text:
\section{Introduction}
For $n$-dimensional hypersurfaces in Euclidean space, it is natural to understand the relations between the intrinsic curvature and the extrinsic curvature.
In 1897, Hadamard~\cite{H} proved that a closed (i.e., compact without boundary) surface embedded in $\mathbb{R}^3$ of positive Gaussian curvature is a convex surface.
Hadamard's result was extended by Stoker~\cite{St} to the complete non-compact case.

In contrast to the strict inequality assumption on the curvature, the non-strict inequality case is more subtle.
About sixty years later, Chern--Lashof~\cite{CL} was able to prove that Hadamard's theorem remains valid if the positive curvature is replaced by the nonnegative curvature.

For $n \ge 2$, the statement that a hypersurface with nonnegative sectional curvature whose second fundamental form is semi-positive definite was completely proved by Sacksteder~\cite{S}, combined with the earlier results of van~Heijenoort~\cite{V} and Hartman--Nirenberg~\cite{HN}. A simpler proof was later provided by do Carmo--Lima~\cite{DL}. The further study of convex hypersurfaces can be found in, for example, H. Wu~\cite{HWu} and the references therein.

Among various notions of the intrinsic curvature, the sectional curvature is the strongest (pointwise) curvature condition, while the scalar curvature is the weakest. In this paper, we consider the condition only on the scalar curvature without imposing any condition on the sectional curvature. We would like to know what kind of convexity can be implied by the nonnegative scalar curvature.

Our another motivation comes from the study of the $k$-th mean curvature. For an $n$-dimensional hypersurface, its $k$-th mean curvature ($1 \le k \le n$), denoted by $\sigma_k$, is defined to be the $k$-th symmetric polynomial of its principal curvatures. It is known that $\sigma_{2k}$ is intrinsically defined, while $\sigma_{2k-1}$ is not, for each $k$ (see~\cite{Reilly}, for example). In particular, $\sigma_1$ is the mean curvature, $2 \sigma_2$ is the scalar curvature, and $\sigma_n$ is the Gauss--Kronecker curvature.

If a closed smooth hypersurface has positive $k$-th mean curvature, then its $l$-th mean curvature is positive for each $1 \le l \le k$ (see, for example, \cite[Proof of Proposition 3.3]{HS} and \cite{CNSIII}). In particular, when $k=2$, the result follows from the Gauss equation. A conjecture proposed by Huisken and Sinestrari~\cite[Proposition 3.3]{HS} asks whether the analogous result holds when one replaces the condition $\sigma_k > 0$ by $\sigma_k \ge 0$. However, this conjecture turns out not true, for all $k \ge 3$. We construct in Section~\ref{se:example} a family of examples with $\sigma_k \ge 0$ but $\sigma_1 < 0$ somewhere, for $k \ge 3$. These examples are inspired by Chern--Lashof~\cite{CL}.

In contrast to the counter-examples for all $k\ge 3$,  the conjecture holds for $k = 2$. More precisely, we have the following result. Throughout this article, we assume that hypersurfaces are embedded and orientable.
\begin{THM} \label{thm:R}
Let $n\ge 2$ and $M$ a closed embedded $n$-dimensional  $C^{n+1}$ hypersurface in $\mathbb{R}^{n+1}$. If the scalar curvature of $M$ is nonnegative, then the mean curvature $H$ has a sign, i.e., either $H\ge 0$ or $H\le 0$ everywhere on $M$.
\end{THM}

The main analytic difficulty is that several natural geometric differential equations, including the linearized scalar curvature equation and scalar curvature flow, may be fully degenerate at points of vanishing mean curvature and cease to be globally elliptic. A new ingredient in the proof of Theorem~\ref{thm:R} is that we consider the level sets of the height function on the hypersurface. We derive a geometric inequality which relates the mean curvature and scalar curvature of $M$ to the mean curvature of the level sets  (see Theorem \ref{th:HHR}). Therefore, the geometry of $M$ has some quantitative influence on the geometry of its level sets. This inequality is rather general and holds at each point. We then carefully investigate the geometry of the level sets, and establish a key lemma (Lemma~\ref{le:Hsign}). This enables us to obtain the following more general result for a complete hypersurface (not necessarily closed), which may have interests of its own.

\begin{THM} \label{th:unbounded}
Let $n\ge 2$ and $M$ a complete embedded $n$-dimensional $C^{n+1}$ hypersurface in $\mathbb{R}^{n+1}$ with nonnegative scalar curvature. Let $H$ be the mean curvature of $M$. If there exists a connected component of $\{p\in M: H(p) =0 \}$ separating the subsets $\{p\in M: H(p) > 0 \}$ and $\{p\in M: H(p) < 0 \}$ in $M$, then the component must be unbounded.
\end{THM}

As an application of Theorem~\ref{thm:R}, we show that nonnegative scalar curvature is preserved by the mean curvature flow. More precisely, we have the following result.

\begin{THM} \label{th:MCF}
Let $n\ge 2$ and $M$ a closed embedded $n$-dimensional $C^{n+1}$ hypersurface in $\mathbb{R}^{n+1}$ with nonnegative scalar curvature. Let $\{ M_t\}$ be a solution to the mean curvature flow with initial hypersurface $M$. Then, the scalar curvature of $M_t$ is strictly positive for all $t>0$.
\end{THM}

 Moreover, by using Theorem~\ref{th:unbounded}, we can provide a simple proof to Sacksteder's theorem for the case of closed hypersurfaces (see Theorem~\ref{th:Sa}). We expect the geometric inequality to have further applications. For example, in a forthcoming work~\cite{HW3}
 we shall prove some rigidity results for hypersurfaces in sphere, parallel to our previous work~\cite{HW1} in non-positive space form. (We refer readers to the excellent survey by Brendle~\cite{Brendle} and the references therein, for the recent rigidity results involving scalar curvature.)

For a general non-closed hypersurface with nonnegative scalar curvature, the mean curvature may change signs locally. For example, consider the $n$-dimensional graph in $\mathbb{R}^{n+1}$ defined by the function $f(x^1,\dots, x^n) = (x^n)^3$ over the unit ball centered at the origin in $\mathbb{R}^n$. The scalar curvature of the graph is zero, but its mean curvature is strictly positive when $x^n>0$ and strictly negative when $x^n<0$.  Theorem~\ref{thm:R} implies that the piece of hypersurface cannot be extended to a closed hypersurface with nonnegative scalar curvature.

By extending the argument in the proof of Theorem~\ref{th:unbounded}, we generalize Theorem \ref{thm:R} to complete non-compact hypersurfaces of finitely many regular ends (see Definition~\ref{de:waf}).
\begin{THM} \label{th:halfspace}
Let $n\ge 2$ and $M$ a complete embedded $n$-dimensional $C^{n+1}$ hypersurface in $\mathbb{R}^{n+1}$ of finitely many regular ends with scalar curvature $R\ge 0$. Then the mean curvature $H$ of $M$ has a sign, i.e. either $H\ge 0$ or $H\le 0$ on $M$.
\end{THM}

Furthermore, we prove a positive mass theorem for such hypersurfaces for all $n\ge 2$. For $n=3$ and for asymptotically flat manifolds with nonnegative scalar curvature, the positive mass theorem was proved by Schoen--Yau~\cite{SY79, SY81} and Witten~\cite{W}. The proofs have been generalized to asymptotically flat manifolds of dimension $3 \le n \le 7$ or to spin manifolds of dimension $n\ge 3$. For higher dimensional non-spin manifolds, some approaches  have been announced by Lockhamp~\cite{Lo} and by Schoen~\cite{Schoen}, which involve very delicate analysis. Recently, Lam~\cite{L} proved the positive mass inequality for graphical asymptotically flat hypersurfaces for all $n\ge 2$. See Bray~\cite{Bray} for a thorough and up-to-date survey article on Riemannian positive mass theorem. Our result is as follows:

\begin{THM} \label{thm:PMT}
Let $n\ge 2$ and $M$ a complete embedded $n$-dimensional $C^{n+1}$ hypersurface in $\mathbb{R}^{n+1}$ of finitely many regular ends with nonnegative scalar curvature. Then, the mass on each end is nonnegative. Moreover, if $M$ is connected and the mass of one end is zero, then $M$ is identical to a hyperplane.
\end{THM}

Notice that in Theorem~\ref{thm:PMT} we impose \emph{no} decay condition on the induced metric of the ends. Generally, the mass defined by \eqref{eq:mass} may be $+\infty$. In Lemma~\ref{le:mass}, we show that the mass is finite and coincides with the classical definition of the ADM mass if the growth rate of the end is controlled. Our conditions are rather general and include interesting examples such as $n$-dimensional Schwarzschild manifolds embedded in $\mathbb{R}^{n+1}$ with two ends (see  Example~\ref{ex:schwarzschild}). We remark that although hypersurfaces in $\mathbb{R}^{n+1}$ are spin, Theorem~\ref{thm:PMT} holds under more general asymptotics and does not seem to be a special case of the positive mass theorem for spin manifolds.

The article is organized as follows. In Section~\ref{se:level}, we prove the geometric inequality (Theorem~\ref{th:HHR}).
In Section \ref{se:thm1}, we prove Theorem~\ref{thm:R} and Theorem~\ref{th:unbounded}, and apply the results to study the mean curvature flow. In addition, we give a simple proof to Sacksteder's theorem for closed hypersurfaces.
In Section~\ref{se:example}, we provide the examples of non-mean convex hypersurfaces satisfying $\sigma_k \ge 0$, for all $k\ge 3$. Finally, Theorem~\ref{th:halfspace} and Theorem~\ref{thm:PMT} are proven in Section~\ref{se:thm2}.

\begin{ack}
We thank Brian White for discussions in an early stage of this work, and Huai-Dong Cao, Dan Lee and Xiao Zhang for  valuable comments after the first version appeared on arXiv. The first author thanks Rick Schoen and Mu-Tao Wang for their encouragement and acknowledges NSF grant DMS-$1005560$ for partial support. The second author would like to thank The Ohio State University for support.
\end{ack}

\section{The mean curvature of the level sets} \label{se:level}
Let us begin with a linear algebra identity, which applies to a real matrix not necessarily being symmetric.
\begin{prop} \label{pr:id}
  Let $A = (a_{ij})$ be an $n \times n$ matrix with $n \ge 2$. Denote
\[
   \sigma_1(A) = \sum_{i=1}^n a_{ii}, \quad \sigma_1(A|1) = \sum_{i=2}^n a_{ii}, \quad \sigma_2(A) = \sum_{1 \le i < j \le n} (a_{ii} a_{jj} - a_{ij}a_{ji}).
\]
  Then, we have
  \begin{equation} \label{eq:id1}
    \begin{split}
    \sigma_1(A) \sigma_1(A|1)
    & = \sigma_2(A) + \frac{n}{2(n-1)} [\sigma_1(A|1)]^2 + \sum_{1 \le i < j \le n} a_{ij} a_{ji} \\
    & \quad + \frac{1}{2(n-1)} \sum_{2 \le i < j \le n} (a_{ii} - a_{jj})^2,
    \end{split}
  \end{equation}
  where the last term is zero when $n = 2$.
  In particular, if $A$ is real and $a_{ij}a_{ji} \ge 0$ for all $1 \le i< j \le n$, then
  \[
     \sigma_1(A) \sigma_1(A|1) \ge \sigma_2(A) + \frac{n}{2(n-1)} [\sigma_1(A|1)]^2,
  \]
  where ``='' holds if and only if $a_{22} = \cdots = a_{nn}$ and $a_{ij} a_{ji} = 0$ for all $i,j = 1,\ldots, n$ and $i \ne j$
\end{prop}
\begin{proof}

   Note that
   \[
      \sigma_2(A) = a_{11}\sigma_1(A|1) + \sum_{2 \le i < j \le n} a_{ii} a_{jj} - \sum_{1 \le i <j \le n} a_{ij} a_{ji}.
   \]
   Then,
   \begin{equation} \label{eq:id2}
     \begin{split}
      \sigma_1(A) \sigma_1(A|1)
     & = a_{11} \sigma_1(A|1) + \sigma_1(A|1)^2 \\
     & = \sigma_2(A) + \sum_{1\le i <j \le n} a_{ij} a_{ji} + \sigma_1(A|1)^2 - \sum_{2 \le i < j \le n} a_{ii} a_{jj}.
   \end{split}
  \end{equation}
   Now \eqref{eq:id1} follows from applying
   \[
   		\frac{n-2}{2(n-1)}\Big(\sum_{j=2}^n a_{jj}\Big)^2 - \sum_{2 \le i < j \le n} a_{ii} a_{jj} \\
     = \frac{1}{2(n-1)} \sum_{2 \le i < j \le n} (a_{ii} - a_{jj})^2
   \]
   to the last term on the right hand side of \eqref{eq:id2}.
\end{proof}

Let $M$ be a complete $C^{2}$ hypersurface in $\mathbb{R}^{n+1}$. Assume that $M$ is embedded and orientable. Consider the height function $h : M \rightarrow \mathbb{R}$ given by
\[
   h(x^1, \dots, x^{n+1})= x^{n+1}.
\]
By Morse--Sard theorem (see~\cite{Sard} for example), for almost every height value, the level set of $h$ is an $(n-1)$-dimensional $C^2$ manifold in $\mathbb{R}^n$. Without loss of generality, we assume that $0$ is a regular value of $h$, i.e.,
\[
	\Sigma = M \cap \{ x^{n+1} = 0\}
\]
is a $C^{2}$ hypersurface in $\mathbb{R}^n$, and $|\nabla^M h| \ne 0$ for every point in $\Sigma$.

