## A characterization of constant higher mean curvature hypersurfaces in space forms

In this post I will talk about a characterization of constant higher mean curvature hypersurfaces in space forms, the main result is a Schur-type lemma.

Let ${(N^{n+1},h)}$ be a space form and let ${\Sigma^n\subset N^{n+1}}$ be a closed hypersurface. Let ${\nu}$ be a global unit normal field on ${\Sigma}$, we define the second fundamental form ${A: T \Sigma \rightarrow T\Sigma}$ by ${A= \overline \nabla \nu}$, where ${\overline \nabla }$ is the connection in ${N}$.

We define the ${r}$-th mean curvature of ${\Sigma}$ in ${N}$ as follows. Suppose ${k_1, \cdots, k_n}$ are the principal curvatures of ${\Sigma}$ (eigenvalues of ${A}$), we define ${H_r}$ by the identity

$\displaystyle \prod_{i=1}^{n} (1+k_i t)= H_0+ H_1 t +H_2 t^2 +\cdots + H_{n}t^{n}.$

We define the ${r}$-th Newton transformation of ${A}$ (which can be regarded as a ${(0,2)}$ tensor) for ${r=0, \cdots, n}$ as the self-adjoint linear map

$\displaystyle T^{r}= H_rI - H_{r-1}A+\cdots + (-1)^r A^r. \ \ \ \ \ (1)$

Alternatively, they can be defined inductively by

$\displaystyle T^{0}=I, \quad T^{r+1}= H_{r+1}I -A T^{r}.$

Clearly, ${AT^{r}=T^{r}A}$ and so they can be simultaneously diagonalized, i.e. ${T^{r}}$ has the same eigenvectors as ${A}$.

In a local frame, the matrix entries of ${{T^{r}}}$ are given by ([Reilly] Proposition A)

$\displaystyle {T^r}_i^{\,j}=\frac 1{r!} \delta_{i \, i_1 \ldots i_r}^{j \,j_1 \ldots j_r}h_{j_1}^{i_1}\cdots h_{j_r}^{i_r} \ \ \ \ \ (2)$

where ${h_i^j}$ are the matrix entries of ${A}$ and ${\delta _{i_1 \ldots i_r}^{j_1 \ldots j_r} }$ are the generalized Kronecker delta symbols. This can be seen by comparing the two sides using a local frame which is a basis consisting of the eigenvectors of ${A}$.

 Lemma 1 Let ${\nabla }$ and ${\mathrm{tr}}$ be the connection and trace on ${\Sigma}$ respectively, then for all ${r}$, $\displaystyle \nabla ^i T^{r}_{ij}=0 \textrm{\quad and \quad}\mathrm{tr}(T^{r})= (n-r)H_r.$

Proof:

The first equation follows from

$\displaystyle \begin{array}{rcl} r! \nabla ^i T^{r}_{ij} &=& \delta_{i \, i_1 \ldots i_r}^{j \,j_1 \ldots j_r}(\nabla^ih_{j_1}^{i_1}h_{j_2}^{i_2}\cdots h_{j_r}^{i_r}+\cdots +h_{j_1}^{i_1}h_{j_2}^{i_2}\cdots \nabla^ih_{j_r}^{i_r})\\ &=& r\delta_{i \, i_1 \ldots i_r}^{j \,j_1 \ldots j_r}\nabla^ih_{j_1}^{i_1}h_{j_2}^{i_2}\cdots h_{j_r}^{i_r}\\ &=&r\delta_{i \, i_1 \ldots i_r}^{j \,j_1 \ldots j_r}\nabla^{i_1}h_{j_1}^{i}h_{j_2}^{i_2}\cdots h_{j_r}^{i_r}\\ &=&-r\delta_{i \, i_1 \ldots i_r}^{j \,j_1 \ldots j_r}\nabla^{i}h_{j_1}^{i_1}h_{j_2}^{i_2}\cdots h_{j_r}^{i_r}\\ &=&-r! \nabla ^i T^{r}_{ij} \end{array}$

where we have used the Codazzi equation ${\nabla ^i h_{j}^k=\nabla ^k h_{j}^i}$ in the third line (as ${N}$ has constant curvature). The second assertion follows by applying the Newton’s identities (hence the name Newton tensors) ${rH_r= H_{r-1} p_1-H_{r-2} p_2+\cdots + (-1)^{r-1}p_r}$ to (1), where ${p_j= \sum_{i=1}^n k_i^j=\mathrm{tr}(A^j)}$. $\Box$

 Theorem 2 Let ${\Sigma}$ be a closed hypersurface in a space form, ${\overline H_r=\frac 1{ \mathrm{Area}(\Sigma)}\int_\Sigma H_r}$ be the average of ${H_r}$ and ${\stackrel{\circ}{T^{r}}= T^{r}- \frac{(n-r)}n H_r g}$ be the traceless part of ${T^{r}}$. Suppose the Ricci curvature of ${\Sigma }$ is non-negative, then for ${r=1,\cdots, n-1}$, we have $\displaystyle \int_\Sigma (H_r-\overline H_r)^2 \le \frac{ n(n-1)}{(n-r)^2} \int_\Sigma |\stackrel \circ {T^{r}}|^2, \ \ \ \ \ (3)$ or equivalently, $\displaystyle \int_\Sigma |T^{r} - \frac{n-r}n \overline H_r g|^2 \le n \int_\Sigma |T^{r} - \frac{n-r}n H_r g|^2.$

Proof:

