A note on Hsiung-Minkowski formulas

In this note, we will give a generalization of Hsiung-Minkowski formulas for hypersurfaces in space forms. This will generalize the results in some of our previous posts. For example, as a special case of Theorem 6, we have the following result:

 Theorem 1 Suppose ${(M^n,g)}$ is a space form, ${\Sigma^{n-1}}$ is a closed oriented hypersurface. Assume ${X\in \Gamma(\phi^*(TM))}$ is a conformal vector field along ${\Sigma}$, and ${f}$ is a smooth function on ${\Sigma}$. $\displaystyle \begin{array}{rl} \int_\Sigma \alpha f \sigma_k =\int_\Sigma f\sigma_{k+1}\nu\cdot X- \frac{1}{(n-1-k){{n-1}\choose k}}\int_\Sigma \langle T_k(\nabla f), X^T\rangle . \end{array}$ Here ${\sigma_k}$ is the normalized ${k}$-th mean curvature, ${\alpha}$ is defined by ${\mathcal{L}_Xg =\alpha g}$ and ${\nu}$ is a unit normal vector field.

The definition of ${T_k}$ will be given in the next section. We remark that the classical Hsiung-Minkowski formulas [Hsiung] can be recovered by putting ${f=1}$ in the above equation. By choosing suitable ${f}$ in Theorem 1, we can obtain the following corollary:

 Corollary 2 (Corollary 13) Suppose ${\Sigma\subset \mathbb R^n}$ is a closed hypersurface with ${\sigma_{k+1}>0}$ and ${0\le k\le n-2}$. Then $\displaystyle \mathrm{Area}(\Sigma)\le \int_\Sigma \sigma_1 r\le \cdots\le \int_\Sigma\sigma_{k+1} r^{k+1}$ and $\displaystyle n\mathrm{Vol}(\Omega)\le \int_\Sigma r\le \int_\Sigma \sigma_1 r^2 \le \cdots \le \int_\Sigma \sigma_{k+1}r^{k+2}.$ Here ${r=|X|}$ and ${\Omega}$ is the region enclosed by ${\Sigma}$. The equality occurs if and only if ${\Sigma}$ is a sphere centered at ${O}$.

As another corollary, we have the following extension of Alexandrov’s theorem:

 Corollary 3 (Corollary 11) Suppose ${\Sigma}$ is a closed embedded hypersurface in ${\mathbb{R}^n}$. Assume ${f>0}$, ${f'\ge 0}$ and there exists ${0\le k \le m}$ such that ${\sigma_k f(r)}$ is constant where ${r}$ is the distance from ${O}$. Then ${\Sigma}$ is a sphere.

The rest of this note is organized as follows. In Section 1 we will give the necessary definitions and preliminary results. In Section 2, we will prove the main results, and several corollaries will be given in Section 3.

1. Preliminaries

Let ${\phi}$ be an isometric immersion of an ${m}$-dimensional semi-Riemannian manifold ${\Sigma}$ into an ${n}$-dimensional semi-Riemannian manifold ${(M,g)}$. We will use ${\overline \nabla }$ and ${\nabla }$ to denote the connection on ${(M,g)}$ and ${\Sigma}$ respectively. The second fundamental form of ${\Sigma }$ in ${M}$ is defined by ${A(X,Y)=-({\overline \nabla _XY})^\perp}$ and is normal-valued. We denote ${A(e_i, e_j)}$ by ${A_{ij}}$, where ${\lbrace e_i\rbrace_{i=1}^m}$ is a local orthonormal frame on ${\Sigma}$. For simplicity, we will write ${g(X, Y)}$ as ${X\cdot Y}$. For any normal vector field ${\nu}$ of ${\Sigma}$ in ${M}$, we define the scalar second fundamental form ${A^\nu\in \mathrm{End}(TM)}$ by ${\langle A^\nu(X), Y \rangle= A(X, Y)\cdot \nu}$, and let ${A^\nu(e_i)= \sum_{j=1}^m (A^\nu)_i ^j e_j}$.

We define the ${k}$-th mean curvature as follows. If ${k}$ is even,

$\displaystyle H _k = \frac 1{k!}\sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon_{j_1\cdots j_k}^{i_1\cdots i_k}(A_{i_1j_1}\cdot A_{i_2j_2})\cdots (A_{i_{k-1}j_{k-1}}\cdot A_{i_kj_k}).$

If ${k}$ is odd, the ${k}$-th mean curvature is a normal vector field defined by

$\displaystyle H _k =\frac 1{k!}\sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon _{j_1\cdots j_k}^{i_1\cdots i_k} (A_{i_1j_1}\cdot A_{i_2j_2})\cdots (A_{i_{k-2}j_{k-2}}\cdot A_{i_{k-1}j_{k-1}})A_{i_k j_k}.$