We shall adopt the following convention for the mean curvature. Let $N$ be a (piece) of hypersurface in Euclidean space. Let $\mu$ be a non-vanishing unit normal vector field to $N$. The mean curvature of $N$ defined by $\mu$ is given by
\[
	H_N = - \mbox{div}_0 \mu,
\]
where $\mbox{div}_0$ is the Euclidean divergence operator. The $n$-dimensional sphere of radius $r$ has positive mean curvature $n/r$ with respect to the inward unit normal vector by this convention.

Moreover, we denote by $\langle \cdot, \cdot \rangle$ the standard metric on $\mathbb{R}^{n+1}$, and denote by $\partial_1,\ldots, \partial_{n+1}$ the tangent vectors with respect to $(\mathbb{R}^{n+1}; x^1,\ldots, x^{n+1})$. For a $C^2$ function $f$, we abbreviate $f_i = \partial f/\partial x^i$, $f_{ij} = \partial^2 f/ \partial x^i \partial x^j$, and denote $Df = (f_1,\ldots, f_n)$.

\begin{thm} \label{th:HHR}
Let $M$ and $h$ be given as before, and $\Sigma = h^{-1}(0)$ with $| \nabla^M h | > 0$ on $\Sigma$. Denote by $\nu$ and $\eta$ the unit normal vector fields to $M \subset \mathbb{R}^{n+1}$ and $\Sigma \subset \mathbb{R}^n$, respectively; and denote by $H$ and $H_{\Sigma}$ the mean curvatures of $M\subset \mathbb{R}^{n+1}$ and $\Sigma \subset \mathbb{R}^n$ defined by $\nu$ and $\eta$, respectively. Let $R$ be the induced scalar curvature of $M$. Then,
\begin{equation} \label{eq:HHR}
	  \langle \nu, \eta \rangle H H_{\Sigma} \ge \frac{R}{2} + \frac{n}{2(n-1)} \langle \nu, \eta \rangle^2 H_{\Sigma}^2 \qquad \textup{on $\Sigma$},
\end{equation}
where ``='' holds at a point in $\Sigma$ if and only if
$(M, \Sigma)$ satisfies the following two conditions at the point:
\begin{enumeratei}
  \item \label{it:S} $\Sigma \subset \mathbb{R}^n$ is umbilic, with the principal curvature $\kappa$;
  \item \label{it:M} $M \subset \mathbb{R}^{n+1}$ has at most two distinct principal curvatures, and one of them is equal to $\langle \nu, \eta \rangle \kappa$, with multiplicity at least $n-1$.
\end{enumeratei}
\end{thm}
\begin{pf}
It suffices to show \eqref{eq:HHR} at a point $p \in \Sigma$. We may assume $\langle \nu, \eta \rangle \ge 0$ at $p$. Otherwise, we can replace $\eta$ by $-\eta$.

Let us divide the proof into two cases:

  \textbf{Case 1}: Assume that $\langle \nu, \eta \rangle < 1$ at $p$.  Since $\Sigma = M \cap \{x^{n+1} = 0\}$, we have in this case that $\langle \nu, \partial_{n+1} \rangle \ne 0$ at $p$. Then, a neighborhood $V$ of $p$ in $M$ can be represented by
   \[
      x^{n+1} = f(x), \qquad \textup{for all $x = (x^1,\ldots, x^n) \in \Omega$},
   \]
   in which $\Omega \subset \{x^{n+1} = 0 \}$ is a small domain containing $p$, and $f \in C^{2}(\Omega)$. It follows that
   \begin{equation} \label{eq:S1}
      \Sigma \cap V = \{ x \in \Omega \mid f(x) = 0 \}.
   \end{equation}
 We assume, without loss of generality, that $\langle \nu, \partial_{n+1} \rangle > 0$ at $p$; then
  \begin{equation} \label{eq:nu1}
     \nu = \frac{(-Df, 1)}{\sqrt{1+ |Df|^2} }  \qquad \textup{at $p$}.
  \end{equation}

   We remark that $|Df| \equiv \sqrt{f_1^2 + \cdots + f_n^2} > 0$ at $p$, for, by the construction we have
  \[
      h(x, f(x) ) = f(x) \qquad \textup{for all $x \in \Omega$};
  \]
  thus, $|\nabla^M h| > 0$ on $\Sigma$ implies that $|Df|> 0$ on $\Sigma \cap V$.

  We can rotate the coordinates $(x^1,\ldots, x^n)$ in $\Omega$ so that at $p$
  \[
      f_1 = |Df|, \qquad \textup{and \quad $f_{\alpha} = 0$ \quad for all $2 \le \alpha \le n$}.
  \]
  Then, the shape operator $A = (A^i_j)$ on $M \subset \mathbb{R}^{n+1}$ at $p$ is given by
  \[
     A^i_j = \frac{\partial}{\partial x^j}\Big(\frac{f_i}{w} \Big) = \frac{1}{w} \Big( f_{ij} - \frac{\delta_{i1} f_{1j} |Df|^2}{w^2} \Big),
  \]
  where $w \equiv \sqrt{1 + |Df|^2}$. Hence, at the point $p$,
  \begin{align}
     \sigma_1(A) & = H = \frac{f_{11}}{w^3} + \frac{1}{w}\sum_{\alpha \ge 2} f_{\alpha \alpha}, \notag\\
     \sigma_1(A|1) & =\sum_{\alpha \ge 2} A^{\alpha}_{\alpha} = \frac{1}{w}\sum_{\alpha \ge 2} f_{\alpha \alpha}. \label{eq:A1}
  \end{align}
   On the other hand, by \eqref{eq:S1} and \eqref{eq:nu1} we have $\eta = - Df/|Df|$ at $p \in \Sigma$; hence,
   \[
      \langle \nu, \eta \rangle = \frac{|Df|}{w} > 0 \qquad \textup{at $p$}.
   \]
   Furthermore, the shape operator $A_{\Sigma}$ on $\Sigma \subset \mathbb{R}^n$ is given by
   \begin{equation} \label{eq:AS}
				(A_{\Sigma})^i_j = \frac{\partial}{\partial x^j} \Big( \frac{f_i}{|Df|}\Big) = \frac{f_{ij}}{|Df|}, \qquad 2 \le i, j \le n.
   \end{equation}
   In particular, the mean curvature
   \begin{equation} \label{eq:HS}
       H_{\Sigma} = \sum_{i=2}^n (A_{\Sigma})^i_i = \frac{1}{|Df|}\sum_{\alpha =2}^n f_{\alpha \alpha}.
   \end{equation}
   Comparing \eqref{eq:A1} and \eqref{eq:HS} we have
   \[
     \sigma_1(A|1) = \frac{|Df|}{w} H_{\Sigma} = \langle \nu, \eta \rangle H_{\Sigma}.
   \]
   Now applying Proposition~\ref{pr:id} with $\sigma_2(A) = R/2$ yields
   \[
      \langle \nu, \eta \rangle H H_{\Sigma}  \ge \frac{R}{2} + \frac{n}{2(n-1)} (\langle \nu, \eta \rangle H_{\Sigma})^2.
   \]
   Here the equality holds if and only if
   \[
      f_{22} = \cdots = f_{nn}, \qquad \textup{and \quad $f_{ij} = 0$ \quad for all $i \ne j$},
   \]
   which, by \eqref{eq:AS}, is the same as that $(M,\Sigma)$ satisfies conditions \eqref{it:S} and \eqref{it:M} at $p$. This proves the result for Case 1.

   \textbf{Case 2}: Assume that $\langle \nu, \eta \rangle = 1$ at $p$. Then, $\nu = \eta$ at $p$; equivalently,
   \[
      \langle \nu, \partial_{n+1} \rangle = 0.
   \]
   Let us assume, without loss of generality, that $\langle \nu, \partial_1 \rangle \ne 0$. We can furthermore rotate the coordinates $(x^1,\ldots, x^n)$ so that
   \[
      \nu = \partial_1 \qquad \textup{at $p$}.
   \]
   Then, by the implicit function theorem, we can represent a neighborhood $U$ of $p$ in $M$ by
   \[
      x^1 = \psi(x^2, \ldots, x^n, x^{n+1}), \qquad \textup{for all $(x^2,\ldots, x^{n+1}) \in \Omega_1$,}
   \]
   where $\Omega_1 \subset \{x^1 = 0\}$ is a small domain containing $p$, and $\psi \in C^{2}(\Omega_1)$ satisfies that
   \begin{equation} \label{eq:dpsi}
      \psi_i(p) = 0 \qquad \textup{for all $2 \le i \le n+1$}.
   \end{equation}
   Since $\Sigma \subset \{x^{n+1} = 0\}$, $\Sigma \cap U$ is given by
   \[
      x^1 = \psi(x', 0) \qquad \textup{for all $(x',0) \equiv (x^2,\ldots, x^{n}, 0) \in \Omega_1$}.
   \]
    By construction above, we have
    \[
       \nu   = \frac{(1, - D' \psi, -\psi_{n+1})}{\sqrt{1 + |D' \psi|^2 + \psi_{n+1}^2}}, \quad \textup{and \quad $\eta  = \frac{(1, - D' \psi, 0)}{\sqrt{1 + |D' \psi|^2}}$},
    \]
    where $D' \psi = (\psi_2,\ldots, \psi_n)$. Using \eqref{eq:dpsi} we obtain the shape operator $A$ for $M \subset \mathbb{R}^{n+1}$ at $p$
    \[
       A^i_j = \frac{\partial}{\partial x^j} \left(\frac{\psi_i}{\sqrt{1 + |D'\psi|^2 + \psi_{n+1}^2 }} \right) = \psi_{ij}, \qquad 2 \le i,j \le n+1,
    \]
    while the shape operator $A_{\Sigma}$ for $\Sigma \subset \mathbb{R}^n$ at $p$ is
    \begin{equation} \label{eq:AS2}
       (A_{\Sigma})^i_j = \frac{\partial}{\partial x^{j}}\left(\frac{\psi_i}{\sqrt{1 + |D'\psi|^2 }} \right) = \psi_{ij}, \qquad \textup{for all $2 \le i, j \le n$}.
    \end{equation}
    Hence, at $p$ the matrix $A_{\Sigma}$ is exactly the first $(n-1)\times (n-1)$ principal minor of the matrix $A$. It then follows from Proposition~\ref{pr:id} that
    \[
       H_{\Sigma} H \ge \frac{R}{2} + \frac{n}{2(n-1)} H_{\Sigma}^2,
    \]
    where ``='' holds if and only if
    \[
       \psi_{22} = \cdots = \psi_{nn}, \qquad \textup{and \quad $\psi_{ij} = 0$ for all $i \ne j$},
    \]
    which is the same as that $(M,\Sigma)$ satisfies conditions \eqref{it:S} and \eqref{it:M} at $p$, by \eqref{eq:AS2}. This proves the result for Case 2. Combining the two cases, we finish the proof. \qed
\end{pf}

\begin{cor} \label{co:HHR}
  With the notations in Theorem~\ref{th:HHR}, if $R\ge 0$ on $M$, then
  \[
     \langle \nu, \eta \rangle H H_{\Sigma} \ge  \frac{n}{2(n-1)} \langle \nu, \eta \rangle^2 H_{\Sigma}^2 \qquad \textup{on $\Sigma$}.
  \]
  In particular, $\langle \nu, \eta \rangle  H H_{\Sigma} \ge 0$ at the point; in addition, if $H= 0$, then $H_{\Sigma} = 0$, and both $M$ and $\Sigma$ are geodesic at the point.
\end{cor}

\section{Complete hypersurfaces with nonnegative scalar curvature} \label{se:thm1}