We can assume ${H_r-\overline H_r\neq 0}$. Let ${f}$ be the solution to

$\displaystyle \begin{cases} \Delta f = H_r- \overline H_r\\ \int_\Sigma f =0. \end{cases}$

The solution exists because ${\int_\Sigma H- \overline H_r=0}$. As ${\stackrel \circ {T^r} = T^r- \frac{(n-r)}n H_r g}$ and ${\textrm{div}(T^r)=0}$,

$\displaystyle \mathrm{div}(\stackrel\circ {T^r})=-\frac {n-r}nd H_r.$

$\displaystyle \begin{array}{rcl} \int_\Sigma (\Delta f)^2 = \int_\Sigma (H_r- \overline H_r)\Delta f &=& -\int_\Sigma \langle dH_r,df\rangle\\ &=&-\frac n{n-r} \int_\Sigma \langle \mathrm{div}(\stackrel \circ {T^r}),df\rangle\\ &=& \frac n{n-r}\int_\Sigma \langle \stackrel \circ {T^r} , \nabla ^2f\rangle\\ &=& \frac n{n-r}\int_\Sigma \langle \stackrel \circ {T^r} , \nabla ^2f- \frac {\Delta f }ng\rangle\\ &\le&\frac n{n-r} \|\stackrel \circ {T^r} \|_{L^2}\|\nabla ^2 f- \frac{\Delta f}n g\|_{L^2}. \end{array} \ \ \ \ \ (4)$

We have

$\displaystyle \begin{array}{rcl} \|\nabla ^2 f -\frac {\Delta f}ng\|_{L^2}^2 &= &\int_\Sigma |\nabla ^2 f|^2 +\frac 1n \int_\Sigma (\Delta f)^2 -\frac 2n \int_\Sigma (\Delta f)^2\\ &= &\int_\Sigma |\nabla ^2 f|^2 -\frac 1n \int_\Sigma (\Delta f)^2. \end{array} \ \ \ \ \ (5)$

Consider

$\displaystyle \begin{array}{rcl} \int_\Sigma |\nabla ^2f|^2 = \int_\Sigma f_{ij}f_{ij} = \int_\Sigma - f_{jij}f_i &=& \int_\Sigma (-f_{jji}f_i - R_{il}f_i f_l)\\ &=& \int_ M f_{jj}f_{ii} -\int_\Sigma Rc(\nabla f, \nabla f)\\ &=& \int_\Sigma (\Delta f)^2 -\int_\Sigma Rc (\nabla f, \nabla f). \end{array}$

Thus (5) becomes

$\displaystyle \begin{array}{rcl} \|\nabla ^2 f -\frac {\Delta f}ng\|_{L^2}^2 &=&\frac{n-1}n \int_\Sigma (\Delta f)^2 - \int_\Sigma Rc(\nabla f,\nabla f)\\ &\le& \frac{n-1}n \int_\Sigma (\Delta f)^2 \end{array}$

given ${Rc \ge 0}$. Substitute this into (4), we have

$\displaystyle \int_\Sigma (H_r-\overline H_r)^2 \le \frac{ n(n-1)}{(n-r)^2} \int_\Sigma |\stackrel \circ {T^{r}}|^2. \ \ \ \ \ (6)$

As ${T^{r}- \frac{n-r}n \overline H_r g = \stackrel\circ {T^{r}}+\frac{n-r}n (H_r-\overline H_r)g}$, we have

$\displaystyle |T^{r}- \frac{n-r}n \overline H_r g |^2= |\stackrel \circ {T^{r}}|^2 +\frac {(n-r)^2}n(H-\overline H_r)^2.$

Thus (6) can be rephrased as

$\displaystyle \int_\Sigma |T^{r} - \frac{n-r}n \overline H_r g|^2 \le n \int_\Sigma |T^{r} - \frac{n-r}n H_r g|^2.$

$\Box$

 Remark 1 When ${r=1}$, as ${T^1= H_1 I-A}$, it is easy to see that ${\stackrel \circ {T^{1}}= -\stackrel \circ A}$ where ${\stackrel \circ A}$ is the traceless part of the second fundamental form. Note also that ${H=H_1=k_1+\cdots+k_n}$ is the mean curvature. Thus (3) can be rewritten as $\displaystyle \int_\Sigma (H-\overline H)^2 \le \frac{ n}{(n-1)} \int_\Sigma |\stackrel \circ {A}|^2,$

 Remark 2 For a hypersurface in Euclidean space, having nonnegative Ricci curvature is equivalent to ${A\ge 0}$ (i.e. convex), see \cite{Perez} p. 48. So the curvature assumptions in Theorem 2 can be replaced by ${\Sigma}$ being convex.

Theorem 2 immediately implies the following Schur-type lemma:

 Corollary 3 With the same assumptions as in Theorem 2, if ${T^r= \frac{n-r}nH_r g}$, then ${H_r}$ is constant.

By the way, this is called a Schur-type lemma because in [Lellis-Topping], it was proved in a similar way that if ${(M^n,g)}$ (${n\ge 3}$)is a closed Riemannian manifold with nonnegative Ricci curvature, then

$\displaystyle \int_M (R-\overline R) ^2 \le \frac{4n(n-1)}{(n-2)^2}\int_M|\stackrel\circ {Ric}|^2.$

Clearly this implies that under the stated assumption, the Schur’s lemma holds: if ${\stackrel\circ {Ric}=0}$, i.e. ${M}$ is Einstein (${Ric=\frac R ng}$), then ${R}$ is constant (the original version has no restriction on ${Ric}$ and for ${M}$ being closed).