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I have two questions about tensor calculation.

First question : In the book, Lectures on mean curvature flows written by Xi-Ping Zhu, there exists the equaility $g^{mn} \nabla_m \nabla_n h_{ij} = g^{mn} \nabla_m \nabla_i h_{jn}$. I do not understand this.

The situation is as follows : $X(\cdot, t) : M^n \rightarrow {\bf R}^{n+1}$ is a one-parameter family of smooth hypersurface immersions in ${\bf R}^{n+1}$, and $ X_t = H \nu$ where $H$ and $\nu$ is the mean curvature and unit normal to $X$. $g_{ij} = (X_i,X_j)$, $h_{ij} = (\nu, X_{ij})$

The question is found in the proof of Lemma 2.3 in 19 page. Please help me.

Second question : In the same book, there exists the equality $\Delta h_{ij} -\epsilon \Delta H g_{ij} = \Delta( h_{ij} - \epsilon H g_{ij}) $ (See the proof of Proposition 2.6 in 22 page)

I cannot understand the equality. Please help me.

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up vote 3 down vote accepted

For your first question, it follows from the identity $$\nabla_n h_{ij} = \nabla_i h_{jn},$$ which is Codazzi equation in $\mathbb{R}^{n+1}$. You can find its proof in P.137, Proposition 3.4 of Riemannian Geometry by Do Carmo. See also remark 3.5.

For your second question, it follows from the fact that $\Delta g=0$. To see this, by definition, we have $\Delta g=\sum_{i=1}^n\nabla_i\nabla_i g$. Note that $g$ is compatible with the connection, we have $\nabla_ig\equiv 0$. So we have $\Delta g=0$.

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