Let us first prove some results on the mean curvature operator. Let $f$ be a $C^2$-function defined over an open set in $\mathbb{R}^n$. The upward unit normal vector of the graph of $f$ is
\begin{align} \label{eq:nu}
	\nu = \frac{( -Df, 1) }{\sqrt{1+ |Df|^2}}.
\end{align}
The mean curvature operator is defined by
\[
	H(f) : = - \mbox{div}_0 \nu= \sum_{i,j=1}^n \left( \delta_{ij} - \frac{f_i f_j}{ 1+ |Df|^2} \right) \frac{f_{ij} }{ \sqrt{1+|Df|^2}}.
\]

\begin{prop} \label{pr:nonempty}
Let $W$ be an open subset in $\mathbb{R}^n$, not necessarily bounded. Let $p\in \partial W$, and denote by $B(p)$ a small open ball in $\mathbb{R}^n$ centered at $p$. Suppose that $f \in C^2(  W\cap B(p)) \cap C^1( \overline{W} \cap B(p))$ satisfies
\begin{align*}
	& H(f) \ge 0\quad &&\mbox{in } W  \cap B(p)   \\
	&f= c, \; |Df|=0\quad &&\mbox{on } \partial W  \cap B(p),
\end{align*}
for some constant $c$. Then either $f\equiv c$ in $W  \cap B(p)$, or
\[
	\{ x \in  W  \cap B(p): f(x) > c \} \neq \emptyset.
\]
\end{prop}
\begin{remark}
Notice that we impose no hypothesis on regularity of $\partial W$ here and below.
\end{remark}
\begin{pf}
Without loss of generality, we can assume $c=0$. We prove by contradiction. Suppose that $f \le 0$ and $f$ is not identically zero on $W\cap B(p)$. If $f = 0$ at  $x_0 \in W \cap B(p)$, then $x_0$ is a local maximum of $f$. This contradicts the strong maximum principle applied to $H(f)\ge 0$. Thus, $f$ must be strictly negative in $W\cap B(p)$. Because $W\cap B(p)$ is open,
we can take a smaller open ball $B$ contained in $ W\cap B(p)$ such that $\partial B$ touches $\partial W \cap B(p)$ at a point $q$; in other words,
$\partial W$ satisfies an interior sphere condition at $q \in \partial W$. Furthermore, $f < 0$ on $B$ and $f(q) = 0$.  Applying the Hopf boundary lemma to the mean curvature operator $H(f)$ yields that $|Df|(q) \neq 0$, which contradicts with the assumption that $|Df|(q) = 0$ for $q \in \partial W \cap B(p)$.
\qed
\end{pf}

\begin{defi}
Let $W$ be a subset in $\mathbb{R}^n$. A point $p\in \partial W$ is called a convex point of $W$, if there exists an $(n-1)$-sphere $S$ in $\mathbb{R}^n$ passing through $p$ so that  $\overline{W}\setminus \{ p \}$ is contained in the open ball enclosed by $S$. Here $\overline{W}$ denotes the closure of $W$ in $\mathbb{R}^n$.
\end{defi}

\begin{remark}
A bounded subset in $\mathbb{R}^n$ has at least one convex point.
\end{remark}

\begin{lemma} \label{le:Hsign}
Let $W$ be an open subset in $\mathbb{R}^n$, not necessarily bounded. Suppose that $p \in \partial W$ is a convex point of $W$. Denote by $B(p)$ an open ball in $\mathbb{R}^n$ centered at $p$. Suppose that $f\in C^2 (W\cap B(p))\cap C^1 ( \overline{W} \cap B(p))$  and $f = c, |Df|=0$ on $\partial W \cap B(p)$ for some constant $c$. If the scalar curvature of the graph of $f$ is nonnegative, then $H(f)$ must change signs in $W\cap B(p)$, unless $f\equiv c$ on $W\cap B(p)$.
\end{lemma}
\begin{figure}[top]
   \centering
   \includegraphics[width=0.5\textwidth]{level_set.png}
   \caption{As in the proof of Lemma~\ref{le:Hsign}, $W$ is an open subset in $\mathbb{R}^n$, $p$ is a convex point of $W$, and $S$ is an $(n-1)$-sphere through $p$. For any small ball $B(p)$ centered at $p$, $\{ x \in W \cap B(p): f(x) > c \} \neq \emptyset$, unless $f\equiv c$. Hence the level set $\Sigma_{c+\epsilon}$ of $f$ is nonempty for $\epsilon > 0$ small. The shaded region represents the set $\{x\in W \cap B(p): f<c\}$ (possibly empty). }
\end{figure}
\begin{pf}
It suffices to prove the lemma for $c=0$. Suppose on the contrary that $H(f)$ does not change signs over  $W\cap B(p)$. Then, we may assume that $H(f) \ge 0$ over $ W\cap B(p)$, for, otherwise, we can replace $f$ by $-f$.

Because $p$ is a convex point, there exists an $(n-1)$-sphere $S$ through $p$ so that $\overline{W} \setminus \{ p \}$ is contained in the open ball enclosed by $S$.
Consider the level sets of $f$ inside $W\cap B(p)$
\[
	\Sigma_{\epsilon} = \{ x \in W \cap B(p): f(x) =\epsilon \}.
\]
By Proposition \ref{pr:nonempty}, $\Sigma_{\epsilon}$ is non-empty for all $\epsilon>0$ small enough.   By Morse--Sard theorem, for almost every $\epsilon$, $\Sigma_{\epsilon}$ is a piece of $C^2$ hypersurface in $\mathbb{R}^n$. Note that either $\Sigma_{\epsilon}$ is closed, or  $\partial \Sigma_{\epsilon}$ is nonempty and is contained in $W \cap \partial B(p)$.

We pick a sufficiently small $\epsilon$ so that $|Df|$ never vanishes on $\Sigma_{\epsilon}$, and the distance from the level set $\Sigma_{\epsilon}$ to the $(n-1)$-sphere $S$ is less than the distance from $\overline{W}\cap \partial B(p)$ to $S$. This can be done because $p$ is a convex point of $W$. Now we continuously translate the $(n-1)$-sphere $S$ toward $\Sigma_{\epsilon}$ along its inward normal at $p$, until ultimately it begins to intersect $\Sigma_{\epsilon}$. Denote by $S'$ the resulting $(n-1)$-sphere. Then, $S'$ must be tangent to $\Sigma_{\epsilon}$ at an interior point $x_0$, and $\Sigma_{\epsilon}$ lies in the ball enclosed by $S'$. Let $\eta =  Df/ |Df|$ be the normal vector to $\Sigma_{\epsilon}$ in $\mathbb{R}^n$ and $H_{\Sigma_{\epsilon}}$ the mean curvature of $\Sigma_{\epsilon}$ with respect to $\eta$. Note that $\eta$ at $x_0$ is pointing inward to $W$, because $f = 0$ on $\partial W \cap B(p)$ and $x_0$ is the intersection of $\Sigma_{\epsilon}$ and the $(n-1)$-sphere $S'$ for the first time. By comparison principle, $H_{\Sigma_{\epsilon}} > 0$ at the tangent point $x_0$.  On the other hand, since the scalar curvature of the graph of $f$ is nonnegative, it follows from Corollary \ref{co:HHR} that
\[
	\langle \nu, \eta \rangle H H_{\Sigma_{\epsilon}} \ge 0.
\]
By \eqref{eq:nu}, $\langle \nu, \eta \rangle < 0$ on $\Sigma_{\epsilon}$. Therefore, $H_{\Sigma_{\epsilon}} \le 0$ at $x_0$. This leads to a contradiction.
\qed
\end{pf}

\begin{lemma} \label{le:maximum-principle}
Denote by $B_r$ the open ball in $\mathbb{R}^n$ centered at the origin of radius $r$. Let $f\in C^2 (B_{r_2}\setminus \overline{B}_{r_1})\cap C^1 (\overline{B_{r_2}\setminus B_{r_1}})$ for some $r_2 > r_1 > 0$. Suppose that $f$ satisfies $H(f) \ge 0$ and the scalar curvature of the graph of $f$ is nonnegative. Then
\[
	 \max_{\overline{B}_{r_2} \setminus B_{r_1}} f =\max_{\partial B_{r_2}} f .
\]
Moreover, if $\displaystyle f(x) = \max_{\partial B_{r_2}} f $ for some interior point $x \in B_{r_2}\setminus \overline{B}_{r_1}$, then $f \equiv  \sup_{\partial B_{r_2}} f$ in $\overline{B_{r_2}\setminus B_{r_1}}$.
\end{lemma}
\begin{remark}
Lemma~\ref{le:maximum-principle} does not follow directly from applying the standard maximum principle to $H(f) \ge 0$, since we impose no hypothesis on $\max_{\partial B_{r_1}} f$. This result may have interests of its own.
\end{remark}
\begin{pf}
By subtracting $\max_{\partial B_{r_2}} f$ from $f$,  we may assume $\max_{\partial B_{r_2}} f  =0$. Suppose on the contrary that $f$ is not identically zero and $f>0$ somewhere in $B_{r_2} \setminus B_{r_1}$. Then the level set
\[
	\Sigma_{\epsilon} = \{ x \in B_{r_2} \setminus B_{r_1}: f(x) =\epsilon \}
\]
is non-empty for $\epsilon>0$ sufficiently small.   By Morse--Sard theorem, for almost every small $\epsilon$, $\Sigma_{\epsilon}$ is a piece of $C^{2}$ hypersurface in $\mathbb{R}^n$. Note that either $\Sigma_{\epsilon}$ has no boundary, or $\partial \Sigma_{\epsilon}$ is contained in $\partial B_{r_1}$. Fix $\epsilon>0$ so that $|Df|$ does not vanish on $\Sigma_{\epsilon}$. Let $p \in \partial B_{r_2}$ be a point that is closest to $\Sigma_{\epsilon}$. Now we continuously translate the $(n-1)$-sphere $\partial B_{r_2}$ toward $\Sigma_{\epsilon}$, along its inward normal at $p$. Denote by $S'$ the $(n-1)$-sphere that touches $\Sigma_{\epsilon}$ for the first time. Then, $S'$ must be tangent to $\Sigma_{\epsilon}$ at an interior point $x_0$, and $\Sigma_{\epsilon}$ lies in the ball enclosed by $S'$.  The rest of proof is similar to the proof of Lemma~\ref{le:Hsign}, so we only highlight the key argument. By comparison principle, the mean curvature $H_{\Sigma_{\epsilon}}$ of $\Sigma_{\epsilon}$ with respect to the inward unit normal vector $Df/ |Df|$ is positive at $x_0$. However, it contradicts $H_{\Sigma_{\epsilon}} \le 0$ at $x_0$ from Corollary~\ref{co:HHR}.

Moreover, if $f(x) =  \max_{\partial B_{r_2}} f $ for some interior point $x$, then by strong maximum principle and $H(f) \ge 0$ we prove the second statement.
\qed
\end{pf}

In the rest of this section, we let $M$ be a complete, embedded and orientable $C^{n+1}$-smooth hypersurface in $\mathbb{R}^{n+1}$, unless otherwise indicated.  We denote by $R$ and $H$ the scalar curvature and mean curvature of $M$, respectively, and denote by $A$ the second fundamental form of $M$. We also abbreviate $\{ H>0 \} = \{ q \in M: H(q) > 0 \}$ and $\{ H< 0 \} = \{ q \in M: H(q) < 0 \}$.

Let $M_0$ be the set of geodesic points in $M$, i.e.,
\[
	M_0 = \{ q\in M : A(q) = 0\}.
\]
Suppose that $M$ has nonnegative scalar curvature.  Then, by Gauss equation,
  \[
    H^2 - |A|^2 = R \ge 0
  \]
implies that a point in $M$ with $H=0$ is also a geodesic point; that is,
\[
	M_0 = \{ q \in M: H(q) = 0 \}.
\]

We recall the following characterization of the set of geodesic points, due to Sacksteder~\cite{S}. For the sake of completeness, we include his proof. This is the only place that requires the hypersurface $M$ to be $C^{n+1}$, instead of $C^2$, in our main theorems (Theorems~\ref{thm:R} to \ref{thm:PMT}).
Notice that the assumption that $R \ge 0$ is not needed in the proof of Lemma~\ref{le:geodesic}.
\begin{lemma}\label{le:geodesic}
Suppose that $M$ is a $C^{n+1}$ hypersurface in $\mathbb{R}^{n+1}$. Let $M_0'$ be a connected component of $M_0$. Then $M_0'$ lies in a hyperplane which is tangent to $M$ at every point in $M_0'$.
\end{lemma}
\begin{pf}
We consider the Gauss map $\nu: M \rightarrow \mathbb{S}^n$. Since $M$ is of $C^{n+1}$, the Gauss map is of $C^n$. Note that the Gauss map $\nu$ has rank zero at any geodesic point. We can then apply a theorem of Sard~\cite[p. 888, Theorem 6.1]{Sard} to obtain that the image $\nu(M_0)$ has Hausdorff dimension zero in $\mathbb{S}^n$. It follows that $\nu(M_0)$ is totally disconnected in $\mathbb{S}^n$. Thus, $\nu(M_0')$ consists of a single point in $\mathbb{S}^n$, denoted by $\nu_0$.

