 2}\phi'F^{-3}F_1^2+2\frac{b_{n, 2} F^{-1}(\phi^{\prime}_1)^2}{\phi'}\end{align*}$$
(4.19) becomes
$$ \begin{align} \begin{aligned} 0&\leq\hat LG\leq\lambda C_1+\left(-b_{n, 2}\phi'F^{-2}\sum f^k-b_{n, 2}\phi'F^{-2}\sum f^k\kappa_k^2\right)\\ &\quad -\frac{3}{2\kappa_1}b_{n, 2}\phi'F^{-3}F_1^2+\frac{b_{n, 2}}{\kappa_1}\phi'F^{-2}F^{pq, rs}h_{pq1}h_{rs1}\\ &\quad +2b_{n, 2}\phi'F^{-1}\kappa_1-\frac{2b_{n, 2}}{\kappa_1}\phi'F^{-2}\sum\limits_i\sum\limits_{p>m}F^{ii}\frac{h^2_{1pi}}{\kappa_1-\kappa_p}\\ &\quad +\frac{b_{n, 2}\phi'F^{-2}}{\kappa_1^2}F^{ii}h^2_{11i}-\lambda b_{n, 2}(\phi')^2F^{-2}\sum f^k. \end{aligned} \end{align} $$
Here and in the rest of this proof, we will use
$C_1, C_2, \ldots $
and
$c_0, c_1, c_2, \ldots $
to denote universal positive constants that only depend on
$n, \rho , u,$
and
$F.$
Equation (4.20) yields
$$ \begin{align} \begin{aligned} 0&\leq\hat LG\leq\lambda C_1-\left(b_{n, 2}\phi'F^{-2}+\lambda b_{n, 2}(\phi')^2F^{-2}\right)\sum f^k\\ &\quad -b_{n, 2}\phi'F^{-2}\sum f^k\kappa_k^2+2b_{n, 2}\phi'F^{-1}\kappa_1\\ &\quad +\frac{b_{n, 2}\phi'}{\kappa_1F^2}\left(-\frac{3}{2}F^{-1}F_1^2+F^{pq, rs}h_{pq1}h_{rs1}-2\sum\limits_i\sum\limits_{p>m}\frac{F^{ii}h^2_{1pi}}{\kappa_1-\kappa_p} +\sum\limits_i\frac{F^{ii}h_{11i}^2}{\kappa_1}\right). \end{aligned} \end{align} $$
By virtue of (2.4) we can see that
$$ \begin{align*} \sigma_2^{pq, rs}h_{pq1}h_{rs1}&=\sum\limits_{p\neq q}\sigma_2^{pp, qq}h_{pp1}h_{qq1}+2\sum\limits_{p>q}\sigma_2^{pq, qp}h^2_{pq1}\\ &=\sum\limits_{p\neq q}h_{pp1}h_{qq1}-2\sum\limits_{p>q}h^2_{pq1}\\ &\leq\sum\limits_{p\neq q}h_{pp1}h_{qq1}-2\sum\limits_{p>m}h^2_{11p}. \end{align*} $$
Note that
$F^2=\sigma _2,$
thus we have
$$\begin{align*}2FF^{pq, rs}h_{pq1}h_{rs1}+2(F_1)^2\leq\sum\limits_{p\neq q}h_{pp1}h_{qq1}-2\sum\limits_{p>m}h^2_{11p}.\end{align*}$$
This gives
