I am reading the book Lecture on mean curvature flow by Xi-Ping Zhu. Suppose $M^n$ is an n-dim smooth manifold and $X(x,t):M^n \rightarrow R^{n+1}$ be a one-parameter family of smooth immersion. Metric and the second fundamental form on $X(x,t)$ is defined as $$g_{ij} = (\frac{\partial X }{\partial x_i},\frac{\partial X }{\partial x_j})\,\,\,\,h_{ij}= (n(x,t),\frac{\partial^2 X }{\partial x_i\partial x_j})$$ And the covariant derivative of vector $v$ is defined by $$\nabla_j v_i = \frac{\partial v_i }{\partial x_j} + \Gamma^i_{jk}v_k$$ 1. How can i define the mean curvature $H$ by using the metric and second fundamental form ?
By setting $$\frac{\partial X }{\partial t}=H n$$ Using the Gauss equation and the Weingarten equation, we have $$g^{ij}\nabla_i\nabla_j X=g^{ij}(\frac{\partial^2 X }{\partial x_i\partial x_j}-\Gamma_{ij}^k\frac{\partial X }{\partial x_k})$$ 2.How can i get the above relation?
3.Using the Gauss equation and the Weingarten equation, the author further calculate that $$g^{ij}(\frac{\partial^2 X }{\partial x_i\partial x_j}-\Gamma_{ij}^k\frac{\partial X }{\partial x_k})=g^{ij}h_{ij}n=Hn=\frac{\partial X }{\partial t}$$ Later he used the De turck trick to make this equation become strictly parabolic. But I think it is already strictly parabolic. Thank you so much.
I am not sure I understand you well. If not , tell me. Besides, if there is wrong in my answer, tell me too. Thanks.
First, the mean curvature can be written $$ H = g^{ij} h_{ij} $$
Second, consider component $X^\alpha$ , treat it as function, so we have $$ \nabla _i \nabla _ j X^\alpha = \nabla_i (\nabla _j X^\alpha) - \nabla _{\nabla _i j} X^\alpha \\ = \frac{\partial ^2 X^\alpha}{ \partial x_i \partial x_j} -\nabla _{\Gamma_{ij}^k \partial _k} X^\alpha \\ = \frac{\partial ^2 X^\alpha}{ \partial x_i \partial x_j} -\Gamma_{ij}^k\frac{\partial X^\alpha }{\partial x_k} $$ In fact, I don't know how connection act on a non-tangential vector.
Third, you can see Anthony Carapetis' comment of this question.