Self-shrinkers for the mean curvature flow in arbitrary codimension

Abstract

In this paper, we generalize Colding–Minicozzi’s recent results about codimension-1 self-shrinkers for the mean curvature flow to higher codimension. In particular, we prove that the sphere ${bf S}^{n}(\sqrt{2n})$ is the only complete embedded connected $F$-stable self-shrinker in $\mathbf{R}^{n+k}$ with $\mathbf{H}\ne 0$, polynomial volume growth, flat normal bundle and bounded geometry. We also discuss some properties of symplectic self-shrinkers, proving that any complete symplectic self-shrinker in $\mathbf{R}^4$ with polynomial volume growth and bounded second fundamental form is a plane. As a corollary, we show that there is no finite time Type I singularity for symplectic mean curvature flow, which has been proved by Chen–Li using different method. We also study Lagrangian self-shrinkers and prove that for Lagrangian mean curvature flow, the blow-up limit of the singularity may be not $F$-stable.

Appendices

Appendix A

In this appendix, we will prove the variations of normal vector field and mean curvature we need in Sect. 3. The proof is standard. When the variation vector $\mathbf{V }$ is the mean curvature vector $\mathbf{H }$, they are proved in [12]. We will follow the computations in [12].

We begin with fixing our notation. In a normal coordinate around some point in $\Sigma$, the induced metric on $\Sigma$ is given by

$$\begin{aligned} g_{ij}=\langle \partial _i F,\partial _j F\rangle , \end{aligned}$$

(8.1)

where $\partial _i(i=1,\ldots ,n)$ are the partial derivatives with respect to the local coordinate. Here, $\langle \cdot ,\cdot \rangle$ is the inner product of $\mathbf{R }^{n+k}$.

We choose a local field of orthonormal frames $e_1, \cdots , e_n, e_{n+1}, \cdots , e_{n+k}$ of $\mathbf{R }^{n+k}$ along $\Sigma _s$ such that $e_1,\ldots , e_n$ are tangent vectors of $\Sigma _s$ and $e_{n+1},\ldots ,e_{n+k}$ are in the normal bundle over $\Sigma _s$. From now on, we will agree on the following index ranges:

$$\begin{aligned} 1\le i,j,k,l \le n, \ \ n+1\le \alpha ,\beta ,\gamma \le n+k, \ \ 1\le A,B,C \le n+k. \end{aligned}$$

We can write

$$\begin{aligned} \mathbf{A }=\mathbf{A }^{\alpha }e_{\alpha }, \ \ \ \mathbf{H }=-H^{\alpha }e_{\alpha }. \end{aligned}$$

Let $\mathbf{A }^{\alpha }=(h^{\alpha }_{ij})$, where $(h^{\alpha }_{ij})$ is a matrix. By the Weingarten equation, we have

$$\begin{aligned} h^{\alpha }_{ij}=\langle \bar{\nabla }_{e_i}e_{\alpha },e_j\rangle =-\langle e_{\alpha },\bar{\nabla }_{e_i}e_j\rangle =h^{\alpha }_{ji}, \end{aligned}$$

where $\bar{\nabla }$ is the Levi-Civita connection on $\mathbf{R }^{n+k}$. Furthermore,

$$\begin{aligned} H^{\alpha }=g^{ij}h^{\alpha }_{ij}=h^{\alpha }_{ii}. \end{aligned}$$

Suppose the variation vector filed is $\mathbf{V }=V^{\alpha }e_{\alpha }$, i.e.,

$$\begin{aligned} F(\cdot ,s):\Sigma \rightarrow \mathbf{R }^{n+k} \end{aligned}$$

satisfies

$$\begin{aligned} \frac{\partial F}{\partial s}=\mathbf{V }. \end{aligned}$$

Then we have

Lemma 8.1

The induced metric satisfies

$$\begin{aligned} \frac{\partial }{\partial s}g_{ij}=2V^{\alpha }h^{\alpha }_{ij}, \end{aligned}$$

(8.2)

and

$$\begin{aligned} \frac{\partial }{\partial s}g^{ij}=-2V^{\alpha }h^{\alpha }_{ij}. \end{aligned}$$

(8.3)