We also define ${H_0=1}$. Here ${\epsilon_{i_1 \cdots i_k}^{j_1\cdots j_k}}$ is zero if ${i_k=i_l}$ or ${j_k=j_l}$ for some ${k\ne l}$, or if ${\lbrace i_1, \cdots, i_k\rbrace \ne \lbrace j_1, \cdots, j_k\rbrace}$ as sets, otherwise it is defined as the sign of the permutation ${(i_1, \cdots, i_k)\mapsto (j_1, \cdots, j_k)}$. We also define the normalized ${k}$-th mean curvature as

$\displaystyle \sigma_k =\frac{1}{{m\choose k}}H_k.$

In the codimension one case, i.e. ${\Sigma}$ is a hypersurface, by taking the inner product with a unit normal if necessary, we can assume ${H_k}$ is scalar valued. In this case the value of ${H_k}$ is given by

$\displaystyle H_k =\pm\sum_{i_1<\cdots< i_k}\lambda_{i_1}\cdots \lambda_{i_k} \ \ \ \ \ (1)$

where ${\{\lambda_i\}_{i=1}^n}$ are the principal curvatures. This definition of ${H_k}$ will be used whenever ${\Sigma }$ is a hypersurface.

Following [Grosjean] and [Reilly1], we define the (generalized) ${k}$-th Newton transformation ${T_k}$ of ${A}$ (as a ${(1,1)}$ tensor, possibly vector-valued) as follows.
If ${k}$ is even,

$\displaystyle {(T_k)}_j^{\,i}= \frac 1 {k!} \sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon^{i i_1 \ldots i_k}_{j j_1 \ldots j_k} (A_{i_1j_1}\cdot A_{i_2j_2})\cdots (A_{i_{k-1}j_{k-1}}\cdot A_{i_kj_k}).$

If ${k}$ is odd,

$\displaystyle {(T_k)}_j^{\,i}= \frac 1 {k!} \sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon^{i i_1 \ldots i_k}_{j j_1 \ldots j_k} (A_{i_1j_1}\cdot A_{i_2j_2})\cdots (A_{i_{k-2}j_{k-2}}\cdot A_{i_{k-1}j_{k-1}})A_{i_kj_k}.$

We also define ${T_0= \mathrm{id}}$. Again, in the codimension one case, by taking the inner product with a unit normal if necessary, we can assume ${T_k}$ is an ordinary ${(1,1)}$ tensor and if ${\lbrace e_i\rbrace_{i=1}^m}$ are the eigenvectors of ${A}$, then

$\displaystyle T_k(e_i)=\pm\frac 1{{m\choose k}}\sum_{\substack{i_1<\cdots

This definition of ${T_k}$ will be used whenever ${\Sigma}$ is a hypersurface.

We collect some basic properties of ${T_k}$ and ${H_k}$:

 Lemma 4 We have (${\mathrm{tr}}$ denotes the trace on ${\Sigma}$): ${\mathrm{tr}(T_k)= (m-k)H_k}$. If ${M}$ is a space form, then ${\mathrm{div}(T_k)=0}$. i.e. ${\sum_{i=1}^m\nabla _{e_i}(T_{k})^i_{j}=0.}$ (Here ${\nabla }$ is the normal connection if ${k}$ is odd. ) If ${k=1}$ and ${m=n-1}$, we can assume ${M}$ is Einstein instead. If ${k=0}$, we can remove any assumption on ${M}$. If ${k}$ is even, then ${\sum_{i,j=1}^m (T_k)_i^jA_{ij}=(k+1) H_{k+1}. }$ If ${k}$ is odd, then ${\sum_{i,j=1}^m (T_k)_i^j\cdot A_{ij}=(k+1) H_{k+1}. }$

Proof: These equations are well-known, at least in the codimension one case (e.g. [Mar] Lemma 2.1). They can be found e.g. in [Grosjean] Lemma 2.1, 2.2 and [K1] Lemma 2.1. For (4), if ${k=1}$ and ${m=n-1}$, then by Codazzi equation, we have ${\mathrm{div}(A)= dH_1}$, which is equivalent to ${\mathrm{div}(T_1)=0}$. The assertions are trivial for ${k=0}$. $\Box$

2. Main results

In this section, we will first derive a simple integral formula and draw a few direct consequences. We will use the notations in Section 1. Throughout this section, we will also assume that ${\Sigma}$ is a closed and oriented semi-Riemannian manifold isometrically immersed in ${(M,g)}$, and denote the induced metric on ${\Sigma}$ by ${\langle \cdot, \cdot\rangle}$. We will omit the area element in the integrals when there is no confusion.

 Proposition 5 Let ${T}$ be a symmetric ${(1,1)}$ tensor on ${\Sigma}$, ${f}$ be a smooth function on ${\Sigma}$ and ${X}$ be a vector field on a neighborhood of ${\Sigma}$ in ${M}$. Then $\displaystyle \int_\Sigma \left ( \langle T(\nabla f), X^T\rangle+ f(\mathrm{div}\;T)(X^T)+ \frac{1}{2}f\langle T^\flat, \phi^* (\mathcal{L}_Xg )\rangle -f\langle T^\flat, A^{X^\perp }\rangle\right )=0.$ Here ${T^\flat}$ is the ${(0,2)}$-tensor defined by ${T^\flat (Y, Z)= \langle T(Y), Z\rangle}$, ${X^T}$ (resp. ${X^\perp}$) is the tangential (resp. perpendicular) component of ${X}$ and ${\mathcal{L}_Xg }$ is the Lie derivative of ${g}$.