$$ \begin{align} F^{pq, rs}h_{pq1}h_{rs1}\leq\frac{1}{2}F^{-1}\left(\sum\limits_{p\neq q}h_{pp1}h_{qq1}-2\sum\limits_{p>m}h^2_{11p}-2(F_1)^2\right). \end{align} $$
Therefore, we get
$$ \begin{align*} &-\frac{3}{2}F^{-1}F_1^2+F^{pq, rs}h_{pq1}h_{rs1}-2\sum\limits_{i}\sum\limits_{p>m}\frac{F^{ii}h^2_{1pi}}{\kappa_1-\kappa_p}+\sum\limits_i\frac{F^{ii}h_{11i}^2}{\kappa_1}\\ &\leq-\frac{5}{2}F^{-1}F_1^2+\frac{1}{2}F^{-1}\sum\limits_{p\neq q}h_{pp1}h_{qq1}-F^{-1}\sum\limits_{p>m}h^2_{11p}-2\sum\limits_{p>m}\frac{F^{11}h^2_{11p}}{\kappa_1-\kappa_p}\\ &\quad -2\sum\limits_{p>m}\frac{F^{pp}h^2_{pp1}}{\kappa_1-\kappa_p}+\sum\limits_{p>m}\frac{F^{pp}h^2_{11p}}{\kappa_1}+\frac{F^{11}h^2_{111}}{\kappa_1}, \end{align*} $$
where we have used
$h_{11i}=0$
for
$1<i\leq m.$
By a straightforward calculation we can see that for each fixed
$p>m,$
$$ \begin{align*} &\left(-F^{-1}-2\frac{F^{11}}{\kappa_1-\kappa_p}+\frac{F^{pp}}{\kappa_1}\right)h^2_{11p}\\ &=F^{-1}\left(-1-\frac{\sigma_1-\kappa_1}{\kappa_1-\kappa_p}+\frac{\sigma_1-\kappa_p}{2\kappa_1}\right)h^2_{11p}\\ &=F^{-1}\frac{(\kappa_1+\kappa_p)(\kappa_p-\sigma_1)}{2(\kappa_1-\kappa_p)\kappa_1}h^2_{11p}\leq 0, \end{align*} $$
here we have used equality (2.3) and the first inequality in Lemma 2.2. Thus we conclude
$$ \begin{align} \begin{aligned} &-\frac{3}{2}F^{-1}F_1^2+F^{pq, rs}h_{pq1}h_{rs1}-2\sum\limits_{i}\sum\limits_{p>m}\frac{F^{ii}h^2_{1pi}}{\kappa_1-\kappa_p}+\sum\limits_i\frac{F^{ii}h_{11i}^2}{\kappa_1}\\ &\leq\frac{1}{2}F^{-1}\sum\limits_{p\neq q}h_{pp1}h_{qq1}-\frac{5}{2}F^{-1}F_1^2-2\sum\limits_{p>m}\frac{F^{pp}h_{pp1}^2}{\kappa_1-\kappa_p}+\frac{F^{11}h^2_{111}}{\kappa_1}. \end{aligned} \end{align} $$
Now let
$\epsilon>0$
be a small constant that will be determined later,
$\delta =\delta _0=\frac {1}{2},$
and let
$\delta '=\delta '(\epsilon , \delta , \delta _0)=O(\epsilon )>0$
be a constant determined by Lemma 2.4. We will divide the rest of this proof into two cases.