Proof

We prove it at a fixed point. We have

$$\begin{aligned} \frac{\partial }{\partial s}g_{ij}&= \bar{\nabla }_{\mathbf{V }}\langle \partial _i F,\partial _j F\rangle = \langle \bar{\nabla }_{\mathbf{V }} \partial _i F,e_j \rangle +\langle e_i,\bar{\nabla }_{\mathbf{V }} \partial _j F\rangle \nonumber \\&= \langle \bar{\nabla }_{e_i} \mathbf{V },e_j \rangle +\langle e_i,\bar{\nabla }_{e_j} \mathbf{V }\rangle \nonumber \\&= \langle \bar{\nabla }_{e_i} (V^{\alpha }e_{\alpha }),e_j \rangle +\langle e_i,\bar{\nabla }_{e_j} (V^{\alpha }e_{\alpha })\rangle \nonumber \\&= V^{\alpha }\langle \bar{\nabla }_{e_i} e_{\alpha },e_j \rangle +V^{\alpha }\langle e_i,\bar{\nabla }_{e_j} e_{\alpha }\rangle \nonumber \\&= 2V^{\alpha }h^{\alpha }_{ij}. \end{aligned}$$

Here we have used the fact that

$$\begin{aligned} \bar{\nabla }_{\mathbf{V }} \partial _i F=\frac{\partial ^2}{\partial s\partial x_i}F=\bar{\nabla }_{\partial _i F}\mathbf{V }. \end{aligned}$$

As at the fixed point $p, g_{ij}(p)=\delta _{ij}$, we know that

$$\begin{aligned} \frac{\partial }{\partial s}g^{ij}=-2V^{\alpha }h^{\alpha }_{ij}. \end{aligned}$$

$\square$

Lemma 8.2

Denote $\langle \frac{\partial }{\partial s}e_{\alpha },e_{\beta } \rangle =\langle \bar{\nabla }_{\mathbf{V }}e_{\alpha },e_{\beta } \rangle =b^{\beta }_{\alpha }$, then $b^{\beta }_{\alpha }=-b_{\beta }^{\alpha }$, and we have

$$\begin{aligned} \frac{\partial }{\partial s}e_{\alpha }=-\nabla V^{\alpha }-V^{\beta }\langle \bar{\nabla }_{e_i}e_{\beta },e_{\alpha } \rangle e_i+b^{\beta }_{\alpha }e_{\beta }. \end{aligned}$$

(8.4)

Here, $\nabla V^{\alpha }$ is the covariant differential for the induced metric on $\Sigma _s$.

Proof

We have

$$\begin{aligned} \frac{\partial }{\partial s}e_{\alpha }&= \langle \frac{\partial }{\partial s}e_{\alpha },e_i \rangle e_i +\langle \frac{\partial }{\partial s}e_{\alpha },e_{\beta } \rangle e_{\beta } \nonumber \\&= \langle \bar{\nabla }_{\mathbf{V }} e_{\alpha },e_i \rangle e_i +b^{\beta }_{\alpha }e_{\beta }\nonumber \\&= -\langle e_{\alpha },\bar{\nabla }_{\mathbf{V }}\partial _i F \rangle e_i +b^{\beta }_{\alpha }e_{\beta }\nonumber \\&= -\langle e_{\alpha },\bar{\nabla }_{e_i}\mathbf{V }\rangle e_i +b^{\beta }_{\alpha }e_{\beta }\nonumber \\&= -\langle e_{\alpha },\bar{\nabla }_{e_i}(V^{\beta }e_{\beta })\rangle e_i +b^{\beta }_{\alpha }e_{\beta }\nonumber \\&= -\langle e_{\alpha },(\bar{\nabla }_{e_i}V^{\beta })e_{\beta } +V^{\beta }\bar{\nabla }_{e_i}e_{\beta }\rangle e_i +b^{\beta }_{\alpha }e_{\beta }\nonumber \\&= -(\bar{\nabla }_{e_i}V^{\alpha })e_i- V^{\beta } \langle \bar{\nabla }_{e_i}e_{\beta },e_{\alpha }\rangle e_i +b^{\beta }_{\alpha }e_{\beta }\nonumber \\&= -\nabla V^{\alpha }-V^{\beta }\langle \bar{\nabla }_{e_i} e_{\beta },e_{\alpha } \rangle e_i+b^{\beta }_{\alpha }e_{\beta }. \end{aligned}$$

$\square$

Lemma 8.3

The second fundamental form satisfies

$$\begin{aligned} \frac{\partial }{\partial s}h_{ij}^{\alpha }=-V^{\alpha }_{,ji}+V^{\beta }h_{ik}^{\alpha }h_{jk}^{\beta } +h_{ij}^{\beta }\langle e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle . \end{aligned}$$

(8.5)

Here, $V^{\alpha }_{,ji}$ denotes the second covariant derivative for the connection on the normal bundle.

