Mean curvature vector in do Carmo's Riemannian Geometry

Question

Pictures below are from do Carmo's Riemannian Geometry. The third picture is from the 133th page.

First, I want to show the mean curvature vector does not depend on the chosen frame $E_i$. But I don't how to do it.

Second, when $f$ is minimal, we have trace of $S_\eta$ $=0$ for all $\eta\in (T_pM)^\perp$. Therefore, $H(p)=0$. But, why $H(p)=0$ means $f$ is minimal ?

Answer 1

First of all, notice that for another orthonormal frame $\{F_j\}$, where $F_j=\sum_ ia_{ji}E_i$, by linearity we have $S_{F_j}=\sum_i a_{ji} S_{E_i}$. We write $A=(a_{ji})$. Also recall that trace is independent on basis chosen, \begin{align}H=\frac 1n \sum_j (\operatorname{trace} S_{F_j})F_j &=\frac 1n \sum_{j,i,k}a_{ji}(\operatorname{trace} S_{E_i})a_{jk}E_k\\&=\frac 1n \sum_{i,k}(\operatorname{trace}S_{E_i})\underbrace{(A^TA)_{ik}}_{\delta_{ik}} E_k\\&=\frac 1n\sum_i(\operatorname{trace} S_{E_i}) E_i\end{align} If $H=0$, then by linear indepedence all trace of $S_i$, for some choice of orthonormal basis, but then $S_\eta$ is always given by linear combinations of $S_i$, so trace of $S_\eta$ is always zero.

Answer 2

To avoid additional check the independence issues, one could put these definitions in a different order.

As an alternative, define $H$ as the trace of $B$ (divided by $n$): $$ H := \frac 1n \operatorname{tr} B. $$ If you already know what a trace of a symmetric bilinear form is, then the above definition doesn't depend on any choices (except for the choice of the dot product). Picking an orthonormal frame $V_1,\ldots,V_n$ of $T_p M$, we can represent the trace as $\frac 1n \sum_j B(V_j,V_j)$, but the result doesn't depend on $V_j$'s. And even more importantly, it doesn't depend on the choice of $E_i$'s (the orthonormal frame of $(T_p M)^\perp$), as they haven't even appeared.

But why is this the same as $H$ introduced in the text? Choosing orthonormal frame $E_i$, we can express $H$ in this basis: $$ H = \sum_i \langle H, E_i \rangle E_i $$ It remains to compare the corresponding coefficients: $\frac 1n \langle \operatorname{tr} B, E_i \rangle$ (as above) and $\frac 1n \operatorname{tr} S_i$ (as in the text).

One can do this e.g. by using the characterization of trace (of $B$ and $S_i$) via $V_j$'s. So: \begin{align*} \langle \operatorname{tr} B, E_i \rangle & = \langle \sum_j B(V_j,V_j), E_i \rangle \\ & = \sum_j \langle B(V_j,V_j), E_i \rangle \\ & = \sum_j \langle S_i(V_j), V_j \rangle \\ & = \operatorname{tr} S_i, \end{align*} which shows that the coefficients match and thus, that the two definitions are equivalent. In particular, the outcome doesn't depend on the choice of $E_i$'s