# Restriction of a smooth vector bundle is a smooth bundle?

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 $$E$$?
• 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. – Jack Lee Sep 28 '18 at 22:08

## 1 Answer

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: $$p \colon f^*E \to S, \qquad p(x,u) := x.$$ It 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 be constructin 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 diffeomorphism that is linear in each fiber; it's inverse can be written down directly: $$\psi^{-1}(x,v) = (x, \varphi^{-1}(f(x),v)).$$ That 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: $$d(\varphi \circ F)(x,u) = \begin{pmatrix} \mathrm{d}_x f & 0 \\ ?? & \operatorname{pr}_{\mathbb{R}^m} \varphi(f(x), \cdot)\end{pmatrix}.$$ If $$f \colon S \to M$$ is an immersion, no matter what $$??$$ is, this matrix as 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.