Proof

We compute at a fixed point $p\in \Sigma$. We can choose a frame $e_i$ so that $\bar{\nabla }^T_{e_i}e_j(p)=0$, i.e., at $p, \bar{\nabla }_{e_i}e_j=-h^{\beta }_{ij}e_{\beta }$. From

$$\begin{aligned} h^{\alpha }_{ij}=-\langle \bar{\nabla }_{e_i}e_j, e_{\alpha }\rangle , \end{aligned}$$

and the fact that $\mathbf{R }^{n+k}$ is flat, we have

$$\begin{aligned} \frac{\partial }{\partial s}h^{\alpha }_{ij}&= -\langle \bar{\nabla }_{\mathbf{V }}\bar{\nabla }_{e_i}e_j, e_{\alpha }\rangle -\langle \bar{\nabla }_{e_i}e_j, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= -\langle \bar{\nabla }_{e_i}\bar{\nabla }_{\mathbf{V }}e_j, e_{\alpha }\rangle -\langle \bar{\nabla }_{[\mathbf{V },e_i]}e_j, e_{\alpha }\rangle -\langle \bar{\nabla }_{e_i}e_j, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= -\langle \bar{\nabla }_{e_i}\bar{\nabla }_{e_j}\mathbf{V }, e_{\alpha }\rangle -\langle -h^{\beta }_{ij}e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= -\langle \bar{\nabla }_{e_i}(\bar{\nabla }_{e_j}^T\mathbf{V }+\bar{\nabla }_{e_j}^{\perp }\mathbf{V }), e_{\alpha }\rangle +\langle h^{\beta }_{ij}e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= -\langle \bar{\nabla }_{e_i}(\bar{\nabla }_{e_j}^T\mathbf{V }), e_{\alpha }\rangle -\langle \bar{\nabla }_{e_i}^{\perp }\bar{\nabla }_{e_j}^{\perp }\mathbf{V }, e_{\alpha }\rangle +\langle h^{\beta }_{ij}e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= \langle \bar{\nabla }_{e_j}^T\mathbf{V }, \bar{\nabla }_{e_i}e_{\alpha }\rangle -\langle V^{\beta }_{,ji}e_{\beta }, e_{\alpha }\rangle +\langle h^{\beta }_{ij}e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= \langle \bar{\nabla }_{e_j}^T(V^{\beta }e_{\beta }), h^{\alpha }_{ik}e_k\rangle -V^{\beta }_{,ji}\langle e_{\beta }, e_{\alpha }\rangle +\langle h^{\beta }_{ij}e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= -V^{\alpha }_{,ji}+V^{\beta }\langle h^{\beta }_{jl}e_l, h^{\alpha }_{ik}e_k\rangle +\langle h^{\beta }_{ij}e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle \nonumber \\&= -V^{\alpha }_{,ji}+V^{\beta }h^{\alpha }_{ik}h^{\beta }_{jk} +h^{\beta }_{ij}\langle e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle . \end{aligned}$$

$\square$

Lemma 8.4

The mean curvature vector satisfies

$$\begin{aligned} \frac{\partial }{\partial s}\mathbf{H }=(\Delta V^{\alpha }+V^{\beta }h^{\beta }_{ij}h^{\alpha }_{ij})e_{\alpha }+H^{\alpha }\nabla V^{\alpha } +H^{\alpha }V^{\beta }\langle \bar{\nabla }_{e_i}e_{\beta },e_{\alpha }\rangle e_i. \end{aligned}$$

(8.6)

Proof

By Lemma 8.1 and Lemma 8.3, we have

$$\begin{aligned} \frac{\partial }{\partial s}H^{\alpha }&= \frac{\partial }{\partial s}\left(g^{ij}h^{\alpha }_{ij}\right) = \left(\frac{\partial }{\partial s}g^{ij}\right)h^{\alpha }_{ij}+g^{ij}\frac{\partial }{\partial s}h^{\alpha }_{ij}\nonumber \\&= -2V^{\beta }h^{\beta }_{ij}h^{\alpha }_{ij}+(-\Delta V^{\alpha }+V^{\beta }h^{\beta }_{ij}h^{\alpha }_{ij} +H^{\beta }\langle e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle )\nonumber \\&= -\Delta V^{\alpha }-V^{\beta }h^{\beta }_{ij}h^{\alpha }_{ij} +H^{\beta }\langle e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle . \end{aligned}$$

