TITLE: Normal Bundle is Trivial
QUESTION [7 upvotes]: 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

REPLY [5 votes]: 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 $\langle \cdot, \cdot\rangle$.