Proof: Let ${Y}$ be the vector field defined by ${Y=f T(X^T)}$. Locally, let ${\{e_i\}_{i=1}^m}$ be a local orthonormal frame on ${\Sigma}$ such that ${\langle e_i , e_j\rangle= \mu_i \delta_{ij}}$, ${\mu=\pm 1}$. Then ${Y= \sum_{j=1}^m Y^j e_j}$, where ${Y^j= f \sum_{i=1}^m \mu_i T_{i}^j X\cdot e_i}$ and that ${ \mu_i T_i^j = \mu_j T_j^i}$. We can assume that ${\nabla _{e_i}e_j(p)=0}$ for all ${i,j}$. We compute the divergence of ${Y}$ at ${p}$:

$\displaystyle \begin{array}{rl} &\mathrm{div}(Y)\\ =& \langle \nabla f, T(X^T)\rangle+ f(\mathrm{div}\;T)(X^T)+ f\sum_{i,j=1}^m \mu_iT_{i}^j (\overline \nabla _{e_j}X\cdot e_i+X\cdot\overline \nabla _{e_j}e_i)\\ =& \langle T(\nabla f), X^T\rangle+ f(\mathrm{div}\;T)(X^T)+ \frac{1}{2}f\sum_{i,j=1}^m \mu_i T_{i}^j (\overline \nabla _{e_j}X\cdot e_i+ \overline \nabla _{e_i}X\cdot e_j)\\ &-f\sum_{i,j=1}^m \mu_i T_i^j (X\cdot A_{ji})\\ =& \langle T(\nabla f), X^T\rangle+ f(\mathrm{div}\;T)(X^T)+ \frac{1}{2}f\langle T^\flat, \phi^* (\mathcal{L}_Xg )\rangle -f\langle T^\flat, A^{X^\perp}\rangle \end{array}$

The result follows by applying divergence theorem. $\Box$

To proceed, let us recall that a vector field ${X}$ on ${M}$ is said to be a conformal (Killing) vector field if it satisfies

$\displaystyle \mathcal{L}_X g = 2 \alpha g \ \ \ \ \ (2)$

for some function ${\alpha}$ on ${M}$, and in this case, it is easy to see that ${\alpha= \frac{1}{n}\overline {\mathrm{div}}(X)}$. Here ${\overline {\mathrm{div}}}$ is the divergence on ${M}$. More generally, for an immersion ${\phi}$ of ${\Sigma}$ into ${(M,g)}$, a vector field ${X\in \Gamma (\phi^*( TM)) }$ is conformal along ${\phi}$ if ${\langle \overline \nabla _YX, Z\rangle+ \langle \overline \nabla _ZX,Y \rangle=2\alpha \langle X, Y\rangle}$ for any ${\Sigma}$-vector fields ${X, Y\in \Gamma(T\Sigma)}$.

We now state and prove our first main result.

 Theorem 6 Suppose ${M}$ is a space form. Assume ${X\in \Gamma(\phi^*(TM))}$ is a conformal vector field along ${\Sigma}$ with ${\alpha}$ given by (2), and ${f}$ is a smooth function on ${\Sigma}$. If ${0\le k\le m-1}$ is even, then $\displaystyle \int_\Sigma \alpha f \sigma_k =\int_\Sigma f\sigma_{k+1}\cdot X- \frac{1}{(m-k){m\choose k}}\int_\Sigma \langle T_k(\nabla f), X^T\rangle . \ \ \ \ \ (3)$ If ${m=n-1}$ (i.e. hypersurface), then $\displaystyle \int_\Sigma \alpha f \sigma_k =\int_\Sigma f\sigma_{k+1}\nu\cdot X- \frac{1}{(m-k){m\choose k}}\int_\Sigma \langle T_k(\nabla f), X^T\rangle . \ \ \ \ \ (4)$ Here ${\sigma_k}$ and ${ \sigma_{k+1}}$ are scalars, ${T_k}$ is understood to be an ordinary ${2}$-tensor, and ${\nu}$ is a unit normal vector field. If ${k=1}$, we can assume ${M}$ is Einstein instead. If ${k=0}$, we can remove any assumption on ${M}$.

Proof: Recall that ${\sigma_k = \frac{H_k}{{m\choose k}}}$. The result follows by applying Proposition 5 to ${T=T_k}$, and using Lemma 4. $\Box$

In general, it does not make sense to talk about ${\int_\Sigma \sigma_k}$ if ${k}$ is odd. Even when the normal bundle ${NM}$ is parallel so that ${\int_\Sigma \sigma_k}$ makes sense, our approach does not seem to produce a result similar to that of Theorem 6, as we cannot produce the term ${\sum_{i,j=1}^m(T_k)_i^j \cdot A_{ij}}$ and apply Lemma 4.