(8.7)

Combining with Lemma 8.2, we obtain

$$\begin{aligned} \frac{\partial }{\partial s}\mathbf{H }&= \frac{\partial }{\partial s}\left(-H^{\alpha }e_{\alpha }\right) =-\left(\frac{\partial }{\partial s}H^{\alpha }\right) e_{\alpha }-H^{\alpha }\frac{\partial }{\partial s}e_{\alpha } \nonumber \\&= \left(\Delta V^{\alpha }+V^{\beta }h^{\beta }_{ij}h^{\alpha }_{ij} -H^{\beta }\langle e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha } \rangle \right)e_{\alpha } \nonumber \\&+H^{\alpha }\nabla V^{\alpha }+H^{\alpha }V^{\beta }\langle \bar{\nabla }_{e_i}e_{\beta },e_{\alpha } \rangle e_i -H^{\alpha }b^{\beta }_{\alpha }e_{\beta }. \end{aligned}$$

Note that

$$\begin{aligned} -H^{\beta }\langle e_{\beta }, \bar{\nabla }_{\mathbf{V }}e_{\alpha }\rangle e_{\alpha } =-H^{\beta }b^{\beta }_{\alpha } e_{\alpha }=-H^{\alpha }b^{\alpha }_{\beta } e_{\beta }=H^{\alpha }b^{\beta }_{\alpha }e_{\beta }. \end{aligned}$$

Thus we have

$$\begin{aligned} \frac{\partial }{\partial s}\mathbf{H }=(\Delta V^{\alpha }+V^{\beta }h^{\beta }_{ij}h^{\alpha }_{ij})e_{\alpha }+H^{\alpha }\nabla V^{\alpha } +H^{\alpha }V^{\beta }\langle \bar{\nabla }_{e_i}e_{\beta },e_{\alpha }\rangle e_i. \end{aligned}$$

$\square$

Appendix B

In this appendix, we will give another two geometric identities satisfied on self-shrinkers with arbitrary dimension and codimension. These results generalized Theorem 5.2 and Lemma 10.8 of [8].

Suppose $\Sigma ^n\subset \mathbf{R }^{n+k}$ is a self-shrinker. We choose a frame $\{e_A\}_{A=1}^{n+k}$ on $\mathbf{R }^{n+k}$ along $\Sigma$ such that $\{e_i\}_{i=1}^{n}$ are tangent to $\Sigma$ and $\{e_{\alpha }\}_{\alpha =n+1}^{n+p}$ are in the normal bundle. We will compute pointwise. So we will always choose the frame $\{e_i\}_{i=1}^{n}$ such that $\bar{\nabla }_{e_i}^Te_j(p)=0$, i.e., at $p, \bar{\nabla }_{e_i}e_j=-h^{\alpha }_{ij}e_{\alpha }$.

Lemma 9.1

Let $L$ be defined by (3.9). Suppose $\mathbf{w }\in \mathbf{R }^{n+k}$ is a fixed vector. Then on a self-shrinker $\Sigma ^n$ in $\mathbf{R }^{n+k}$, we have

$$\begin{aligned} L\mathbf{w }^{\perp }=\frac{1}{2}\mathbf{w }^{\perp }. \end{aligned}$$

(9.1)

Proof

By definition,

$$\begin{aligned} \mathbf{w }^{\perp }=\langle \mathbf{w },e_{\alpha }\rangle e_{\alpha }\equiv f^{\alpha }e_{\alpha }, \end{aligned}$$

where

$$\begin{aligned} f^{\alpha }=\langle \mathbf{w },e_{\alpha }\rangle . \end{aligned}$$

(9.2)

