Lemma $1.30$ - A course in minimal surfaces by Colding and Minicozzi This is a lemma which appears in A course in minimal surfaces by Colding and Minicozzi, section $8.1$ The second variation formula on page $39$. Before the lemma, I put some notations of this section.

Suppose now that $\Sigma^k \subset M^n$ is a minimal submanifold; we want to compute the second derivative of the area functional for a variation $\Sigma$. Therefore, again let $F$ be a variation of $\Sigma$ with compact support. In fact, we will assume that $F$ is a normal variation, that is, on $\Sigma$ we have 
$$F^T_t \equiv 0.$$
As before, let $x_i$ be local coordinates on $\Sigma$ and set
$$g_{ij}(t) = g(F_{x_i},F_{x_j}).$$
$\textbf{Lemma 1.30.}$ At the point $x$, we get
$$|g'(0)|^2 = 4 |\langle A(\cdot,\cdot),F_t \rangle|^2,$$
$$\text{Tr} (g''(0)) = 2 |\langle A(\cdot,\cdot),F_t \rangle|^2 + 2 |\nabla^N_{\Sigma} F_t|^2 + 2 \text{Tr} \ \langle R_M(\cdot,F_t),F_t, \cdot \rangle^2 + 2 \text{div}_{\Sigma}(F_{tt}).$$

My doubt is concerning just to the second equality, so I will put only the proof of this equality.

$\textbf{Proof.}$ To get the equality, we compute
$$(1.140) \text{Tr} g''(0) = 2 \sum\limits_{i=1}^k \langle F_{x_itt},F_{x_i} \rangle + 2 \sum\limits_{i=1}^k \langle F_{x_it},F_{x_it} \rangle.$$
Next, use the definition of the Riemann curvature tensor $R_M$ of $M$ ($R_M(U,V)W = \nabla_V \nabla_U W - \nabla_U \nabla_V W + \nabla_{[U,V]} W$) to get
$$\begin{align*}
(1.141) \sum\limits_{i=1}^k \langle F_{x_itt},F_{x_i} \rangle &= \sum\limits_{i=1}^k \langle \nabla_{F_t} \nabla_{F_t} F_{x_i},F_{x_i} \rangle = \sum\limits_{i=1}^k \langle \nabla_{F_t} \nabla_{F_{x_i}} F_t,F_{x_i} \rangle\\
&= \sum\limits_{i=1}^k \langle R_M(F_{x_i},F_t) F_t,F_{x_i} \rangle + \sum\limits_{i=1}^k \langle \nabla_{F_{x_i}} \nabla_{F_t} F_t,F_{x_i} \rangle\\
&= \sum\limits_{i=1}^k \langle R_M(F_{x_i},F_t) F_t,F_{x_i} \rangle + \text{div}_{\Sigma}(F_{tt}).
\end{align*}$$
Combining these and using again that $F_{x_i}$ is an orthonormal frame gives
$$\begin{align*}
\text{Tr} g''(0) &= 2 \sum\limits_{i=1}^k \langle F_{x_it}^T,F_{x_it}^T \rangle + 2 \sum\limits_{i=1}^k \langle F_{x_it}^N,F_{x_it}^N \rangle + 2 \sum\limits_{i=1}^k \langle R_M(F_{x_i},F_t) F_t,F_{x_i} \rangle 2 + 2 \text{div}_{\Sigma}(F_{tt})\\
&= 2 |\langle A(\cdot,\cdot),F_t \rangle|^2 + 2 |\nabla^N_{\Sigma} F_t|^2 + 2 \text{Tr}_{\Sigma} \langle R_M(\cdot,F_t),F_t,\cdot \rangle + 2 \text{div}_{\Sigma}(F_{tt})
\end{align*}$$

My doubts are
$1.$ $\sum\limits_{i=1}^k \langle \nabla_{F_t} \nabla_{F_t} F_{x_i},F_{x_i} \rangle = \sum\limits_{i=1}^k \langle \nabla_{F_t} \nabla_{F_{x_i}} F_t,F_{x_i} \rangle$ is true because the connection is symmetric and $[F_t,F_{x_i}] = 0$ since $\{ F_{x_i} \}_{i=1}^n \cup \{ F_t \}$ are a local coordinate vector fields?
$2.$ Why $\sum\limits_{i=1}^k \langle F_{x_it}^T,F_{x_it}^T \rangle = |\langle A(\cdot,\cdot),F_t \rangle|^2$? I imagine that I need to use the definition of the norm of the tensors, but I can't see how to obtain this equality.
Thanks in advance!
 A: I tried to verify $\sum\limits_{i=1}^k \langle F_{x_it}^T,F_{x_it}^T \rangle = |\langle A(\cdot,\cdot),F_t \rangle|^2$ again and I got it, I will leave here if someone has the same doubt I had.
Observe that
\begin{align*}
    F_{x_it}^T &= \sum\limits_{j=1}^n \langle F_{x_it}^T,F_{x_j} \rangle F_{x_j}\\
    &= \sum\limits_{j=1}^n \langle F_{x_it},F_{x_j} \rangle F_{x_j}\\
    &= \sum\limits_{j=1}^n - \langle A(F_{x_i},F_{x_j}), F_t \rangle F_{x_j},
\end{align*}
where the last equality is valid because
\begin{align*}
    F_{x_i} \langle F_t, F_{x_j} \rangle = 0 &\Longrightarrow \langle F_{sx_i}, F_{x_j} \rangle + \langle F_t, F_{x_jx_i} \rangle = 0\\
    &\Longrightarrow \langle F_{sx_i}, F_{x_j} \rangle = - \langle F_t, F_{x_jx_i} \rangle\\
    &\Longrightarrow \langle F_{sx_i}, F_{x_j} \rangle = - \langle F_t, \nabla_{F_{x_j}} F_{x_i} \rangle\\
    &\Longrightarrow \langle F_{sx_i}, F_{x_j} \rangle = - \left\langle F_t, \left( \nabla_{F_{x_j}} F_{x_i} \right)^T + \left( \nabla_{F_{x_j}} F_{x_i} \right)^N \right\rangle\\
    &\Longrightarrow \langle F_{sx_i}, F_{x_j} \rangle = - \left\langle F_t, \left( \nabla_{F_{x_j}} F_{x_i} \right)^N \right\rangle\\
    &\Longrightarrow \langle F_{sx_i}, F_{x_j} \rangle = - \langle F_t, A(F_{x_i}, F_{x_j}) \rangle.
\end{align*}
Thus,
\begin{align*}
    \langle \nabla_{F_{x_i}}  F_t^T,\nabla_{F_{x_i}} F_t^T \rangle &= \langle F_{x_it}^T, F_{x_it}^T \rangle\\
    &= \sum\limits_{j=1}^n \langle A(F_{x_i},F_{x_j}), F_t \rangle^2,
\end{align*}
therefore
$$g^{ii} (\langle \nabla_{F_{x_i}}  F_t^T,\nabla_{F_{x_i}} F_t^T \rangle = |\langle A(\cdot,\cdot),F_t \rangle|^2,$$
where we used the definition of the norm of tensors.
