## A generalization of Hsiung-Minkowski formulas for hypersurfaces in the Euclidean space

In this short note, we will prove the following generalization of Hsiung-Minkowski formulas for hypersurfaces in the Euclidean space.

 Theorem 1 Let ${\Sigma\subset \mathbb R^n}$ be a closed hypersurface, then for any ${p\ge0}$ and ${0 \le k\le n-2}$, we have $\displaystyle \int_\Sigma u^p \sigma_k=\int_\Sigma u^{p+1}\sigma_{k+1}- \frac{p}{m{{m-1}\choose k}}\int_\Sigma u^{p-1}AT_k(X^T, X^T).$ Here ${\sigma_k}$ is the normalized ${k}$-th mean curvature, $u=X\cdot \nu$, ${X^T}$ is the tangential component of the position vector onto ${T\Sigma}$. (If ${p=0}$, we use the convention that ${u^p\equiv 1}$ and the second term on the RHS of this equation is understood to be zero. )

The classical Hsiung-Minkowski formulas [Hsiung] can be recovered by putting ${p=0}$ in the above equation. This also generalizes the results in one of my previous posts.

0.1. Preliminaries

Let us fix some notations. Let ${\Sigma\subset \mathbb R^n}$ be a hypersurface and ${m=n-1}$. Let ${\lbrace e_i\rbrace_{i=1}^{m}}$ be a local orthonormal frame on ${\Sigma}$ and let ${\nu=e_n}$ be the unit outward normal of ${\Sigma}$. Let ${\overline \nabla }$ and ${\nabla }$ be the connections on ${\mathbb R^n}$ and ${\Sigma}$ respectively. We define the the shape operator by ${A=\overline \nabla \nu: T\Sigma \rightarrow T\Sigma}$ and ${A_i^j}$ is defined by ${A(e_i)=\sum_{j=1}^m A_i^j e_j}$. By abusing of notation, we will also denote the second fundamental form ${\langle A(u), v\rangle}$ by ${A(u,v)}$.

We define the ${k}$-th mean curvature ${H_k}$ and the normalized ${k}$-th mean curvature ${\sigma_k}$ of ${\Sigma}$ by

$\displaystyle H _k = \frac 1{k!}\sum_{\substack{i_1,\cdots, i_k\\ j_1, \cdots, j_k}} \epsilon_{j_1\cdots j_k}^{i_1\cdots i_r}A_{i_1}^{j_1}\cdots A_{i_k}^{j_k}= \frac 1{k!}\sum_{i_1,\cdots, i_k} A_{i_1}^{[i_1}\cdots A_{i_k}^{i_k]} \textrm{\; and\;} \sigma_k= \frac{H_k}{{{n-1}\choose k}}$

respectively. We use the convention that ${H_0=\sigma_0=1}$.

Following [Reilly], we define the ${k}$-th Newton transformation ${T_k: T\Sigma\rightarrow T\Sigma}$ of ${A}$ as

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

Alternatively, ${T_k}$ can be defined recursively by (see e.g. [Reilly])

$\displaystyle T_0=\mathrm{Id}, \quad T_{k}=H_k I - AT_{k-1}\textrm{ for }k\ge 1. \ \ \ \ \ (1)$

We use the convention that ${A_i^{[i}A_j^{j]}=A_i^i A_j^j-A_i^jA_j^i}$, for example. Unless otherwise stated, repeated indices will be summed over ${1, \cdots, m}$.

We will need the following two lemmas.

 Lemma 2 For all ${1\le k\le m}$, we have the following identities: ${A_{[i_1}^{i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{i]}^{l}=(k-1)! (T_{k-1})^l_q A^q_i}$. ${\mathrm{div}(T_k)=0}$, i.e. ${(T_{k})^i_{j,i}=0.}$ ${(T_{k-1})^l_q A^q_{i.l}=(H_k)_i}$.

Proof: The equations (2) and (3) require the Codazzi equation ${A^i_{j,k}=A^i_{k,j}}$ which holds for any hypersurface in a space form. (2) is well-known and can be found e.g. in [Reilly]. To prove (1), let us assume that ${\lbrace e_i\rbrace_{i=1}^m}$ are the principal directions with principal curvatures ${\lbrace \lambda_i\rbrace_{i=1}^m}$.

