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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!

George
  • 3,957

1 Answers1

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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.

George
  • 3,957