In John Lee's Smooth manifolds, pg255, he wrote

(Restriction of a vector bundle.) Suppose $\pi:E \rightarrow M$ is a rank $k$ vector bundle and $S \subseteq M$ is any subset. We define the restriction of $E$ to $S$ to be the set $E|_S = \bigcup_{p \in S}E_p$. with projection $E|S \rightarrow S$ obtained by restriciting $\pi$. A local trivilization $\Phi: \pi^{-1}(U) \rightarrow U \times \Bbb R^k$ restricts to a bijection $$ \Phi|_S: (\pi|S)^{-1}(U \cap S) \rightarrow (U \cap S) \times \Bbb R^k.$$ If $E$ is a smooth vector bundle, and $S \subseteq M$ is an immersed or embedded submanifold it follows from chart lemma that $E|_S$ is a smooth vector bundle.

  1. Where have we used the fact that $S$ is an immersed submanifold?

i.e. that $i$ is of constant rank and at each point $d_pi:T_pS \rightarrow T_pM$ is an injective map.

  1. Is $E|_S$ an immersed submanifold of $E$?
  • 4
    $\begingroup$ The short answer to your question 1 is that you need to assume $S$ is an immersed submanifold for the statement "$E|_S$ is a smooth vector bundle" even to make sense. The only kinds of smooth submanifold are embedded ones and immersed ones, and the latter category encompasses the former. $\endgroup$
    – Jack Lee
    Sep 28 '18 at 22:08

At 1: Not really, if one actually interprets $S \subset M$ as a mapping $f \colon S \to M$ and instead of $E|_S$ the the construction of the pullback $f^*E$ of a (vector) bundle $\pi \colon E \to M$ along a map $f \colon S \to M$. I find this construction much easier to grasp, so I sketch it here. Just think of $f$ as the immersion; but it can be any other smooth mapping.

The total space of the pullback bundle $f^*E$ is defined by $$ f^*E : = \{ (x,u) \in S \times E \;|\; f(x) = \pi(u) \} \subset S \times E. $$ Its footpoint map is just the projection onto the first factor: $$ p \colon f^*E \to S, \qquad p(x,u) := x. $$ Its addition and multiplication are defined by $$ (x,u) + (x,v) := (x, u+v) \quad \text{and} \quad \lambda (x,u) := (x, \lambda u). $$ We have to show that it is locally trivial. We do so by constructing a bundle atlas as follows: For each $x\in S$ choose an open neighborhood $V \subset M$ of $f(x)$ with local trivialization $\varphi \colon E|_V \to V \times \mathbb{R}^m$. Then $U = f^{-1}(V)$ is an open neighborhood of $x$. A local trivialization can be defined as follows $$ \psi \colon (f^*E)|_U \to U \times \mathbb{R}^m, \quad \psi(x,u) = (x, \operatorname{pr}_{\mathbb{R}^m}\varphi(f(x),u)), $$ where $\operatorname{pr}_{\mathbb{R}^m} \colon V \times \mathbb{R}^m \to \mathbb{R}^m$ is the projection onto the second factor. Indeed, this is a diffeomorphism that is linear in each fiber; it's inverse can be written down directly: $$ \psi^{-1}(x,v) = (x, \varphi^{-1}(f(x),v)). $$ These local trivializations are compatible with each other so that they form a bundle atlas follows by construction. It is a bit tediuous to show (but not hard, really), so I skip it here.

At 2.: We can show that the mapping $$ F \colon f^*E \to E, \quad F(x,u) = u$$ is an immersion. With the local trivialization $\varphi$ from above, we have $\varphi \circ F(x,u) = (f(x), \operatorname{pr}_{\mathbb{R}^m} \varphi(f(x),u)$ (locally of course). Then the differential looks like this in block matrix form: $$ d(\varphi \circ F)(x,u) = \begin{pmatrix} \mathrm{d}_x f & 0 \\ ?? & \operatorname{pr}_{\mathbb{R}^m} \varphi(f(x), \cdot)\end{pmatrix}. $$ Since $f \colon S \to M$ is an immersion, no matter what $??$ is, this matrix has rank equal to $$ \operatorname{rank}(\mathrm{d}_x f) + \operatorname{rank}(\varphi(f(x), \cdot)) = \operatorname{dim}(M)+ (\operatorname{dim}(E|_{f(x)})). $$ Since $\varphi$ is a diffeomorphism, $F$ has maximal rank and is thus an immersion.


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