$\displaystyle \begin{array}{rcl} (k-1)! (T_{k-1})^l_q A^q_{i} &=&\sum_{\substack{q, i_1,\cdots, i_{k-1}\\j_1, \cdots, j_{k-1}}} \epsilon^{li_1 \cdots i_{k-1}}_{qj_1\cdots j_{k-1}}A_{i_1}^{j_1}\cdots A_{i_{k-1}}^{j_{k-1}}A^q_i\\ &=&\sum_{\substack{q, i_1,\cdots, i_{k-1}\\j_1, \cdots, j_{k-1}}} \epsilon^{li_1 \cdots i_{k-1}}_{qj_1\cdots j_{k-1}}\lambda_{i_1}\cdots \lambda_{i_{k-1}}\lambda_{i}\delta_{i_1}^{j_1}\cdots \delta_{i_{k-1}}^{j_{k-1}}\delta^q_i\\ &=&\sum_{\substack{ i_1,\cdots, i_{k-1}\\j_1, \cdots, j_{k-1}}} \epsilon^{li_1 \cdots i_{k-1}}_{i j_1\cdots j_{k-1}}\lambda_{i_1}\cdots \lambda_{i_{k-1}}\lambda_{i}\delta_{i_1}^{j_1}\cdots \delta_{i_{k-1}}^{j_{k-1}}\\ &=&\sum_{i_1,\cdots, i_{k-1}} \lambda_{i_1}\cdots \lambda_{i_{k-1}}\lambda_{i}\delta_{i_1}^{[i_1}\cdots \delta_{i_{k-1}}^{j_{k-1}}\delta^{l]}_i\\ &=&A_{[i_1}^{i_1}\cdots A_{i_{k-1}}^{j_{k-1}}A^{l}_{i]}. \end{array}$

We now prove (3). Note that by (1) and (2), we have

$\displaystyle \begin{array}{rcl} T^l_q A^q_{i,l}&=&(T^l_q A^q_{i})_l =\frac{1}{(k-1)!}(A_{i_1}^{[i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{i}^{l]})_l\\ &=&\frac{1}{(k-1)!}\left((k-1)A_{i_1,l}^{[i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{i}^{l]}+A_{i_1}^{[i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{i,l}^{l]}\right)\\ &=&\frac{1}{(k-1)!}A_{i_1}^{[i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{i,l}^{l]}\\ &=&\frac{1}{(k-1)!}A_{i_1}^{[i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{l.i}^{l]}\\ &=&\frac{1}{k!}(A_{i_1}^{[i_1}\cdots A_{i_{k-1}}^{i_{k-1}}A_{l}^{l]})_i =(H_k)_i. \end{array}$

The third line follows from the fact that ${A_{i, l}^{i_1}}$ is symmetric in the indices ${i}$, ${l}$. $\Box$

 Lemma 3 For ${k\ge 1}$, we have $\displaystyle (k+1)H_{k+1}= H_k H_1-(T_{k-1})^l_q A^q_{i}A^{i}_l .$

Proof: See e.g. [LWX] Lemma 2. $\Box$

0.2. Main result

We can now state and prove our main result.

 Theorem 4 Let ${\Sigma\subset \mathbb R^n}$ be a closed hypersurface, then for any ${p\ge0}$ and ${k\ge 0}$, we have $\displaystyle \int_\Sigma u^p \sigma_k=\int_\Sigma u^{p+1}\sigma_{k+1}- \frac{p}{m{{m-1}\choose k}}\int_\Sigma u^{p-1}AT_k(X^T, X^T).$ Here ${X^T}$ is the tangential component of the position vector onto ${T\Sigma}$.

Proof: We first assume ${k\ge 1}$. In the following computation, we will omit ${\Sigma}$ in the integral symbol. Using Lemma 2 and integration by parts, we compute

$\displaystyle \begin{array}{rcl} -k \int u ^p H_k &=&-\frac{1}{(k-1)!}\int u^p A_{[i_1}^{i_1}\cdots A_{i_k]}^{i_k}\\ &=& \frac{1}{(k-1)!}\int u^p A_{[i_1}^{i_1}\cdots A_{i_{k-1}}^{i_{k-1}} X_{i_k]}^{i_k}\cdot \nu\\ &=& -\frac{1}{(k-1)!}\int u^p A_{[i_1}^{i_1}\cdots A_{i_{k-1}}^{i_{k-1}} \nu_{i_k]}\cdot X^{i_k}\\ &=& -\frac{1}{(k-1)!}\int u^p A_{[i_1}^{i_1}\cdots A_{i_{k-1}}^{i_{k-1}} A_{i_k]}^lX_l\cdot X^{i_k}\\ &=& -\int u^p (T_{k-1})^l_q A^q_{i}X_l\cdot X^{i}\\ &=& \int [(u^p )_l(T_{k-1})^l_q A^q_{i}X\cdot X^{i}+ u^p (T_{k-1})^l_q A^q_{i,l}X\cdot X^{i} + u^p (T_{k-1})^l_q A^q_{i}X\cdot X^{i}_l]\\ &=& \int [(u^p )_l(T_{k-1})^l_q A^q_{i}X\cdot X^{i}+ u^p (H_k)_{l}X\cdot X^{l} + u^p (T_{k-1})^l_q A^q_{i} X\cdot X^{i}_l]\\ &=& \int [(u^p )_l(T_{k-1})^l_q A^q_{i} X\cdot X^{i}-(u^p)_l H_k X\cdot X^{l}-u^pH_k X_l\cdot X^{l}\\ &&-u^p H_k X\cdot X^{l}_l-u^p (T_{k-1})^l_q A^q_{i}A^{i}_l X\cdot \nu]\\ &=& \int [(u^p )_l(T_{k-1})^l_q A^q_{i} X\cdot X^{i}-(u^p)_l H_k X\cdot X^{l}-mu^pH_k \\ &&+u^{p+1} H_k H_1-u^{p+1} (T_{k-1})^l_q A^q_{i}A^{i}_l ]. \end{array}$