Instead, we will now take a different approach to derive a formula similar to (3) for odd ${k}$, which is due to Strobing [S]. Similar to Section 1, for a family of normal vector fields ${\nu_1, \nu_2, \cdots}$ (not necessarily distinct), we define

$\displaystyle H_k (\nu_1, \cdots, \nu_k)= \frac 1{k!}\sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon_{j_1\cdots j_k}^{i_1\cdots i_k}(A^{\nu_1})_{i_1}^{j_1} \cdots(A^{\nu_k})_{i_k}^{j_k},$

$\displaystyle \sigma_k (\nu_1, \cdots, \nu_k)=\frac{1}{{m\choose k}}H_k(\nu_1, \cdots, \nu_k),$

and

$\displaystyle {(T_k(\nu_1, \cdots, \nu_k))}_j^{\,i}= \frac 1 {k!} \sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon^{i i_1 \ldots i_k}_{j j_1 \ldots j_k} (A^{\nu_1})_{i_1}^{j_1} \cdots (A^{\nu_k})_{i_k}^{j_k}.$

Similar to Lemma 4, we have

 Lemma 7 For ${k\ge 1}$, we have (${\mathrm{tr}}$ denotes the trace on ${\Sigma}$): ${\mathrm{tr}(T_k(n_1, \cdots, \nu_k))= (m-k)H_k(\nu_1, \cdots, \nu_k)}$. If ${M}$ is a space form, and ${\nu_1,\cdots,\nu_k}$ are parallel in the normal bundle, then ${\mathrm{div}(T_k(\nu_1, \cdots, \nu_k))=0}$. i.e. ${\sum_{i=1}^m\nabla _{e_i}(T_{k}(\nu_1, \cdots, \nu_k))^i_{j}=0.}$ ${\sum_{i,j=1}^m (T_k(\nu_1, \cdots, \nu_k))_i^j (A^{\nu_{k+1}})_j^i=(k+1) H_{k+1}(\nu_1, \cdots, \nu_{k+1}). }$

Proof: The proof is exactly the same as in the codimension one case of Lemma 4, see e.g. [Mar] Lemma 2.1, except that in (7), we need the fact that ${\nabla _{e_i}(A^\nu)_j^k= \nabla _{e_j}(A^\nu)_i^k}$ if ${\nu}$ is parallel. $\Box$

By applying Proposition 5 to ${T_k(\nu_1, \cdots, \nu_k)}$ and using Lemma 7, we obtain the following

 Theorem 8 Suppose ${M}$ is a space form. Assume ${X\in \Gamma(\phi^*(TM))}$ is a conformal vector field along ${\Sigma}$ with ${\alpha}$ given by (2), ${f}$ is a smooth function on ${\Sigma}$ and ${\nu_1, \cdots, \nu_k}$ are (not necessarily distinct) normal fields to ${\Sigma}$ which are parallel in the normal bundle. Then $\displaystyle \begin{array}{rl} \int_\Sigma \alpha f \sigma_k(\nu_1, \cdots, \nu_k) =& \int_\Sigma f\sigma_{k+1}(\nu_1, \cdots, \nu_k, X^\perp)\\ &- \frac{1}{(m-k){m\choose k}}\int_\Sigma \langle T_k(\nu_1,\cdots, \nu_k)(\nabla f), X^T\rangle . \end{array}$
 Remark 1 If ${m=n-1}$, then Theorem 8 is reduced to (4) in Theorem 6.

3. Some examples and applications

By substituting different functions ${f}$ and ${X}$ in Theorem 6, we have several corollaries.

 Corollary 9 Suppose ${\Sigma}$ is immersed in ${\mathbb{R}^n}$. Then For all odd ${1\le k \le m}$, we have ${\int_\Sigma \sigma_k=0.}$ Here we regard ${\sigma_k}$ as a vector valued function. If ${\Sigma}$ is a hypersurface, then for all ${0\le k\le m=n-1}$, we have ${\int_\Sigma \sigma_k \nu=0.}$ Here we regard ${\sigma_k}$ as a scalar.

Proof: This follows from Theorem 6 by putting ${f=1}$ and ${X= E_i}$, ${i=1,\cdots, n}$, where ${E_i}$ are the standard orthonormal basis of ${\mathbb{R}^n}$. As ${E_i}$ are Killing vector fields, we have ${\alpha=0}$ and the result follows. (Alternatively, this also follows from integrating the divergence of the vector field (more appropriately, an ${n}$-tuple of vector fields) ${\sum_{i,j=1}^m(T_{k-1})^i_j X^je_i}$ on ${\Sigma}$, where ${X}$ is the position vector and ${X^j = \nabla _{e_j}X}$ regarded as an ${n}$-tuple.) $\Box$

 Corollary 10 Suppose ${\Sigma}$ is a closed hypersurface immersed in ${\mathbb{R}^n}$. Let ${X}$ be the position vector, ${r=|X|}$, ${0\le k\le m-1}$ and ${f}$ is a smooth function on ${\mathbb{R}}$. Assume that ${O\notin \Sigma}$, then we have $\displaystyle \int_\Sigma f(r) \sigma_k=\int_\Sigma f(r)\sigma_{k+1}X\cdot \nu- \frac{1}{(m-k){m\choose k}}\int_\Sigma \frac{f'(r)}{r}\langle T_k(X^T), X^T\rangle$ and $\displaystyle \int_\Sigma f(u) \sigma_k=\int_\Sigma u f(u)\sigma_{k+1}- \frac{1}{(m-k){m\choose k}}\int_\Sigma f'(u)\langle T_kA(X^T), X^T\rangle$ where ${u=X\cdot \nu}$.