Then computing at $p$ using the above chosen frame, we have

$$\begin{aligned} f^{\alpha }_{,i}&= \langle \bar{\nabla }^N_{e_i}(f^{\gamma }e_{\gamma }),e_{\alpha }\rangle =e_i(f^{\gamma })\langle e_{\gamma },e_{\alpha }\rangle +f^{\gamma }\langle \bar{\nabla }^N_{e_i}e_{\gamma },e_{\alpha }\rangle \nonumber \\&= \bar{\nabla }_{e_i}f^{\alpha }+\langle \mathbf{w } ,e_{\gamma }\rangle \langle \bar{\nabla }^N_{e_i}e_{\gamma },e_{\alpha }\rangle \nonumber \\&= \bar{\nabla }_{e_i}\langle \mathbf{w } ,e_{\alpha }\rangle -\langle \mathbf{w } ,e_{\gamma }\rangle \langle e_{\gamma },\bar{\nabla }^N_{e_i}e_{\alpha }\rangle \nonumber \\&= \langle \mathbf{w },\bar{\nabla }_{e_i}e_{\alpha }\rangle -\langle \mathbf{w },\bar{\nabla }^N_{e_i}e_{\alpha }\rangle =h^{\alpha }_{ij}\langle \mathbf{w } ,e_j\rangle \end{aligned}$$

(9.3)

and

$$\begin{aligned} f^{\alpha }_{,ik}&= \langle \bar{\nabla }^N_{e_k}\bar{\nabla }^N_{e_i}(f^{\gamma }e_{\gamma }), e_{\alpha }\rangle =\langle \bar{\nabla }^N_{e_k}(f^{\gamma }_{,i}e_{\gamma }),e_{\alpha }\rangle \nonumber \\&= e_k(f^{\gamma }_{,i})\langle e_{\gamma },e_{\alpha }\rangle +f^{\gamma }_{,i}\langle \bar{\nabla }^N_{e_k}e_{\gamma },e_{\alpha } \rangle \nonumber \\&= e_k(f^{\alpha }_{,i}) +h^{\gamma }_{ij}\langle \mathbf{w } ,e_j\rangle \langle \bar{\nabla }^N_{e_k}e_{\gamma },e_{\alpha }\rangle \nonumber \\&= e_{k}(h^{\alpha }_{ij})\langle \mathbf{w } ,e_j\rangle +h^{\alpha }_{ij}\bar{\nabla }_{e_k}\langle \mathbf{w } ,e_j\rangle +h^{\gamma }_{ij}\langle \mathbf{w } ,e_j\rangle \langle \bar{\nabla }^N_{e_k}e_{\gamma },e_{\alpha }\rangle \nonumber \\&= \left(e_{k}(h^{\alpha }_{ij}) +h^{\gamma }_{ij}\langle \bar{\nabla }^N_{e_k}e_{\gamma },e_{\alpha }\rangle \right)\langle \mathbf{w } ,e_j\rangle -h^{\alpha }_{ij}\langle \mathbf{w } ,\bar{\nabla }_{e_k}e_j\rangle \nonumber \\&= h^{\alpha }_{ik,j}\langle \mathbf{w } ,e_j\rangle -h^{\alpha }_{ij}h^{\beta }_{kj}\langle \mathbf{w },e_{\beta }\rangle . \end{aligned}$$

(9.4)

Here, we have used (6.23) and Codazzi equation. Taking trace of (9.4) and using (6.25), we obtain

$$\begin{aligned} \Delta f^{\alpha } = H^{\alpha }_{,j}\langle \mathbf{w } ,e_j\rangle -h^{\alpha }_{ij}h^{\beta }_{ij}\langle \mathbf{w },e_{\beta }\rangle = \langle \mathbf{w } ,\nabla H^{\alpha }\rangle -f^{\beta }h^{\alpha }_{ij}h^{\beta }_{ij}. \end{aligned}$$

(9.5)

By (6.21) and (9.3), we have

$$\begin{aligned} \langle \mathbf{w } ,\nabla H^{\alpha }\rangle = H^{\alpha }_{,i}\langle \mathbf{w } ,e_i\rangle =\frac{1}{2}h^{\alpha }_{ij}\langle \mathbf{x } ,e_j\rangle \langle \mathbf{w } ,e_i\rangle = \frac{1}{2}f^{\alpha }_{,j}\langle \mathbf{x } ,e_j\rangle =\frac{1}{2}\langle \mathbf{x }, \nabla f^{\alpha }\rangle . \end{aligned}$$

(9.6)

Putting (9.6) into (9.5), we obtain

$$\begin{aligned} \Delta f^{\alpha }+f^{\beta }h^{\alpha }_{ij}h^{\beta }_{ij}-\frac{1}{2}\langle \mathbf{x }, \nabla f^{\alpha }\rangle +\frac{1}{2}f^{\alpha } =\frac{1}{2}f^{\alpha }. \end{aligned}$$

(9.7)

By definition of the operator $L$, this is equivalent to (6.24). $\square$

The following result needs “flat normal bundle” assumption on the self-shrinker.

