Convergence of vectors in the tangent bundle after an embedding implies convergence at the source

Question

$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Vol}{\operatorname{Vol}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\Volm}{\operatorname{Vol}_{\M}}$ $\newcommand{\Voln}{\operatorname{Vol}_{\N}}$

This is a self-answered question, which I put here since it wasn't obvious for me. (Tangent bundle stuff can be confusing.)

Let $\N$ be a smooth compact $n$-dimensional Riemannian manifold, and let $i:\N \to \R^D$ be an isometric embedding. Let $q_k,q \in \N$, $w_k \in T_{q_k}\N,w \in T_q\N$.

Claim: Suppose that $di_{q_k}(w_k) \to di_{q}(w)$. (this is a convergence of a sequence of vectors in $\R^D$). Then $w_k \to w$ w.r.t the topology of $T\N$.

Of course, I would be happy to see other approaches.

Answer 1

I think the general result is, that the tangent bundle functor preserves embeddings, i.e. if $f:M\to N$ is an embedding (smooth immersion+ homeomorphism onto its image) then the differential $Tf :TM\to TN$ is also an embedding.

If im not mistaken, this can be shown by taking adapted charts for the submanifold $f(M)$ or using the constant rank theorem.

Answer 2

$\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Vol}{\operatorname{Vol}}$ $\newcommand{\Det}{\operatorname{Det}}$ $\newcommand{\Volm}{\operatorname{Vol}_{\M}}$ $\newcommand{\Voln}{\operatorname{Vol}_{\N}}$

First, we note that $|w_k|=|di_{q_k}(w_k)|$ is bounded. Thus the sequence $(q_k,w_k)$ has a convergent subsequence in $T\N$ which converges to some $(q,\tilde w)$. It suffices to show that $\tilde w=w$.

Since $di:T\N \to T\R^D$ is continuous, $(q_k,w_k) \to (q,\tilde w)$ implies $di_{q_k}(w_k) \to di_q(\tilde w)$. By our assumption, we have $di_{q_k}(w_k) \to di_{q}(w)$, so $di_{q}(w)=di_{q}(\tilde w)$. The injectivity of $di_q$ now implies that $w=\tilde w$.