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 <i_k\\ i\ne i_l}}\lambda_{i_1}\cdots \lambda_{i_k}e_i.

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}):

  1. {\mathrm{tr}(T_k)= (m-k)H_k}.
  2. 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}.
  3. 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}.

  1. 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)

  2. 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}):

  1. {\mathrm{tr}(T_k(n_1, \cdots, \nu_k))= (m-k)H_k(\nu_1, \cdots, \nu_k)}.
  2. 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.}
  3. {\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

  1. 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.
  2. 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

  1. {\sigma_k f(r)} is constant, or
  2. {\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<k}. 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).

  1. 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}

  2. 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)}.

  3. 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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