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).

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One Response to A characterization of constant higher mean curvature hypersurfaces in space forms

  1. Charles Li says:

    I added <!–more–> to break the long page (same for some of your other posts). Hope you don’t mind.

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