It remains to show that $M_0'$ lies in a hyperplane which is orthogonal to $\nu_0$. Pick a point $p_0 \in M_0'$. Let $(V; y^1,\ldots,y^n)$ be a local coordinate chart centered at $p_0$ in $M$, and define
\[
   \varphi(y) = \langle \nu_0, x(y) - x(0) \rangle, \qquad \textup{for each $y = (y^1,\ldots,y^n) \in V$}.
\]
Here $x = (x^1,\ldots, x^{n+1})$ is the coordinates in $\mathbb{R}^{n+1}$, and $\langle \cdot, \cdot \rangle$ is the Euclidean metric on $\mathbb{R}^{n+1}$.
Then, the function $\varphi \in C^{n+1}(V)$, and by our construction,
\[
   M_0' \cap V \subset \{ y \in V \mid \frac{\partial \varphi}{\partial y^i}(y) = 0, \; i = 1,\ldots,n \}.
\]
It follows from a theorem of A. P. Morse~\cite[p. 70, Theorem 4.4]{APM} that $\varphi$ is a constant on $M_0' \cap V$; thus, $\varphi \equiv 0$ on $M_0' \cap V$. Since $M_0'$ is connected,
\[
   \langle \nu_0, x(p) - x(p_0) \rangle = 0 \qquad \textup{for all $p \in M_0'$},
\]
namely, $M_0'$ lies in the hyperplane orthogonal to $\nu_0$.
\qed
\end{pf}

\begin{defi}
Let $N$ be a complete manifold and $N_1$ and $N_2$ be two nonempty disconnected open subsets of $N$. For a closed subset $E$ in $N$, we say that $E$ separates $N_1$ and $N_2$ in $N$, if there exists an open neighborhood $V$ of $E$ in $N$ so that $V \setminus E = N_1 \cup N_2$.
\end{defi}

Now we are in a position to prove Theorem~\ref{th:unbounded}.
\begin{pfthunbounded}
Suppose on the contrary that $M_0'$ is a bounded connected component of $M_0$ and that $M_0'$ separates the subsets $\{ H > 0 \}$ and $\{ H < 0 \}$ in $M$. By Lemma \ref{le:geodesic}, $M_0'$ lies in a tangent plane to $M$. Assume that the tangent plane is $\{ x^{n+1} = 0 \}$. Then, near $M_0'$ the hypersurface $M$ can be represented as the graph of a $C^{n+1}$-function $f$ in a neighborhood of $M_0'$ in $\{ x^{n+1} = 0\}$, which satisfies $f=0, |Df|=0$ on $M_0'$.
Because $M$ is complete and $M_0'$ separates $\{ H > 0 \}$ and $\{ H < 0 \}$ in $M$, $M_0'$ also separates regions in $\{ x^{n+1} = 0 \}$. We denote by $W$ one of the connected open subsets in $\{x^{n+1}=0\}$ enclosed by $M_0'$. Then, $\partial W \subset M_0'$, and $W$ has a convex point $p \in \partial W$. For a small open ball $B(p)$, we have either $H(f)>0$, or $H(f)<0$ in $W \cap B(p)$. However, it contradicts Lemma \ref{le:Hsign}.
\qed
\end{pfthunbounded}

The proof of Theorem~\ref{thm:R} follows directly from Theorem~\ref{th:unbounded}.
\begin{pf1}
Because $M$ has nonnegative scalar curvature, the set of points where $H=0$ is equal to $M_0$, the set of geodesic points. Suppose on the contrary that $H$ changes signs. Then, a connected component of $M_0$ would separate the sets $\{ H > 0 \} $ and $\{H<0\}$. The component is bounded because $M$ is closed. This contradicts Theorem~\ref{th:unbounded}.
\qed
\end{pf1}

Using Theorem~\ref{th:unbounded}, we provide another proof to Sacksteder's theorem for closed hypersurfaces.
\begin{thm} \label{th:Sa}
Suppose that $M$ is a $C^{n+1}$-smooth closed hypersurface in $\mathbb{R}^{n+1}$. If the sectional curvature of $M$ is nonnegative, then the induced second fundamental form is semi-positive definite. As a consequence, $M$ is the boundary of a convex body in $\mathbb{R}^{n+1}$.
\end{thm}
\begin{pf}
Let $(A_{ij})$ be the second fundamental form of $M$. Then, at a point in $M$, the principal curvatures are either all nonnegative or all non-positive, because the sectional curvature of $M$ is nonnegative. Suppose on the contrary that $(A_{ij})$ is not semi-positive definite. Then, there exists a connected component $M_0'$ of $M_0$, which separates the sets $\{ (A_{ij} ) > 0) \} $ and $\{ (A_{ij} ) < 0\}$ in $M$. In particular, $M_0'$ separates the sets $\{ H > 0 \}$ and $\{ H < 0 \}$ in $M$. Since $M$ is compact, $M_0'$ is necessarily bounded. This contradicts with Theorem~\ref{th:unbounded}.
\qed
\end{pf}

Theorem~\ref{thm:R} can be applied to study the mean curvature flow, especially when the initial hypersurface is closed and has nonnegative scalar curvature. Let us briefly recall the setting of the mean curvature flow, and then prove Theorem \ref{th:MCF}. Let $M$ be a closed hypersurface in $\mathbb{R}^{n+1}$ represented by a diffeomorphism: For  an open subset $U \subset \mathbb{R}^n$,
\[
	F_0 : U \rightarrow F_0(U) \subset M \subset \mathbb{R}^{n+1}.
\]
Let $F(x,t)$ be a family of maps satisfying
\begin{align*}
	& \frac{\partial}{\partial t} F(x,t) = H(x,t)\nu(x,t), \quad x \in U,\\
	& F(\cdot, 0) = F_0,\notag
\end{align*}
where $\nu(\cdot,t)$ is the inward unit normal vector to $M_t:= F(M, t)$ and $H(\cdot, t)$ is the mean curvature with respect to $\nu$. The family of closed hypersurfaces $\{ M_t\}$ for $t > 0$ is called a solution to the mean curvature flow.

\begin{pfmcf}
By Theorem \ref{thm:R}, the mean curvature of $M$ is nonnegative. It is well known that
if $M$ has nonnegative mean curvature, then $M_t$ has positive mean curvature for all $t>0$. We then consider
\[
	q_2 : = \frac{R}{2H},
\]
where $H$ and $R$ are the mean curvature and scalar curvature of $M_t$, respectively.
A result of Huisken--Sinestrari~\cite[p. 61, Corollary 3.2]{HS} shows that the evolution equation of $q_2$ satisfies the parabolic  strong  maximum principle. It follows that $q_2 > 0$ on $M_t$ for all $t>0$, because $q_2 \ge 0$ on $M$. Thus, we conclude that $R>0$ on $M_t$ for all $t>0$.
\qed
\end{pfmcf}

\section{Examples of nonnegative $k$-th mean curvature} \label{se:example}
Let $M$ be a smooth closed hypersurface in $\mathbb{R}^{n+1}$. Denote by $\kappa_i$, $i=1,\ldots,n$, the principal curvatures of $M$. We define, for each $1 \le k \le n$, the \emph{$k$-th mean curvature} of $M$ to be
\[
   \sigma_k(A) = \sum_{1 \le i_1 < \cdots < i_k \le n} \kappa_{i_1}\cdots \kappa_{i_k}.
\]
In particular, $\sigma_1(A)$, $2 \sigma_2(A)$, and $\sigma_n(A)$ are the mean curvature, the scalar curvature, and the Gauss--Kronecker curvature of $M$, respectively.

It is well known that if $\sigma_k(A) > 0$ then $\sigma_l(A) > 0$ for all $1 \le l \le k$ (see, for example, \cite{CNSIII} and \cite[p. 51]{HS}). We are interested in the non-strict inequality case: Namely, the question is, \emph{whether $\sigma_k(A) \ge 0$ would imply $\sigma_l(A) \ge 0$ for all $1 \le l \le k$}?

For $k = 2$, Theorem~\ref{thm:R} tells us that $\sigma_2(A) \ge 0$ implies $\sigma_1(A) \ge 0$. However, it is no longer true for $k \ge 3$.
In fact, for any $n \ge 3$ and $k \ge 3$, we construct below a family of smooth closed hypersurfaces, which satisfy $\sigma_k(A) \ge 0$ but are not mean convex, i.e., $\sigma_1(A)$ changes signs.

\begin{ex} \label{ex:odd}
  Let $n \ge 3$ and $k$ be an odd integer such that $3 \le k \le n$. Consider the hypersurface in $\mathbb{R}^{n+1}$ given by
  \begin{equation} \label{eq:odd}
     (r - a)^2 + (x^{n+1})^2 = 1,
  \end{equation}
  where $a > 1$ is a constant, and
  \[
     r = \sqrt{(x^1)^2 + \cdots + (x^n)^2}.
  \]
  This hypersurface is homeomorphic to $\mathbb{S}^1 \times \mathbb{S}^{n-1}$, for it can be obtained by rotating a unit circle about the $x^{n+1}$-axis.

 We shall show that, for each $a \in (n/k, n)$,  $\sigma_k(A) \ge 0$ but $\sigma_1(A)$ change signs. (When $k<n$, the same conclusion holds if $a = n/k$.)
In particular, letting $a = n/2$, the hypersurface given by \eqref{eq:odd} has nonnegative $k$-th mean curvature while its first mean curvature changes signs.

   Note that the hypersurface is symmetric about $\{ x^{n+1} = 0\}$. Let us consider the lower half portion $x^{n+1} = \phi(r)$,
  where
  \[
     \phi (r) = -[ 1 - (r -a)^2 ]^{1/2}, \qquad \textup{for all $a-1 \le r \le a+1$}.
  \]
By a direct computation, the $k$-th mean curvature of the graph of $x^{n+1} = \phi(r)$ is
\begin{align*}
   \sigma_k(A)
   & = \binom{n-1}{k-1} \frac{(\phi')^{k-1}}{r^{k-1}[ 1 + (\phi')^2]^{k/2}} \left(\frac{\phi''}{1 + (\phi')^2} + \frac{n - k}{k} \frac{\phi'}{r} \right) \\
   & = \binom{n-1}{k-1} \big(1 - \frac{a}{r}\big)^{k-1} \left[1 + \frac{n-k}{k}\big(1 - \frac{a}{r}\big) \right].
\end{align*}
Since $k$ is odd, to get $\sigma_k(A)\ge 0$ it suffices to consider
\begin{equation*}

   0 \le 1 + \frac{n-k}{k} \big( 1 - \frac{a}{r}\big).
\end{equation*}
That is, when $1 \le k < n$, $\sigma_k(A) \ge 0$ if
\begin{equation*}
    1 - \frac{a}{r} \ge - \frac{k}{n - k}, \qquad \mbox{ for all } a -1 < r < a+1;
\end{equation*}
and when $k =n$ (then $n$ is odd), $\sigma_n(A)$ is always nonnegative.
On the other hand,
\[
	\frac{1}{a + 1} > 1 - \frac{a}{r} >  -\frac{1}{a-1}, \qquad \mbox{for $a -1 < r < a+1$.}
\]
Therefore, for all the real numbers $a$ satisfying
\[
   -\frac{1}{n-1} >  - \frac{1}{a - 1} > - \frac{k}{n - k}, \qquad \textup{i.e.}, \quad n > a > \frac{n}{k},
\]
we have $\sigma_k (A) \ge 0$, but $\sigma_1(A)$ changes signs. More precisely, let us fix any real number $a \in (n/k, n)$; then
\begin{align*}
	\sigma_1(A) & > 0, \quad \mbox{ for }  \frac{n-1}{n}a < r <  a + 1, \quad \textup{ and}\\
	\sigma_1(A) & < 0, \quad \mbox{ for }   a - 1 < r < \frac{n-1}{n}a.
\end{align*}
\end{ex}

\begin{ex} \label{ex:even}
Let $n \ge 4$, and $k$ an even integer satisfying $4 \le k \le n$. We consider the smooth embedded hypersurface in $\mathbb{R}^{n+1}$ given by the equation
\begin{equation} \label{eq:CLa}
   (r - a)^2 + (x^n)^2 + (x^{n+1})^2 = 1,
\end{equation}
where $a>1$ is a constant to be determined.  This hypersurface is obtained by rotating a $2$-dimensional unit sphere about a $2$-dimensional coordinate plane, and is therefore homeomorphic to $\mathbb{S}^2 \times \mathbb{S}^{n-2}$.