Case 1. When
$\kappa _2\leq \delta '\kappa _1$
at
$P_0,$
by Lemma 2.4, we get
$$ \begin{align} \sum\limits_{p\neq q}\frac{h_{pp1}h_{qq1}}{\sigma_2}-\frac{(\sigma_2^{ii}h_{ii1})^2}{\sigma_2^2} \leq(\epsilon-1)\frac{h^2_{111}}{\kappa_1^2}+\frac{1}{2}\sum\limits_{p>1}\frac{\sigma_2^{pp}h^2_{pp1}}{\kappa_1\sigma_2}. \end{align} $$
Moreover, since
$\kappa \in \Gamma _2$
we have
$\sigma _2^{11}=\sum \limits _{i=2}^{n-1}\kappa _i+\kappa _n>0,$
which implies
$$ \begin{align}\kappa_n>-(n-2)\delta'\kappa_1.\end{align} $$
When
$\delta '>0$
small we obtain
$$\begin{align*}2\sum\limits_{p>1}\frac{\sigma_2^{pp}h^2_{pp1}}{\kappa_1-\kappa_p}>2\sum\limits_{p>1}\frac{\sigma_2^{pp}h^2_{pp1}}{(1+(n-2)\delta')\kappa_1} >\sum\limits_{p>1}\frac{\sigma_2^{pp}h^2_{pp1}}{\kappa_1}.\end{align*}$$
Therefore, in this case (4.23) becomes
$$ \begin{align} \begin{aligned} &-\frac{3}{2}F^{-1}F_1^2+F^{pq, rs}h_{pq1}h_{rs1}-2\sum\limits_{i}\sum\limits_{p>1}\frac{F^{ii}h^2_{1pi}}{\kappa_1-\kappa_p}+\sum\limits_i\frac{F^{ii}h_{11i}^2}{\kappa_1}\\ &\leq\frac{1}{2}F^{-1} \left\{\sum\limits_{p\neq q}h_{pp1}h_{qq1}-\frac{5}{4}\sigma_2^{-1}(\sigma_2^{ii}h_{ii1})^2-2\sum\limits_{p>1}\frac{\sigma_2^{pp}h_{pp1}^2}{\kappa_1-\kappa_p}+\frac{\sigma_2^{11}h^2_{111}}{\kappa_1}\right\}\\ &\leq\frac{1}{2}F^{-1}\left\{(\epsilon-1)\sigma_2\frac{h_{111}^2}{\kappa_1^2}+\frac{1}{2}\sum\limits_{p>1}\frac{\sigma_2^{pp}h^2_{pp1}}{\kappa_1}-\frac{1}{4}\sigma_2^{-1}(\sigma_2)^2_1 -2\sum\limits_{p>1}\frac{\sigma_2^{pp}h^2_{pp1}}{\kappa_1-\kappa_p}+\frac{\sigma_2^{11}h_{111}^2}{\kappa_1}\right\}\\ &\leq\frac{1}{2}F^{-1}\left[(\epsilon-1)\sigma_2+\sigma_2^{11}\kappa_1\right]\frac{h_{111}^2}{\kappa_1^2}. \end{aligned} \end{align} $$
Plugging (4.26) into (4.21) and applying the first equality in Lemma 2.1 we obtain
$$ \begin{align} \begin{aligned} 0&\leq\hat LG\leq\lambda C_1-\left(b_{n, 2}\phi'F^{-2}+\lambda b_{n, 2}(\phi')^2F^{-2}\right)\sum f^k\\ &\quad -b_{n, 2}\phi'F^{-2}\sum f^k\kappa_k^2+2b_{n, 2}\phi'F^{-1}\kappa_1+\frac{b_{n, 2}\phi'}{2F^3\kappa_1}[\epsilon\sigma_2-\sigma_2(\kappa|1)]\frac{h^2_{111}}{\kappa_1^2}. \end{aligned} \end{align} $$
In view of our assumption that
$\kappa _2\leq \delta '\kappa _1,$
we know
$$\begin{align*}|\sigma_2(\kappa|1)|\leq c_0(\delta'\kappa_1)^2.\end{align*}$$
Therefore, we have
$$ \begin{align} \begin{aligned} 0&\leq\hat LG\leq\lambda C_1-\left(b_{n, 2}\phi'F^{-2}+\lambda b_{n, 2}(\phi')^2F^{-2}\right)\sum f^k\\ &\quad -b_{n, 2}\phi'F^{-2}\sum f^k\kappa_k^2+2b_{n, 2}\phi'F^{-1}\kappa_1+\frac{b_{n, 2}\phi'}{2F^3\kappa_1}[c_1\epsilon+c_0(\delta'\kappa_1)^2]\frac{h^2_{111}}{\kappa_1^2}. \end{aligned} \end{align} $$
By (4.18), we get at
$P_0$
$$\begin{align*}\frac{h_{111}}{\kappa_1}=-\lambda\Phi_1=-\lambda\left<V, \tau_1\right>.\end{align*}$$
Combining with (4.28) and (4.12) yields