Proof: The first equation follows from Theorem 6 and the observation that ${r\nabla r=X^T}$. The second equation follows by putting ${f=f(u)}$ noting that ${\nabla (X\cdot \nu)= A(X^T)}$.

$\Box$

We have the following extension of Alexandrov’s theorem.

 Corollary 11 Suppose ${\Sigma}$ is a closed hypersurface embedded in ${\mathbb{R}^n}$. Assume ${f>0}$, ${f'\ge 0}$ and there exists ${0\le k \le m}$ such that ${\sigma_k f(r)}$ is constant, or ${\sigma_k f(u)}$ is constant, and ${A\ge 0}$. Assume also that ${f}$ is injective if ${k=0}$. Then ${\Sigma}$ is a sphere, which is centered at ${O}$ if ${f}$ is injective.

Proof: Assume ${k\ge 1}$ and ${\sigma_k f(u)}$ is constant. Since ${\Sigma}$ has an elliptic point (i.e. point at which ${A>0}$), ${\sigma_kf(u)}$ must be positive and hence ${\sigma_k>0}$. By [Mar] Proposition 3.2, then ${T_j}$ is positive definite and ${\sigma_j>0}$ for ${0\le j. So by Corollary 10,

$\displaystyle \int_\Sigma \sigma_{k-1}f(u)\le \int_\Sigma \sigma_k f(u) X\cdot \nu=\sigma_k f(u) \int_\Sigma X\cdot \nu= \sigma_k f(u)n \mathrm{Vol}(\Omega)$

where ${\Omega}$ is the region bounded by ${\Sigma}$. Therefore

$\displaystyle n \mathrm{Vol}(\Omega)\ge \int_\Sigma\frac{\sigma_{k-1}}{\sigma_k}.$

By Proposition 12 below, ${\Sigma}$ must be a sphere. If ${f}$ is injective, then ${\langle X, \nu\rangle}$ is a positive constant as ${\sigma_k}$ is constant. In particular, ${\Sigma}$ is star-shaped w.r.t. ${O}$. Now, consider the furthest point ${p_1}$ and the nearest point ${p_2}$ on ${\Sigma}$ from ${O}$, we have ${r(p_1)=\langle X, \nu\rangle= r(p_2)}$. We conclude that ${\Sigma}$ is centered at ${O}$. The remaining cases can be proved similarly. $\Box$

 Proposition 12 Suppose ${\Sigma}$ is a closed hypersurface embedded in ${\mathbb{R}^n}$ such that ${\sigma_{k}>0}$, then $\displaystyle \int_\Sigma \frac{\sigma_{k-1}}{\sigma_{k}}\ge\int_\Sigma \frac{\sigma_{k-2}}{\sigma_{k-1}} \ge\cdots\ge \int_\Sigma\frac{1}{\sigma_1}\ge n \mathrm{Vol}(\Omega) \ \ \ \ \ (5)$ where ${\Omega}$ is the region bounded by ${\Sigma}$. The equality holds if and only if ${\Sigma}$ is a sphere.

Proof: By [Mar] Proposition 3.2, we have ${\sigma_{l}>0}$ for all ${l\le k}$. In particular, by Newton’s inequalities, we have

$\displaystyle \frac{1}{\sigma_1}\le \frac{\sigma_1}{\sigma_2}\le \cdots \le\frac{\sigma_{k-1}}{\sigma_k}.$

By a result of Ros ([Ros] Theorem 1), we have ${\int_\Sigma \frac{1}{\sigma_1}\ge n \mathrm{Vol}(\Omega).}$ The result follows.

If one of the equality ${\int_\Sigma \frac{\sigma_{j-1}}{\sigma_{j}}= \int_\Sigma \frac{\sigma_{j-2}}{\sigma_{j-1}}}$ holds then we conclude from the equality case of the Newton’s inequalities that ${\Sigma}$ is umbilical, and therefore is a sphere. If ${\int_\Sigma \frac{1}{\sigma_1}=n\mathrm{Vol}(\Omega)}$, then ${\Sigma}$ is also a sphere by [Ros] Theorem 1 again.