We would like to prove that, for any $a \in ( 1+ b(n,k) , n/2)$ the hypersurface defined by \eqref{eq:CLa} satisfies that $\sigma_k(A) \ge 0$ and $\sigma_1(A)$ changes signs, where
\[
   b(n,k) = \frac{n-k}{k-1} + \frac{1}{k-1} \sqrt{\frac{(n-1)(n-k)}{k}} \ge 0.
\]
That the interval $( 1 + b(n,k) , n/2)$ is nonempty for $n \ge k \ge 4$ is justified by Proposition~\ref{pr:a} below. (However, the interval is empty for $k =2$ and all $n \ge 2$.)

We first derive a formula for $\sigma_k(A)$ for all $n \ge 3$ and $1 \le k \le n$. Let
\[
   \psi(r, x^n) = - \sqrt{1 - (r -a )^2 - (x^n)^2}, \qquad 0 < a -1 \le r \le a + 1.
\]
By rotation symmetry, it suffices to carry out the calculation at a point where $x^1 = \cdots = x^{n-2} = 0$ and $x^{n-1} = r$. Unless otherwise indicated, we let Greek letters such as $\alpha, \beta$ range from $1$ to $n$, and English letters such as $i, j$ range from $1$ to $n-1$. We denote $\psi_{\alpha}= \partial \psi/\partial x^{\alpha}$ and $\psi_{\alpha\beta} = \partial^2 \psi/\partial x^{\alpha} \partial x^{\beta}$.

Note that the induced metric is given by
\[
   g_{\alpha \beta} = \delta_{\alpha \beta} + \psi_{\alpha} \psi_{\beta}.
\]
It follows that the matrix, at the point $(0, \dots, 0, r, x^n)$,
\[
   (g_{\alpha \beta}) = \begin{bmatrix}
   I_{n-2} & 0 \\
   0 & T_2
   \end{bmatrix}.
\]
Here $I_m$ denotes the $m \times m$ identity matrix, and $T_2$ is a $2 \times 2$ matrix defined by
\begin{align}
   T_2 & = \left[ \begin{matrix}
         1 + (\psi_{r})^2 & \psi_r \psi_n \\
         \psi_r \psi_n & 1 + (\psi_n)^2
   \end{matrix}
   \right] \notag \\
   & = (-\psi)^{-2} \left[ \begin{matrix}
         1 - (x^n)^2 & x^n (r - a) \\
         x^n (r - a) & 1 - (r - a)^2
   \end{matrix}
   \right], \label{eq:T2}
\end{align}
in which
\[
  \psi_r = \frac{\partial \psi}{ \partial r} = \frac{r - a}{- \psi}, \qquad \psi_n = \frac{\partial \psi}{ \partial x^n} = \frac{x^n}{- \psi}.
\]
On the other hand, the second fundamental form
\[
   A_{\alpha \beta} = \frac{\psi_{\alpha \beta}}{\sqrt{1 + |D \psi|^2}} = (-\psi) \psi_{\alpha \beta}.
\]
Then, at the point $(0,\ldots, 0, r, x^n)$, the second fundamental form matrix
\[
   (A_{\alpha \beta}) = \left[ \begin{matrix}
   (1 - a/r) I_{n-2} & 0 \\
   0 & T_2
   \end{matrix}
   \right],
\]
where $T_2$ is the $2 \times 2$ matrix given by \eqref{eq:T2}. Hence, the matrix of the shape operator is given by
\begin{align*}
   (A^{\beta}_{\alpha})
   = (A_{\alpha \gamma}) (g_{\beta \gamma})^{-1}
    = \left[ \begin{matrix}
   (1 - a/r) I_{n-2} & 0 \\
   0 & I_2
   \end{matrix}
   \right].
\end{align*}
Therefore, we have
\begin{equation}\label{eq:H}
   H = \sigma_1(A) = (n-2)t + 2,
\end{equation}
and for any $2 \le k \le n$,
\begin{equation} \label{eq:eg2k}
  \sigma_k(A) = \binom{n - 2}{k} t^k + 2 \binom{n-2}{k-1} t^{k-1} + \binom{n-2}{k-2} t^{k-2}.
\end{equation}
Here we denote $t = 1 - a/r$, which satisfies that
\begin{align} \label{eq:ranget}
    - \frac{1}{a - 1} <  t < \frac{1}{a + 1}, \qquad \mbox{for all $a - 1 < r < a + 1$}.
\end{align}
Moreover, in \eqref{eq:eg2k}, we use the combinatoric convention so that
  \begin{equation*}
     \sigma_n (A) = t^{n-2}, \qquad \sigma_{n-1}(A) = 2t^{n-2} + (n-2) t^{n-3}.
  \end{equation*}

Now return to our setting $n \ge k \ge 4$ and $k$ being even.
Clearly, for $k = n$ (thus $n$ is even), we always have $\sigma_n(A) \ge 0$.
For $k < n$ and $k$ being even, $\sigma_k(A) \ge 0$ if
\begin{equation} \label{eq:quad}
		\binom{n - 2}{k} t^2 + 2 \binom{n-2}{k-1} t + \binom{n-2}{k-2} \ge 0.
\end{equation}
Note that the inequality \eqref{eq:quad} holds for all $t \ge t_1$, where
\[
   t_1 = - \frac{k-1}{n - k} \left[\sqrt{\frac{n - 1}{k(n-k)}} + 1\right]^{-1} = - \frac{1}{b(n,k)}.
\]
On the other hand, by \eqref{eq:H} we have $H < 0$ for
\[
		t < - \frac{2}{n - 2}.
\]
Thus, if we can show that $t_1 < - 2/(n-2)$, i.e.,
\begin{equation} \label{eq:nt1}
   \frac{n}{2} > 1 - t_1^{-1}

   = 1 + b(n,k),
\end{equation}
then by \eqref{eq:ranget} for any real number $a$ satisfying
\begin{equation} \label{eq:defa}
   - \frac{2}{n-2} > - \frac{1}{a-1} > t_1, \qquad \textup{namely, \quad $\frac{n}{2} > a > 1 + b(n,k)$},
\end{equation}
we have that $\sigma_k(A) \ge 0$ and that $\sigma_1(A)$ changes signs; more precisely, for a fixed $a \in (1 + b(n,k),n/2)$,
\begin{align*}
   \sigma_1(A) & > 0, \qquad \textup{for $\frac{n-2}{n} a < r < a + 1$, \quad and} \\
   \sigma_1(A) & < 0, \qquad \textup{for $a - 1 < r < \frac{n-2}{n}a $}.
\end{align*}
Now notice that \eqref{eq:nt1} is assured by Proposition~\ref{pr:a} below. This finishes the proof.
We remark that, when $k<n$, the same result holds if $a = 1 + b(n,k)$.
\end{ex}

\begin{prop} \label{pr:a}
   For each $n \ge k \ge 4$,
   \[
       \frac{n}{2} >  1 + b(n,k) = 1 + \frac{n-k}{k-1} + \frac{1}{k-1} \sqrt{\frac{(n-1)(n-k)}{k}}.
   \]
\end{prop}
\begin{proof}
Observe that
  \begin{equation*}
     \frac{n}{2} - 1 - b(n,k) = c(n,k) \left\{\Big[\frac{(k-3)^2}{4} - \frac{1}{k}\Big] n + \big(k - 2 + \frac{1}{k} \big) \right\} > 0,
  \end{equation*}
  in which
  \[
     c(n,k) = \frac{n}{k-1} \left( \frac{(k-3)n}{2} + 1 + \sqrt{\frac{(n-1)(n-k)}{k}}\right)^{-1} > 0,
  \]
  for $n \ge k \ge 4$.
\end{proof}

\begin{remark}
By varying the parameter $a$ in Example~\ref{ex:odd} and Example~\ref{ex:even}, we can also provide examples of closed hypersurfaces in $\mathbb{R}^{n+1}$, which satisfy that $\sigma_k(A) \ge 0$ but $\sigma_{k-1}(A)$ changes signs, for each $3 \le k \le n$. (This in particular answers a question raised by H. D. Cao, for $k$ being even.) Therefore, Theorem~\ref{thm:R} is optimal for the Huisken--Sinestrari conjecture.
\end{remark}

\section{Positive mass theorem for hypersurfaces} \label{se:thm2}

Throughout this section, we denote by $M$ a complete non-compact,  embedded, and orientable $C^{n+1}$-smooth  hypersurface in $\mathbb{R}^{n+1}$, unless otherwise indicated. Let $M_0 \subset M$ be the set of geodesic points, i.e.,
\[
	M_0 = \{ p \in M: A(p) = 0 \},
\]
where $A $ is the induced second fundamental form on $M$. Let $H$ and $R$ be, respectively, the mean curvature and scalar curvature of $M$.
We adopt the convention that $H = - \mbox{div}_0 \nu$, where $\nu$ is a smooth unit normal vector field to $M$ and $\textup{div}_0$ is the Euclidean divergence operator. In the following, we often abbreviate $\{ H>0\} = \{ q \in M: H(q) > 0\}$ and $\{ H< 0\} = \{ q \in M: H(q) < 0\}$. For a function $f(x) = f(x^1, \dots, x^n)$, we denote  $f_i = \partial f/\partial x^i$, $f_{ij} = \partial^2 f/\partial x^i \partial x^j$ for all $i,j= 1,\ldots,n$, and $Df = (f_1,\ldots,f_n)$.

\begin{defi} \label{de:waf}
We say that $M \subset \mathbb{R}^{n+1}$ is a $C^{k}$  hypersurface of finitely many regular ends if  $M$ satisfies the following conditions:
\begin{itemize}
\item[(1)] $M$ is a complete non-compact, embedded, and orientable $C^{k}$ hypersurface in $\mathbb{R}^{n+1}$;
\item[(2)] There is a compact subset $K \subset M$ so that $M\setminus K$ consists of finitely many components $N_i$, where each $N_i$ is the graph of a function $f_{(i)}$ over the exterior of a bounded region in some hyperplane $\Pi_i$;
\item[(3)] If $\{x^1, \ldots, x^n\}$ are coordinates in $\Pi_i$, we require $\lim_{|x| \rightarrow \infty} f_{(i)}(x) = a_i$, where $a_i$ is either a bounded constant, $a_i = \infty$, or $a_i = -\infty$.
\end{itemize}
We refer $N_i$ the ends of $M$. We say that an end $N_i$ is asymptotic to the hyperplane $\Pi_i$ if $a_i=0$. By translation, whenever $|a_i| < \infty$, $N_i$ is asymptotic to a hyperplane.
\end{defi}

\begin{ex}[\cite{Bray}] \label{ex:schwarzschild}
The Schwarzschild manifolds of mass $m>0$ in dimension $n\ge 3$ have the expression
\[
	\left(\mathbb{R}^n \setminus \bar{B},  \left( 1+ \frac{m}{2r^{n-2}}\right)^{4/(n-2) } \delta \right).
\]
Here $\bar{B}$ denotes a closed ball of radius $(m/2)^{1/(n-2)}$. An $n$-dimensional Schwarzschild manifold can be isometrically embedded into $\mathbb{R}^{n+1}$, as a spherically symmetric  $C^{\infty}$ hypersurface of two regular ends.
\end{ex}

In Section \ref{se:thm1}, we have proved that a \emph{closed} hypersurface with nonnegative scalar curvature is weakly mean convex (up to an orientation). Below, we generalize the result to complete non-compact hypersurfaces of finitely many regular ends.

\begin{pf-complete}
Suppose on the contrary that $H$ changes signs. The condition $R\ge 0$ implies that the set of geodesic points, denoted by $M_0$, is equal to $\{ q\in M: H(q) = 0\}$. By Theorem~\ref{th:unbounded}, the connected component $M_0'$ of $M_0$ which separates $\{ H > 0\}$ and $\{ H < 0\}$ must be unbounded.  Lemma \ref{le:geodesic} yields that  $M_0'$ lies in a hyperplane, say $\{ x^{n+1} = 0 \}$, which is tangent to $M$ at every point of $M_0'$. Because $M_0'$ is unbounded, we can conclude that an end $N$ of $M$ is asymptotic to $\{ x^{n+1} = 0 \}$. That is, $N$ is the graph of a function $f$ over the exterior of a bounded region in $\{x^{n+1}=0\}$ and $ |f(x)| \to 0$ as $|x| \to \infty$. Notice that $f$ is not identically zero. Denote by $h = x^{n+1}\big|_M$ the height function on $M$. By Morse--Sard theorem, the level set $h^{-1}(\epsilon)$ is a $C^{n+1}$  submanifold for almost every $\epsilon$.

