how can I show the normal bundle $$\nu(\mathbb S^n):=\bigcup_{p\in\mathbb S^n} T_p\mathbb R^{n+1}/T_p\mathbb S^n, $$ is a trivial bundle? Any help will be valuable.. Thanks


Solution 1:

The normal bundle of the standard embedding of $S^n$ in $\Bbb R^{n+1}$ has a nowhere vanishing section (the unit outward normal to the sphere, $\displaystyle \frac{\mathbf{x}}{\|\mathbf{x}\|}$). So you just need to show that a nowhere vanishing section of a rank $1$ vector bundle can be used to construct an isomorphism of the bundle with the trivial bundle.

If you already know this fact, then you're done. If you don't know it, the proof is a simple exercise.

This is a special case of a more general fact:

If $E \longrightarrow M$ is a rank $k$ vector bundle and $s_1, \dots, s_k$ are sections of $E$ such that $\{s_1(p), \dots, s_k(p)\}$ is a basis for that fiber $E_p$ for all $p \in M$, then $E$ is isomorphic to the trivial rank $k$ bundle over $M$.

Solution 2:

We will construct a vector bundle isomorphism between $\mathcal{V}(\mathbb S^n)$ and the trivial vector bundle $\mathbb S^n\times \mathbb R$.

Define $\Phi:\nu(\mathbb S^n)\rightarrow \mathbb S^n\times \mathbb R$ setting $$[v]\in \nu_p(\mathbb S^n)\mapsto (p, \langle v, p\rangle).$$ Notice $\Phi$ is well defined. In fact, if $u\in [v]$ then $u-v\in T_p\mathbb S^n$ so that, \begin{align*} \displaystyle (p, 0)=( p, \langle u-v, p\rangle)=(p, \langle u, p\rangle-\langle v, p\rangle), \end{align*} hence, \begin{align*} \displaystyle \Phi([u])=(p, \langle u, p\rangle)=(p, 0)+(p, \langle v, p\rangle)=(p, \langle v, p\rangle)=\phi([v]). \end{align*} Now define $\Psi: \mathbb{S}^n\times \mathbb R\rightarrow \nu(\mathbb S^n)$ setting $$(q, t)\mapsto [tq]\in \nu_q(\mathbb S^n).$$ Then, \begin{align*} \displaystyle (\Phi\circ \Psi)(p, t)=\Phi([tp])=(p, \langle tp, p\rangle)=(p, t\langle p, p\rangle)=(p, t), \end{align*} whereas, \begin{align*} \displaystyle (\Psi\circ \Phi)([v])=\Psi(p, \langle v, p\rangle)=[\langle v, p\rangle p]=[v], \end{align*} for $v-\langle v, p\rangle p\in T_p\mathbb S^n$ since, \begin{align*} \displaystyle \langle p, v-\langle v, p\rangle p\rangle=\langle p, v\rangle-\langle v, p\rangle \langle p, p\rangle=\langle p, v\rangle-\langle v, p\rangle=0. \end{align*} It is an exercise to check $\Phi$ is vector bundle morphism. The linearity on the fibers follows from the linearity of the scalar product $\langle \cdot, \cdot\rangle$.