$\Box$

 Corollary 13 Suppose ${\Sigma}$ is a closed hypersurface immersed in ${\mathbb{R}^n}$ such that ${\sigma_{k}>0}$ for some ${1\le k\le n-1}$. Assume that ${p\ge0}$ and ${O\notin \Sigma}$. Then we have $\displaystyle \int_\Sigma r^{p}\le \int_\Sigma \sigma_1 r^{p+1}\le \cdots \le \int_\Sigma \sigma_k r^{p+k}$ where ${r=|X|}$. The equality occurs if and only if ${\Sigma}$ is a sphere centered at ${O}$. In particular, we have $\displaystyle \begin{array}{rl} \mathrm{Area}(\Sigma)\le \int_\Sigma \sigma_1 r\le \cdots\le \int_\Sigma\sigma_{k} r^{k} \end{array}$ and $\displaystyle n\mathrm{Vol}(\Omega)\le \int_\Sigma r\le \int_\Sigma \sigma_1 r^2 \le \cdots \le \int_\Sigma \sigma_{k}r^{k+1} \ \ \ \ \ (6)$ if ${\Sigma}$ is embedded. Here ${\Omega}$ is the region enclosed by ${\Sigma}$. The equality holds if and only if ${\Sigma}$ is a sphere centered at ${O}$.

Proof: By [Mar] Proposition 3.2, if ${\sigma_{k}>0}$ on ${\Sigma}$, then ${T_j}$ is positive for ${0\le j< k}$. By applying Corollary 10 with ${f=r^l}$ and the Cauchy-Schwarz inequality, we can get the inequalities. If the equality holds, then ${X^T=0}$ as ${T_j>0}$ for ${0\le j< k}$, but then ${\nabla (|X|^2)=0}$, which implies ${\Sigma}$ is a sphere centered at ${O}$. The converse is easy. The inequality (6) follows from the fact that ${n\mathrm{Vol}(\Omega)=\int_\Sigma X\cdot \nu\le \int_\Sigma r}$. $\Box$

 Remark 2 Corollary 13 generalizes [K1] Theorem 3.2 (1) and also [K2] Theorem 2.

To state our next result, we first set up the notations. We define ${\mathbb R^{p,q}}$, ${p+q=n+1}$, to be the vector space ${\mathbb R^{p+q}}$ equipped with the semi-Riemannian metric ${dx_1^2+\cdots + dx_p^2-dx_{p+1}^2-\cdots - dx_{p+q}^2}$. The position vector in ${\mathbb R^{p,q}}$ will be denoted by ${X}$ and the inner product on ${\mathbb R^{p,q}}$ by ${"\cdot"}$. Let ${\mu=\pm1}$ and ${M_{p,q}(\mu)=\lbrace X\in \mathbb R^{p,q}: X\cdot X=\mu\rbrace}$ be a pseudo-sphere in ${\mathbb R^{p,q}}$. It is easy to see that ${M_{p,q}(\mu)}$ is totally umbilic in ${\mathbb R^{p,q}}$ and in particular has constant curvature.

Let us recall that the classical Hsiung-Minkowski formulas [Hsiung]: if ${(M^n, g)}$ is a space form and ${\Sigma}$ is a closed oriented hypersurface in ${N}$ with a unit normal vector field ${\nu}$. Suppose ${M}$ possesses a conformal vector field ${Y}$, i.e. the Lie derivative of ${g}$ satisfies ${\mathcal L_Yg= 2 \alpha g}$ for some function ${\alpha}$, then we have

$\displaystyle \int_\Sigma \alpha\sigma_k= \int_\Sigma \sigma_{k+1}\langle Y,\nu\rangle.$

It is a nice observation that in general, if ${(M, g)}$ is a semi-Riemannian manifold which is isometrically embedded as a totally umbilic hypersuface in another semi-Riemannian manifold ${(N, h)}$, and such that there exists a conformal vector field ${Z}$ on ${N}$, then the orthogonal projection ${Z^T}$ of that vector field on ${M}$ is a conformal vector field on ${M}$. Indeed, a simple calculation shows that on ${ TM}$, if ${\mathcal{L}_Zh= \alpha h}$, then

$\displaystyle \mathcal{L}_{Z^T}g= 2 \alpha g -2 A^{Z^\perp}. \ \ \ \ \ (7)$

Therefore ${Z^T}$ is conformal on ${M}$ if ${M}$ is totally umbilic. In particular, we can construct a conformal vector field on ${M_{p,q}(\mu)}$ by projecting any conformal vector field on ${\mathbb R^{p,q}}$ onto ${M_{p,q}(\mu)}$.