Let $\nu$ be the unit normal vector field on $M$ which is pointing upward on $N$, i.e.,
\begin{equation} \label{eq:afnu}
	\nu = \frac{(-Df, 1)}{\sqrt{1+ |Df|^2}} \quad \mbox{at } (x, f(x))\in N.
\end{equation}
Let $H$ be the mean curvature with respect to $\nu$. By Proposition~\ref{pr:nonempty}, for $\epsilon > 0$ sufficiently small, $h^{-1}(\epsilon) \cap \{H> 0\} \ne \emptyset$.
More importantly,  we have $H \ge 0$ on each component of $h^{-1}(\epsilon)$ which has non-empty intersection with $\{ H > 0\}$, because $H$ can only change signs through an unbounded subset of $M$ contained in a hyperplane that is tangent to $M$ by Theorem~\ref{th:unbounded}.

Notice that $h^{-1}(\epsilon)$ is closed for $0< \epsilon \ll 1$. Let $\Sigma_{\epsilon}$ be the \emph{outermost} connected component of $h^{-1}(\epsilon)$ such that $\Sigma_{\epsilon} \cap \{H>0\} \ne \emptyset$, i.e., it is not enclosed by any other component of $h^{-1}(\epsilon)$ which has nonempty intersection with $\{H > 0\}$.

Now we fix a sufficiently small $\epsilon >0$ so that $\Sigma_{\epsilon}$ has nonempty intersection with $N$ and $|\nabla^M h|\neq 0$ on every point in $\Sigma_{\epsilon}$. We caution readers that $\Sigma_{\epsilon}$ may not completely lie in $N$.  Denote by $\eta$ the unit normal vector on $\Sigma_{\epsilon}$, pointing inward to the region enclosed by $\Sigma_{\epsilon}$. Then, $\eta =  Df/ |Df|$ on $\Sigma_{\epsilon} \cap N$ because $f$ tends to zero at infinity and $\Sigma_{\epsilon}$ is outermost.  Thus, $\langle \nu, \eta\rangle < 0$ on $\Sigma_{\epsilon} \cap N$. Because $|\nabla^M h | \neq 0$ on every point in $\Sigma_{\epsilon}$,  $\langle \nu, \eta \rangle$ remains strictly negative on the rest part of $\Sigma_{\epsilon}$. Denote by $H_{\Sigma_{\epsilon}}$ the mean curvature with respect to $\eta$.  Apply Corollary~\ref{co:HHR} to obtain $H_{\Sigma_{\epsilon}} \le 0$. This contradicts the compactness of $\Sigma_{\epsilon}$ (for,  a compact set has at least one convex point at which $H_{\Sigma_{\epsilon} }>0$). Therefore, $H$ has a sign on $M$.
\qed
\end{pf-complete}

\begin{cor} \label{co:half-space}
Let $M$ be a connected $C^{n+1}$  hypersurface in $\mathbb{R}^{n+1}$ of finitely many regular ends with nonnegative scalar curvature. Suppose that an end $N$ of $M$ is asymptotic to the hyperplane $\Pi$, and $N$ is the graph of a function over the exterior of an open ball  in $\Pi$. Then $N$ strictly lies in one side of $\Pi$, unless $M$ is identical to $\Pi$.
\end{cor}
\begin{pf}
Suppose that $N$ is the graph of a function $f$ over $\{x^{n+1}=0\}\setminus B_{r_1}$ for some $r_1>0$ and $ |f(x)| \to 0 $ as $|x| \to \infty$. We would like to prove that $N$ is contained  in either $\{x^{n+1} > 0\}$ or $\{ x^{n+1} < 0\}$, unless $N$ is identical to $\{ x^{n+1} = 0\}$.

By Theorem~\ref{th:halfspace}, the mean curvature of $M$ has a sign. Suppose that $H \ge 0$ with respect to $\nu$, where $\nu$ is the upward pointing unit normal vector on $N$ given by \eqref{eq:afnu}. (Otherwise, we reflect $M$ about $\{ x^{n+1} = 0\}$.) By Lemma~\ref{le:maximum-principle},
\[
	\max_{\overline{B}_{r_2} \setminus B_{r_1}} f = \max_{\partial B_{r_2}} f \qquad \mbox{for all } r_2 > r_1 .
\]
Because $|f(x)| \to 0$ as $|x|\to \infty$, we then conclude that $f \le 0$ outside $ B_{r_1}$. Moreover, by applying the strong maximum principle to $H(f)\ge 0$, we have $f<0$ outside $B_{r_1}$, unless $f \equiv 0$. In the latter case, we can further conclude that $M$ is identical to $\{ x^{n+1} = 0\}$ by repeating the argument over $B_{r_2 } \setminus B_{r_0}$ for $0\le r_0 < r_1$.

\qed
\end{pf}

Next, we need a nice observation, due to Lam~\cite{L}, that the scalar curvature has a divergence form for graphical hypersurfaces. This was originally proved by using the intrinsic definition of the scalar curvature~\cite{L}. We provide below a different proof which makes use of the extrinsic geometry.
\begin{prop} If $\Omega$ is an open subset in $\mathbb{R}^n$. Let $f\in C^{2}(\Omega)$. Then the scalar curvature of the graph of $f$ is
  \begin{align}
    R(f) =  \sum_j \partial_j \sum_i \left(\frac{f_{ii} f_j - f_{ij} f_i}{1+|Df|^2} \right). \label{eq:div}
  \end{align}
\end{prop}
\begin{pf}
By Gauss equation,
\[
     R = 2 \sigma_2 (A) = \sum_{i,j} (A_i^i A_j^j - A_i^j A_j^i),
\]
  in which $A = (A_j^i)$ is the shape operator with
\[
     A_j^i = \partial_j \big(\frac{f_i}{w}\big) = \Big(\frac{f_i}{w}\Big)_j, \qquad \textup{where $w = \sqrt{1 + |D f|^2}$.}
\]
  Observe that
  \begin{align*}
     A_i^i A_j^j
     & = \partial_j \big[\Big(\frac{f_i}{w} \Big)_i \frac{f_j}{w} \big] - \frac{f_j}{w} \Big(\frac{f_i}{w}\Big)_{ij}, \\
     A_i^j A_j^i & =  \partial_i \big[\Big(\frac{f_i}{w} \Big)_j \frac{f_j}{w} \big] - \frac{f_j}{w} \Big(\frac{f_i}{w}\Big)_{ij}.
  \end{align*}
  It follows that
  \begin{align*}
    R (f)
    & = \sum_{j} \partial_j \sum_i \left[ \Big(\frac{f_i}{w} \Big)_i \frac{f_j}{w} -  \Big(\frac{f_j}{w} \Big)_i \frac{f_i}{w}\right] \\
    & = \sum_j \partial_j \sum_i \left(\frac{f_{ii} f_j - f_{ij} f_i}{w^2} \right).
  \end{align*}
\qed
\end{pf}

\begin{defi}[cf. \cite{L}]
Let $M$ be a $C^2$  hypersurface of finitely many regular ends. Let $N$ be one end of $M$, which is the graph of $f$ over the exterior of a bounded region in the hyperplane $\Pi$.  The mass of $N$ is  defined by
\begin{align} \label{eq:mass}
	m &= \frac{1}{2(n-1)  \omega_{n-1}}  \lim_{r\rightarrow \infty}\int_{S_r} \frac{1}{ 1 + |Df|^2 } \sum_{i,j}  (f_{ii} f_j - f_{ij} f_i)  \frac{x^j}{|x|}  \, d\sigma,
\end{align}
where  $S_r = \{ (x^1, \dots, x^n) \in \Pi: |x|=r\}$, $d\sigma$ is the standard spherical volume measure of $S_r$, and $\omega_{n-1} = \textup{vol}(\mathbb{S}^{n-1})$.
\end{defi}

\begin{lemma} \label{le:mass-level-set}
Let $M$ be a  $C^2$  hypersurface of finitely many regular ends.  Let $N$ be an end of $M$, which is the graph of $f$ over the exterior of a bounded region in the hyperplane $\Pi$.  If there exists a bounded region $\Omega_r$ in $\Pi$ such that $\partial \Omega_r$ is the disjoint union of  $S_r $ and $\Sigma = \{ x\in \Pi: f(x) = c\}$ for some constant $c$, and that $|Df|$ does not vanish on $\Sigma$, then
\[
	m= \frac{1}{2(n-1)  \omega_{n-1}}\left( \int_{\Sigma} \frac{|Df|^2}{ 1 + |Df|^2} H_{\Sigma}\, d\sigma + \lim_{r \to \infty} \int_{\Omega_r} R(f) \, dx \right),
\]
where $R(f)$ is the scalar curvature of the graph of $f$,  $\eta$ is the unit normal vector on $\Sigma$ pointing away from $\Omega_r$, and $H_{\Sigma}$ is the mean curvature of $\Sigma$ with respect to $\eta$.
\end{lemma}
\begin{pf}
Applying the divergence theorem to \eqref{eq:div} over $\Omega_r$ yields
\begin{align*}
	&\int_{S_r} \frac{1}{ 1 + |Df|^2 } \sum_{i,j=1}^n  (f_{ii} f_j - f_{ij} f_i)  \frac{x^j}{|x|}  \, d\sigma \\
	&=  \int_{\Omega_r} R(f) \, dx- \int_{\Sigma} \frac{1}{ 1 + |Df|^2 } \sum_{i,j=1}^n  (f_{ii} f_j - f_{ij} f_i) \eta^j  \, d\sigma.
\end{align*}
Because $\Sigma$ is a level set of $f$, $\eta$ equals either $Df/ |Df|$ or $- Df/ |Df|$. If $\eta = - Df/ |Df|$,
\begin{align*}
 	 H_{\Sigma} &= - \mbox{div}_0 \eta= \sum_{i=1}^n \frac{\partial}{ \partial x^i}\left( \frac{f_i}{ |Df|}\right)\notag\\
	& = \frac{ 1}{ |Df|^3} \sum_{i,j=1}^n( f_{ii}f_{j}f_j-  f_{ij} f_i f_j ).
\end{align*}
We then derive
\begin{align} \label{eq:pmt}
	\begin{split}
		&\int_{S_r} \frac{1}{ 1 + |Df|^2 } \sum_{i,j=1}^n  (f_{ii} f_j - f_{ij} f_i)  \frac{x^j}{|x|}  \, d\sigma  \\
		&= \int_{\Omega_r} R(f) \, dx+ \int_{\Sigma} \frac{|Df|^2}{ 1 + |Df|^2} H_{\Sigma}\, d\sigma.
	\end{split}
\end{align}
If $\eta = Df/ |Df|$, we also derive the same identity. Letting $r\to \infty$, we prove the lemma.
\qed
\end{pf}

Generally,  the set $\Omega_r$  may not exist. We shall prove that if $M$ has nonnegative scalar curvature, then such $\Omega_r$ exists, and moreover  $H_{\Sigma} \ge 0$.

\begin{pf2}
We assume that $M$ is not a hyperplane; otherwise the theorem trivially holds. Consider an end $N$ of $M$, and suppose that $N$ is the graph of $f$ over $\{x^{n+1} = 0\} \setminus B_{r_1}$ for some $r_1>0$. By Theorem~\ref{th:halfspace}, $H$ has a sign on $M$. We may without loss of generality assume that $H\ge 0$ with respect to $\nu$, where $\nu$ is the upward unit normal to $N$, given by
\[
		\nu = \frac{(-Df, 1)}{\sqrt{1+ |Df|^2}}.
\]
(Otherwise, we may replace $f$ by $-f$.) We divide the proof  into the following cases. \vspace{10pt}

\noindent{\bf Case 1:} $\lim_{|x| \to \infty} f(x) = a$ for some bounded constant $a$. By translation, we may assume that $|f(x)| \to 0$ as $|x| \to \infty$; namely, $N$ is asymptotic to the hyperplane $\{ x^{n+1} = 0 \}$. By proof of Corollary \ref{co:half-space}, $N \subset \{ x^{n+1} < 0\}$. Denote by $h =x^{n+1}\big|_M$ the height function on $M$. For $\epsilon>0$ small enough, a connected component  of the level set $h^{-1}(-\epsilon)$ is contained in $N$. We define $\Sigma_{-\epsilon}$ to be an  outermost connected component of  the subset $\{x\in \{ x^{n+1} = 0\} : f(x) = - \epsilon \}$. By Morse--Sard theorem, $\Sigma_{-\epsilon}$ is  $C^{n+1}$ for almost every $\epsilon$.  Moreover, because $f$ tends to zero,  for some small $\epsilon > 0$,  $\eta = - Df/ |Df|$ is the unit vector on $\Sigma_{-\epsilon}$, pointing inward to the bounded region in $\{x^{n+1} = 0\}$ enclosed by $\Sigma_{-\epsilon}$. Let $H_{\Sigma_{-\epsilon}}$ be the mean curvature of $\Sigma_{-\epsilon}$ defined by $\eta$. Then, $H_{\Sigma_{-\epsilon}} \ge 0$ by Corollary \ref{co:HHR} and by $H\ge 0$.  Applying Lemma~\ref{le:mass-level-set}, we have $m\ge 0$.

