Technical Details in do Carmo's Riemannian Geometry I am confused about a proof in do Carmo's Riemannian Geometry. The following side-to-side screenshots are pages 227 - 228 in Section 4 of Chapter 10. I have included them so as to keep the question self-contained. The main post is directly below the pictures. If it helps, we may assume that $N$ is an embedded submanifold (the case I care about).

For reference's sake, I will also include a screenshot of the "Weingarten equation" theorem from Chapter 6 of the same book.


I am confused about do Carmo's proof of property (ii). do Carmo writes
$$
\frac{DA}{ds}(0) = \left.\overline{\nabla}_{J(0)}A(s)\right|_{s = 0}.
$$
The right-hand side doesn't make a lot of sense to me. $A(s)$ is a vector field defined along a curve and not on an open subset of $M$. If $A(s)$ is extendible to a vector field $X$ in a neighborhood of $p$ with $X(\alpha(s)) = A(s)$ for small enough $s$, then the definition of $DA/ds(0)$ implies that $DA/ds(0) = \overline{\nabla}_{J(0)}X$, and then we could continue the proof just fine. I cannot think of any other possible interpretation for the right-hand side of the above equation. But what if $A(s)$ is not extendible? Is there something that guarantees it in this case, or is this a missing assumption?

Question 1: How do we choose such a vector field $X$ on $M$ near $p$, such that for small enough $s$, $X(\alpha(s)) = A(s)$? (Are we necessarily able to in this case, or is do Carmo omitting an important hypothesis?)

If, for example, $\alpha'(0) = J(0) \neq 0$, then $\alpha$ is locally an embedding, and so I can find local coordinates $(x^1,\dots,x^n)$ for $M$ with $\alpha(s) = (0, \dots, 0, s)$, and then extend $A$ in the obvious way, But what if $J(0) = 0$? I do not see what to do in this case. It suffices in this case to prove that $J'(0) \in (T_pM)^\perp$, but I don't know how to show this.
Now, let's assume that we've chosen such an $X$. I interpret the last equation as saying that
$$
\left.\left\langle\left(\overline{\nabla}_{J(0)}A(s)\right)^\top,v\right\rangle\right|_{s = 0} = \left\langle \left(\overline{\nabla}_{J(0)}X\right)^\top,v\right\rangle = \left\langle -S_{\gamma'(0)}(J(0)), v \right\rangle,
$$
where the last equality supposedly follows from the Weingarten equation (reference above). In order to apply this, we have to choose our local extension $X$ in such a way that $X$ is normal to $N$.

Question 2: As a follow-up to Question 1, how can we choose $X$ normal to $N$?


I apologize for the very long post. Any insights or help would be appreciated. I haven't been able to find anything on this online or in other textbooks, which is why I've decided to post this. If anyone could share any references with proofs of the fact do Carmo is proving, that would also be very helpful. (Of course, if I am making a fundamental misunderstanding and overcomplicating something, I would also welcome that as an answer.)

Edit: Inspired by this post, I noticed that Question 1 can be whittled down to a specific case. If $\alpha'(0) \neq 0$, then, as I remarked above, we can extend $A(s)$. (The question of how to choose the extension normal to $N$ still stands.) If $\alpha'(s) = 0$ for all $s$ sufficiently close to $0$, then $\alpha(s) \equiv p$ for small $s$. In this case, $A(s)$ is a curve in the vector space $(T_pN)^\perp$, so it follows that $J'(0) = DA/ds(0) = dA/ds(0) \in (T_pN)^\perp$, which proves (ii) directly. It only remains to figure out what happens when we have a sequence $s_n \to 0$ with $\alpha'(s_n) \neq 0$ but $\alpha'(0) = 0$. This, I have made no progress on.
 A: If I haven't made any mistakes in the following, no extensions are necessary. It boils down to a computation in the right frame. (I'm curious to know if there's a reasonable coordinate/frame-free proof; Deane's comment leads me to believe there might be one, in terms of pullback connections.) Let $E_1,\dots,E_m,\dots,E_n$ be a local orthonormal frame near $p$ adapated to $N$ (i.e., $(E_1,\dots,E_m)$ is an orthonormal frame for $N$ along $N$). Write
$$
A(s) = \sum_{i=m+1}^n A^i(s) E_i(f(s, 0))
$$
for some functions $A^{m+1},\dots,A^n$. The covariant derivative $DA/ds(0)$ is easily computed:
\begin{align*}
\frac{DA}{ds}(0) &= \sum_{i=m+1}^n \left.\frac{D}{ds} A^i(s)E_i(f(s,0))\right|_{s=0} \\
&= \sum_{i=m+1}^n \left[ \frac{d A^i}{d s}(0) E_i(p) + A^i(0)\left.\frac{D}{ds}E_i(f(s,0))\right|_{s=0} \right] \\
&= \sum_{i=m+1}^n \left[ \frac{d A^i}{d s}(0) E_i(p) + A^i(0) \overline{\nabla}_{J(0)}E_i\right] \\
&= \sum_{i=m+1}^n \left[ \frac{d A^i}{d s}(0) E_i(p) + \sum_{j=1}^n A^i(0) \omega_i{}^j(J(0))E_j(p)\right],
\end{align*}
where $(\omega_i{}^j)$ is the matrix of connection forms for $\overline\nabla$ with respect to $E_1,\dots,E_n$. Taking the tangential part gives
$$
\tag{1}
\left(\frac{DA}{ds}(0)\right)^\top = \sum_{j=1}^m \left(\sum_{i=m+1}^n A^i(0)\omega_i{}^j(J(0))\right)E_j(p).
$$
Now let's compute $-S_{\gamma'(0)}(J(0))$:
\begin{align*}
-S_{\gamma'(0)}(J(0)) &= -\sum_{j=1}^m \langle S_{\gamma'(0)}(J(0)), E_j(p) \rangle E_j(p) \\
&= -\sum_{j=1}^m \langle \gamma'(0), (\overline{\nabla}_{J(0)}E_j)^\perp \rangle E_j(p) \\
&= -\sum_{j=1}^m \langle \gamma'(0), \overline{\nabla}_{J(0)}E_j \rangle E_j(p) \\
&= -\sum_{j=1}^m \left(\sum_{i=1}^n \langle \gamma'(0), E_i(p)\rangle \omega_j{}^i(J(0))\right) E_j(p) \\
&= -\sum_{j=1}^m \left(\sum_{i=m+1}^n A^i(0) \omega_j{}^i(J(0))\right) E_j(p).
\end{align*}
Since the matrix of connection forms in an orthonormal frame is skew-symmetric, this equals (1), and we conclude
$$
\left(\frac{DA}{ds}(0)\right)^\top = -S_{\gamma'(0)}(J(0)),
$$
what we wanted to prove.
A: A(s) is a vector field along $\alpha(s)$ and $\alpha'(0)=J(0)$. Then by definition of the covariant derivative $(DA/ds)(0)$, one has the first equation.