So we have

$\displaystyle \begin{array}{rcl} (m-k)\int_\Sigma u^pH_k &=& \int_\Sigma [(u^p )_l(T_{k-1})^l_q A^q_{i} X\cdot X^{i}-(u^p)_l H_k X\cdot X^{l}\\ &&+u^{p+1} H_k H_1-u^{p+1} (T_{k-1})^l_q A^q_{i}A^{i}_l ]\\ &=& \int_\Sigma \left(u^p )_l(T_{k-1})^l_q A^q_{i} X\cdot X^{i}-(u^p)_l H_k X\cdot X^{l}\right)\\ & &+(k+1)\int_\Sigma u^{p+1}H_{k+1} \end{array} \ \ \ \ \ (2)$

where we have used Lemma 3. Using ${u_l=A_l^j(X\cdot X_j)}$, we have

$\displaystyle \begin{array}{rcl} &&(u^p )_l((T_{k-1})^l_q A^q_{i} X\cdot X^{i}-H_k X\cdot X^{l})\\ &=&pu^{p-1}A_l^j A_i^q(T_{k-1})_{q}^{l}(X\cdot X_j)(X\cdot X^{i})-pu^{p-1}H_kA_l^j (X\cdot X^{l})(X\cdot X_j)\\ &=& -pu ^{p-1} A(H_k I - AT_{k-1})(X^T, X^T)\\ &=& -pu ^{p-1} AT_{k}(X^T, X^T). \end{array}$

Here we have used (1) in the last line. Substitute this into (2), we obtain

$\displaystyle (m-k)\int_\Sigma u^p H_k = (k+1)\int_\Sigma u^{p+1}H_{k+1}- p \int_\Sigma u^{p-1}AT_k (X^T, X^T).$

This is equivalent to

$\displaystyle \int_\Sigma u^p \sigma_k=\int_\Sigma u^{p+1}\sigma_{k+1}- \frac{p}{m{{m-1}\choose k}}\int_\Sigma u^{p-1}AT_k(X^T, X^T).$

We now consider the case where ${k=0}$. Define the function ${f=\frac{r^2}{2}=\frac{|X|^2}{2}}$ on ${\mathbb R^n}$. Then it is easy to check that ${\overline \nabla ^2 f=g_0}$ where ${g_0}$ is the Euclidean metric. Let ${\lbrace e_i\rbrace_{i=1}^m}$ be a local orthonormal frame on ${\Sigma}$, ${e_n=\nu}$ and ${z}$ be the restriction of ${f}$ on ${\Sigma}$. Then we have ${\nabla z=X^T}$ and ${u=\frac{\partial f}{\partial \nu}=X\cdot \nu}$. From this we see that

$\displaystyle \delta_{ij}= \overline \nabla ^2_{ij} f= \nabla ^2_{ij}z+ u A_{ij}\textrm{\quad and \quad}0=\delta_{in}=\overline \nabla ^2 _{in}f= u_i - A_{i}^jz_j.$

So we have

$\displaystyle \begin{array}{rcl} \int_\Sigma pu^{p-1}A(X^T, X^T) =\int_\Sigma pu^{p-1}A_{j}^iz_iz_j &=&\int_\Sigma pu^{p-1}u_j z_j\\ &=&-\int_\Sigma u^pz_{jj}\\ &=&-\int_\Sigma u^p(\delta_{jj}-u A_{jj})\\ &=&\int_\Sigma (mu^{p+1} \sigma_1-mu^p ). \end{array}$

Therefore

$\displaystyle \int_\Sigma u^p= \int_\Sigma u^{p+1}\sigma_1-\frac p m \int_\Sigma u^{p-1}A(X^T, X^T).$

This completes the proof. $\Box$

 Corollary 5 For ${0\le k \le n-2}$, suppose ${\Sigma}$ is convex (i.e. ${A\ge 0}$), then we have $\displaystyle \int_\Sigma \sigma_k = \int_\Sigma \sigma_{k+1}u \le \int_\Sigma \sigma_{k+1}u^2 \le \int_\Sigma \sigma_{k+2}u^3 \le \cdots$ In particular, $\displaystyle n\mathrm{Vol}(\Omega)\le \int_\Sigma \sigma_1 u^2 \le \int_\Sigma \sigma_2 u^3\le \cdots$ and $\displaystyle \mathrm{Area}(\Sigma)\le \int_\Sigma \sigma_1 u \le \int_\Sigma \sigma_2 u^2\le \cdots$ Here ${\Omega}$ is the region bounded by ${\Sigma}$.

This entry was posted in Geometry, Inequalities. Bookmark the permalink.