If $m=0$, then $M$ must be identical to $\{ x^{n+1} = 0\}$. For, otherwise, there exists some positive $\epsilon$ so that $\Sigma_{-\epsilon}$ is contained in $\{ x^{n+1} = 0\}$ with $H_{\Sigma_{-\epsilon}} \equiv 0$ by \eqref{eq:pmt}. This contradicts compactness of $\Sigma_{-\epsilon}$. \vspace{10pt}

\noindent{\bf Case 2:}  $\lim_{|x| \to \infty} f(x) = \infty$. The set $\{ x\in \{ x^{n+1} = 0 \}: f(x) = \Lambda\}$ lies in $\{ x^{n+1} \} \setminus B_{r_1}$ for $\Lambda \gg 1$. Let $\Sigma_{\Lambda}$ be the outmost component of the above set. For $\Lambda$ sufficiently large, $\eta = -Df/ |Df|$ is the normal vector to $\Sigma_{\Lambda}$ pointing inward to the bounded region enclosed by $\Sigma_{\Lambda}$. Let $H_{\Lambda}$ be the mean curvature with respect to $\eta$. Hence, $H_{\Sigma_{\Lambda}} \ge 0$ by Corollary \ref{co:HHR} and by $H\ge 0$.  Applying Lemma~\ref{le:mass-level-set}, we have $m\ge 0$. If $m=0$, we can show that $M$ is identical to a hyperplane as in Case 1.\vspace{10pt}

\noindent{\bf Case 3:} $\lim_{|x| \to \infty} f(x) = -\infty$. This case cannot happen. Otherwise, for some $\Lambda\gg 1 $, there is a closed submanifold $\Sigma_{-\Lambda} \subset \{ x\in \{x^{n+1}=0\} \setminus B_{r_1} : f(x) =  - \Lambda\}$ so that the unit normal vector $\eta = Df/|Df|$ is pointing inward to the region enclosed by $\Sigma_{-\Lambda}$. Let $H_{\Sigma_{-\Lambda}}$ be the mean curvature with respect to $\eta$. Then, $H_{\Sigma_{-\Lambda}} \le 0$ by Corollary~\ref{co:HHR}. This contradicts compactness of $\Sigma_{-\Lambda}$.

\qed
\end{pf2}

Last, we verify below that our definition of the mass~\eqref{eq:mass} coincides with the classical definition of the ADM mass, if we assume certain decay condition on the derivatives of $f$. Let us recall the definition of the ADM mass (see, for example, \cite[Equation (4.1)]{Bartnik}).
\begin{defi}
We say that an $n$-dimensional manifold $(M, g)$ has an asymptotically flat end $N$ if $N\subset M$ is diffeomorphic to $\mathbb{R}^n \setminus B_1$ and $N$ has a coordinate chart $\{y\}$ so that $g_{ij} (y)= \delta_{ij} + O_2(|y|^{-q})$  for some $q> (n-2)/2$. The $O_2$ indicates that first and second derivatives also decay at rates one and two orders faster, respectively.

For $n\ge 3$, the ADM mass of the asymptotically flat end $N$ is defined by
\begin{align}\label{eq:classical-mass}
	 \frac{1}{2(n-1)  \omega_{n-1}} \lim_{r\rightarrow \infty} \int_{|y|=r} \sum_{i,j} \left(\frac{\partial g_{ij}}{\partial y^i} - \frac{\partial g_{ii} }{\partial y^j} \right) \tau^j\, d\sigma_g,
\end{align}
where $\tau$ is the outward unit normal to $\{ |y|=r\}$ with respect to $g$, $d\sigma_g$ is the volume measure of $\{ |y| = r\}$ with respect to $g$, and $\omega_{n-1} = \textup{vol} (\mathbb{S}^{n-1})$.
\end{defi}

\begin{lemma}  [cf. \cite{L}] \label{le:mass}
Let $n\ge 2$ and $M$  a complete $n$-dimensional $C^3$ hypersurface of finitely many regular ends.  Let $N$ be an end of $M$ which is the graph of $f$. Suppose that $R\in L^1(N)$  and $|D f(x)|=O(1)$ as $|x| \to \infty$. Then, the mass of $N$ defined by \eqref{eq:mass} is  finite.

If in addition $n\ge 3$, $|Df(x)|^2 = O_2(|x|^{-q})$ for some $q> (n-2)/2$, and $|Df(x)|^2 |D^2 f(x)| = o(|x|^{1-n})$ as $|x| \to \infty$, then \eqref{eq:mass} equals \eqref{eq:classical-mass}.
\end{lemma}
\begin{pf}
Applying the divergence theorem to \eqref{eq:div} yields
\begin{align*}
	 & 2(n-1)  \omega_{n-1}  m\\
	 & =  \int_{S_{r_0}} \frac{1}{ 1 + |Df|^2 } \sum_{i,j =1}^n  (f_{ii} f_j - f_{ij} f_i)  \frac{x^j}{|x|}  \, d\sigma+ \lim_{r\rightarrow \infty} \int_{r_0 \le |x| \le r} R(f) \, dx.
\end{align*}
Because $R$ is integrable over $N$ and $|Df(x)|=O(1)$ as $|x| \to \infty$, the second term on the right hand side is bounded.
Therefore, $m$ is bounded.

To prove the second statement, we consider the coordinate chart $\{y\}$ of $N$, where
\[
	y^i =(0, \dots, \underbrace{x^i}_{i-\mbox{th}}, \dots, 0, f(0, \dots, \underbrace{x^i}_{i-\mbox{th}}, \dots, 0)).
\]
Then
\[
	\frac{\partial}{\partial y^i} = \partial_i + f_i \partial_{n+1},
\]
where we denote $\partial_i = \frac{\partial}{\partial x^i}$. Moreover, at the point $(x, f(x)) \in N$,
\begin{align*}
	g_{ij} &= \langle\frac{\partial}{\partial y^i}, \frac{\partial}{\partial y^j}\rangle = \delta_{ij} + f_i f_j,\\
	\frac{\partial g_{ij} }{\partial y^k} &= \frac{\partial (f_i f_j) }{\partial x^k} = f_{ik} f_j + f_i f_{jk}.
\end{align*}
Therefore, $N$ is an asymptotically flat end of $M$ by hypothesis of $|Df|$. Denote by $\mu =\sum_{i=1}^n \frac{x^i}{r} \partial_i$ the outer unit normal to $S_r$ in $\{x^{n+1} = 0\}$. Let $\tau$ be  the outer unit normal to the graph of $f$ over $S_r$ in $M$. Then,
\[
	\tau = \frac{ \mu + \mu(f) \partial_{n+1} }{\sqrt{1 +  |\mu(f) |^2}},
\]
and hence,
\[
	\tau^j = g(\tau,\frac{\partial}{\partial y^j}  ) = \langle\tau,\frac{\partial}{\partial y^j}  \rangle= \frac{\mu^j + \mu (f) f_i}{ \sqrt{1+ | \mu(f)|^2}}.
\]
It follows that
\begin{align}\label{eq:Sr}
\begin{split}
	 & \int_{f( S_r)} \sum_{i,j} (\frac{\partial g_{ij}}{\partial y^i} - \frac{\partial g_{ii} }{\partial y^j} ) \tau^j\, d\sigma_g \\
	 &= \int_{S_r} \sum_{i,j}  (f_{ii} f_j - f_{ij} f_i)  \frac{\mu^j + \mu (f)  f_j}{ \sqrt{1+ |\mu(f)|^2}} \sqrt{1+ |D^T f|^2}\, d\sigma,
\end{split}
\end{align}
where $d\sigma_g$ and $d\sigma$ are the $(n-1)$-Hausdorff measures on $f(S_r)$ and $S_r$ respectively, and $D^T f$ is the derivative along the directions tangent to $S_r$.  By \cite[Proposition 4.1]{Bartnik} and hypotheses on the derivatives of $f$, the left hand side of \eqref{eq:Sr} converges to \eqref{eq:classical-mass} as $r\to\infty$, and the right hand side of \eqref{eq:Sr} converges to \eqref{eq:mass}.
\qed
\end{pf}

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Facing a myriad of security regulatory requirements and evolving threats, critical infrastructure owners and operators in all sectors – from telecommunications, energy, transportation, financial, defense industrial base, critical manufacturing, and water – need a way to manage risk efficiently across their enterprises. Crowell & Moring recognizes comprehensive risk assessments as good tools in helping companies manage and mitigate these risks.
Our representative experience includes:
Security Compliance Reviews: Assisting companies undertake comprehensive and coordinated cyber and physical security risk assessments and compliance reviews led by security personnel and legal counsel whose efforts can help direct compliance efforts and preserve privilege and confidentiality for business and proprietary information and data.
Security Plans: Working with companies to develop cyber and physical security policies and procedures, operations, and incident response and business continuity plans (including restoration, mitigation, and contingency plans) and testing and exercise regimes.
Risk Mitigation: Helping companies reduce the risk profile and regulatory burdens associated with critical infrastructure and key assets.
Despite preparation and best efforts, incidents and crises will occur. Our Privacy & Cybersecurity and Government & Internal Investigations groups help clients conduct and manage privileged post-incident internal and external investigations in coordination with forensic investigators and represent clients before agencies and other government entities.
Advising companies in a multitude of sectors on hundreds of data breaches, including advising on government and contractual reporting obligations and potential mitigation responsibilities.
Representing Los Alamos National Security, LLC, the operator and manager of Los Alamos National Laboratory for the National Nuclear Security Administration, in connection with a February 2014 radiological release incident at the Waste Isolation Pilot Plant and compliance with applicable laws, regulations, contracts, and policies.
Providing representation before the Chemical Products Safety Board, Federal Trade Commission, Securities and Exchange Commission, offices of the U.S. Attorney General, and other relevant authorities.
Opportunities to Mitigate Risk
Given the potentially significant risks that critical infrastructure owners and operators face in the event of a security incident, our Insurance Group advises companies on identifying appropriate coverage.
Oftentimes, insurance isn't enough. Through the DHS Support Anti-Terrorism by Fostering Effective Technologies Act (the SAFETY Act) program, we help a broad array of clients seek and obtain liability protections, including immunity from tort claims for damages arising out of an act of terrorism and limits on potential liability for loss when approved anti-terrorism products and services are in use. Most critical infrastructure and service providers are eligible for SAFETY Act protection.
Cybersecurity and privacy concerns have played an increasingly critical role in influencing how government and the private sector think about information collection, use, and disclosure. Information sharing — a core Homeland Security mission — poses unprecedented challenges in the information arena, such as privacy, cybersecurity, and international data exchanges. Concerns about cybersecurity and privacy have caused delays, restructuring, and even termination of Homeland Security programs and challenges for the private sector. Domestic laws, like the U.S. Patriot Act, may also clash with international privacy requirements.
U.S. and foreign regulators have responded with a patchwork of privacy principles (such as FTC and state Attorney General "recommendations"), new laws (including federal laws and different security breach statutes in 47 states and in Washington, D.C.), and aggressive enforcement. Together, these varying requirements and best practices create a complex and uncertain regulatory environment. We guide clients through these myriad laws, regulations, and best practices to design and implement efficient and effective privacy programs.
Compliance Counseling: We work with companies to adopt or improve efficient and effective privacy programs that are tailored to the particular threats and vulnerabilities as well as industry-specific issues each organization faces.
Regulatory: Our team identifies and helps organizations comply with all state, federal, and international laws, regulations, and best practices that apply to the organization.
Transactional: We help companies perform due diligence — particularly relating to privacy concerns – in advance of a transaction and develop key deal terms.
Litigation: Our team's national reputation for successful litigation, settlement, and thought-leadership make us a clear choice when a client faces significant class action litigation, such as under the Telephone Consumer Protection Act (TCPA), California Confidentiality of Medical Information Act (CMIA), and other state and federal laws.
Energy Sector Security and Reliability
Managing the evolving risks faced by the energy sector requires both a comprehensive understanding of how this sector works and is regulated, and a comprehensive knowledge of the cyber and physical security environment and its overlapping regulatory framework. Our firm's experienced Energy Group and security professionals work together to help our clients comply with legal requirements and manage the material risks associated with security and reliability concerns.
Representative experience includes:
Recently assisted a generator operator client in handling a potential violation of NERC Critical Infrastructure Protection standards and analyzed whether any such violation occurred.
Advised companies on implementation of the TSA Pipeline Security Guidelines, including providing advice on developing compliant corporate security programs, plans and risk analyses, conducting site reviews, and preparing for government inspections.
We help our clients create, implement, and maintain security policies and procedures to effectively manage compliance with the evolving regulations confronting members of the defense industry. Recognizing that critical infrastructure owners and operators cannot always rely on regulatory compliance to manage risk, we also help our clients assess security vulnerabilities and develop policies and procedures to mitigate risk.
Because of the deep knowledge and experience of our Government Contracts Group, we are able to help clients navigate and address regulatory and contractual issues that may be affected in the event of a cyber or physical attack.
Advising companies on compliance with the National Industrial Security Program Operating Manual (NISPOM) obligations including development of security training and incident response programs and policies.
Advising companies on Defense Federal Acquisition Regulations Supplement (DFARS) rules on safeguarding classified and other controlled or sensitive government information entrusted to them.
Advising and counseling energy companies on NERC compliance issues, with a particular focus on Critical Infrastructure Protection (CIP) Standards.
Conducting a comprehensive physical security assessment and providing a gap analysis for major government contractor, in coordination with technical expert, and assisting on developing best practice compliance program.
Chemical Security
Crowell & Moring's professionals have broad experience helping industrial sectors, including pipelines, pulp and paper mills, petroleum facilities, metal production and manufacturing plants, hospitals, universities and colleges, and industrial cleaning facilities cope with chemical security regulatory programs, including the evolving DHS Chemical Facility Anti-Terrorism Standards (CFATS), the Coast Guard Maritime Safety and Transportation Act (MTSA) regulations, and the EPA Chemical Accident Prevention provisions.
Helping companies develop hundreds of CFATS and MTSA surveys, vulnerability assessments and site and facility security plans.
Assisting companies prepare for CFATS and MTSA field inspections and technical assistance visits.
Advising companies on internal and third-party audits, investigations, and compliance reviews.
Counseling companies on avoiding regulatory risks by modifying chemical holdings.
A Crowell & Moring interdisciplinary team, which includes a broad base of knowledge in the areas of government contracting, energy, transportation and mining, provides advice and counsel to our clients on unique and complex national security regulatory programs, including the export and other controls over sensitive technologies and issues related to Committee on Foreign Investment in the United States (CFIUS), an inter-agency committee empowered to review all foreign investments in U.S. companies (including of a U.S. business from one foreign owner to another) that may affect U.S. national security. US national security is interpreted very broadly by CFIUS and includes not only mergers and acquisitions involving US defense contractors or businesses but also those that could result in foreign control of US critical infrastructure, such as communications, energy, transportation, chemical and biopharmaceutical manufacturing and mining assets.
The Government Contracts and International Trade groups at Crowell & Moring include lawyers who know the CFIUS process inside and out — through guiding both foreign acquirers and domestic businesses through the CFIUS process and participating in CFIUS reviews through prior government service — and understand what companies need to do to gain approval for their transactions. We are also deeply familiar with the myriad of export control regulations (ITAR, EAR, NRC and DOE) that control the export and import of defense-related articles and services, nuclear power and other sensitive technologies, including access by foreign owners of U.S. businesses.
Counseling companies on identifying U.S. national security implications early in the consideration of potential transactions, on determining whether to submit a voluntary notice to CFIUS, on undertaking the filing process, and, ultimately, on managing the many formal and informal considerations involved in obtaining CFIUS clearance.
Counseling clients concerning US laws and regulations applicable to foreign investment and ownership in the United States, such as ITAR, EAR, NRC and DOE.