Lemma 9.2

If we extend the operator $L$ to tensors, then on a self-shrinker $\Sigma ^n$ in ${{\varvec{R}}}^{n+k}$ with flat normal bundle, we have

$$\begin{aligned} L{{\varvec{A}}}={{\varvec{A}}}. \end{aligned}$$

(9.8)

Proof

We will show that

$$\begin{aligned} (L\mathbf{A })^{\alpha }_{ij}=h^{\alpha }_{ij}. \end{aligned}$$

(9.9)

In general, we have the following Simons’ equality for the second fundamental form [12, 26]:

$$\begin{aligned} \Delta h^{\alpha }_{ij} = \nabla _i\nabla _jH^{\alpha }+H^{\beta }h^{\beta }_{il} h^{\alpha }_{lj}-h^{\beta }_{ik}h^{\beta }_{kl}h^{\alpha }_{lj} +2h^{\beta }_{ik}h^{\alpha }_{kl}h^{\beta }_{lj} -h^{\beta }_{ij}h^{\beta }_{kl}h^{\alpha }_{kl}-h^{\alpha }_{ik} h^{\beta }_{kl}h^{\beta }_{lj}.\qquad \quad \end{aligned}$$

(9.10)

Combining with (9.1), we have

$$\begin{aligned} (L\mathbf{A })^{\alpha }_{ij}&= \Delta h^{\alpha }_{ij}+h^{\beta }_{ij}h^{\beta }_{kl}h^{\alpha }_{kl} -\frac{1}{2}\langle \mathbf{x },\nabla h^{\alpha }_{ij}\rangle +\frac{1}{2}h^{\alpha }_{ij}\nonumber \\&= \frac{1}{2}\left(h^{\alpha }_{ij,k}\langle \mathbf{x } ,e_k\rangle +h^{\alpha }_{ij} -h^{\alpha }_{jk}h^{\beta }_{ki}\langle \mathbf{x },e_{\beta }\rangle \right) +H^{\beta }h^{\beta }_{il}h^{\alpha }_{lj}\nonumber \\&-h^{\beta }_{ik}h^{\beta }_{kl}h^{\alpha }_{lj}\ +2h^{\beta }_{ik}h^{\alpha }_{kl}h^{\beta }_{lj} -h^{\beta }_{ij}h^{\beta }_{kl}h^{\alpha }_{kl}-h^{\alpha }_{ik} h^{\beta }_{kl}h^{\beta }_{lj}\nonumber \\&+h^{\beta }_{ij}h^{\beta }_{kl}h^{\alpha }_{kl} -\frac{1}{2}\langle \mathbf{x },\nabla h^{\alpha }_{ij}\rangle +\frac{1}{2}h^{\alpha }_{ij}\nonumber \\&= h^{\alpha }_{ij}+2h^{\beta }_{ik}h^{\alpha }_{kl}h^{\beta }_{lj} -h^{\beta }_{ik}h^{\beta }_{kl}h^{\alpha }_{lj} -h^{\alpha }_{ik}h^{\beta }_{kl}h^{\beta }_{lj}. \end{aligned}$$

(9.11)

By Ricci equation,

$$\begin{aligned} 2h^{\beta }_{ik}h^{\alpha }_{kl}h^{\beta }_{lj}-h^{\beta }_{ik} h^{\beta }_{kl}h^{\alpha }_{lj}-h^{\alpha }_{ik}h^{\beta }_{kl}h^{\beta }_{lj}&= h^{\beta }_{ik}(h^{\alpha }_{kl}h^{\beta }_{lj}-h^{\beta }_{kl}h^{\alpha }_{lj}) +(h^{\beta }_{ik}h^{\alpha }_{kl}-h^{\alpha }_{ik}h^{\beta }_{kl}) h^{\beta }_{lj}\nonumber \\&= h^{\beta }_{ik}R_{\alpha \beta kj}+R_{\beta \alpha il}h^{\beta }_{lj}=0. \end{aligned}$$

The last equality follows from our assumption that the normal curvature is zero. Thus we obtain (9.9) from (9.11), This proves the lemma. $\square$