In the following, we will consider the special case where the conformal vector field ${Y}$ on ${M_{p,q}(\mu)}$ is the orthogonal projection of a constant vector field on ${\mathbb R^{p,q}}$. More precisely, fix ${Z_0\in \mathbb R^{p,q}}$, considered as a parallel vector field on ${\mathbb R^{p,q}}$. The orthogonal projection ${Y}$ of ${-\mu Z_0}$ (this choice will make the conformal factor looks neater) on ${M_{p,q}(\mu)}$ is then given by ${-\mu Z_0=(- \mu Z_0)^T+ (-\mu Z_0)^\perp= Y(X)-(Z_0\cdot X)X}$, or equivalently,

$\displaystyle Y(X)=-\mu Z_0+(Z_0\cdot X)X\quad \textrm{ for }X\in M_{p,q}(\mu). \ \ \ \ \ (8)$

It is easily shown that the second fundamental form of ${M_{p,q}(\mu) }$ in ${\mathbb{R}^{p,q}}$ is

$\displaystyle A(U,V)=\mu g(U,V)X\quad \textrm{for }X\in M\textrm{ and }U,V \in T_XM_{p,q}(\mu).$

In particular, for ${Y}$ defined in (8), in view of (7), we have

$\displaystyle \mathcal{L}_Yg = 2 (Z_0 \cdot X )g \quad \textrm{at }X\in M_{p,q}(\mu). \ \ \ \ \ (9)$

By Theorem 6, Theorem 8, and in view of (9), we have the following result:

 Theorem 14 Let ${\Sigma}$ be an ${m}$-dimensional closed oriented semi-Riemannian manifold isometrically immersed in ${M_{p,q}(\mu)}$. Let ${f}$ be a smooth function on ${\Sigma}$, ${Z_0\in \mathbb{R}^{p,q}}$ be fixed and ${Y(X)}$ be given by (8). If ${0\le k\le m-1}$ is even, then $\displaystyle \begin{array}{rl} \int_\Sigma (Z_0\cdot X) f \sigma_k dS(X) =&\int_\Sigma f\sigma_{k+1}\cdot Y(X) dS(X)\\ &- \frac{1}{(m-k){m\choose k}}\int_\Sigma \langle T_k(\nabla f), Y^T\rangle dS(X). \end{array}$ If ${m=n-1}$ (i.e. hypersurface), then $\displaystyle \begin{array}{rl} \int_\Sigma (Z_0\cdot X)f \sigma_k dS(X) =&\int_\Sigma f\sigma_{k+1}\nu\cdot Y(X)dS(X)\\ &- \frac{1}{(n-1-k){{n-1}\choose k}}\int_\Sigma \langle T_k(\nabla f), Y^T\rangle dS(X). \end{array}$ Here ${\sigma_k}$ and ${ \sigma_{k+1}}$ are scalars, ${T_k}$ is understood to be an ordinary ${2}$-tensor, and ${\nu}$ is a unit normal vector field of ${\Sigma}$ in ${M_{p,q}(\mu)}$. If there exists (not necessarily distinct) normal fields ${\nu_1, \cdots, \nu_k}$ to ${\Sigma}$ which are parallel in the normal bundle. Then $\displaystyle \begin{array}{rl} &\int_\Sigma (Z_0 \cdot X) f \sigma_k(\nu_1, \cdots, \nu_k) dS(X)\\ =&\int_\Sigma f\sigma_{k+1}(\nu_1, \cdots, \nu_k, Y^\perp)dS(X) - \frac{1}{(m-k){m\choose k}}\int_\Sigma \langle T_k(\nu_1,\cdots, \nu_k)(\nabla f), Y^T\rangle dS(X). \end{array}$

In the following, we will apply Theorem 14 to ${M_{p,q}(\mu)}$ for different ${(p,q)}$ and ${\mu}$. For simplicity, we will only give the result when ${\Sigma}$ is a hypersurface in ${M_{p,q}(\mu)}$ (and consequently ${\sigma_k}$ are scalars).

Let us consider the case where ${(p,q)=(n+1, 0)}$ and ${\mu=1}$ so that ${M_{p, q}(\mu)=\mathbb S^n}$. Choose ${Z_0\in \mathbb S^n}$ and ${(r, \theta)}$ be the geodesic polar coordinates around ${Z_0}$ on ${\mathbb S^n}$, where ${\theta\in \mathbb S^{n-1}}$. Then

$\displaystyle Y=\sin r\partial _r\quad \textrm{and}\quad Z_0\cdot X=\cos r.$

By Theorem 14 and the above, we have

 Corollary 15 With the notations above, let ${\Sigma}$ be a closed hypersurface in ${\mathbb{S}^n}$ and ${\nu}$ be its unit normal. Suppose ${f}$ is a smooth function on ${\Sigma}$. Then for ${0\le k\le n-2}$, $\displaystyle \begin{array}{rl} \int_\Sigma \cos r f \sigma_k =&\int_\Sigma f\sigma_{k+1}\nu\cdot \sin r \partial _r - \frac{1}{(n-1-k){{n-1}\choose k}}\int_\Sigma \langle T_k(\nabla f), (\sin r \partial _r) ^T\rangle. \end{array}$

By substituting different functions ${f}$ in Corollary 15, we have the following corollary:

 Corollary 16 With the same assumptions as in Corollary 15, suppose ${0\le k\le n-2}$ and ${\Sigma}$ is contained in the open hemisphere centered at ${Z_0}$. Assume that ${Z_0\notin \Sigma}$, then we have $\displaystyle \int_\Sigma \tan^p r \cos r\sigma_k=\int_\Sigma \tan^{p+1}r\cos r \sigma_{k+1}\nu\cdot \partial _r- \frac{p}{(n-1-k){{n-1}\choose k}}\int_\Sigma \frac{\tan ^{p}r}{\cos r}\langle T_k(\partial _r^T), \partial _r^T\rangle$ and $\displaystyle \int_\Sigma u^p \cos r\sigma_k=\int_\Sigma u^{p+1}\sigma_{k+1}- \frac{p}{(n-1-k){{n-1}\choose k}}\int_\Sigma u^{p-1}\langle T_kA((\sin r \partial _r)^T), (\sin r \partial _r)^T\rangle$ where ${u=Y\cdot\nu=\sin r \partial _r\cdot \nu}$.