redpajama_set_name: RedPajamaCommonCrawl


text:
Interspecific interactions, movement patterns and habitat use in a diverse coastal shark assemblage
Heupel, Michelle R., Munroe, Samantha E. M., Ledee, Elodie J. I., Chin, Andrew, and Simpfendorfer, Colin A. (2019) Interspecific interactions, movement patterns and habitat use in a diverse coastal shark assemblage. Marine Biology, 166 (6). 68.
PDF (Published version) - Published Version
View at Publisher Website: https://doi.org/10.1007/s00227-019-3511-...
Sharks are a highly diverse predatory taxon and are regularly found in large, potentially competitive, assemblages. However, the mechanisms that enable long-term coexistence and factors that drive complementary movement are poorly understood. As interspecific interactions can have a large influence on survival and trophic linkages, research on shark assemblages could substantially increase our understanding of marine community dynamics. In this study, we used passive acoustic telemetry to compare the activity space size, spatial overlap and habitat use patterns of six co-occurring shark species from the same family in a tropical nearshore embayment. Our results indicated that all sizes of Rhizoprionodon taylori (a small-bodied, highly productive species) used significantly larger amounts of space (e.g., mean 95% KUD = 85.9 km2) than juveniles of large-bodied, less productive species (e.g., Carcharhinus amboinensis; 62.3 km2) that use nearshore areas as nursery areas. Most large, less productive species appeared risk averse by using less space, while the smaller more productive species took greater risk by roaming broadly. These movement strategies are likely a means of avoiding predation or gaining access to new or additional resources. Spatial overlap patterns varied substantially between species with overlap in core use areas ranging from 1.2 to 27.6%, but were consistent over time. Most species exhibited low spatial overlap, suggesting spatial partitioning to reduce interspecific competition. While a few species exhibited a high degree of spatial overlap (up to 60% of activity space extent), dietary diversity may reduce competition to support co-occurrence. These data suggest that complex interactions occur in communal nurseries in nearshore waters where species are in direct competition for resources at vulnerable life stages.
Article (Research - C1)
shark, behaviour, movement, resource, ecology
© Springer-Verlag GmbH Germany, part of Springer Nature 2019
Marine Tropical Sciences Research Facility (MTSRF), Great Barrier Reef Marine Park Authority (GBRMPA), Australian Research Council (ARC)
06 BIOLOGICAL SCIENCES > 0602 Ecology > 060205 Marine and Estuarine Ecology (incl Marine Ichthyology) @ 75%
06 BIOLOGICAL SCIENCES > 0608 Zoology > 060801 Animal Behaviour @ 25%
96 ENVIRONMENT > 9608 Flora, Fauna and Biodiversity > 960808 Marine Flora, Fauna and Biodiversity @ 50%
96 ENVIRONMENT > 9608 Flora, Fauna and Biodiversity > 960802 Coastal and Estuarine Flora, Fauna and Biodiversity @ 50%


text:
12 People Share the Secrets They Won't Share with Their SO
Alisha Richardson
I think we all have secrets we keep from people, but (hopefully) there aren't too many secrets we keep from our significant others. Take a look at these secrets some AskReddit users have kept from theirs.
1. Fart = 7 year relationship
The first time my SO and I slept together I woke up during the night to use the bathroom. I came back in the room and she farted one of the loudest farts I've ever heard. I started laughing like crazy and she almost woke up. Almost. We've been together for seven years.
Photo Credit: BBC Films
2. Caesar Salad
Her Caesar salad dressing. It's actually the best thing ever but even after two years she refuses to tell me what's in it. What she doesn't know is that I have secretly been watching her make it over the past few months. I have finally pieced it together and now I have the power!!
Edit: Everyone wants the secret recipe. But she has reddit, and I prefer life.
Photo Credit: Instagram, @g3n1j4l4c
3. Reflection
My ex, and the cheating. She thinks I found out when I looked at her phone, but I new about two months earlier when she sent me a pic of herself watching Netflix in bed. I don't think she realized that the dude's face was 100% visible reflected in the laptop. After that it was a matter of me being in denial and trying to fix things.
Photo Credit: Instagram, @maci_welch
4. Steamer Pack
A girl I dated a few years back had just graduated with her bachelors so I took her out for a night on the town. Her and our friends got absolutely demolished and since this was before Uber, I volunteered to be DD. So around 2am she's tanked, we head out and she wants krystal burgers. Very adamant about that, so I stop by Krystals and order a steamer pack so I can have some too and then have leftovers. She eats somewhere around 8, I get her home, get her into her bed and she immediately passes out. I am sitting next to her watching some TV when I smell something. I notice that she has just shit herself. She is one of those people who would be so ashamed of herself if anyone found out so I just… left. I called her the next day and told her I dropped her off, got her some water and headed home. Never mentioned her shitting herself or anything so to this day she thinks she did it in her sleep after I left. I could have stayed and helped her clean it up and I probably should have, but she would have cried over that and avoided me sporadically for weeks.
Photo Credit: Universal Pictures
5. Frozen Croutons
Everybody likes snacks, right? Some of us like chips, others like fruit, or yogurt. Well, my wife likes croutons. Frozen croutons to be exact and she tries to hide it like an alcoholic hides bottles of vodka. Where does she hide the croutons? In the freezer. At first I'd put them in the pantry when we get home from the grocery store, but they'd always end up in the freezer. I'd be like, "Uh, hon, what's up with the croutons in the freezer?" and she'd act like she had no idea what I was talking about. Now here's the thing, I've tried them and now I'm addicted to them too. So now I openly eat cold croutons out of the freezer as a snack, and she tells me what a weirdo I am while she eats half the bag with me. No idea why she's embarassed to admit what a genius she is for discovering such an awesome snack. BTW, the best kind is Chatham Village garlic and cheese flavor.
Photo Credit: Instagram, @koaticmamii
6. Upgrade
My wife dropped my iPhone in a Port-A-Potty toilet. Luckily it has been so used there was a large pile of shit which gently held the phone above the liquid. She cleaned it off and gave it back to me. A year later I upgraded and gave her my old phone.
She insisted on purchasing a new cover even though I had a high-end OtterBox. I never understood why until one of the kids told on her.
Photo Credit: Instagram, @geekcommander
7. Mysterious moving router
She smokes when I'm not home, I know this because I always move the router to the window where the signals best, when I come home the router is in a different position away from the window, I put it back without her knowing and call her out for smoking, she has no idea how I keep finding out!
Photo Credit: Instagram, @fullpinkmoon
8. Never trust a fart
I am not sure what events led up to it but he pooped the bed. I am guessing he trusted a fart he shouldn't have but it happened. He didnt hide it veru well, he just put the sheets in the washing machine but didnt start it so i smelled it. He claims the dog sharted but the boxers he attempted to hide( pretty haphazardly i might add) in the trash tell a different story. I have attempted to bring it up but i don't want to break the illusion he has of himself being the james bond of bed shitting.
Photo Credit: Pixabay, stevepb
9. Work History
She had sex with my boss before we knew each other, before I had been hired into the company. I'm not mad about it or anything, everyone has a past but she carefully dances around ever coming into my work now and I'm careful not to bring him up.
Photo Credit: 20th Century Fox
10. Secret Clogging
we were long distance when we first started dating. The one time she came up to where I lived for 4 days, but I had to work during one of the days. When I was gone she had to take a dump and clogged the toilet. She couldn't get it to plunge and was freaking out so she went to the kitchen and got a fork and a plastic bag and dug the turds out of the bowl and threw everything in the dumpster out back. She called her friend panicking and the friend thought it was hilarious. Friend told me about it one night when the girl had left the room after we moved in together. I find it funny but don't want to embarrass her!
Photo Credit: Pixabay, orzalaga
11. Sleep Secrets
When he's about to fall asleep he's completely honest. it was about 7 months in when he grabbed me and whispered into my ear "You're gonna be the mother of my children."
I told him and he was thoroughly embarrassed and he apologized. however I neglected to tell him that pretty much every night he would tell me that he loved me starting a month in.
I don't mind and I'm not trying to rush things and neither of us have any plans to jump into that sort of stuff right away but it's nice to know
Photo Credit: Alcon Entertainment
12. Escape Plan
Before my ex-wife and I had first split up, I had found out she was hiding a storage unit with an entire apartment set aside in it, just in case the breakup went bad.
She worked graveyard, and I was using her laptop while watching The Walking Dead to use the story sync. And because she uses all Apple products, her iMessage was tied to her MacBook. I don't normally read her messages, but considering the amount of notifications that were chiming in, I could tell she was having a rather lengthy conversation. I opened it and started reading and it turns out she had sent a picture message of a fully packed storage unit to her best friend out in South Carolina.
It had Furniture, appliances, essentially everything she needed to start out with an apartment on her own
I sat on that for several months, before finally bringing it up out of nowhere. We were literally sitting on the couch one night and I just turned to her and said, "so I know about your storage unit". She about shit a brick, it was the last thing she ever expected to hear because she thought she was being sneaky.
Conversation actually went quite well though, and we ended up deciding that we were going to file for divorce. That divorce just finalized March 3rd of this year.
Photo Credit: Instagram, @jamiedoesntknow
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