$$ \begin{align} 0\leq\hat LG\leq\lambda C_1+C_2(\delta')^2\lambda^2\kappa_1+C_3\kappa_1-\frac{n-1}{2}\lambda b_{n, 2}(\phi')^2F^{-3}\sigma_1. \end{align} $$
Without loss of generality, we will always assume
$\delta '\leq \frac {1}{2n^2},$
then by (4.25) we have,
$$ \begin{align*}\sigma_1>\kappa_1+(n-1)\kappa_n>\frac{\kappa_1}{2}.\end{align*} $$
Therefore, (4.29) implies
$$ \begin{align} 0\leq\hat LG\leq\lambda C_1+C_2(\delta')^2\lambda^2\kappa_1+C_3\kappa_1-\frac{n-1}{4}\lambda b_{n, 2}(\phi')^2F^{-3}\kappa_1. \end{align} $$
We can choose
$\lambda =\lambda (n, \rho , F, C_3)>0$
large such that
$\frac {n-1}{4}\lambda b_{n, 2}(\phi ')^2F^{-3}>2C_3+1$
, then choose
$\delta '=\delta '(\lambda , C_2)>0$
small (this can be achieved by choosing a small
$\epsilon $
) such that
$C_2(\delta ')^2\lambda ^2<\frac {1}{2}$
. Then if at
$P_0$
we have
$\kappa _1>N_0=N_0(\lambda , C_1)>0$
large, the right hand side of (4.30) would be strictly negative. This leads to a contradiction.
Case 2. When
$\kappa _2\geq \delta '\kappa _1$
at
$P_0.$
By Lemma 4.4 and Lemma 4.5 we know
$f\left (1, \frac {\kappa _2}{\kappa _1}, \frac {\kappa _3}{\kappa _1}, \ldots , \frac {\kappa _n}{\kappa _1}\right )=O\left (\frac {1}{\kappa _1}\right ).$
In view of our assumption that
$\frac {\kappa _2}{\kappa _1}\geq \delta '$
, when
$\kappa _1$
is very large we have
$\kappa _n<0,$
and
$$ \begin{align*} \frac{C_4}{\kappa_1}&>f\left(1, \frac{\kappa_2}{\kappa_1}, \frac{\kappa_3}{\kappa_1}, \cdots, \frac{\kappa_n}{\kappa_1}\right)>\sigma_2\left(1, \frac{\kappa_2}{\kappa_1}, \frac{\kappa_3}{\kappa_1}, \cdots, \frac{\kappa_n}{\kappa_1}\right)\\ &=\sum\limits_{i\geq 2}\frac{\kappa_i}{\kappa_1}+\sigma_2\left(0, \frac{\kappa_2}{\kappa_1}, \frac{\kappa_3}{\kappa_1}, \cdots,\frac{\kappa_n}{\kappa_1}\right)\\ &>\delta'+c_2\frac{\kappa_n}{\kappa_1} \end{align*} $$
for some
$c_2=c_2(n)>0.$
Therefore, if
$\kappa _1>N_1=N_1(1/\delta ', C_4)$
very large at
$P_0$
, then there exists
$\eta _0=\eta _0(\delta ', c_2)>0$
such that
$\frac {\kappa _n}{\kappa _1}\leq -\eta _0<0.$
This implies
$$ \begin{align*}\sum f^k\kappa_k^2\geq f^n\kappa_n^2\geq\frac{1}{n}\left(\sum f^i\right)\eta_0^2\kappa_1^2\geq c_3\kappa_1^2,\end{align*} $$
where
$c_3=c_3(n, \eta _0)>0.$
Combining with (4.21), we obtain
$$ \begin{align} 0\leq\hat LG\leq\lambda C_1-b_{n, 2}\phi'F^{-2}c_3\kappa_1^2+2b_{n, 2}\phi'F^{-1}\kappa_1+\frac{b_{n, 2}\phi'}{F^2}\sum F^{ii}\lambda^2\left<V, \tau_i\right>^2, \end{align} $$
where we have used (4.18) and the fact that
$F=\sigma _2^{1/2}$
is concave. We can see that when
$\kappa _1>N_2=N_2(N_1, n, \rho , u, F, \lambda , c_3)>0$
very large
$$\begin{align*}\lambda C_1-b_{n, 2}\phi'F^{-2}c_3\kappa_1^2+2b_{n, 2}\phi'F^{-1}\kappa_1+\frac{b_{n, 2}\phi'}{F^2}\sum F^{ii}\lambda^2\left<V, \tau_i\right>^2<0.\end{align*}$$
This leads to a contradiction. It follows that if G achieves its maximum at an interior point, then G is uniformly bounded. Therefore, the lemma is proved.