Proof: The first equation follows by putting ${f=\tan^p r }$ into Corollary 15 and observing that ${\sin r\nabla r=Y^T}$. The second equation follows by putting ${f=(Y\cdot\nu)^p}$ and noting that ${\nabla (Y\cdot \nu)= A(Y^T)}$.

$\Box$

 Corollary 17 With the same assumptions as in Corollary 15, suppose ${1\le k\le n-1}$ and ${\Sigma}$ is contained in the open hemisphere centered at ${Z_0}$ with ${\sigma_{k}>0}$. $\displaystyle \mathrm{Area}(\Sigma)=\int_\Sigma \sigma_0\le \int_\Sigma \sigma_1\tan r\le \int_\Sigma \sigma_2 \tan^2 r\le \cdots \le \int_\Sigma \sigma_{k}\tan ^{k}r. \ \ \ \ \ (10)$ The equality occurs if and only if ${\Sigma}$ is a sphere centered at ${Z_0}$.

Proof: By [Mar] Proposition 3.2, if ${\sigma_{k}>0}$ on ${\Sigma}$, then ${T_j}$ and ${\sigma_j}$ are both positive for ${0\le j< k}$. Applying Corollary 15, we have

$\displaystyle \begin{array}{rl} &\int_\Sigma \sigma_j \tan ^j r =\int_\Sigma \sigma_j\frac{\tan ^j r}{\cos r}\cos r \\ =& \int_\Sigma \sigma_{j+1} \tan^{j+1} r\partial _r\cdot\nu-\frac{1}{(n-1-j){{n-1}\choose j}}\int_\Sigma (\frac{j \tan ^{j}r}{\cos ^2 r}+\tan^{j+2} r)\langle T_j( \nabla r),\nabla r\rangle\\ \le& \int_\Sigma \sigma_{j+1} \tan^{j+1} r. \end{array}$

The equality (10) then follows by induction. If the equality case holds, then ${\nabla r=0}$ and so ${\Sigma}$ is a sphere centered at ${O}$. The converse is easy. $\Box$

For the case where ${(p,q)=(n,1)}$ and ${\mu=-1}$ so that ${M_{p, q}(\mu)=\mathbb H^n\sqcup \mathbb H^n}$. We can choose ${Z_0\in \mathbb H^n}$ and ${(r, \theta)}$ be the geodesic polar coordinates around ${Z_0}$ on ${\mathbb H^n}$, where ${\theta\in \mathbb S^{n-1}}$. Then

$\displaystyle Y=-\sinh r\partial _r\quad \textrm{and}\quad Z_0\cdot X=-\cosh r.$

By Theorem 14, we have

 Corollary 18 With the notations above, let ${\Sigma}$ be a closed hypersurface in ${\mathbb H^n}$ with unit normal vector ${\nu}$. Suppose ${f}$ is a smooth function on ${\Sigma}$, then for ${0\le k \le n-2}$, we have $\displaystyle \int_\Sigma f \cosh r \sigma_k= \int_\Sigma f\sigma_{k+1}\nu\cdot \sinh r \partial _r- \frac{1}{(n-1-k){{n-1}\choose k}}\int_\Sigma \langle T_k(\nabla f), (\sinh r \partial _r)^T\rangle.$

For the case where ${(p,q)=(n,1)}$ and ${\mu=1}$ so that ${M_{p, q}(\mu)=dS_n}$, the de Sitter space. We choose ${Z=(0, \cdots, 0, 1)}$ and parametrize ${dS_n}$ by ${X=(\cosh r \theta, \sinh r)}$, where ${\theta\in \mathbb S^{n-1}}$. Then

$\displaystyle Y=-\cosh r\partial _r\quad \textrm{and}\quad Z\cdot X=-\sinh r.$

By Theorem 14, we have

 Corollary 19 With the notations above, let ${\Sigma^{m}}$ (${m=n-1}$) be a closed spacelike hypersurface in ${d S_n}$ with unit normal vector ${\nu}$. Suppose ${f}$ is a smooth function on ${\Sigma}$, then for ${0\le k \le m-1}$, we have $\displaystyle \int_\Sigma f \sinh r \sigma_k= -\int_\Sigma f\sigma_{k+1}\langle \cosh r \partial _r, \nu\rangle- \frac{1}{m{{m-1}\choose k}}\int_\Sigma \langle T_k(\nabla f), (\cosh r \partial _r)^T\rangle.$

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