# Normal Bundle is Trivial

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

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 $M$.

• You mean I must construct an isomorphism between $\nu(\mathbb S^n)$ and $\mathbb S^{n}\times \mathbb R$ with the aid of the nowhere vanishing section? – PtF Sep 23 '13 at 21:56
• Yes. A specific isomorphism $f: S^n \times \Bbb R \longrightarrow \nu(S^n)$ is given by $f(p, t) = t \frac{\mathbf{x}}{\|\mathbf{x}\|}$. – Henry T. Horton Sep 23 '13 at 22:18
• what I thought was defining $\phi:\nu(\mathbb S^n)\rightarrow \mathbb S^n\times \mathbb R$ by $$v+T_p\mathbb S^n\mapsto (p, \langle p, v\rangle)$$ I guess it works, what you think @Henry? – PtF Sep 26 '13 at 2:15
• @PtF: You want to map $v \in T_p S^n$ to $(p, \langle p, v \rangle)$, where $\langle p, v \rangle$ is the dot product of $p$ and $v$? This won't work, because for any $p \in S^n$, considered as a vector in $\Bbb R^{n+1}$, and any $v \in T_p S^n$, also considered as a vector in $\Bbb R^{n+1}$, $\langle p, v \rangle = 0$. – Henry T. Horton Sep 27 '13 at 21:45
• no, $\nu(\mathbb S^n)$ is an union of quotient spaces so I want to map the classe $[v]\in \nu_p(\mathbb S^n)$ to $(p, \langle v, p\rangle)$ and it works fine, I wrote a formal proof for this.. Later I'll post here.. – PtF Sep 27 '13 at 22:12

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 $\langle \cdot, \cdot\rangle$.