So far we have obtained uniform
$C^2$
bounds for the flow hypersurfaces
$M_t.$
This implies the uniform parabolicity of the operator
$L.$
Due to the concavity of the operator, we can apply the Krylov and Safonov regularity theorem (see [Reference Krylov14]) to deduce
$C^{2, \alpha }$
bounds. The
$C^{\infty }$
bounds follow from the Schauder theory. We conclude:
Proposition 4.7 Let
$M_0\subset \mathbb S^{n+1}_1$
be a spacelike, compact, star-shaped, and strictly
$2$
-convex hypersurface. Then the flow (4.6) exists for all time with uniform
$C^\infty $
-estimates.
5 Convergence and inequality
In this section, we complete the proof of the geometric inequality, that is, Corollary 1.3.
Proof of Corollary 1.3
Recall Lemma 3.1, we know that
$\mathcal A_0$
is decreasing. By the
$C^2$
estimates obtained in the section 4.3, we have
$\mathcal A_2$
is uniformly bounded from above. Moreover, Lemma 3.1 also tells us that
$\mathcal A_2$
is increasing. Therefore, we have
$$\begin{align*}\int_0^\infty\partial_t\mathcal A_2<\infty,\end{align*}$$
which implies that
$$ \begin{align} \partial_t\mathcal A_2=3\int_{M_t}C_{n, 2}\phi'\sigma_2-b_{n, 2}\phi'\sigma_2^{-1/2}\sigma_3d\mu_g\rightarrow 0, \end{align} $$
as
$t\rightarrow \infty .$
By Lemma 4.1, we know
$\mathcal A_0=\int _{M_t}d\mu _g$
is bounded away from
$0.$
Thus by (5.1) we have
$C_{n, 2}\phi '\sigma _2-b_{n, 2}\phi '\sigma _2^{-1/2}\sigma _3\rightarrow 0$
as
$t\rightarrow \infty .$
In view of the Newton–Maclaurin inequality we conclude that any convergent subsequence
$\{M_{t_i}\}$
must converge to a totally umbilical hypersurface, that is, a radial coordinate slice as
$t_i\rightarrow \infty .$
Note that by the proof of Lemma 4.1, we can see that both
$\rho _{\min }$
and
$\rho _{\max }$
are monotonic. Therefore, we conclude that the limiting radial coordinate slice is unique.
In sum, we have the flow hypersurfaces
$M_t$
converge to a radial coordinate slice smoothly as
$t\rightarrow \infty ,$
and the inequality
$$ \begin{align} \int_M\sigma_2d\mu_g-(n-1)\mathcal A_0\leq\xi_{2, 0}(\mathcal A_0),\end{align} $$
follows from Lemma 3.1 easily. Moreover, the equality holds if and only if M is a radial coordinate slice. Following the argument in [Reference Guan and Li8], when M is
$2$
-convex instead of strictly
$2$
-convex, we may approximate it by strictly
$2$
-convex star-shaped hypersurfaces. The inequality (5.2) follows from the approximation. Therefore, we complete the proof of Corollary 1.3.
